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Transcript
KVS OLYMPIAD
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KVS – CHANDIGARH REGION
MATHEMATICS OLYMPIAD
In association with
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{Celebrating International Mathematics and KVS golden jubilee}
Study Booklet
{For Senior classes}
“It’s not enough to have good mind, the main thing is use it well”
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{Mathematics is 99% Commonsense & 1% Syllabus by: L.K.Gupta}
KVS-OLYMPIAD Sample booklet Contents
Index {Syllabus} for Round-I {Vidyalaya level}
Section A : SAT (Subject Aptitude Test)
Chapter
N0.:
Chapter Name
Classes
{Theory+Problems+answers}
eligible
Page No.:
0.
Common Sense.
6th TO 12th
6-7
1
Number System
6th TO 12th
8-20
2
Simple Arithmetic
6th TO 12th
21-29
3
Square root & cube root
6th TO 12th
30-35
4
Polynomials
7th to 12th
36-41
5
Quadratic Equation
9th to 12th
42-47
6
Mensuration-1{Area}
7th to 12th
48-79
7
Mensuration-2{volume}
8th to 12th
80-97
8
Geometry-I
8th to 12th
98-114
9
Geometry-II
9th to 12th
115-133
10
Coordinate Geometry
9th to 12th
134-142
11
Algebra-I
9th to 12th
143-156
12
Combinatorics
12th only
157-166
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Section B : MAT (Mental Aptitude Test )
Chapter N0.:
Chapter Name
Classes
{Theory+Problems+answers}
eligible
Page No.:
(A) Verbal
1
Blood Relation
6th TO 12th
167-170
2
Classification
6th TO 12th
171-173
3
Directions Sense
6th TO 12th
174-177
4
Coding decoding.
6th to 12th
178-181
( B) Non-Verbal
5
In complete Figures
6th TO 12th
182-185
6
Mirror images
6th TO 12th
186-188
7
Paper Cutting
6th TO 12th
189-193
8
Analytical Reasoning
6th TO 12th
194-199
{For more chapters visit www.pioneermathematics.com}
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Index {Syllabus} of Free Sample booklet for Round-II
{Cluster level}
Contents
1. Number Theory
1.1 Divisibility of Integers
1.2 Congruences
1.3 Fermat, Euler, Wilson and Lagrange's Theorems
1.4 Greatest Integer Function
1.5 Arithmetic Functions
1.6 Pythagorean Triples
1.7 Representation of a positive integer
2. Algebra
2.1 Polynomials
2.2 Inequalities
2.3 Functional Equations
3. Geometry
3.1 Some Important Theorems
3.2 Concurrency and collinearity
3.3 Pythagoras Theorem
3.4 Properties of triangles
3.5 Constructions
4. Combinatorics
4.1 Basic Counting Principles
4.2 Permutations - Combinations
4.3 Permutations with repetitions
4.4 The Pigeonhole Principle
5. Miscellaneous topics
1 Time and work
3. Profit –loss
5.Mixture And Allegation
2.Percentage.
4.Speed and Time
6.Average and partnership
7.Calender and clock
6. Logical Reasoning & Analytical Ability
7. Decision Making & Problem solving.
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Index {Syllabus} of Free Sample booklet
for Round-III {Regional level}
1. Number Theory
1.1 Divisibility of Integers
1.2 Congruences
1.3 Fermat, Euler, Wilson and Lagrange's Theorems
1.4 Greatest Integer Function
1.5 Arithmetic Functions
1.6 Pythagorean Triples
1.7 Representation of a positive integer
2. Algebra
2.4 Polynomials
2.5 Inequalities
2.6 Functional Equations
3. Geometry
3.6 Some Important Theorems
3.7 Concurrency and collinearity
3.8 Pythagoras Theorem
|
3.9 Properties of triangles
3.10 Constructions
4. Combinatorics
4.3 Basic Counting Principles
4.4 Permutations - Combinations
4.3 Permutations with repetitions
4.4 The Pigeonhole Principle
5. Miscellaneous topics
1 Time and work
3. Profit –loss
5.Mixture And Allegation
2.Percentage.
4.Speed and Time
6.Average and partnership
7.Calendar and clock
6. Logical Reasoning & Analytical Ability
6. Decision Making & Problem solving.
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Chapter: 0 Common Sense
1. A number of bacteria are placed in a glass. One second later each bacterium divides in two, the next
second each of the resulting bacteria divides in two again, et cetera. After one minute the glass in full. When
was the glass half-full?
2. Jack tore out several successive pages from a book. The number of the first page he tore out was 183, and
it is known that the number of the last page is written with the same digits in some order. How many pages
did Jack tear out of the book?
3. Cut the figure shown in below figure into four figures, each similar to the original with dimensions twice
as small.
4. Matches are arranged to form the figure shown in below fig.. Move two matches to change this fig. into
four squares with sides equal in length to one match.
5. Ten coins are arranged as shown in below Fig. What is the minimum number of coins we must remove so
that no three of the remaining coins lie on the vertices of an equilateral triangle?
6. The pendulum of a clock takes 7s to strike 4 o’ clock. How much time will it take to strike 11 o’ clock?
Here Dr. Dim helps you that 14 is not right Answer.
7. Amit Mittal buys 80 tubes of fair & Lovely. All tubes except one weigh 1000 gms. To determine the tube
with lesser weight he uses a pan balance. How many weightings are required to find out the defected tube?
8. A ladder leans against a vertical wall. The top of the ladder is 8m above the ground. When the bottom of
the ladder is moved 2 m farther away from the wall, the top of the ladder rests against the foot of the wall.
Can you guess the length of the ladder?
9. M milkman mixes 20 Litres of water with 80 litres of milk. After selling one-fourth of this mixture he adds
water to replenish the quantity that he has sold. What is the current proportion of water to milk?
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10. In nuts and bolts factory, on machine produces only nuts at the rate of 100 nuts per minute and needs to
be cleaned for 5 minutes after production of every 1000 nuts. Another machine produces only bolts at the
rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If
both the machines start production at the same time, what is the minimum duration required for producing
9000 pairs of nuts and bolts.
11. A solid cube of each side 12 cm has been painted yellow pink and white on pairs of opposite faces. It is
then cut into cubical block of each of cms. Can you define how many cubes have only two faces painted, also
count the cube having no face painted?
12. Step 1 – Put A = 0, B = 1, C = 1
Step 2- Replace A by C.
Step 3- Replace B by 2A + 1.
Step 4- Replace C by A + B.
Step 5-If C = 100 go to step – 7 otherwise go to step – 6.
Step 6- Go to step – 2.
Step 7- Stop.
What is the value of A.?
13. How many rectangles are there in the diagram given below?
14. Peter said: “The day before yesterday I was 10, but I will turn 13 in the next year” Is this possible?
15. The son of a professor’s father is talking to the father of the professor sor’s son, and the professor does
not take part in the conversation. In this possible?
Answers
1. After 59 seconds
2. 136
5. 4
6. 23.335
7. 7
8. 17
9. 20
10. 170 minutes
11. 8 cubes
12. 40
13. 14
14. Yes
15. Yes
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Chapter: 1 NUMBER THEORY
In Hindu Arabic System, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number. This
is the decimal system where we use the numbers 0 to 9. 0 is called insignificant digit whereas 1, 2, 3, 4, 5, 6,
7, 8, 9 are called significant digits.
A group of figures, denoting a number is called a numeral. For a given numeral, we start from extreme right as
Unit's place, Ten's place, Hundred's place and so on.
Illustration 1: We represent the number 309872546 as shown below:
We read it as "Thirty crores, ninety-eight lacs, seventy-two thousands five hundred and forty-six."
In this numeral:
The Place value of 6 is 6 × 1 = 6
The place value of 4 is 4 × 10 = 40
The place value of 5 is 5 × 100 = 500
The place value of 2 is 2 × 1000 = 2000 and so on.
The face value of a digit in a number is the value itself wherever it may be.
Thus, the face value of 7 in the above numeral is 7. The face value of 6 in the above numeral is 6 and so on.
NATURAL NUMBERS
Counting numbers 1, 2, 3, 4, 5, ... are known as natural numbers.
The set of all natural numbers can be represented by
N = {1, 2, 3, 4, 5, ...}.
WHOLE NUMBERS
If we include 0 among the natural numbers, then the numbers 0, 1, 2, 3, 4, 5, ... are called whole
numbers.
The set of whole numbers can be represented by
W = {0,1,2,3,4,5, ...}
Clearly, every natural number is a whole number but 0 is a whole number which is not a natural number.
INTEGERS
All counting numbers and their negatives including zero are known as integers.
The set of integers can be represented by
Z or I = {... 4, ―3, ―2, ―1, 0, 1, 2, 3, 4 ......}
POSITIVE INTEGERS
The set I+ = {1, 2, 3, 4, ....} is the set of all positive integers. Clearly, positive integers and natural numbers are
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synonyms.
NEGATIVE INTEGERS
The set {0, ―1, ―2, ― 3, ...} is the set of all negative integers. 0 is neither positive nor negative.
NON-NEGATIVE INTEGERS
The set {0, 1, 2, 3, ... } is the set of all non-negative integers.
RATIONAL NUMBERS
p
The numbers of the form where p and q are integers and q 0 , are known as rational numbers,
q
4 3 5 0 2
, , , , , etc.
7 2 8 1 3
The set of all rational numbers is denoted by Q.
e.g.
i.e. Q {x : x
p
; p, q I, q 0} .
q
a
, every natural number is a rational number. Since 0 can be
1
0
a
written as and every non-zero integer ‘a’ can be written as , every integer is a rational number.
1
1
Every rational number has a peculiar characteristic that when expressed in decimal form is expressible either in
terminating decimals or in non-terminating repeating decimals.
Since every natural number ‘a’ can be written as
1
1
22
8
0.2, 0.333...,
=3.1428714287, 0.181818 ....,etc
5
3
7
44
The recurring decimals have been given a short notation as
For example,
0.333... 0.3
4.1555... 4.05
0.323232... 0.32.
IRRATIONAL NUMBERS
Those numbers which when expressed in decimal form are neither terminating nor repeating decimals are
known as irrational numbers,
e.g. 2 , 3 , 5 , , etc.
22 22
22
,
is rational while is irrational number.
is approximate
7 7
7
value of . Similarly, 3.14 is not an exact value of it.
REAL NUMBERS
The rational and irrational numbers combined together are called real numbers,
13 2
3
, ,
, 3 , 4 2 , etc. are real numbers.
e.g.
21 5
7
Note that the exact value of is not
The set of all real numbers is denoted by R.
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Note that the sum, difference or product of a rational and irrational number is irrational, e.g.
2
3 2, 4 3 , 5 , 4 3 , 7 5 are all irrational.
3
EVEN NUMBERS
All those numbers which are exactly divisible by 2 are called even numbers, e.g. 2, 6, 8,10, etc., are even
numbers.
ODD NUMBERS
All those numbers which are not exactly divisible by 2 are called odd numbers, e.g. 1, 3, 5, 7 etc., are odd numbers.
PRIME NUMBERS
A natural number other than 1, is a prime number if it is divisible by 1 and itself only.
For example, each of the numbers 2, 3, 5, 7 etc., are prime numbers.
COMPOSITE NUMBERS
Natural numbers greater than 1 which are not prime, are known as composite numbers.
For example, each of the numbers 4, 6, 8, 9, 12, etc., are composite numbers.
Note:
1. The number 1 is neither a prime number nor a composite number.
2. 2 is the only even number which is prime.
3. Prime numbers up to 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, i.e. 25 prime
numbers between 1 and 100.
4. Two numbers which have only 1 as the common factor are called co-primes or relatively prime to
each other, e.g. 3 and 5 are co-primes.
Note that the numbers which are relatively prime need not necessarily be prime numbers, e.g. 16 and 17 are
relatively prime although 16 is not a prime number.
ADDITION AND SUBTRACTION (SHORT-CUT METHODS)
The method is best illustrated with the help of following example :
Illustration 2: 54321 (9876 + 8967 + 7689) = ?
Step 1: Add 1st column:
6 + 7 + 9 = 22
To obtain 1 at unit's place add 9 to make 31. In the answer, write 9 at unit's place and carry over 3.
Step 2: Add 2nd column:
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3 + 7 + 6 + 8 = 24
To obtain 2 at ten's place, add 8 to make 32. In the answer, write 8 at ten's place and carry over 3.
Step 3: Add 3rd column:
3 + 8 + 9 + 6 = 26
To obtain 3 at hundred's place, add 7 to make
33. In the answer, write 7 at hundred's place and carry over 3.
Step 4: Add 4th column:
3 + 9 + 8 + 7= 27
To obtain 4 at thousand's place add 7 to make
34. In the answer, write 7 at thousand's place and carry over 3.
Step 5: 5th column:
To obtain 5 at ten-thousand's place add 2 to it to make 5. In the answer, write 2 at the ten-thousand's
place.
54321 ― (9876 + 8967 + 7689) = 27789.
MULTIPLICATION (SHORT-CUT METHODS)
1. Multiplication of a given number by 9, 99,999, etc., that is by 10n ― 1
Method: Put as many zeros to the right of the multiplicand as there are nines in the multiplier and from the result
subtract the multiplicand and get the answer.
Illustration 3: Multiply:
(a) 3893 by 99
(b) 4327 by 999
(c) 5863 by 9999.
Sol:
(a) 3893 × 99 = 389300 ― 3893 = 385407.
(b) 4327 × 999 = 4327000 ― 4327 = 4322673
(c) 5863 × 9999 = 58630000 ― 5863 = 58624137.
2. Multiplication of a given number by 11, 101, 1001, etc., that is by, 10n + 1.
Method: Place n zeros of the right of the multiplicand and then adds the multiplicand to the number so
obtained.
Illustration 4: Multiply:
(a) 4782 × 11
(b) 9836 × 101
(c) 6538 × 1001.
Sol:
(a) 4782 × 11 = 47820 + 4782 = 52602
(b) 9836 × 101= 983600 + 9836 = 993436
(c) 6538 × 1001= 6538000 + 6538 = 6544538
3. Multiplication of a given number by 15, 25, 35 etc.
Method: Double the multiplier and then multiply the multiplicand by this new number and finally divide the
product by 2.
Illustrations 5: Multiply:
(a) 7054 × 15
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(b) 3897 × 25
(c) 4563 × 35
Sol:
1
1
7054 30 = 211620 105810 .
2
2
1
(b) 3897 × 25 = (3897 × 50) =(194850) = 97425.
2
1
1
(c) 4536 × 35 = (4563 × 70) = (319410) = 159705.
2
2
4. Multiplication of a given number by 5, 25, 125, 625, etc., that is, by number which is some power of 5.
Method: Place as many zeros to the right of the multiplicand as is the power of 5 in the multiplier, then
divide the number so obtained by 2 raised to the same power as is the power of 5.
Illustration 6: Multiply:
(a) 3982 × 5
(b) 4739 × 25
(c) 7894 × 125
(d) 4863 × 625
Sol:
39820
(a) 3982 × 5 =
= 19910.
2
473900 473900
118475 .
(b) 4739 × 25 =
22
4
7894000 7894000
(c) 7894 125
= 986750
23
8
48630000 48630000
(d) 4863 × 625 =
= 3039375.
24
16
DISTRIBUTIVE LAWS:
For any three numbers a, b, c, we have
(a) a × b + a × c = a × (b + c)
(b) a × b ― a ×c = a × (b ― c)
Illustration 7:
438 × 637 + 438 × 367 = ?
Sol:
438 × 637 + 438 + 367 = 438 × (637 + 367) = 430 ×1000 = 438000.
Illustration 8:
674 × 832 ― 674 × 632 =?
Sol:
674 × 832 ― 674 × 632 = 674 × (832 ― 632) = 674 200 = 134800.
SQUARES (SHORT-CUT METHODS)
1. To square any number ending with 5.
Method: (A5)2 = A (A + 1)/25
Illustration 9:
(a) (25)2 = 2 (2 + l)/25 = 6/25 = 625
(b) (45)2 = 4 (4 + i)/25 = 20/25 = 2025
(a) 7054 × 15 =
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(c) (85)2 = 8 (8 + l)/25 = 72/25 = 7225.
2. To square a number in which every digit is one.
Method: Count the number of digits in the given number and start writing numbers in ascending order
from one to his number and then in descending order up to one.
Illustration 10:
(a) 11 2 = 121
(b) 1112 =12321
(c) 11112 =1234321
(d) 2222 = 22 (111)2 = 4 (12321) = 49284
(e) 33332 = 32 (1111)2 = 9 (1234321) = 11108889
3. To square a number which is nearer to 10x.
Method: Use the formula:
x2 x2 y 2 y 2 x y x y y 2
Illustration 11:
(a) (97)2 = (97 + 3) (97 3) + 32 = 9400 + 9 = 9409.
(b) (102)2 = (102 ― 2) (102 + 2) + 22 = 10400 + 4 = 10404.
(c) (994)2 = (994 + 6) (994 ― 6) + 62 = 988000 + 36 = 988036.
(d) (1005)2 = (1005 ― 5) (1005 + 5) + 52 = 1010000 + 25 = 1010025.
DIVISION:
Division is repeated subtraction.
For example, when we divide 63289 by 43, it means 43 can be repeatedly subtracted 1471 times from
63289 and the remainder 36 is left.
Dividend = (Divisor × Quotient) + Remainder
Dividend Remainder
or, Divisor =
Quotient
Illustration 12: On dividing 7865321 by a certain number, the quotient is 33612 and the remainder is 113.
Find the divisor.
Sol:
Dividend Remainder
7865321 113 7865208
234 .
Divisor =
=
Quotient
33612
33612
Illustration 13: A number when divided by 315 leaves remainder 46 and the value of quotient is 7. Find the
number.
Sol:
Number = (Divisor × Quotient) + Remainder = (315 × 7) + 46 = 2205 + 46 = 2251.
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Illustration 14: Find the least number of 5 digits which is exactly divisible by 632.
Sol:
The least number of 5 digits is 10000. Dividing this number by 632, the remainder is 520. So, the required number
= 10000 + (632 + 520) = 10112.
Illustration 15: Find the greatest number of 5 digits which is exactly divisible by 463.
Sol:
The greatest number of 5 digits is 99999. Dividing this number by 463, the remainder is 454. So, the required
number = 99999 ― 454 = 99545.
Illustration 16: Find the number nearest to 13700 which is exactly divisible by 235.
Sol:
On dividing the number 13700 by 235, the remainder is 70. Therefore, the nearest number to 13700, which is
exactly divisible by 235 = 13700 ― 70 = 13630.
TESTS OF DIVISIBILITY
o Divisibility by 2: A number is divisible by 2 if the unit’s digit is zero or divisible by 2.
For example, 4, 12, 30, 18, 102, etc., are all divisible by 2.
o Divisibility by 3: A number is divisible by 3 if the sum of digits in the number is divisible by 3.
For example, the number 3792 is divisible by 3 since 3 + 7 + 9 + 2 = 21, which is divisible by 3.
o Divisibility by 4: A number is divisible by 4 if the number formed by the last two digits (ten’s digit and unit’s
digit) is divisible by 4 or are both zero.
For example, the number 2616 is divisible by 4 since 16 is divisible by 4.
o Divisibility by 5: A number is divisible by 5 if the unit’s digit in the number is 0 or 5.
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For example, 13520, 7805, 640, 745, etc., are all divisible by 5.
o Divisibility by 6: A number is divisible by 6 if the number is even and sum of its digits is divisible by 3.
For example, the number 4518 is divisible by 6 since it is even and sum of its digits 4 + 5 + 1 + 8 = 18 is divisible by 3.
o Divisibility by 7: The unit digit of the given number is doubled and then it is subtracted from the number
obtained after omitting the unit digit. If the remainder is divisible by 7, then the given number is also divisible by 7.
For example, consider the number 448. On doubling the unit digit 8 of 448 we get 16.
Then, 44 ― 16 = 28.
Since 28 is divisible by 7, 448 is divisible by 7.
o Divisibility by 8: A number is divisible by 8, if the number formed by last 3 digits in divisible by 8.
For example, the number 41784 is divisible by 8 as the number formed by last three digits, i.e. 784 is divisible by 8.
o Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
For example, the number 19044 is divisible by 9 as the sum of its digits 1 + 9 + 0 + 4 + 4 = 18 is divisible by 9.
o Divisibility by 10: A number is divisible by 10, if it ends in zero.
For example, the last digit of 580 is zero, therefore, 580 is divisible by 10.
o Divisibility by 11: A number is divisible by 11 if the difference of the sum of the digits at odd places and sum of
the digits at even places is either zero or divisible by 11.
For example, in the number 38797, the sum of the digits at odd places is 3 + 7 + 7 = 17 and the sum of the digits at
even places is 8 = 9 =17. The difference is 17 ― 17 = 0, so the number is divisible by 11.
o Divisibility by 12: A number is divisible by 12 if it is divisible by 3 and 4.
o Divisibility by 25: A number is divisible by 25 if the number formed by the last two digits is divisible by 25 or
the last two digits are zero.
For example, the number 13675 is divisible by 25 as the number formed by the last two digits is 75 which is divisible
by 25.
o Divisibility by 125: A number is divisible by 125 if the number formed by the last three digits is divisible by 125
or the last three digits are zero.
For example, the number 5250 is divisible by 125 as 250 is divisible by 125.
o Divisibility by 18: An even number satisfying the divisibility test of 9 is divisible by 18.
o Divisibility by 88: A number is divisible by 88 if it is divisible by 11 and 8.
SOME USEFUL SHORT-CUT METHODS:
1. Test to find whether a given number is a prime
Step 1: Select a least positive integer n such that n2 > given number
Step 2: Test the divisibility of given number by every prime number less than n.
Step 3: The given number is prime only if it is not divisible by any of these primes.
Illustration 17: Investigate whether 571 is a prime number
Sol:
Since (23)2 = 529 < 571 and (24)2 = 576 > 571
n = 24.
Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. Since 24 is divisible by 2, 571 is not a prime number.
Illustration 18: Investigate whether 923 is a prime number.
Sol:
Since (30)2 = 900 < 923 and (31)2 = 961 > 923
n = 31.
Prime numbers less than 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Since 923 is not divisible by any of these primes,
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therefore 923 is a prime number.
2. The least number, which when divided by d1, d2 and d3 leaves the remainders r1, r2 and r3 respectively
such that
(d1 ― r1) = (d2 ― r2) = (d3 ― r3), is = (L.C.M. of d1, d2 and d3) ― (d1 ― r1) or (d2 ― r2) or (d3 ― r3).
Illustration 19: Find the least number which when dividend by 9, 10 and 15 leaves the remainders 4, 5 and 10,
respectively.
Sol:
Here, 9 ― 4 = 10 ― 5 = 15 ― 10 = 5.
Also, L.C.M. (9, 10, 15) = 90,
The required least number = 90 ― 5 = 85.
3. A number on being divided by d1 and d2 successively leaves the remainders r1 and r2, respectively. If the number
is divided by d1 × d2, then the remainder is = (d1 × r2 + r1).
4. Illustration 20: A number on being divided by 10 and 11 successively leaves the remainders 5 and 7,
respectively. Find the remainder when the same number is divided by 110.
Sol:
The required remainder = d1 × r2 + r1 = 10 × 7 + 5 = 75.
5. To find the number of numbers divisible by a certain integer. The method is best illustrated with the help of
following example.
Illustration 21: How many numbers up to 532 are divisible by 15?
Sol:
We divide 532 by 15.
532 35 15 7
The quotient obtained is the required number of numbers. Thus, there are 35 such numbers.
Illustration 22: How many numbers up to 300 are divisible by 5 and 7 together?
Sol:
L.C.M. of 5 and 7 = 35.
We divided 300 by 35.
300 = 8 35 20
Thus, there are 8 such numbers.
6. Two numbers when divided by a certain divisor give remainders r1 and r2. When their sum is divided by
the same divisor, the remainder is r3. The divisor is given by r1 + r 2 r3.
Illustration 23: Two numbers when divided by a certain divisor give remainders 473 and 298,
respectively. When their sum is divided by the same divisor, the remainder is 236. Find the divisor.
Sol:
The required divisor
=437 + 298 236 = 499.
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Sample Questions
In each of the following questions a number of possible answers are given, out of which one answer is
correct. Find out the correct answer.
1. 3149 × 1? 5 = 425115
(a) 3
(c) 4
(b) 2
(d) 6
2. 24 is divided into two parts such that 7 times the first part added to 5 times the second part makes 146.
The first part is
(a) 11
(b) 13
(c) 16
(d) 17
1
1
of a number subtracted from of the number gives 12. The number is
4
3
(a) 144
(b) 120
(c) 72
(d) 63
3.
4. A fraction becomes 4 when 1 is added to both the numerator and denominator; and it becomes 7 when 1
is subtracted from both the numerator and denominator. The numerator of the given fraction is
(a) 2
(b) 3
(c) 7
(d) 15
5. The quotient arising from a division of a number by 62 is 463 and the remainder is 60, what is the
number?
(a) 28666
(b) 28766
(c) 28576
(d) 28676
6. Which one of the following is the least number of four digits divisible by 71?
(a) 1006
(b) 1065
(c) 1094
(d) 1056
7. How many numbers between 100 and 300 are divisible by 11?
(a) 11
(b) 10
(c) 12
(d) 18
8. The least value to be given to * so that the number 5 * 3457 is divisible by 11 is
(a) 2
(b) 3
(c) 0
(d) 4
9. Which one of the following is the greatest number of five digits divisible by 231?
(a) 99792
(b) 99892
(c) 99692
(d) 99972
10. What is the number just more than 5000 which is exactly divisible by 73?
(a) 5001
(b) 5009
(c) 5037
(d) 5027
11. Three-fourths of one-fifth of a number is 60. The number is
(a) 300
(b) 400
(c) 450
(d) 1200
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12. The product of two numbers is 120. The sum of their squares is 289. The sum of the two numbers is
(a) 20
(b) 23
(c) 169
(d) None of these
13. A number when divided by a certain divisor left remainder 241, when twice the number was divided by
the same divisor, the remainder was 112. Find
the divisor.
(a) 370
(b) 365
(c) 380
(d) 456
14. Two numbers when divided by a certain divisor give remainders 43 and 37 respectively, when their sum
is divided by the same divisor, the remainder
is 13. Find the divisor.
(a) 71
(b) 67
(c) 57
(d) 77
15. For what value of K, the number 7236K2 is divisible by 8?
(a) 7
(b) 5
(c) 4
(d) 9
16. Sum of three numbers is 132. First number is twice the second and third number is one-third of the first.
Find the second number.
(a) 18
(b) 36
(c) 20
(d) 16
17. The sum of the digits of a two-digit number is 8. If the digits are reversed the number is increased by 54.
Find the number.
(a) 17
(b) 19
(c) 21
(d) 23
18. The smallest number by which 3600 can be divided to make it a perfect cube is
(a) 9
(b) 50
(c) 300
(d) 450
19. The remainder when 784 is divided by 342 is
(a) 0
(b) 1
(c) 49
(d) 341
20. A two-digit number is such that the product of the digits is 14. When 45 is added to the number, then the
digits interchange their places. Find the number,
(a) 72
(b) 27
(c) 37
(d) 14
21. In three coloured boxes-Red, Green and Blue, 108 balls are placed. There are twice as many in the Green
and Red boxes combined as they are in the Blue box and twice as many in the Blue box as they are in the Red
box. How many balls are there in the Green box?
(a) 18
(b) 36
(c) 45
(d) None of these
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22. Which of the following is a prime number?
(a) 889
(b) 997
(c) 899
(d) 1147
23. If a and b are two integers and b > 0, then there exist two integers q and r such that
(a) b = aq + r, where 0 < r < b
(b) b = rq + a, where 0 < r < b
(c) a = bq + r, where 0 < r < b
(d) None of these
24. The denominator of a rational number is 3 more than its numerator. If the numerator is increased by 7
and the denominator is decreased by 2, we obtain 2. The rational number is
1
5
(b)
(a)
4
8
7
8
(c)
(d)
10
11
25. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and
1
stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave of the
2
diamonds he had then, and 2 more besides. He escaped with one diamond. How many did he steal originally?
(a) 40
(b) 36
(c) 25
(d) None of these
26. A child was asked to add first few natural numbers (that is 1 + 2 + 3 + ...) so long his patience permitted.
As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he
had missed one number in the sequence during addition. The number he missed was:
(a) less than 10
(b) 10
(c) 15
(d) more than 15
27. After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4
respectively. What will be the remainder if 84 divide the same number?
(a) 80
(b) 75
(c) 41
(d) 53
28. Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. The number of boxes
containing the same number of oranges is at least:
(a) 5
(b) 103
(c) 6
(d) None of these
29. Consider a sequence of seven consecutive integers. The average of the first integers is n. The average of
all the seven integers is :
(a) n
(b) n + 1
2
(c) k × n, where k is a function of n (d) n
7
30. Which of the following integers has most number of divisors?
(a) 176
(b) 182
(c) 99
(d) 101
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Answers
1. (a)
2. (b)
3. (a)
4. (d)
5. (b)
6. (b)
7. (d)
8. (a)
9. (a)
10. (c)
11. (b)
12. (b)
13. (a)
14. (b)
15. (a)
16. (b)
17. (a)
18. (d)
19. (b)
20. (b)
21. (d)
22. (b)
23. (c)
24. (b)
25. (b)
26. (d)
27. (d)
28. (a)
29. (b)
30. (a)
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Chapter: 2 SIMPLE ARITHMETIC OPERATIONS
It is a common need to simplify the expressions formulated according to the statements of the problems
relating to practical life. To do this, it is essential to follow in sequence the mathematical operations given by
the term “BODMAS”.
BODMAS
Each letter of the word BODMAS stands as follows:
B for Bracket: [{(−)}]
There are four brackets, namely, − bar, ( ), { } and [ ]. They are removed, strictly in the order -, ( ), { } and [ ].
O for Of
:
of
D for Division
:
M for Multiplication :
×
A for Addition
:
+
S for Subtraction
:
−
The order of various operations in exercises involving brackets and fractions must be performed strictly
according to the order of the letters of the word BODMAS.
Note: Here, 5 8 3 3 .
Illustration 1: Simplify:
1 1
1
1
1 5
8 3 4 of 5 11 3 1
2 5
2
3
4 8
Sol:
Given expression
17 16 9 16
5 5
=
of
11 3
2 5 2
3
4 8
17 16 9 16
5
of
11 3
2 5 2
3
8
17 16 9 16
19
of
11
2 5 2
3
8
17 16 9 16 69
of
2 5 2
3
8
17 16 9 16 69
2 5 2 3
8
17 16 24 69
2 5
1
8
17 16 1 69 17 16 69
2 5 24 8 2 120 8
17 16 1035 17 1051
2 120 2
120
1020 1051
31
.
120
120
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Illustration 2: Simplify:
1 1 1
1 1
5 4 3 2
3 3 3
3 3
Sol:
Given expression
16 13 10 7 1
3 3 3 3 3
16 13 10 6 16 13 4
3 3 3 3 3 3 3
16 9 16 9 7
1
=
2 .
3 3 3 3 3
3
USE OF ALGEBRAIC FORMULAE
The following important formulae are sometimes found useful, in dealing with the simplifications:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
2
a b a2 2ab b2
2
a b a2 2ab b2
2
2
a b a b 2 a2 b2
2
2
a b a b 4ab
a 2 b2 a b a b
3
a b a3 3a2b 3ab2 b3 = a3 b3 3ab a b
3
a b a3 3a2b 3ab2 b3 = a3 b3 3ab a b
a3 b3 a b a2 ab b2
a3 b3 a b a2 ab b2
a3 b3 c3 3abc
a b c .
a 2 b2 c2 ab bc ca
a 4 b4 a2 b2 a b a b .
Illustration 3: Simplify the following:
(i) 0.32 × 0.32 + 0.64 × 0.68 + 0.68 × 0.68
Sol:
Given expression
= 0.32 × 0.32 + 2 × 0.32 × 0.68 + 0.68 × 0.68
2
2
= 0.32 20.32 0.68 0.68
2
= 0.32 0.68
2
[ a2 2ab b2 a b ]
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= 12 1 .
(ii) 2.45 × 2.45 – 0.9 × 2.45 + 0.45 × 0.45
Sol:
Given expression
= 2.45 × 2.45 – 2 × 2.45 × 0.45 + 0.45 × 0.45
2
2
= 2.45 2 2.450.45 0.45
= 2.45 0.45
2
2
2
2
[ a 2ab b2 a b ] = 2 4 .
7 {(146 92)2 (146 92)2 }
(iii)
(146)2 (92)2
Sol:
7 2{(146)2 (92)2 }
Given expression =
(146)2 (92)2
2
2
[ a b a b 2 a 2 b2 ] 14.
2
(iv)
2
0.345 0.255 0.345 0.255
0.345 1.02
Sol:
Given expression
2
2
0.345 0.255 0.345 0.255
=
4 0.345 0.255
2
2
4 0.345 0.255
[ a b a b 4ab] = 1.
=
4 0.345 0.255
0.682 0.682 0.318 0.318
(v)
0.682 0.318
Sol:
Given expression
2
2
0.682 0.318
=
0.682 0.318
= 0.682 0.318
a 2 b2
a b
ab
= 1.
2
(vi)
3.29 0.81
2
4
Sol:
2
Given expression =
2
3.29 0.81
3.29 0.81
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a 2 b2
= 3.29 0.81
a b
ab
= 2.48.
3
2
2
3
(vii) 2.35 1.95 2.35 7.05 0.65 0.65
Sol:
Given expression
3
2
3
= 2.35 3 0.65 2.35 2 + 3 × 2.35 × 0.65 0.65
= (2.35 + 0.65)3
3
[ a3 3a 2b 3ab2 b3 a b ]
3
= 3 27 .
3
(viii)
4.32
2
2
3
0.96 4.32 12.96 0.32 0.32
44 4
Sol:
Given expression
3
2
2
3
4.32 3 0.32 4.32 3 4.32 0.32 0.32
=
44 4
3
4.32 0.32 [ a3 3a2b 3ab2 b3 a b 3 ]
=
43
3
4
= 1.
4
885 885 885 115 115 115
(ix)
885 885 115 115 885 115
Sol:
Given expression
3
3
885 115
2
2
885 115 885 115
a 3 b3
= (885 + 115) 2
a b
2
a ab b
= 1000.
0.62 0.62 0.62 0.41 0.41 0.41
(x)
0.62 0.62 0.62 0.41 0.41 0.41
Sol:
Given expression
3
3
0.62 0.41
=
2
2
0.62 0.62 0.41 0.41
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a3 b3
= 0.62 0.41 2
a b
2
a ab b
= 0.21.
3
3
3
2.3 1.5 1.2
(xi)
3 2.3 1.5 1.2
2.3 2.3 1.5 1.5 1.2 1.2 2.3 1.5 2.3 1.2 1.5 1.2
Sol:
Given expression
3
3
3
2.3 1.5 1.2 3 2.3 1.5 1.2
=
2
2
2
2.3 1.5 1.2 2.3 1.5 2.3 1.2 1.5 1.2
= (2.3 + 1.5 + 1.2)
a3 b3 c3 3abc
a b c = 5.
2
2
2
a b c ab ac bc
Samples Questions
4 3
2
9
1. 48 12 of of ?
3 4
3
8
(a) 9
(b) 12
(c) 15
(d) None of these
8
2. 3 (8 5) (4 2) 2 ?
13
(a)
33
71
(b)
(c)
13
17
(d) None of these
3. Evaluate
55
17
0.530.53 2 0.53 0.41 0.410.41
0.53 0.41
(a) 0.16
(b) 0.8
(c) 0.12
(d) None of these
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4. Which of the following fractions is less than
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7
1
and greater than ?
8
3
(a)
1
4
(b)
23
24
(c)
11
12
(d)
17
24
5.
272 32124 176 ?
17 15 15
(a) 0
(b) 2.25
(c) 300
(d) None of these
a 1
3a 2b
, then
is equal to
b 3
3a 2b
6. If
(a) 3
(b) – 3
(c) – 5
(d) −1
7.
20 5 2 16 8 2 10 5 3 2 ?
(a) 9
(b) 12
(c) 15
(d) 18
8. If we multiply a fraction by itself and divide the product by its reciprocal, the fraction thus obtained is
18
(a)
26
. Te fraction is
27
8
27
(c) 1
1
3
(b) 2
2
3
(d) None of these
9. What fraction must be subtracted from the sum of
1
1
1
and to have an average of
of all the three
4
6
12
fractions?
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
6
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10.
In association with
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? 12
2
0.23.6
(a) 17.82
(b) 17.22
(c) 17.28
(d) 17.12
11. If
a
17
ab
, what is
equal to?
a b 23
ab
(a)
11
23
(b)
17
32
(c)
23
11
(d)
23
17
12. If
a 2 b2 ab
ab
, then find the value of
in terms of c and d only
2
2
ab
c d
cd
(a)
cd
cd
(b)
cd
cd
(c)
c d
cd
(d)
cd
c d
2
13. 1.06 0.04 ? 4 1.06 0.04
(a) 1.04
(b) 1.4
(c) 1.5
(d) Cannot be determined
14. If a =
(a)
x
y
ab
and b
then
is equal to
xy
xy
ab
xy
2
x y2
x
(c)
xy
(b)
x2 y 2
xy
2
y
(d)
xy
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1 1 1
of
15. The value of 2 2 2 is
1 1 1
of
2 2 2
(a) 2
2
3
(b) 1
(c) 1
1
3
(d) 3
16. Find the positive integer, which when added to the numerator and denominator of 2/3 will result in a
fraction nearest to 13/15
(a) 6
(b) 5
(c) 4
(d) 3
17.
5
15
of 24 is equal to
of what?
8
7
(a) 15
(c)
(b) 8
7
225
(d) 7
18. More than half of the members of a club are ladies. If
4
7
of the ladies and
of the gents in the club
7
11
attended the meeting, then what is the smallest number of members that the club could have?
(a) 25
(b) 18
(c) 39
(d) 22
1
1
1
1
19. The value of 1
1
x 1
x 2
x 3
1
1 x 4 is
(a) 1
1
x 5
(b)
1
x 5
(c) x
1
x 5
(d)
x 5
x 1
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20. The expression:
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In association with
x 1 x 2 x2 9x 4
x 7 x2 3x 2
simplifies to
(a) (x – 1)
(b) (x – 2)
(c) (x – 7)
(d) 1/(x – 7)
Answers
1. (b)
2. (c)
3. (c)
4. (d)
5. (c)
6. (b)
7. (a)
8. (b)
9. (d)
10. (c)
11. (c)
12. (d)
13. (a)
14. (a)
15. (a)
16. (b)
17. (d)
18. (a)
19. (d)
20. (b)
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Chapter: 3 SQUARE ROOT AND CUBE ROOT
SQUARE
A number multiplied by itself is known as the square of the given number. For example, square of 6 is 6 × 6 =
36.
SQUARE ROOT
Square root of a given number is that number which when multiplied by it is equal to the given number.
For example, square root of 81 is 9 because 92 = 9 × 9 = 81.
The square root of a number is denoted by the symbol √ or , called the radical sign.
Thus,
81 9, 64 8 and so on.
Note that 1 1 .
METHODS OF FINDING THE SQUARE ROOT
I. Prime Factorization Method
1. Find the prime factors of the given number.
2. Group the factors in pairs.
3. Take one number from each pair of factors and then multiply them together.
This product is the square root of the given number.
Illustration 1: Find the square root of:
(i) 4761
(ii) 207025
Sol:
3
(i) 4761 = 23 23 3
4761 23 3 69 .
5 7
7 13 13
(ii) 207025 = 5
207025 5 7 13 455 .
Note: The above method is used when the given number is a perfect square or when every prime factor of
that number is repeated twice.
II. Method of Division
This method is used when the number is large and the factors cannot be easily determined.
The working rule is explained with the help of following example:
Step 1: The digits of a number, whose square root is required, are separated into periods of two beginning
from the right. The last period may be either single digit or pair.
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Step 2: Find a number (here, 4) whose square may be equal to or less than the first period (here, 22).
Step 3: Find out the remainder (here, 6) and bring down the next period (here, 65).
Step 4: Double the quotient (here, 4) and write to the left (here, 8).
Step 5: The divisor of this state will be equal to the above sum (here, 8) with the quotient of this stage (here,
7) suffixed to it (here, 87).
Step 6: Repeat this process (step 4 and step 5) till all the periods get exhausted.
The quotient (here, 476) is equal to the square root of the given number (here, 226576).
Illustration 2: Find the square root of:
(i) 180625
(ii) 1498176
Sol:
(i)
Thus, 180625 425 .
SQUARE ROOT OF A DECIMAL:
If the given number is having decimal, we separate the digits of that number into periods of two to the right
and left starting from the decimal point and then proceed as in the following illustration:
Illustration 3: Find the square root of :
(i) 12.1801
(ii) 127.0129
Sol:
(i)
12.1801 3.49 .
(ii)
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127.0129 11.27
SQUARE ROOT OF A FRACTION:
(a) If the denominator is a perfect Square:
The square root is found by taking the square root of the numerator and denominator separately.
(b) If the denominator is not a perfect square :
The fraction is converted into decimal and then square root is obtained or the denominator is made perfect
square by multiplying and dividing by a suitable and then its square root is obtained.
Illustration 4: Find the square root of:
2704
(i)
49
44
(ii)
25
Sol:
52 52 52
2704
2704
3
(i)
=7 .
7
49
7
49
77
44
44
44
44
6.6332
= 1.3266 (nearly).
25
5
25
55
55
CUBE
Cube of a number is obtained by multiplying the number itself thrice.
For example, 27 is the cube of 3 as 27 = 3 × 3 × 3.
CUBE ROOT
The cube root of a given number is that number which when raised to the third power produces the given
number, that is the cube root of a number x is the number whose cube is x.
The cube root of x is written as 3 x .
For example, cube root of 64 is 4 as 4 × 4 × 4 = 64.
METHODS TO FIND CUBE ROOT
I. Method of Factorization
1. Write the given number as product of prime factors.
2. Take the product of prime numbers, choosing one out of three of each type.
This product given the cube root of the given number.
Illustration 5: Find the cube root of 42875.
Sol:
Resolving 42875 into prime factors, we get
(ii)
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42875= 5
5 5
7 7
7
3 42875 5 7 35 .
II. Short-cut Method to Find Cube Roots of Exact Cubes consisting of up to 6 Digits:
Before we discuss the method to find the cube roots of exact cubes, the following two remarks are very
useful and must be remembered by heart.
1. 13 1; 23 9; 33 27; 43 64; 53 125; 63 216; 73 343; 83 512; 93 729; 103 1000 .
2. If the cube ends in 1, then its cube root ends in 1
If the cube ends in 2, then its cube root ends in 8
If the cube ends in 3, then its cube root ends in 7
If the cube ends in 4, then its cube root ends in 4
If the cube ends in 5, then its cube root ends in 5
If the cube ends in 6, then its cube root end in 6
If the cub e ends in 7, then its cube root ends in 3
If the cube ends in 8, then its cube root ends in 2
If the cube ends in 9, then it cube root ends in 9
If the cube ends in 0, then its cube root ends in 0
Clearly, from above
1 1, 4 4, 5 5, 6 6, 9 9, 0 0 2 8, 3 7 .
The method of finding the cube root of a number up to 6 digits which is actually a cube of some number
consisting of 2 digits is best illustrated with the help of following examples:
Illustration 6: Find the cube roots of the following:
(i) 2744
(ii) 9261
Sol:
(i) Make groups of 3 digits from the right side. 2 744 2 lies between 13 and 23, so left digit is 1.
744 end in 4, so right digit is 4. Thus, cube root of 2744 is 14.
(ii) 9261
9 lies between 23 and 33, so left digit is 2. 261 ends in 1, so right digit is 1.Thus, cube root of 9261 is 21
Sample Questions
1. Find the square root of 104976
(a) 324
(b) 424
(c) 326
(d) None of these
2. If
27
x
1
1 , then x equals
13
169
(a) 1
(b) 3
(c) 5
(d) 7
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1296
?
?
2.25
3.
(a) 6
(b) 7
(c) 8
(d) 9
4.
10 15 ?
(a) 5 6
(b) 6 5
(c) 5
(d) 30
4
3
?
3
4
5.
(a)
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1
2 3
(c) 1
6.
(b) −
(d)
1
2 3
5 3
6
248 52 144 1
(a) 14
(b) 16
(c) 16.6
(d) 18.8
7. If
x
54
then x is equal to:
169 39,
(a) 108
(b) 324
(c) 2916
(d) 4800
8. If 0.03 0.3 a 0.03 0.3 b , value of
a
is
b
(a) 0.009
(b) 0.03
(c) 0.09
(d) None of these
9. Find the cube root of 15.625
(a) 3.5
(b) 2.5
(c) 4.5
(d) 5.5
10. The least number by which 14175 be divided to make it a perfect square is
(a)3
(b)5
(c) 7
(d) 15
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11.
5 3
5 3
In association with
is equal to
(a) 4 15
(b) 4 15
(c)1/2
(d) 1
12.
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24 216
?
96
(a) 2 6
(b) 6 2
(c) 2
(d)
2
6
4
to one place of decimal?
12.1
13. Given that 10 3.16, what is the value of
(a) 0.16
(b) 0.06
(c) 0.6
(d) 0.016
14. A general wishing to draw up his 16160 men in the form of a solid square found that he had 31 men over.
The number of men in the front row is
(a) 127
(b) 123
(c) 137
(d) 129
15. A general wishing to draw up his 5180 men in the form of a solid square found that he had 4 men less.If
he could get four more men and form the solid ' square, the number of men in the front row is
(a) 68
(b) 72
(c) 78
(d) 82
Answers
1. (a)
8. (a)
2. (a)
9. ( b )
3. (d)
10. (C)
4. (a)
1 1 . (b)
5. (a)
12. (c)
6. (b)
13. (c)
7. (b)
14. (a)
15. (b)
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Chapter: 4 POLYNOMIALS
A function p (x) of the form
p x a 0 a1 x a 2x 2 ... a n x n
where, a 0 , a 1 ,a 2 ,...,a n are real numbers, a n 0 and n is a non-negative integer is called a polynomial in x
over reals.
The real number a 0 , a 1 ,...,a n are called the coefficients of the polynomial.
If a0 ,a1 ,a 2 ,....,a n Integers, we call it a polynomial over integers.
If they are rational numbers, we call it a polynomial over rationales.
Illustration 1:
(a) 4x 2 7x 8 is a polynomial over integers.
7
2
8
(b) x3 x2 x 5 is a polynomial over rationales.
4
3
7
2
(c) 4x 3 x 5 is a polynomial over reals.
Monomial
A polynomial having only one term is called a monomial. For example, 7, 2x, 8x3 are monomials.
Binomial
A polynomial having two terms is called a binomial. For example, 2x + 3, 7x2 − 4x, x2 + 8 are binomials.
Trinomial
A polynomial having three terms is called a trinomial. For example, 7x 2 − 3x + 8 is a trinomial.
Degree of a Polynomial
The exponent in the term with the highest power is called the degree of the polynomial.
For example, in the polynomial 8x 6 4x 5 7x3 8x 2 3, the term with the highest power is x6. Hence, the
degree of the polynomial is 6.
A polynomial of degree 1 is called a linear polynomial.
It is of the form ax b, a 0.
A polynomial of degree 2 is called a quadratic polynomial.
It is of the form ax2 bx c, a 0.
Division of a Polynomial by a Polynomial
Let p x andf x be two polynomials and f x 0 . Then, if we can find polynomials q(x) and r(x), such that
P x f x . q x r x ,
where, degree r(x) < degree f(x), then we say that; p(x) divided by f(x), gives q(x) as quotient and r(x) as
remainder.
If the remainder r(x) is zero, we say that divisor f(x) is a factor of p(x) and we have
p x f x .q x .
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Illustration 2: Divide f(x) = 5x3 – 70x2 + 153x − 342 by g(x) = x2 - 10x + 16. Find the quotient and the
remainder.
Sol:
Quotient = 5x – 20 and
Remainder = − 127x – 22.
Illustration 3: Determine if (x – 1) is a factor of p(x) = x3 – 3x2 + 4x + 2.
Since the remainder is not zero, (x – 1) is not a factor of p(x).
SOME BASIC THEOREMS
Factor Theorem
Let p(x) be a polynomial of degree n > 0. If p (a) = 0 for a real number a, then (x – a) is a factor of p(x).
Conversely, if (x – a) is a factor of p(x), then p (a) = 0.
Illustration 4: Use factor theorem to determine if (x – 1) is a factor of x 8 x7 x 6 x5 x 4 x 1.
Sol :
Let p x x 8 x7 x 6 x5 x 4 x 1.
8
7
6
5
4
Then, p 1 1 1 1 1 1 1 1 1 0.
Hence, (x – 1) is not a factor of p(x).
Remainder Theorem
Let p(x) be any polynomial of degree 1 and an any number.
If p(x) is divided by x – a, the remainder is p (a).
Illustration 5: Let p(x) = x2 + 5x 4 −3x + 7 be divided by (x – 1). Find the remainder.
Sol:
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4
Remainder = p 1 1 5 1 3 1 7 10.
SOME Useful Results and Formulae
2
A B A2 B2 2AB
2
2
2. A B A 2 B2 2AB A B 4AB
3. A B A B A2 B2
2
2
4. A B A B 2 A2 B2
2
2
5. A B A B 4AB
3
6. A B A3 B3 3AB A B
3
7. A B A3 B3 3AB A B
2
8. A 2 B2 A B 2AB
9. A 3 B3 A B A2 B2 AB
10. A 3 B3 A B A 2 B2 AB
2
11. A B C A2 B2 C2 2 AB BC CA
12. (A3 B3 C3 3ABC A B C A 2 B2 C2 AB CA BC
1.
13. A B C 0 A3 B3 C3 3ABC.
14. A n Bn is divisible by A B for all values of n.
15. An – Bn is divisible by (A + B) only for even values of n.
16. An + Bn is never divisible by (A − B).
17. An + Bn is divisible by (A + B) only when n is odd.
A Useful Short cut Method
When a polynomial f(x) is divided by x – a and x – b, the respective remainders are A and B. Then, if the same
polynomial is divided by (x – a) (x – b), the remainder will be
A B
Ba ab
x
.
ab
ab
When a polynomial f(x) is divided by (x – 1) and (x – 2), the respective remainders are 15 and
Illustration 6: What is the remainder when it is divided by
(x – 1) (x – 2)?
Sol:
A B
Ba Ab
x
Remainder =
ab
ab
9 1 15 2
15 9
x
12
12
x 21 .
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Sample Questions
1. If (x − 2) is a factor of the polynomial x3 2ax2 ax 1, find the value of a.
(a) 5/6
(b) 7/6
(c) 11/6
(d) None of these.
2. If x + a is a factor of the polynomial x3 + ax 2 −2x + a + 4, find the value of a.
(a) −4/3
(b) +2/3
(c) +4/3
(d) None of these.
3. Find the value of k if f x x3 kx 2 11x 6 and (x – 1) is a factor of f(x).
(a) 6
(b) 4
(c) 8
(d) None of these.
4. If 5x2 − 4x − 1 is divided by x − 1, the remainder is
(a) 0
(b) 2
(c) 1
(d) None of these.
5. Find the values of m and n in the polynomial 2x3 + mx 2 + nx − 14 such that (x − 1) and (x + 2) are its
factors.
(a) m = 4, n = 5
(b) m = 9, n = 3
(c) m = 6, n = 7
(d) None of these.
6. What value should a possess so that x + 1 may be a factor of the polynomial,
f x 2x3 ax 2 2a 3 x 2?
(a) 2
(b) −2
(c) 3
(d) None of these.
7. Divide the polynomial 4y3 − 3y2 + 2y − 4 by y + 2 and find the quotient and remainder.
(a) 4y 2 11y 24, 52
(b) 6y2 – 13y + 36, − 64
(c) 4y2 + 13y – 24, + 52
(d) None of these.
2
2
8. Resolve into factors: 16 x y 9 x y .
(a) x 5y 5x y
(b) x 7y 7x y
(c) x 7y 7x y
(d) None of these
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9. Resolve into factors: 4x 2 12xy 9y 2 8x 12y.
(a) (3x + 2y) (4x + 2y – 3)
(b) (2x + 3y) (2x + 3y – 4)
(c) (2x – 3y) (2x + 3y + 4)
(d) None of these.
10. Resolve into factors:
16x 2 72xy 81y 2 12x 27y.
(a) (6x – 7y) (6x – 7y – 5)
(b) (4x – 9y) (4x – 9y – 3)
(c) (4x + 9y) (4x + 9y + 3)
(d) None of these.
11. Resolve into factors: (a + b)2 – 14c (a + b) + 49 c2.
(a) (a – b – 9z)3
(b) (a + b – 7c)2
(c) (a + b + 9c)2
(d) None of these.
12. Resolve into factors: 81x2 y 2 108xyz 36z2 .
(a) (6xy+9z)2
(b) (9xy–7z) 2
(c) (9xy + 6z) 2
(d) None of these.
2
2
13. Factorise: a b c b c a 2 a b c b c a .
(a) 4a2
(b) 6a2
(c) 8a2
(d) None of these.
14. Resolve into factors:
2
2
9 3x 5y 12 3x 5y 2x 3y 4 2x 3y .
(a) (7x + 9y)2
(b) (5x + 9y)2
(c) (5x – 9y)2
(d) None of these.
2
2
15. Factorise: 2x 3y 2 2x 3y 2x 3y 2x 3y .
(a) 16x2
(b) 18x2
(c) 12x2
(d) None of these.
16. Factorise: 45a3b 5ab3 30a2b2 .
2
(a) 5ab 5a b
2
(c) 5ab 3a b
2
(b) 7ab 5a b
(d) None of these.
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3
3
3
17. Find the factors of a b b c c a .
(a) 3 a b b c c a
(b) 5 a b b c c a
(c) 3 a b b c c a
(d) None of these.
18. Factorise a 2
1
2
3 2a .
2
a
a
1
1
(a) a 1 a 1
a
a
1
1
(b) a 1 a 1
a
a
1
1
(c) a 1 a 1
a
a
1
1
(d) a 1 a 1 .
a
a
1
1
19. If x 2, find the value of x 4 4 .
2
x
(a) 2
(b) 4
(c) 6
(d) 8
x y
x3 y 3
6, find the value of 3 3 .
y x
y
x
(a) 176
(b) 198
(c) 184
(d) None of these.
20. If
ANSWER
1. (b)
8. (c)
2. (a)
9. (b)
3. (a)
10. (b)
4. (a)
11. (b)
5. (b)
12. (c)
6. (c)
13. (a)
15. (a)
16. (c)
17. (c)
18. (d)
19. (a)
20. (b)
7. (a)
14. (b)
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Chapter: 5 QUADRATIC EQUATIONS
THEORY
INTRODUCTION
The general form of a quadratic equation in x is ax2+ bx + c = 0, where a, b, c R & a ≠ 0.
The solution of the quadratic equation, ax²+ bx + c = 0 is given by x =
b b2 4ac
2a
The expression b2 – 4ac = D is called the discriminate of the quadratic equation.
If & are the roots of the quadratic equation ax² + bx + c = 0, then;
(i) + = –b/a
(ii) = c/a (iii) – =
D
.
a
NATURE OF ROOTS:
(1) Consider the quadratic equation ax² + bx + c = 0 where a, b, c R & a ≠ 0 then;
(i) D > 0 roots are real & distinct (unequal).
(ii) D = 0 roots are real & coincident (equal).
(iii)
D < 0 roots are imaginary.
(iv) If p + iq is one root of a quadratic equation, then the other must be the conjugate p – iq & vice versa.
(p, q R & i = 1 ).
(2) Consider the quadratic equation ax2 + bx + c = 0 where a, b, c Q & a ≠ 0 then;
(i) If D > 0 & is a perfect square, then roots are rational & unequal.
(ii) If = p + q is one root in this case, (where p is rational & is a surd) then the other root must be the
conjugate of it i.e. = p – q & vice versa.
(3) A quadratic equation whose roots are a & b is (x – a)(x – b) = 0 i.e. x2– (a + b)x + ab = 0 i.e. x2 –(sum of
roots)x + product of roots = 0.
Remember that a quadratic equation cannot have three different roots & if it has, it becomes an identity.
(4) Consider the quadratic expression, y = ax²+ bx + c, a ≠ 0 & a, b, c R then;
(i) The graph between x, y is always a parabola. If a > 0 then the shape of the parabola is concave upwards &
if a < 0 then the shape of the parabola is concave downwards.
(ii)
x R, y > 0 only if a > 0 & b²– 4ac < 0 (figure 3).
(iii)
x R, y < 0 only if a < 0 & b²– 4ac < 0 (figure 6).
SOLUTION OF QUADRATIC INEQUALITIES:
ax2 + bx + c > 0 (a ≠ 0).
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(i) If D > 0, then the equation ax2+ bx + c = 0 has two different roots x1 < x2.
Then a > 0 x (–∞, x1) (x2, ∞) a < 0 x (x1, x2)
(ii) If D = 0, then roots are equal, i.e. x1 = x2.
In that case a > 0 x (–∞, x1) (x1, ∞), a < 0 x
(iii)
Inequalities of the form
P x
Q x
0 can be quickly solved using the method of intervals.
MAXIMUM & MINIMUM VALUE:
Maximum & minimum value of y = ax²+ bx + c occurs at x = –(b/2a) according as;
4ac b2
4ac b2
a < 0 or a > 0. y
, if a > 0 & y ,
if a < 0 .
4a
4a
COMMON ROOTS OF TWO QUADRATIC EQUATIONS [ONLY ONE COMMON ROOT]:
Let be the common root of ax² + bx + c = 0 & a’x2 + b’x + c’ = 0. Therefore
2
1
a² + b + c = 0; a’² + b’ + c’ = 0. By Cramer’s Rule
bc' b'c a'c ac' ab' a'b
Therefore,
ca' c'a bc' b'c
.
ab' a'b a'c ac'
So the condition for a common root is (ca’ – c’a)² = (ab’ – a’b)(bc’ – b’c).
The condition that a quadratic function f(x , y) = ax² + 2hxy + by² + 2gx + 2fy + c may be resolved into
a h g
two linear factors is that abc + 2fgh – af2 – bg2 – ch2 = 0 or h b f = 0.
g f c
THEORY OF EQUATIONS:
If 1, 2, 3, ......n are the roots of the equation;
f(x) = a0xn + a1xn–1 + a2xn–2 + .... + an–1x + an = 0 where a0, a1, .... an are all real & a0 ≠ 0 then,
1
a1
a
a
n a
, 12 2 , 123 3 ,....., 123 ........ n 1 n
a0
a0
a0
a0
Note:
(i) If is a root of the equation f(x) = 0, then the polynomial f(x) is exactly divisible by (x – ) or (x – ) is a
factor of f(x) and conversely.
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(ii) Every equation of nth degree (n 1) has exactly n roots & if the equation has more than n roots, it is an
identity.
(iii) If the coefficients of the equation f(x) = 0 are all real and + i is its root, then – i is also a root. i.e.
imaginary roots occur in conjugate pairs.
(iv) If the coefficients in the equation are all rational & + is one of its roots, then – is also a root
where, Q & is not a perfect square.
(v) If there be any two real numbers 'a' & 'b' such that f(a) & f(b) are of opposite signs, then f(x) = 0
must have atleast one real root between 'a' and 'b'.
(vi) Every equation f(x) = 0 of degree odd has at least one real root of a sign opposite to that of its last
term.
LOCATION OF ROOTS:
Let f(x) = ax2 + bx + c, where a > 0 & a, b, c R.
(i) Conditions for both the roots of f(x) = 0 to be greater than a specified number‘d’ are b2– 4ac 0; f(d)>0 &
(–b/2a) > d.
(ii) Conditions for both roots of f(x) = 0 to lie on either side of the number‘d’ (in other words the number‘d’
lies between the roots of f(x) = 0 is f(d)< 0.
i. Conditions for exactly one root of f(x) = 0 to lie in the interval (d, e) i.e. d < x < e are b2 – 4ac > 0 &
f(d).f(e)< 0.
(iv) Conditions that both roots of f(x)=0 to be confined between the numbers p & q are (p<q).b2– 4ac 0;
f(p)>0; f(q)> 0 & p<(– b/2a) < q.
LOGARITHMIC INEQUALITIES
(i) For a > 1 the inequality 0 < x < y & loga x < loga y are equivalent.
(ii) For 0 < a < 1 the inequality 0 < x < y & loga x > loga y are equivalent.
(iii) If a > 1 then loga x < p 0 < x < ap
(iv) If a > 1 then logax > p x > ap
(v) If 0 < a < 1 then loga x < p x > ap
(vi) If 0 < a < 1 then logax > p 0 < x < ap
LAGRANGE’S IDENTITY:
If a1, a2, a3, b1, b2, b3 R
2
2
2
2
then, a12 a 22 a 32 b12 b22 b32 a 1b1 a 2b2 a3b3 a1b2 a 2b1 a2b3 a3b2 a3b1 a1b3
Equations which can be reduced to linear, Quadratic and Biquadratic equations:
Type I: An equation of the form (x – a)(x – b)(x – c)(x – d) = A,
where a < b < c < d, b – a = d – c, can be solved by change of variable.
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i.e., y =
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x a x b x c x d
= y x
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a b c d
4
4
Type II: An equation of the form (x – a)(x – b)(x – c)(x – d) = Ax2,
where ab = cd, can be reduced to a collection of two quadratic equations by a change of variable y = x +
ab
.
x
Type III: An equation of the form (x – a)4 +(x – b)4 = A can also be solved by a change of variable, i.e.,
Making a substitution y =
x a x b
2
IMPORTANT RESULT
(1) For the quadratic equation ax2 + bx + c = 0.
(i) One root will be reciprocal of the other if a = c
(ii) One root is zero if c = 0
(iii) Roots are equal in magnitude but opposite in sign is b = 0
(iv) Both roots are zero if b = c = 0.
(v) Roots are positive if a and c are of same sign & b is of opposite sign.
(iv) Root are of opposite sign if a and c are of opposite sign.
(vii) Roots are negative if a, b, c are of the same sign.
(2) Let f(x) =a x2 + bx + c, where a > 0. Then
(i) Conditions for both the roots of f(x) = 0 to be greater than a given number k are b2 – 4ac
0; f(k) = 0;
b
2a
> k.
(ii) Conditions for both the roots of f(x) = 0 to be less than a given number k are b2 – 4ac
0; f(k) = 0;
b
<
2a
k.
(iii) The number k lies between the roots of f(x) = 0, if b2 – 4ac > 0; f(k)< 0.
(iv) Conditions for exactly one root of f(x) = 0 to lie between k1 and k2 is f(k1) f(k2) < 0, b2 – 4ac 0.
(v) Conditions for both the roots of f(x) = 0 are confined between k1 and k2 is f(k1) > 0, f(k2) > 0, b2 – 4ac 0
b
< k2 < k2
and k1 <
2a
(vi) Conditions for both the numbers k1 and k2 lie between the roots of f(x)= 0 is b2 – 4ac >0; f(k1) < 0; f(k2) <
0
Sample Questions
1. The roots of the equation
3a 2 x 2 − abx − 2b 2 = 0 are
b 2b
(a) ,
a 3a
b 2b
,
(c)
a 3a
(b)
b 2b
,
a 3a
(d) None of these
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2. The roots of the equation: a 2 x 2 − 3abx + 2b 2 = 0 are
2b b
2b b
,
,
(a)
(b)
a a
a a
2b b
,
(c)
(d) None of these
a a
3. Construct a quadratic equation whose roots are 2 and 2 2
(a) x2 3 2 x 4 0
(b) x2 3 2 x 4 0
(c) x2 3 2 x 4 0
(d) x2 3 2 x 4 0
4. The roots of the equation
ax 2 + (40 2 − 3b)x − 12ab = 0 are
3b
3b
(b) 4a,
(a) 4a,
a
a
3b
3b
(c) 4a,
(d ) 4a,
a
a
5. Construct a quadratic equation whose roots have the sum = 6 and product = −16.
(a) x 2 − 6x −16 = 0
(b) x 2 + 6x − 16 = 0
(c) x 2 − 3 x 6 0
(d) None of these
x2
6. The roots of the equation a + bx + c = 0 will be reciprocal if
(a) a = b
(b) b = c
(c) c = a
(d) None of these
7. Form a quadratic equation whose one root is 3 5 and the sum of roots is 6
(b) x2 + 6x + 4 = 0
(a) x2 − 6x + 4 = 0
(c) x2 − 6x − 4 = 0
(d) None of these
8. The value of k for which the root , of the equation: x2 − 6x + k = 0 satisfy the relation
3 2 20, is
(a) 8
(b) − 8
(c) 16
(d) −16
9. Find two consecutive positive odd integers whose squares have the sum 290.
(a) 11, 13
(b) 13, 15
(c) 9, 11
(d) None of these
x2
10.If , are the roots of the equation + kx + 12 = 0 such that 1 , the value of k is
(a) 0
(b) ± 5
(c) ± 1
(d)±7
2
1 3
1
11.
The value of x in the equation: x x 4 is
x 2
x
1
(a) −2
(b)
2
(c) −1
(d) 0
12. If , are the roots of the quadratic equation x2 – 8x + k = 0, find the value of k such that 2 2 40
(a) 12
(b) 14
(c) 10
(d) 16
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13. Find the value of k so that the sum of the roots of the equation 3x2 + (2x + 1)x − k − 5 = 0 is equal to the
product of the roots
(a) 4
(b) 6
(c) 2
(d) 8
14. Let p and q be the roots of the quadratic equation x2 2 x 1 0 What is the minimum
possible value of p2 + q2?
(a) 0
(c) 4
(b) 3
(d) 5
15. If both a and b belong to the set {1, 2, 3, 4} then the number of equations of the form ax2 + bx + 1 = 0
having teal roots is
(a) 10
(b) 7
(c) 6
(d) 12
Answers
1. (a) 2. (b)
8. (a) 9. (a)
15. (b)
3. (b)
10. (d)
4. (c)
11. (c)
5. (a)
12. (a)
6. (c)
13. (a)
7. (a)
14. (d)
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Chapter:6 MENSURATION-I (Area & Perimeter)
INTRODUCTION
In this chapter, we shall be dealing with plane figures of various shapes finding their sides, perimeters and
areas.
AREA
The area of any figure is the amount of surface enclosed within its bounding lines. Area is always expressed in
square units.
UNITS OF MEASURING AREA
100 sq Millimeters = 1sq centimeter
100 sq centimeter = 1 sq decimeter
100 sq decimeters = 1sq meter
100 sq meters = 1sq decameter or arc
10,000 sq meters =1 hectare
1,000,000 sq meters = 100 hectares = 1 sq kilometer
Perimeter
The perimeter of a geometrical figure is the total length of the sides enclosing the figure.
SOME BASIC FORMULA
A triangle is a closed figure bounded by three sides. ABC is a triangle.
The sides AB, BC and AC are respectively denoted by c, a and b.
Area of a triangle (A)
1
1
(a) A (base height) ah
(b) A s(s a)(s b)(s c),
2
2
where formula is known as Hero’s formula. Perimeter (P) = a +b +c =2s.
2. Right Angled Triangle
A triangle having one of its angles equal to 90° is called a right-angled triangle. The side opposite to the
right angle is called the hypotenuse.
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In a right angled triangle,
(Hypotenuse)2 = sum of squares of sides
h2 a2 b2 .
i.e.
1
Area (a)= (product of the sides containing the right angle)
2
1
i.e. A ab.
2
3. Equilateral Triangle
A triangle whose all sides are equal is called an equilateral triangle.
Area (A)of an equilateral triangle
3
3 2
(side)2
a
4
4
Perimeter (P)of an equilateral triangle
3 (side) 3a
Altitude (h) of an equilateral triangle
3
3
(side)
.
2
4
In an equilateral triangle
∠A= ∠B =∠C =600
Area (a)of an equilateral triangle
(altitude)2 h2
.
3
3
Illustration 1 Length of the side of an equilateral triangle is
Sol: Height of the equilateral triangle
4
cm. . Find its height.
3
;
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(side)
2
3 4
2 cm
2
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Illustration 2: Height of an equilateral triangle is 4 3 cm. Find its area.
Sol: Area of equilateral triangle
2
altitude
3
4 3 4 3
3
16 3 sq cm.
4. Isosceles Triangle
A triangle whose two sides are equal is an isosceles triangle.
Area (a) of an isosceles triangle
b
4a 2 b2
4
Perimeter (P)of an isosceles triangle (2a b)
Height (h) of an isosceles triangle
1
4a2 b2 .
2
5. Isosceles Right-angled Triangle
An isosceles right-angled triangle has two sides equal with equal sides making 90° to each other.
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Hypotenuse (h)= 2a
1
Area (a) a 2
2
Perimeter (P) 2a 2a 2a( 2 1) h( 2 1)
Pb
If the perimeter of an isosceles triangle is P and the base is b, then the length of the equal sides is
.
2
If the perimeter of an isosceles triangle is P and the length of the sides is a, then base is (P−2a).
Illustration 3: If the base of an isosceles triangle is 10 cm and the length of equal sides is 13 cm, find its
area.
Sol: Area of the isosceles triangle
b
4a 2 b2
4
10
4 (13)2 (10)2
4
10
10
676 100 24 = 60 sq cm
4
4
6. Quadrilateral
A closed figure bounded by four sides is called a quadrilateral.
It has four angles included in it.
The sum of these four angles is 360°.
i.e. A + B + C + D = 360°.
1
Area (a) of a quadrilateral ×one diagonal × (sum of perpendiculars to it from opposite vertices)
2
1
d(p1 p2 )
2
Note: If the length of four sides and one of its diagonals are known, then
A=Area of ∆ADC+ Area of ∆ABC. The special cases of quadrilateral are parallelogram, rectangle, square,
rhombus, trapezium, etc., which are discussed below separately.
7. Parallelogram
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A quadrilateral in which opposite sides are equal and parallel is called a parallelogram.
The diagonals of a parallelogram bisect each other.
Area (a)of a parallelogram = base × altitude corresponding to the base = b × h
Area (a)of parallelogram
2 s(s a(s b)(s c)
where a and b are adjacent sides, d is the length of the diagonal connecting the ends of the two
abd
.
sides and s =
2
In a parallelogram, the sum of the squares of the diagonals = 2
(the sum of the squares of the two adjacent sides),
i.e. d12 d22 2(a 2 b2 )
Perimeter (P)of a parallelogram
= 2 (a+b),
where a and b are adjacent sides of the parallelogram.
Illustration 4: In a parallelogram, the lengths of adjacent sides are 11 cm and 13 cm, respectively. If the length of
one diagonal is 20 cm, find the length of the other diagonal.
Sol: We have,
d12 d22 2(a 2 b2 )
(20)2 d22 2(112 132 )
d12 2(121 169) 400 180
d2 180 = 13.4 m (approx.)
Illustration 5: Find the area of a quadrilateral of whose diagonal is 38 cm long and the lengths of
perpendiculars from the other two vertices are 31 cm and 19cm, respectively.
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Sol: Area of the quadrilateral
1
d(p1 p2 )
2
1
38 (31 19)
2
19 50 950 sq cm
Illustration 6: Find the area of a parallelogram whose two adjacent sides are 130 m and 140 m and one of the
diagonals is 150m long.
Sol: Here, a = 130, b = 140 and d = 150.
s
a b d 130 140 150 420
210
2
2
2
Area of parallelogram
2 s(s a)(s b)(s a)
2 210(210 130(210 140)(210 150)
2 210 80 70 60 =2×8400 sq m.
8. Rectangle
A rectangle is a quadrilateral with opposite sides equal and all the four angles equal to 90°.
The diagonals of a rectangle bisect each other and are equal.
(a)Area of rectangle = length × breadth = l × b
OR Area of rectangle (l d2 l2 ), if one sides (l) and diagonal (d)are given.
p2 d2
OR Area of rectangle , if perimeter (P) and diagonal (d) are given.
8 2
(b) Perimeter (P) of rectangle= 2(length + breadth) = 2 (l + b).
OR Perimeter of rectangle 1(l d2 l2 , if one sides (l) and diagonal (d) are given.
(c) Diagonal of a rectangle
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(length)2 (breadth)2 l2 b2
P2
P
(d)If Area (a) and perimeter (P) of a rectangle are given, then length of the rectangle
A and
16
4
P
P2
breadth of the rectangle
A
4
16
Illustration7: Calculate the area of a rectangular field whose one side is 16cm and the diagonal is 20cm.
Sol: Area of the rectangular field
(l d2 l2 ) (16 202 162 ) 16 12 = 192sq cm.
Illustration 8: A rectangular carpet has an area of 120 sq m and perimeter of 46 m. Find the length of its
diagonal.
Sol: We have,
p2 d2
Area of rectangle
8 2
462 d2
120
462 4d2 120 8 4d2 2116 960 1156
8
2
d 289 = 17m
Illustration 9: The perimeter of a rectangle is 82 cm and its area is 400 sq m. Find the length and breadth of
the rectangle.
p2
P
Sol: Length of the rectangle
A
16
4
(82)2
82
400 (4.5 20.5) 25 m.
16
4
P
p2
Breadth of the rectangle
A
4
16
82
(82)2
400 (20.5 4.5) 16 m.
4
16
9. Square
A square is a quadrilateral with all sides equal and all the four angles equal to 90°.
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The diagonals of a square are equal and bisect each other at 90°.
(a) Area (a)of a square
= a2, i.e. (side)2
d2
(diagonal)2
P2
(perimeter)2
,i.e.
,i.e.
2
2
16
16
(b)Perimeter (P) of a square
= 4a, i.e. 4 × side
16 area 2 2a,i.e. 2 diagonal
(c)Length (d) of the diagonal of a square
2a,i.e. 2 side
P
Perimeter
16 area
,i.e.
.
2 2
2 2
Illustration 10: If the area of a square field be 6050 sq m, find the length of its diagonal.
Sol: Length of the diagonal of the square field= 2 area 2 6050 12100 , i.e. 110m.
Illustration 11: Find the area of a square with perimeter 48m.
Sol: Area of the square
Perimeter
16
2
(48)2 48 48
3 48 144 sq m
16
16
Illustration 12: Find the diagonal of a square field whose side is of 6 m length.
Sol: Length of the diagonal
2 side 6 2 m.
Illustration 13: In order to fence a square Ramesh fixed 36 poles. If the distance between two poles is
6 metres, then find the area of the square so formed.
Sol: Perimeter of the square = 36×6 = 216m.
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.'. Area of the square
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2
2
216
54 54 = 2916 sq m.
4
Illustration 14: Perimeter of a square field is 16 2 cm. Find the length of its diagonal.
Sol: We have,
Perimeter of square field = 2 2 × diagonal
16 2 2 2 diagonal
Length of the diagonal
16 2
= 8 cm.
2 2
10. Rhombus
A. rhombus is a quadrilateral whose all sides are equal.
The diagonals of a rhombus bisect each other at 90°
(a)Area. (a) Of a rhombus
= a × h, i.e. base × height
1
1
d1 d2 ,i.e. Product of its diagonals
2
2
2
d
d1 a2 1
2
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2
2
2 d1 2
Perimeter d1
since d 4 a d1
,
4
2
2
2
2
Perimeter 2 d1 2
since d 2 4
2
4
2
(b) Perimeter (P)of a rhombus
= 4a, i.e. 4 × side
2 d21 d22 ,
Where d1 and d2 are two-diagonals.
(c)Side (a) of a rhombus
1
d12 d22 .
2
Illustration 16: Find the area of a rhombus whose one side is 13 cm and one diagonal is 24 cm.
2
d
Sol: Area of rhombus d1 a 1
2
2
24
24 (13)
2
2
2
24 169 144 = 120 sq cm.
Illustration 17: If the perimeter of a rhombus is 73 cm and one of its diagonals is 27.5 cm, find the other
diagonal and the area of the rhombus.
Sol: One side of rhombus (a)
73
18.25 cm
4
d
Other diagonal (d2 ) 2 a 1
2
2
2
2
27.5
2 (18.25)2
24 cm
2
1
1
Area of rhombus d1 d2 24 27.5 = 330 sq cm.
2
2
Illustration 18: In a rhombus, the lengths of two diagonals are 18m and 24m. Find its perimeter.
Sol: Perimeter of the rhombus
2 d12 d22
2 (18)2 (24)2 2 324 576 2 900 60 m.
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Illustration 19: Find the side of a rhombus, one of whose diagonals measure 4 m and the other 3 m.
Sol: Side of the rhombus
1
d12 d22 .
2
1
1
(4)2 (3)2 25 ,i.e. 12.5 m.
2
2
11. Trapezium (Trapezoid)
A trapezium is a quadrilateral whose any two opposite sides are parallel.
Distance between parallel sides of a trapezium is called its height.
(a)Area (a)of a trapezium
1
(sum of parallel sides) × perpendicular
2
Distance between the parallel sides
1
(a b) h
i.e.
2
ab
s(s l)(s c)(s d),
l
Where, l = b − a if b > a = a − b if a > b
And s
cdl
2
(b)Height (h) of the trapezium
2
2A
s(s l)(s c)(s d)
l
ab
Illustration 20: The parallel sides of a trapezium are 24 m and 52 m. If its other two sides are 26 m and 30
m, what is the area of the trapezium?
Sol: Area of trapezium
ab
s(s l)(s c)(s d)
l
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Here, a = 24, b =52, c = 26, d =30, l = b−a =28
c d l 26 30 28
s
42
2
2
.'. Area of trapezium
24 52
76
42(42 28)(42 26)(42 30) 42 14 16 12
28
28
76 336
912 sq m
28
Illustration 21: The two parallel sides of a trapezium of area ISO sq cm measure 28cm and 12cm. What is the
height of the trapezium?
Sol: Height of the trapezium
2A 2 180 360
9 cm
a b 28 12 40
12. Walls of a Room
Area of four walls of a room
= 2(length +breadth × height
12. Circle
A circle is the path travelled by a point which moves in such a way that its distance from a fixed point remains
constant.
The fixed point is known as centre and the fixed distance is called the radius.
(a)Circumference or perimeter of circle 2r d,
where r is radius and d is diameter of circle
(b)Area of circle
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r2 ,r is radius
d2
,d is diameter
2
c2
,c is circumference
4
1
× circumference × radius
2
Area
Perimeter or circumference
2
(d) Ratio of the area of the two circles is
(c)Radius of circle
Area of circle circumscribing the square 2
=
Area of circle inscribed in the square
1
(e) Ratio of the area of the two squares is
Area of square circumscribing the circle 2
=
1
Area of square inscribed in the circle
Illustration 22: Find the length of a rope by which a buffalo must be tethered in order that she may be able
to graze an area of 9856 sq m.
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Sol: The required length of rope
r
Area
9856 7
3136 56m
22
Sector
A sector is a figure enclosed by two radii and an arc lying between them.
For sector AOB,
2r
, where r =radius and ∠AOB =θ
Arc AB
360
1
Area of sectors ACBO (arc AB) radius
2
(radius)2
.
360
Semi-Circle
A semi-circle is a figure enclosed by a diameter and the part of the circumference cut off by it.
Segment
A segment of a circle of a figure enclosed by a chord and an arc which it cuts of.
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Note: Any chord of a circle which is not a diameter (such as AB), divides the circle into two segments, one
greater and one less than a semi-circle.
Area of segment ACB
= area of sector ACBO − area of ∆OAB
And area of segment ADB
= area of circle −area of segment ACB
Illustration 23: Find the area of sector of a circle whose radius is 14 cm and the angle at the centre is 60°.
Sol:
2
Area of sector
radius
360
22 14 14 60 22 2 14
7 360
6
2
102 sq cm.
3
Polygon
A polygon is a plane figure enclosed by four or more straight lines.
Regular Polygon
If all the sides of a polygon are equal, it is called a regular polygon
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All the interior angles of a regular polygon are equal
For a regular polygon:
Sum of exterior angles = 2π
Sum of exterior angles = (n−2)π
n(n 3)
No. of diagonals in a polygon
2
Perimeter (P)= n×a,
where n=number of sides
and , a = length of each side
n 2
Each exterior angle
n
2
Each exterior angle
n
1
1
Area P r n a r,
2
2
where, r is radius of the circle drawn inside the polygon touching its sides.
2
1
a
n a R2 ,
2
2
Where, R is radius of the circle drawn outside the polygon touching its sides.
Area of a regular hexagon
na2
cot .
4
n
3 3
(side)2
2
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Area of a regular octagon 2( 2 1)(side)2 .
Illustration 24: Find the area of a regular octagon whose side measures
2 cm.
Sol: Area of regular octagon 2( 2 1) (side)2 2( 2 1) ( 2)2 4( 2 1) sq cm.
Illustration 25: Find the sum of interior angles of a regular polygon of 10 sides. Also, find the value of each
interior angle.
Sol: Sum of interior angles = (n − 2) ×π
= (10−2) ×π = 8π.
4
n2
10 2
Also, value of each interior angle
.
5
n
10
Illustration 25: Find the sum of all the exterior angles of a regular polygon of 12 sides. Also, find the value of
each exterior angle.
Sol: Sum of exterior angles =2π
Also, value of each exterior angle
2 2
.
n 12 6
Cyclic Quadrilateral
A quadrilateral whose vertices lie on the circumference of the circle is called a cyclic quadrilateral.
For a cyclic quadrilateral
• Area s(s a)(s b)(s c)(s d), where
a bc d
2
• ∠A + ∠B + ∠C + ∠D = 2π
• ∠A + ∠C=∠B + ∠ D=π
s
SOME USEFUL SHORT-CUT MET HODS
1. If the length and the breadth of a rectangle are increased by x% and y% respectively, then the
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xy
Area of rectangle will increase by x y
%.
100
Explana tion
Area of the original rectangle is
A = l × b.
Area of the new rectangle is
100 x 100 y
A' l
b
100 100
100 x 100 y
100 x 100 y
lb
A
100 100
100 100
∴
A'
100 x 100 y
1
1
A
100 100
or
A' A (100 x)(100 y) (100)2
A
(100)2
(100)2 100(x y) xy (100)2
A' A
A 100
(100)
∴ % increase in area =
100(x y) xy
%
100
or,
xy
x y 100 %.
Note: If any of x or y decreases, we put negative sign.
Illustration 1: The length and breadth of a rectangle are increased by 20% and 5%, respectively. Find the
percentage increase in its area.
Sol: % increase in the area of rectangle
xy
x y
%
100
20 5
20 5
% 26%
100
2. If the length of a rectangle is increased by x%, then its breadth will have to be decreased
100x
by
% in order to maintain the some area of rectangle.
100 x
Explanation
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% increase in area of rectangle
xy
x y
%
100
xy
or 0 x y
100
x
100x
or x y 1
or y
100
100 x
− ve sign indicates decrease.
Therefore, breadth must be decreased by
100x
100 x % in order to maintain the same area.
Illustration 2: The length of a rectangle is increased by 25%. By what per cent should its breadth be
decreased so as to maintain the same area?
Sol: The breadth must be decreased by
100x
100 25
%
%, i.e. 20% .
100 x
100 25
3. If the defining dimensions or sides of any two-dimensional figure (triangle, rectangle, square, circle
x
quadrilateral, pentagon, hexagon, etc is changed by x %, its area changes by x 2
%
100
Illustration 3: If the radius of a circle is decreased by 10%, what is the percentage decrease in its area?
Sol: Here, x = −10 (− ve sign indicates decrease)
x
∴ % change in area x 2
%
100
10
10 2
%
100
19
( 10) % 19%.
10
∴ Area of the circle decreases by 19%.
4. If all the sides of a quadrilateral are increased (or decreased) by x%, its diagonals also increase (or
decrease) by x %
Illustration 4: The length and the two diagonals of a rectangle are decreased by 5% each. What is the percentage
decrease in its breadth?
Sol: Since the length and the two diagonals decrease by 5% each, the breadth also must decrease by 5%.
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5. If each of the defining dimensions or sides of any to – dimensional figure are increased (or
decreased) by x%, its
perimeter also increases (or decreased) by x%.
Illustration 5: If all the sides and diagonals of a square are increased by 8% each, then find the percentage
increase in its perimeter.
Sol: The perimeter also increases by 8%.
6. If the ratio of the areas of two squares be a : b, then the ratio of their sides , ratio of their
perimeters and the ratio of their diagonals , each will be in the ratio
a: b
Illustration 6: Ratio of the areas of two squares is 16:9. Find the ratio of their diagonals.
Sol: The ratio of their diagonals =
a : b 16 : 9, i.e. 4:3.
7. If the diagonal of a square increases by x times, then the area of the square becomes x2 times.
Illustration 7: The diagonal of a square is doubled. How many times will the area of the new square
become?
Sol: The area of the new square will become x2 times, i.e. (2)2 = 4 times.
9. Carpeting the Floor of a Room
If the length and breadth of a room are l and b, respectively, and a carpet of width w is used to cover
lb
.
the floor, then the required length of the carpet
w
Illustration 8: How many meters of a carpet 12 cm wide will be required to cover the floor of a room which
is 600 cm long and 420 cm broad? Also, calculate the amount required in carpeting the floor if the cost of
carpet is Rs. 15 per meter.
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Sol: Length of the carpet
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lb
w
600 420
= 21000cm, i.e. 210m.
12
The amount required for carpeting the floor = 15 × 210 = Rs. 3150.
10. Number of Square Tiles Required for Flooring If the length and breadth of a room are l and b
lb
respectively, then the least number of square tiles required to cover the floor
H.C.F(l,b)
Also, the size of the largest tile so that the tiles exactly fit = H.C.F. (l, b).
Illustration 9: A hall of length 24 cm and breadth 20 m is to be paved with equal square tiles. What will be the
size of the largest tile so that the tiles exactly fit and also find the number of tiles required?
Sol: Size of the largest possible square tile = H.C.F. (l, b) = H.C.F. (24, 20) = 4 m.
lb
24 20
120 tiles.
Number of tiles required
H.C.F.(l,b)
4
11. Path around a Rectangular Space
(a)A rectangular garden l m long and b m broad is surrounded by a path w m wide. The area of the path is
given by.
= 2w(l + b + 2w) sq m.
Explanation
Area of part I = Area of part II = (l + 2w)w sq m
Area of part III = Area of part IV = b w sq m.
∴ Total area of the path = 2[(l + 2w)w + bw] = 2w(l + b + 2w) sq m.
(b)A rectangular garden l m long and b m broad is surrounded by a path w m wide constructed inside it
along its boundary. The area of the path is given by
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Explanation
Area of part I = Area of part II
= lw sq m
Area of part III = Area of part IV = (b −2w) w sq m
∴ Total area of the part
= 2[l w + (b −2w) w] = 2w (l + b −2w) sq m.
(c)A rectangular park is l m long and b m board. Two paths w m wide each are perpendicular to each other
inside the park. The area of the path = w (l+b−w) sq m
Also, area of the park minus the paths
=(l−w)(b−w) sq m.
Explanation
Total area of the path
= Area of part I + Area of path II − Area of common central part
= l w + b w −w2 = w (l + b −w) sq m.
.'. Area of the park minus the parts = [l b – w (l + b− w)] = lb – l w −w(b −w)
= l(b− w) − w(b − w) = (l− w) (b −w) sq m.
Notes:
1. Clearly, from the figure, the area of the path does not change on shifting their location as long as they are
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perpendicular to each other.
2. For a square park, take l = b in all the results derived above.
Illustration 10: A rectangular park 18m × 12m, is surrounded by a path 4 m wide. Find the area of the path.
Sol: The 'area of the path
= 2w (l + b + 2w)
= 2 × 4(18 + 12 + 2 × 4) = 304 sq m.
Illustration 11: A park is square in shape with side 18m. Find the area of the pavement 3 m wide to be
laid all around it on its inside.
Sol: Area of the pavement
= 2w (l + b −2w)
= 2 × 3(18 + 18−2×3) (Here, l = b = 18)
= 180 sq m.
Illustration 12: A playground measures 27 m × 13m. From the centre of each side a path 2 m wide goes
across to the centre of the opposite side. Calculate the area of the path and the cost of constructing it at Rs. 4
per sq m.
Sol: Area of the path
= w (l + b − w) = 2(27 +13−2) = 76 sq m.
∴ Cost = 4 × 76 = Rs. 304.
12. Square Room surrounded by a Verandah
(a) A square room of side a is surrounded by a verandah of width w on the outside of the square room. If the
area of the verandah is A, then the area of the room is given by
Explanation
Area of the room = a 2.
Area of the (room + verandah) = (a + 2w)2.
.'. Area (a)of the verandah = (a + 2w)2 −a2
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= (4aw + 4w2)
or
a
A 4w 2
4w
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∴ Area of the room = a 2
A 4w 2
.
4w
(b)A square room of side a is surrounded by a verandah of width w on its inside. If the area of the verandah is
A, then the area of the given room by
Explanation
Area (a) of the verandah = a 2 −(a − 2w) 2
= 4aw − 4w 2 = 4w (a — w)
or,
A
A 4w 2
a
w
4
4w
A 4w 2
∴ Area or the room = a
4w
Illustration 13: A square field is surrounded by a path 2 m wide on its outside. The area of the path is 72 sq
m. What is the area of the field?
2
Sol: Area of the field
2
2
A 4w 2
72 4 22
=49 sq m.
4w
42
Illustration 14: A square room has a verandah of area 24 sq m and width 1 m all round it on its inside.
Find the area of the room.
Sol: Area of the room
2
A 4w 2
4w
2
24 4 12
= 49 sq m
4
1
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13. (a)A circular ground of radius r has a pathway of width w around it on outside. The area of
circular pathway is given by w(2r w).
Explanation
Area of circular ground = π r2
Area of circular ground + pathway
= π(r + w)2 = πr 2 + 2πrw + πw 2.
∴ Area of circular path way = (πr 2 + 2πrw + πw 2 ) −πr 2 = πw(2r + w).
(b)A circular ground of radius r has a pathway of width w around it on inside. The area of circular
pathway is given by w(2r w).
Explanation
Area of circular ground =πr2
Area of circular ground − pathway = π(r - w)2
= πr2 −2πrw + πw2.
∴ Area of circular pathway
= πr 2 −(πr 2 -2πrw + πw 2 )
= π w (2r −w).
Illustration 15: A circular ground of radius 16m has a path of width 2.8 m around it on its outside. Find the
area of the path.
Sol: The area of the circular path = π w (2r + w)
=
22
×2.8 ×(2 ×16 + 2.8)
7
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= 8.8 × (32 + 2.8) = 306.2 sq m.
Illustration 16: A circular park of radius 22 m has a path of width 1.4 m around it on its inside. Find the area
of the path.
Sol: The area of the circular path
= π w (2r + w) =
22
×1.4× (2 ×22− 1.4)
7
= 4.4 × (44 − 1.4) = 187.45 sq m.
14. If the area of a square is a sq cm, then the area of the circle formed by the same perimeter is
4a
sq cm.
Explanation
Area of the square = a.
.'. Side of the square = Area a.
.'. Perimeter of the square = 4 a .
Given: Circumference of the circle = Perimeter of the square
2r 4 a
∴ Radius of circle (r) =
4 a 2 a
.
2
2
2 a 4a
sq cm
∴ Area of circle = π r1 =
Illustration 17: If the area of a square is 33 sq cm, then find the area of the circle formed by the same
perimeter.
Sol: Required area of the circle
4a 4 33 4 33 7
42 sq cm
22
a 2
15. The area of largest the circle that-can be inscribed in a square of side a is
.
4
Explanation
Clearly, from the figure, the diameter of the inscribed circle equals the side of the square i.e. D = a.
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Area of the circle =
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D2
4
∴ Area of the inscribed circle =
a2
.
4
Illustration 18: Find the area of largest circle inscribed in a square of side 112 cm.
a 2
Sol: The area of the largest circle =
4
22 112 112
9856 sq cm
74
16. Area of a square inscribed in a circle of radius
r is 2r2 and the side of the square is 2r.
Explanation
Clearly, from the figure, diagonal of the inscribed square is equal to the diameter of the circle, i.e. 2r.
1
1
∴ Area of square = (diagonal)2 (2r2 ) 2r2 .
2
2
Also, side of the square = Area 2r2 2 r.
Illustration 19: Find the side of the square inscribed in a circle whose circumference is 308 cm.
Sol: Circumference of the circle (2πr) = 308
308 308 7
r
49 cm
2
2 22
∴ Side of the inscribed square = 2r 49 2 cm
17. The area of largest triangle inscribed m a semi-circle of radius r is r2.
Explanation
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Clearly, from the figure, the largest triangle inscribed in a semi-circle is an isosceles triangle with diameter as
its base and radius as its height.
Area of the triangle =
1
× base × height
2
1
= ×2r×r=r2.
2
18. The number of revolutions made by a circular wheel of radius r in travelling distance d is given
d
by
.
2r
Explanation
Circumference of the wheel = 2π r
In travelling a distance 2πr, the wheel makes 1 revolution.
d
∴ In travelling a distance d, the wheel makes
revolutions.
2r
Illustration 20: The diameter of a wheel is 2 cm. If it rolls forward covering 10 revolutions, find the distance
travelled by it.
Sol: Radius of the wheel = 1 cm.
The distance travelled by the wheel in 10 revolutions = 10 ×2π
22
10 2 1 62.8 cm.
7
Sample Questions
1. Three sides of a triangular field are 20 m, 21 m and 29 m long, respectively. The area of the field is
(a) 215 sqm
(b) 230 sqm
(c) 210 sqm
(d) None of these
2. The hypotenuse and the semi-perimeter of right triangle are 20cm and 24cm, respectively. The other
two sides of the triangle are
(a) 16cm, 12cm
(b) 12cm, 16cm
(c) 20 cm, 16 cmn
(d) None of these
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3. The sides of a triangle are in the ratio 3 : 4 : 5 . If its perimeter is 36 cm, area of the triangle is
(a) 57 sqm
(b) 54 sqm
(c) 56.5 sq m
(d) None of these
4. For a triangle whose sides are 50m, 78m and 112m, respectively, the length of perpendicular from
the opposite angle on the side 112m is
(a) 45m
(b) 35m
(c) 30 m
(d) None of these
5. A ladder is resting with one end in contact with the top of a wall of height 12m and the other end on
the ground is at a distance 5 m from the wall. The length of the ladder is
(a) 13m
(b) 17m
(c) 16m
(d) None of these
6. A ladder is placed so as to reach a window 63 cm high. The ladder is then turned over to the opposite side
of the street and is found to reach a point 56 cm high. If the ladder is 65 cm long, the width of the street is
(a) 59cm
(b) 39cm
(c) 49 cm
(d) None of these
7. Find the area of an isosceles right-angled triangle whose hypotenuse is 8 cm.
(a) 32 sq m
(b) 24 sq cm
(c) 16sqcm
(d) None of these
8. The perimeter of an isosceles triangle is equal to 14 cm; the lateral side is to the base in the ratio
5 to 4. The area of the triangle is
(a) 3 21 sqcm
(b) 2 21 sqcm
(c) 4 21 sq cm
(d) None of these
9. If all the sides of a triangle are increased by 200%, the area of the triangle will increase by
(a) 400%
(b) 600%
(c) 800%
(d) None of these
10. A plot of land is in the shape of a right-angled isosceles triangle. The length of hypotenuse is 50V2
m. The cost of fencing is Rs. 3 per metre. The cost of fencing the plot will be
(a) less than Rs. 300
(b) less than Rs. 500
(c) more than Rs. 500
(d) None of these
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11. If the area of an equilateral triangle is equal to the area of an isosceles triangle whose base and equal
sides area 16 cm and 10 cm respectively, then the side of the equilateral triangle is
(a) 10.5cm
(b) 9.5cm
(c) 12.5cm
(d) None of these
12. A rectangular park is 65 m long and 50 m wide. Two cross paths each 2 m wide are to be constructed
parallel to the sides. If these paths pass through the centre of the rectangle and cost of construction is 17.25
per sq m, find the total cost involved in the construction.
(a) Rs. 2265.59
(b) Rs. 1772.45
(c) Rs. 3898.50
(d) Rs. 8452.32
13. The area of a trapezium is 2500 sq m. One of its parallel sides is 75 m. If the distance between the
two parallel sides is 40 m, find the other parallel side.
(a) 20m
(b) 30m
(c) 40m
(d) 50m
14. If the ratio of the areas of two squares is 9 : 1, the ratio of their perimeters is
(a) 9 : 1
(b) 3 : 4
(c) 3 : 1
(d) 1 : 3
15. A square field with side 30 m is surrounded by a path of uniform width. If the area of the path is
256m2, its width is
(a) 16m
(b) 14m
(c) 4m
(d) 2m
16.
Four circular cardboard pieces, each of radius 7 cm are placed in such a way that each piece touches two
other pieces. The area of the space enclosed by the four pieces is
(a) 21 cm2
(b) 42 cm2
(c) 84cm2
(d) 168cm 2
17. A piece of wire 132 cm long is bent successively in the shapes of an equilateral triangle, a square, a
regular hexagon, and a circle. Then, which has the largest surface area?
(a) equilateral triangle
(b) square
(c) circle
(d) regular hexagon
18. If the radius of one circle is twelve times the radius of another, how many times does the area of the
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greater contain the area of the smaller?
(a) 12
(b) 72
(c) 144
(d) 96
19. If the circumference of a circle is equal to the perimeter of a square, what is the ratio of the area of the
circle to the area of the square?
(a) 22 : 7
(b) 14 : 11
(c) 11 : 7
(d) 4 : 1
20.Two poles 1 5 m and 30 m high stand upright in a playground. If their feet be 36 m apart, find the
distance between their tops.
(a) 36m
(b) 39m
(c) 15m
(d) None of these
21.Each side of an equilateral triangle is increased by 1.5%. The percentage increase in its area is
(a) 1.5%
(b) 3%
(c) 4.5%
(d) 5.7%
22.A rope, by which a horse is tied, is increased from 12 to 23 m. How much additional ground will it be
able to graze?
(a) 1315m2
(b) 765m 2
(c) 1210m2
(d) 1012m2
23. If a diagonal of a square is doubled, how does the area of the square change ?
(a) Becomes four-fold
(b) Becomes three-fold
(c) Becomes two-fold
(d) None of these
24.If the sides of a rectangle are increased by 20%, the percentage increase in its perimeter is
(a) 80%
(b) 40%
(c) 20%
(d) None of these
25. If the diagonal of a square is doubled to make another square, the area of the new square will
(a) become four-fold
(b) become two-fold
(c) become six-fold
(d) become eight-fold
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Answer
1. (c) 2. (a)
3. (b)
4. (c)
5. (a)
6. (c)
7. (c)
8. (b) 9. (c)
10. (c)
11. (a)
12. (c)
13. (d)
14. (c)
15. (d) 16. (b)
17. (c)
18. (c)
19. (b)
20. (b)
21. (a)
22. (c) 23. (a)
24. (c)
25. (a)
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Chapter: 7 VOLUME AND SURFACE AREA
INTRODUCTION
Solids
A solid is a figure bounded by one or more surfaces. It has three dimensions namely length, breath or
width and thickness or height. The plane surfaces that bind it are called its faces. The volume of any
solid figure is the amount of space enclosed within its bounding faces. It is measured in cubic units, e.g.,
m3, cm3, etc.
The area of the plane surfaces that bind the solid is called its surface area.
For any regular solid, Number of faces + Number of vertices = Number of edges + 2.
We discuss below some important three dimensional figures and the formulae associated with them.
Some Basic Formulae
Cubic
It is a solid figure which has six rectangular faces. It is also called rectangular parallelepiped.
If l, b and h denote the length, breadth and height of the cuboid and d denotes the body diagonal (AF or BE
or DG or CH),
then
(i) Volume l b h A1 A 2 A3 , where A1 = area of base or top
A2 = area of one side face, and
A3 = area of other sides face
(ii) Total Surface Area 2(lb bh lh) (l b h)2 d2
(iii) Diagonal of cuboid = l2 b2 h2
Note: (i) For painting the surface area of a box or to know how much tin sheet is required for making a box,
we use formula(ii)
(ii) To find how much a box contains or how much space a box shall occupy, we use formula(i). To find the
length of the
longest pole to be placed in a room, we use formula (iii)
Total volume of objects submerged or taken out
(iii) The rise or fall of liquid level in a container
Cross sectional area of container
Illustration 1: Find the length of the longest bamboo that can be placed in a room 12m long, 9 m broad
and 8 m high.
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Solution:
Length of the bamboo
= length of the diagonal of the room
122 92 82
289 17m
Illustration 2: The area of a side of a box is 120 sq cm. The area of the other side of the box is 27 sq cm. If the
area of the upper surface of the box is 60 sq cm, then find the volume of the box.
Solution:
Volume to the box
area of base area of one face x area of the other face
60 120 72 518400 720 cm3 .
Illustration 3: The sum of length, breadth and height of a cuboid is 12 cm long. Find the total surface area of the
cuboid.
Solution:
Total surface area
= (Sum of all three sides)2 - (Diagonal)2 = 122 – 82 = 144 - 64 = 80 sq cm.
Cube
It is a special type of cuboid in which each face is a square.
For a cube, length, breadth and height are equal and is called the edge of the cube.
If a be the edge of a cube, then
(i) Volume of the cube = (edge)2 = a3
(ii) Total surface area of the cube = 6 (edge)2 =6a2
(iii)Diagonal of the cube 3a (edge) 3a
3
2
3
diagonal d Surface area
(iv) Volume of the cube
6
3 3
2
2
(v) Total surface area of the cube 2 = (diagonal) = 2d
(vi) For two cubes
(a) Ratio of volumes = (ratio of sides)2
(b) Ratio of surface areas = (Ratio of sides)2
(c) (Ratio of surface areas)2 = (Ratio of volumes)2.
Illustration 4:
The diagonal of a cube is 8 3 cm. Find its total surface area and volume.
Solution:
We have,
Diagonal of cube = 3 (edge)
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Diagonal of cube 8 3
=
8 cm.
3
3
Total surface area = 6 (edge)2 = 6(8)2 = 384 sq cm.
Volume of cube = (edge)3 = (8)3 = 512 cm3.
Illustration 5: If the volumes of two cubical blocks are in the ratio of 8 : 1, what will be the ratio of their edges?
Solution:
We have,
Ratio of volumes = (Ratio of sides)3
Since, ratio of volumes = 8 : 1 , i.e. 23 : 13
ratio of sides = 2:1.
Illustration 6:
Volumes of the two cubes are in the ratio of I : 9. Find the ratio of their surface areas.
Solution:
(Ratio of the surface areas)3
= (Ratio of volumes)2
Edge of cube
Ratio of surface areas = 3 1 : 81 1 :3 (3)1/3 .
RIGHT CIRCULAR CYLINDER:
A right circular cylinder is a solid with circular ends of equal radius and the line joining their centres
perpendicular to them. This is called axis of the cylinder. The length of the axis is called the height of the cylinder.
Note: Take a rectangular sheet of paper and role it lengthwise or breadth wise in around way, you will get a
cylinder, i.e a
cylinder is generated by rotating a rectangle by fixing one of its sides.
It r is the radius of base and h is the height of the cylinder, then
(i) Volume of cylinder
= Area of the base × height
r2 h r2h cubic units
(ii) Area of the curved surface
= Circumference of the base × height
2 r h 2 rh sq units
(iii) Area of the total surface
= Area of the curved surface + Area of the two circular ends
2 rh 2 r2
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2 r(h r) sq units.
(iv) For two cylinders,
When radii are equal
(a) Ratio of volumes = Ratio of height
(b) Ratio of volumes = Ratio of curved surface areas
(c) Ratio of curved surface areas = Ratio of height
When heights are equal
(a) Ratio of volumes = (Ratio of radii)2
(b) Ratio of volumes = (Ratio of curved surface areas)2
(c) Ratio of curved surface areas = Ratio of radii
When volumes are equal
(a) Ratio of radii =
Inverse ratio of height
(b) Ratio of curved surface areas = Inverse ratio of radii
(c) Radii of curved surface areas = Ratio of height
When curved surface areas are equal
(a) Ratio of radii = Inverse ratio of height
(b) Ratio of volumes = Inverse ratio of height
(c) Ratio of volumes = Ratio of radii
For cylinder
(a) Ratio of radii = (Ratio of curved surfaces) × (Inverse ratio of heights)
(b) Ratio of height = (Ratio of curved surfaces) × (Inverse ratio of heights)
(c) Ratio of curved surfaces =( Ratio of radii) × (Ratio of heights).
Illustration 7: A rectangular piece of paper is 71 cm long and 10 cm wide. A cylinder is formed by rolling
355
the paper along its breadth. Find the volume of the cylinder. Take
113
Solution:
Circumference of the paper = Breadth of the paper
2 r 10
10 10 113 113
cm.
2 2 355
71
As the length of the paper becomes the height of the cylinder,
Volume of the cylinder = r2l
355 113 113
71 565 cm3 .
113 71 71
Illustration 8: Two circular cylinders of equal volume have their heights in the ratio of 9 : 16. Find the ratio
of their radii.
Solution:
r
Ratio of radii = inverse ratio of heights
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16:9 = 4 : 3 .
(vi) If the ratio of height and the ratio of radii of two right circular cylinder are given, then Ratio of
curved surface areas = (ratio of radii) (ratio of height)
Illustration 9: If the heights and the radii of two right circular cylinders are in the ratio 2 : 3 and 4: 5
respectively. Find the ratio of their curved surface areas.
Solution:
Ratio of curved surface areas = (ratio of radii) (ratio of heights)
= (4 : 5) (2 : 3) = 8 : 15.
(vii) If the ratio of height and the ratio of curved surface areas of the two right circular cylinders are
given then
Ratio of radii = (ratio of curves surface areas) (inverse ratio of height).
Illustration 10: The heights and curved surface areas of two right circular cylinders are in the ratio 3: 4
and 5: 8, respectively. Find the ratio of their radii.
Solution:
Ratio of radii = (ratio of curved surface areas) (inverse ratio of heights)
1 1
(5: 8) : (5 :8)(4 :3) 5:6.
3 4
(viii) If the ratio of radii and the ratio of curved surface areas of two right circular cylinder are given,
then
Ratio of heights = (ratio of curved surface areas)(inverse ratio of radii)
Illustration 11: The radii of two right circular cylinders are in the ratio of 3 : 4 and their curved surface
areas are in the ratio of 5 : 6. Find the ratio of their heights.
Solution:
Ratio of heights = (ratio of curved surface areas) (inverse ratio of radii)
1 1
(5:6) :
3 4
(5:6)(4 :3) 10: 9.
Right Circular Cone
A right circular cone is a solid obtained by rotating a right-angled triangle around its height.
If r = radius of base; h = height
l = slant height =
h2 r2 , then
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(i) Volume of cone
1
1
area of the base height r2h cubic units
3
3
(ii) Areas of curved surface = r l
r h2 r2 sq units.
(iii) Total surface area of cone = Area of the base + area of the curved surface
r2 r l r(r 1) sq units.
(iv) For two cones
(a) When volumes are equal
Ratio of radii =
inverse ratio of heights
(b) When radii are equal
Ratio of volumes = Ratio of heights
(c) When height are equal
Ratio of volumes = (ratio of radii)2
(d) When curved surface areas are equal
Ratio of radii = inverse ratio of slant height.
Illustration 12: Two right circular cones of equal curved surface areas have their slant heights in the ratio of
3 : 5 . Find the ratio of their radii.
Solution:
Ratio of radii = inverse ratio of slant heights
1 1
: 5:3.
3 5
Illustration 13: Two right circular cones of equal volumes have their heights in the ratio of 4 : 9. Find the
ratio of their radii.
Solution:
Ratio of radii
inverse ratio of heights
1 1
: 9: 4 3:2.
4 9
(v) If the ratio of volumes and the ratio of heights of two right circular cones (or cylinders) are given,
then Ratio of radii
(ratio of volumes) (inverse ratio of heights)
(3:2)(8 :3) : 4 :1 2:1.
Illustration 14: The volumes of two cones are in the - ratio 3 : 2 and their heights in the ratio 3 : 8. Find
the ratio of their radii.
Solution:
Rat io of r adii
(ratio of volumes) (inverse ratio of heights)
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(3:2)(8 :3) : 4 :1 2:1
(vi) If the ratio of heights and the ratio of diameters (or radii) of two right circular cones (or
cylinders) are given, then Ratio of volumes = (ratio of radii)2 × (ratio of heights).
Illustration 15: The heights of two cones are in the ratio of 5: 3 and their radii in the ratio 2 : 3 . Find the
ratio of their volumes.
Solution:
Ratio of volumes
= (ratio of radii)2 ×(ratio of heights)
= (2 : 3) 2 ×(5 : 3)
4 5
20 :27.
9 3
(vii) If the ratio of radii (or diameter) and the ratio of volumes of two right circular cones are
given, then
Ratio of heights
(inverse ratio of radii)2 (ratio of volumes).
Illustration 16: The volumes of two cones are in the in ratio of 1:4 and their diameters are in the
ratio of j 4 : 5 . Find the ratio of their heights.
Solution:
Ratio of heights
= (inverse ratio of diameters)2 (ratio of volumes)
2
1 1
= : (1 : 4) (5: 4)2(1 : 4)
4 5
25 1
25:64.
16 4
Frustum of a right circular cone
A cone with some of its top portion cut off is called the frustum of the original cone.
If R = Radius of the base of frustum
r = Radius of the top of the frustum
h = Height of the frustum
l = Slant height of the frustum, then
(a) Slant height h2 (R r)2 units
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(b) Area of the curved surface (R r)l sq units
(c) Total surface area of the frustum [(R 2 r2 ) l(R r) sq units
(d) Volume of the frustum
h 2 2
(R r Rr) cu units
3
Illustration 17: A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top
and 4 m across the bottom. It is 6 m deep. Find the area of its curved surface, total surface area and also its
volume.
Solution:
Here, R = 4, r = 2 and h = 6.
.'. Slant height (l) h2 (R r)2
(6)2 (4 2)2 40.
.'. Area of the curved surface = (R + r) l
22
(4 2) 40
7
= 18.8× 6.3 = 118.4 m
Total surface area = [(R2 + r2) + l(R + r)]
22
[(42 22 ) 40(4 2)]
7
22
(20 6 40) 181.6sq m
7
Volume of the frustum =
h 2
(R + r 2 + R r)
3
22 6 2 2
(4 2 4 2)
7 3
44
(20 4 8) 176m3 .
7
SPHERE
A sphere is the solid figure formed by revolving a semicircle on its diameter.
The mid-point of the diameter is called centre c he sphere and the radius of the semi-circle is called the
radius of the sphere. If r = radius of the spheres, then
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4 2
r cubic units
3
(ii) Surface area = 4 r2 sq units.
(i) Volume of sphere
2
(iii) Volume of hemisphere = r 2 cubic units
3
(iv) Area of curved surface 2r2 sq units of hemisphere
(v) Total surface area of hemisphere 3r2 sq units.
(vi) For two spheres
(a) (Ratio of radii)2 = Ratio of surface areas
(b)(Ratio of radii)3 = Ratio of volumes
(c) (Ratio of surface areas)3 = (Ratio of volumes)2.
Illustration 18: The radii of two spheres are in the ratio of 2: 3. What is the ratio of their surface areas?
Solution:
Ratio of surface areas = (ratio of radii)2
= (2 : 3) 2 = 4 : 9.
Illustration 19: The surface areas of two spheres are in the ratio 1 : 2 . Find the ratio of their volumes.
Solution:
We have,
(Ratio of surface areas)3 = (Ratio of volumes)2 (1 : 2)3 = (Ratio of volumes)2
Ratio of volumes 1 :8 1 :2 2.
Illustration 20: The radii of two spheres are in the ratio of 2 : 45. Find the ratio of their volumes.
Solution: Ratio of volumes = (Ratio of radii)3
= (2 : 5)3 = 8 : 125.
Prism:
A solid having top and bottom faces identical and side faces rectangular is a prism.
In a prism with a base of n sides,
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Number of vertices = 2n and Number of faces = n +2.
Volume of the prism = area of base × height
Lateral surface area = perimeter of base × height
Total surface area = 2×Base area + Lateral Surface area.
Illustration 21: Find the volume and the total surface area of a triangular prism whose height is 30 m and
the sides of whose base are 21 m, 20 m and 13m, respectively.
Solution:
Perimeter of base = 21 + 20 + 13 = 54 m. height = 30 m
Area of base s(S a)(S b)(S c)
27(27 21)(27 20)(27 c)
27 6 7 14 126sq m.
Volume of the prism = area of base × height = 126 × 54 = 6804 m3 .
Also, surface area of the prism
= 2× Base area + lateral surface area
= 2 Base area + perimeter of base × height
= 2×126 + 54×30 = 1872 sq m.
SOLID INSRIBED/CIRCUMSCRIBING OTHER SOLIDS
1. If a largest possible sphere is circumscribed by a cube of edge 'a' cm, then the radius of the sphere
= a/2
Illustration 22: Find the volume of largest possible sphere circumscribed by a cube of edge 8 cm.
Solution:
a 8
Radius of the sphere 4cm .
2 2
4
Volume of the sphere = r3
3
4 22
4 4 4 26.81 cm3 .
3 7
2a
.
2. If a largest possible cube is inscribed in a sphere of radius ‘a’ cm, then the edge of the cube
3
Illustration 23: Find the surface area of largest possible cube inscribed in a sphere of radius 4 cm.
Solution:
2a 2 4 8
.
Edge of the cube=
3
3
3
Surface area of the cube = 6 (edge)2
64
128 sq cm.
6
3
3. If a largest possible sphere is inscribed in a cylinder of radius ‘a’ cm and height ‘h’ cm, then
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a for h a
radius of the sphere h
a for a h
Illustration 24: Find the surface area of largest possible sphere inscribed in a cylinder of radius 14 cm and
height 17 cm.
Solution:
Radius of the sphere = 14 cm ( h > a)
Surface area of sphere = 4 r2
22
4 14 14 2464 sq cm.
7
4. If a largest possible sphere is inscribed in a cone of radius 'a' cm and slant height equal to the
a
.
diameter of the base, then radius of the sphere
3
Illustration 25: Find the surface area of largest possible sphere inscribed in a cone of radius 21 cm and slant
height equal to the diameter of the base.
Solution:
a
21
.
. cm.
Radius of the sphere =
3
3
Surface area of the sphere = 4 r2
22 21 21
4
1848 sq cm.
7
3
3
5. If a largest cone is inscribed in a cylinder of radius ‘a’ cm height ‘h’ cm, then radius of the cone = a and
height = h
Illustration 26: Find the volume of largest possible cone inscribed in a cylinder of radius 6 cm and height
14 cm.
Solution:
Radius of the cone (r) = 6 cm. and height of the cone (h) = 1 4 cm.
1
.'. Volume of the cone = r2h
3
1 22
6 6 14 528 sq cm
3 7
6. If a largest possible cube is inscribed in a hemisphere 12of radius V cm, then the edge of the
2
.
3
Illustration 27: Find the length of the diagonal of largest possible cube inscribed in a hemisphere of
cube a
radius 4 2 cm.
Solution:
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2
2 8
cm.
4 2
3
3
2
Diagonal of the cube = 3 (edge)
8
3
8 cm.
3
SOME USEFULL SHORT-CUT METHODS
1. If all three measuring dimensions of a sphere, cuboid, cube, cylinder or cone are increased or decreased
by x%, y% and z% respectively, then the volume of the increase or decrease
xy yz zx xyz
by x y z
%
100
1002
For cuboid, the three measuring dimensions are length, breadth and height.
For cube, all three measuring dimensions are equal, i.e. x = y = z.
For sphere also, (or diameter) all three measuring dimensions are equal and is given by radius, i.e. x = y = z = r.
For cylinder or a cone two measuring dimensions are equal to radius and third measuring dimension is height
i.e. x = y = r and z - h.
Illustration 28: The length, breadth and height of a cuboid are increased by 5%, 10% and 20%, respectively.
Find the percentage increase in its volume.
Solution:
Here, x = 5, y = 10 and z = 20
Percentage increase in volume
xy yz zx xyz
x y z
%
100
1002
(5 10) (5 20) (10 20) 5 10 20
5 10 20
%
100
(100)2
350 1000
35
100 (100)2
(35 3.5 0.1)% 38.6%
Illustration 29: The diameter of a sphere is increased by 20%. What is the percentage increase in its
volume?
Solution:
Percentage increase in volume
3x2
3x2
3x
%
100 (100)2
[Here, x y z]
3(20)2 (20)3
3 20)
%
100 (100)2
(60 12 0.8)% 72%.
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Illustration 30: The radius of a right circular cylinder is decreased by 5% but its height is increased by
10%. What is the percentage change in its volume?
Solution:
Here, x = y = -15 and z = 10.
Percentage change in volume
( 5)( 5) ( 5)(10) ( 5)(10) ( 5)( 5)(10)
5 5 10
%
100
(100)2
(0 0.75 0.025)% 0.725%
Therefore, volume decrease by 0.725%.
Illustration 31: Each of the radius and the height of a cone is increased by 25%. Find the percentage
increase in volume.
Solution:
Here, x = y = 25 and z = 25.
Percentage increase in volume
25 25 25 25 25 25 25 25 25
%
25 25 25
100
(100)2
(75 + 18.75 + 1.56)% = 95.3%.
2. If the two measuring dimensions which are included in the surface area of a sphere, cuboid, cube,
cylinder or cone are increased or decreased by x% and y%, then the surface area of the figure will increase
xy
or decrease by
x y 100 %
Note that in case of percentage increase, values of x, y and z are positive and in case of percentage
decrease, values of
x, y and z are negative.
Illustration 32: The radius of a hemisphere is decreased by 10%. Find the percentage change in its
surface area.
Solution:
Here, x = y = -10.
Percentage change in surface area
xy
x y
%
100
( 10)( 10)
10 10
%
100
( 20 1)% 19%.
Therefore, surface area of hemisphere decreases by 19%.
Illustration 33: The radius of a right circular cone is increased by 25% and slant height is decreased
by 30%. Find the percentage change in curved surface area of the cone.
Solution:
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Here, x = 25 and y = -30
Percentage change in curved surface area
xy
%
x y
100
(25)( 30)
25 30
%
100
( 5 7.5)% 12.5%.
Therefore, curved surface area decreases by 12.5%.
Illustration 34: The radius and height of a cylinder are increased by 10% and 20%, respectively.
Find the percentage increase in its surface area.
Solution:
Here, x = 10 and y = 20.
Percentage increase in surface area
10 20
xy
x y
% 10 10
% 32%.
100
100
3. If a sphere of radius R is melted to form smaller spheres each of radius r, then
The number of smaller spheres
3
Volume of the bigger sphere
R
= .
Volume of the smaller sphere r
Illustration 35: Find the number of lead balls of radius 1 cm each that can be made from a sphere of
radius 4 cm.
Solution:
3
3
R 4
Number of lead balls 64.
r 1
4. If by melting n spheres, each of radius r, a 'big sphere is made, then
Radius of the big sphere r. 3 n
Illustration 36: If by melting 8 spheres, each of radius 5 cm, a big sphere is made, what will be the
radius of the big sphere?
Solution: Radius of the big sphere r. 3 n
5. 3 8 = 5.2 = 19 cm.
5. If a cylinder is melted to form smaller spheres each of radius r,
then
Volume of cylinder
The number of small spheres =
Volume of 1 sphere
Illustration 37: How many bullets can be made out of a loaded cylinder 24 cm high and 5 cm
diameter, each
bullet being 2 cm in diameter?
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Solution:
Number of bullets =
Volume of cylinder
Volume of 1 sphere
6. If a sphere of radius r is melted and a cone of height: h is made, then
Radius of the cone =2
r3
.
h
or,
If a cone of height h is melted and a sphere of radius r is made, then
r3
.
h
Illustration 38: A solid cone of copper of height 3 cm is melted and a solid sphere of radius 3 cm is
made. What is the diameter of the base of the cone?
Solution: Radius of the base of the cone
Radius of the cone =2
r3
33
2
2
6.
h
3
.'. Diameter of the base of the cone = 2×6 = 12 cm.
SAMPLE QUESTIONS
1
3
1
1. Find the cost of the log of wood measuring 15 m by 2 m by 1 m at Rs. 45 per cu m.
2
4
4
(a) Rs. 4257.50
(b) Rs. 4005.00
(c) Rs. 4207.50
(d) Rs. 4357.50
2. Surface area of a cube is 600 sq. cm. Find the length of its diagonal.
(a) 15 3
(b) 12 3
(c) 10 3 cm
(d) None of these
3. A rectangular tank is 30 m long 20 m broad. Water is being flown into it through a square pipe of side 5
cm. What is the speed of water if the level of water in the tank rises by 1 m in 8 hours?
(a) 30 km/hr
(b) 36 km/hr
(c) 40 km/hr
(d) None of these
4. Calculate the number of bricks, each measuring 25 cm ×15 cm× 8 cm, required to construct a wall of
dimension 19 m × 4 m×5 m, when 10% of its volume is occupied by mortar.
(a) 4000
(b) 8000
(c) 7000
(d) 6000
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5. The height of a right circular cylinder is 6 m. Three times the sum of the areas of its two circular faces is
twice the area of its curved surface. The radius of the base is
(a) 4 m
(b) 2 m
(c) 6 m
(d) 1.5 m
6. How many coins 2 mm thick and 1.5 cm. in diameter should be melted in order to form a right circular
cylinder having base diameter 6 cm and height 8 cm?
(a) 640
(b) 540
(c) 740
(d) 840
7. A solid cylinder has a total surface area of 231 sq cm. Its curved surface area is (2/3) of the total surface
area. Find the volume of the cylinder
(a) 270 cu cm
(b) 269.5 cu cm
(c) 256.5 cu cm
(d) 289.5 cu cm
8. The radius of a cylinder is made twice as large. How should the height be changed so that volume remains
the same?
(a)
1
th of original
4
(b)
1
rd of original
3
(c)
1
of original
2
(d)
1
th of original
8
9. A spherical ball of lead, 3 cm in diameter is melted and re-cast into three spherical balls. The diameters of
two of these are 1.5 cm and 2 cm, respectively. The diameter of the third ball is
(a) 2.66 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm
10. A cone and a cylinder having the same area of the base have also the same area of curved surfaces. If the
height of cylinder be 2 m, find the slant height of the cone
(a) 3 m
(b) 3.5 m
(c) 4.5 m
(d) 4 m
11. A hollow cylinder of height 3 cm, is re-casted into a solid cylinder. If the external and internal radii of the
hollow cylinder are 4.3 cm and 1.1 cm respectively. What will be the radius of the solid cylinder?
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(a) 2.8 cm
(b) 2.4 cm
(c) 3.2 cm
(d) 4.8 cm
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12. A solid consists of a circular cylinder with an exact fitting right circular cone placed on the top. The height
of the cone is h. If the total volume of the solid is three times the volume of the cone, then the height of the
cylinder is
(a) 2h
(c)
2h
3
(b) 4h
(d)
3h
2
13. The radius of a cylinder is doubled and the height is halved, what is the ratio between the new volume
and the previous volume?
(a) 3 : 1
(b) 2 : 3
(c) 2 : 1
(d) 1 : 3
14. A circus tent is cylindrical to a height of 3 m and conical above it. If its diameter is 105 m and slant height
of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the tent,
(a) 1857 m
(b) 1647 m
(c) 1947 m
(d) 1847 m
15. If base radius of a cone is increased by 20% and its slant height is made double, then by how much per
cent will the area of its curved surface be increased?
(a) 140%
(b) 160%
(c) 130%
(d) 180%
16. The radius of the base of conical tent is 5 cm. If the tent is 12 m high then area of the canvas required in
making the tent is
(a) 60 m 2
(b) 300 m 2
(c) 90 m 2
(d) None of these
17. A cone of height 7 cm and base radius 3 cm is carved from a rectangular block of wood 10 cm× 5 cm ×2
cm. The percentage % wood wasted is
(a) 34%
(b) 46%
(c) 54%
(d) 66%
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18. The diameter and slant height of a conical tomb are 28 m and 50 m, respectively. The cost of white
washing its curved surface at the rate of 80 paise per sq m is
(a) Rs. 2640
(b) Rs. 1760
(c) Rs. 264
(d) Rs. 176
19. A rectangular sheet of area 264 cm2 and width 11cm is rolled along its breadth to make a hollow cylinder.
The volume of the cylinder is
(a) 231 c.c.
(b) 230 c.c.
(c) 235 c.c.
(d) 234 c.c.
20. A cylinder and a cone have their heights in the ratio 2: 3 and the radii of their bases in the ratio 3 : 4. Find
the ratio of their volumes.
(a) 1 : 9
(b) 2 : 9
(c) 9 : 8
(d) 1 : 8
Answers
1.(c)
8. (a)
15.(a)
2. (c)
9. (b)
16. (d)
3. (a)
10. (d)
17. (a)
4. (d)
11.(b)
18. (b)
5. (a)
12. (c)
19. (a)
6. (a)
13. (c)
20. (c)
7. (b)
14. (c)
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Chapter: 8 PLANE GEOMETRY
LINES AND ANGLES
Line: A geometrical straight line is a set of points that extends endlessly in both the directions.
Axiom-1: A line contains infinitely many points.
Axiom-2: Through a given point, infinitely many lines pass.
Axiom-3: Given two distinct points A and B, there is one and only one line that contains both the points.
Parallel Lines: If two lines have no point in common, they are said to be parallel lines
Intersecting Lines: If two lines have a point in common, they are said to be Intersecting lines. Two lines
can intersect at the most at one point.
Line Segment and Ray: A part (or portion) of a line with two end points is called a line-segment and a
part of a line with one end point is called a ray. A line segment AB and its length is denoted as AB. Ray
AB (i.e. A towards B) is denoted as AB and ray BA (i.e. B towards A) is denoted as BA .
Collinear Points: Three or more than three points are said to be collinear if there is a line which
contains them all.
Concurrent lines: Three or more than three lines are said to be concurrent if there is a point which lies
on all of them.
Angle: An angle is a figure formed by two rays with a common initial point. The two rays forming an
angle are called the arms of the angle and the common initial point is called the vertex of the angle.
Types of Angles:
An angle is said to be:
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(vi) Complete angle: an angle, whose measure is 360°, is called a complete angle.
Complementary angles: Two angles, the sum of whose measures is 90° are called complementary angles e.g.
50° and 40° is a pair of complementary angles.
Supplementary angles: Two angles, the sum of whose measures is 180°, are called supplementary angles, e.g.
72° and 108° is a pair of supplementary angles.
Adjacent Angles: Two angles are called adjacent angles if
(i) They have the same vertex,
(ii) they have a common arm, and
(iii) uncommon arms are on either side of the common arm.
e.g. AOC and BOC are adjacent angles
Linear Pair: Two adjacent angles are said to form a linear pair of angles if their non-common arms are two
opposite rays
e.g. AOC and BOC form a linear pair.
Linear Pair Axiom: If a ray stands on a line, then the sum of the two adjacent angles so formed is 180°.
Conversely, if the sum of two adjacent angles is 180°; then the non-common arms of the angles are two opposite
rays.
Vertically opposite angles: When two lines intersect, four angles are formed. The angles opposite to each
other are called vertically opposite angles
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a and c are vertically opposite angles, a = c
b and d are vertically opposite angles, b = d
Angles Made by a Transversal with two parallel lines
Suppose PQ || RS and a transversal AB cuts them, then
(a) Pair of corresponding angles is
(1 and 5), ( 2 and 6),
( 4 and 8) and ( 3 and 7)
(b) Pair of alternate angles are
( 3 and 6), ( 4 and 5)
(c) Pair of interior angles (consecutive interior angles or cointerior angles) on the same side of the transversal
are ( 3 and 5), ( 4 and 6)
If two parallel lines are intersected by a transversal, then.
Key Point to Remember:
(i) Each pair of corresponding angles are equal.
(ii) Each pair of alternate angles are equal.
(iii) Interior angles on the same side of the transversal arc supplementary.
TRIANGLES
Triangle: A plane figure bounded by three lines in a plane is called a triangle.
TYPES OF TRIANGLES: (ON THE BASIS OF SIDES)
Scalene triangle: A triangle, no two of whose sides are equal is called a scalene triangle.
Isosceles triangle: A triangle, two of whose sides are equal in length is called an isosceles triangle.
Equilateral triangle: A triangle, all of whose sides are equal is called an equilateral triangle.
TYPES OF TRIANGLES: (ON THE BASIS OF ANGLES)
Acute triangle: A triangle, each of whose angles is acute, is called an acute triangle or acute angled triangle.
Right triangle: A triangle with one right angle is called a right triangle or a right angled triangle.
Obtuse triangle: A triangle with one angle and obtuse angle is known as obtuse triangle or obtuse angled
triangle.
SOME IMPORTANT TERMS RELATED TO A TRIANGLE
Median: The median of a triangle corresponding to any side is the line segment joining the midpoint of that
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side with the opposite vertex.
In the figure given below, AD, BE and CF are the medians.
The medians of a triangle are concurrent i.e. they intersect each other at the same point.
Centroid: The point of intersection of all the three medians of a triangle is called its centroid.
In the above figure G is the centroid of ABC.
Note: The centroid divides a median in the ratio 2:1.
Altitudes: The altitude of a triangle corresponding to any side is the length of perpendicular drawn
from the opposite vertex to that side.
In the figure given above, AL, BM and CN are the altitudes.
Note: The altitudes of a triangle are concurrent.
Orthocenter: The point of intersection of all the three altitudes of a triangle is called its orthocenter.
In the figure given above H is the orthocenter of ABC.
Note: The orthocentre of a right angled lies at the vertex containing the right angle.
Incentre of a triangle: The point of intersection of the internal bisectors of the angles of a triangle is
called its incentre.
In the figure given below, the internal bisectors of the angles of ABC intersect at I.
I is the Incentre of ABC.
Let ID BC
Then, a circle with centre I and radius ID is called the in circle of ABC.
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Circumcentre of a triangle: The point of intersection of the perpendicular bisectors of the sides of a
triangle is called its circumcentre. In the figure given below, the right bisectors of the sides of ABC
intersect at O.
O is the circumcentre of ABC with O as centre and radius equal to OA = OB = OC. We draw a circle passing
through the vertices of the given. This circle is called the circumcircle of ABC.
Note: The circumcentre of a triangle is equidistant from its vertices.
CONGRUENT TRIANGLES
Two triangles are congruent if and only if one of them can be superposed on the other, so as to cover it
exactly.
Thus, congruent triangles are exactly identical
For example, If ABC DEF then we have
A D, B E, C F;
and AB DE,BC EF and AC =DF.
Similar Triangles:
Congruent figures: Two geometric figures having the same shape and size are known as congruent
figures
Similar figures: Two figures (plane or solid) are said to be similar if they have the same shape
irrespective of their sizes.
Note: Two similar figures may not be congruent as their size may be different.
For example,
1. Any two lines segments are similar,
2. Any two equilateral triangles are similar,
3. Any two squares are similar,
4. Any two circles are similar,
5. Any two rectangles are similar.
Similar triangles: Two triangles are similar if
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(a) their corresponding angles are equal and
(b) their corresponding sides are proportional
KEY RESULTS TO REMEMBER:
1. The sum of all the angles round a point is equal to 3600.
2. Two lines parallel to the same line are parallel to each other.
3. The sum of three angles of a triangle is 1800.
4. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior
opposite angles.
(Exterior Angle Theorem)
5. If two sides of a triangle are unequal, the longer side has greater angle opposite to it.
6. In a triangle, the greater angle has the longer side opposite to it.
7. The sum of any two sides of a triangle is greater than the third side.
8. If a, b, c denote the sides of a triangle then
(i) If c2 < a2 + b2, triangle is acute angled.
(ii) If c2 = a2 + b2, triangle is right angled.
(iii) If c2 > a2 + b2, triangle is obtuse angles.
9. Two triangles are congruent if:
(i) Any two sides and the included angle of one triangle are equal to any two sides and the included angle of
the other triangle.
(SAS congruence theorem)
(ii) Two angles and the included side of one triangle are equal to the corresponding two angles and the
included side of the other triangle.
(ASA congruence theorem)
(iii) The three sides of one triangle are equal to the corresponding three sides of the other triangle.
(SSS congruence theorem)
Note: Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal
to the hypotenuse and the corresponding side of the other triangle.
(RHS Congruence theorem)
10.The line-segments joining the mid-points of any two sides of a triangle is parallel to the third side and
equal to half of it.
11.Basic Proportionality Theorem: If a line is drawn parallel side one side of a triangle, to intersect the
other two sides in distinct points; the other two sides are divided in the same ratio.
In the figure given below, In a ABC
If DE||BC
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Then,
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AD AE
DB EC
Illustration 1: In the figure given above D and E are the points on the AB and A C respectively such that DE
||BC. If AD = 8 cm, AB = 12 cm and AE = 12 cm. Find CE
Sol:
In ABC, DE || BC
AD AE
(Basic Proportionality Theorem)
DB EC
8
12 8 12
12 8 EC 4 EC
Or EC = 6cm
12. If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.
EXPLANATION: In the figure given in 11, In ABC
AD AE
, then DE|| BC.
if
DB EC
SIMILARITY THEOREMS
13. AAA Similarity: If in two triangles, corresponding angles are equal, then the triangles are similar.
Corollary: (AA- similarity): If two angle of one triangle are respectively equal to two angles of another
triangle then the
two triangles are similar.
Illustration 2: In the figure given below, QA and PB are perpendiculars to AB. If AO = 15 cm, BO = 9 cm, PB =
12 cm, find AQ.
Sol:
If s AOQ and BOP
1 = 2 [vertically opposite angles]
3 = 4 [each 900]
AOQ BOP [AA Similarity Criterion]
AO AQ
(corresponding sides of ~ s)
BO BP
15 AQ
or
9 12
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5 AQ
AQ 20 cm.
1 4
14. SSS –Similarity: If the corresponding sides of two triangles are proportional then they are similar.
EXPLANATION: In s ABC and DEF,
AB BC AC
DE EF DF
Then ABC ~ DEF [SSS Similarity]
if
15. SAS-Similarity: If in the two triangles, one pair of corresponding sides are proportional and the
included angles are equal, then the two triangles are similar.
EXPLANATION: In s ABC and DEF,
AB AC
DE DF
AB BC
or B E and
DE EF
AC BC
or C F and
DF EF
then ABC DEF [SAS-Similarity]
if A D and
16. Internal Bisector Property: The internal bisector of an angle of a triangle divides the opposite side in
the ratio of the
sides containing the angle
EXPLANATION:
In ABC , if 1 2
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AB BD
AC CD
17. If a line-segment drawn from the vertex of an angle of a triangle to its opposite side divides it in the ratio
of the sides
containing the angle, then the line segment bisects the angles.
Illustration 3: In PQR, PQ = 6 cm, PR = 8 cm, QS = 1.5 cm, RS = 2 cm
PQ 6 3
QS 1.5 3
and
PR 8 2
RS 2 2
PQ QS
Thus
PR RS
PS is the bisector of P.
18. Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of
the other
two sides.
EXPLANATION: In a right ABC, right angled at B, AC 2 = AB 2 + BC 2
Illustration 4: A man goes 15m west and then 8 m due north. How far is he from the starting point?
Sol:
Let the initial position of the man be A.
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Let AB = 15m and BC = 8m
AC2 = AB2 + BC2 (Pythagoras Thm.)
= (15)2 + (8)2 = 225 + 64 = 289
AC = 289 = 17m. Hence, the man is 17m away from the starting point.
19. Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of
the other two sides, then the angle opposite the first side is a right angle.
EXPLANATION: In a ABC if AB2 + BC2 = AC2 Then, ABC = 90°
Then, ABC = 90°
20. Area Theorem: The ratio of the areas of two similar s is equal to the ratio of the squares of any two
corresponding
sides.
EXPLANATION: If ABC ~ DEF,
ar( ABC) AB2 AC2 BC2
ar( DEF) DE2 DF2 EF2
Illustration 5: The areas of two similar s ABC and PQR are 64 cm2 and 121 cm2, respectively. If QR =
15.4 cm, find BC
Sol:
Since ABC~ PQR
then
ar( ABC) BC2
(Area Theorem)
ar( PQR) QR 2
64
BC2
8
BC
i.e.
2
121 (15.4)
11 15.4
BC 11.2
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21. The ratio of the areas of two similar triangle is equal to the
(i) ratio of the squares of the corresponding medians
(ii) ratio of the squares of the corresponding altitudes
(iii) ratio of the squares of the corresponding angle bisector segments
22. If two similar triangles have equal areas, then the s are congruent.
23. In two similar triangles, the ratio of two corresponding sides is same as the ratio of their perimeters.
24. Obtuse Angle Property: In a ABC, if B is obtuse then
AC2 AB2 BC2 2 BC.BD
where AD BC
25. Acute Angle Property: In a ABC. if C then AC2 AC2 BC2 2BC CD where AD BC
26. Apollonius theorem: The sum of the squares on any two sides of a triangle is equal to the sum of
twice the square of the median, which bisects the third side and half the square of the third side.
EXPLANATION:
In the given ABC ,
1
AB2 AC2 2AD2 BC2
2
Or
AB2 AC2 2[AD2 BD2 ]
27. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the
triangles on each side of the perpendicular are similar to the whole triangle and to each other. Also the
square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.
EXPLANATION: In the figure given below, ABC is a right triangle, right angled at B and BD AC then
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(i) ADB ~ ABC (AA Similarity)
(ii) BDC ~ ABC (AA Similarity)
(iii) ADB ~ BDC also BD2 = AD.CD
SAMPLE QUESTIONS
In each of the following questions a number of possible answers are given, out of which one
answer is correct. Find out the correct answer.
1. An angle is equal to one-third of its supplement. Its measure is equal to
(a) 40°
(b) 50°
(c) 45°
(d) 55°
2. The complement of 30°20' is
(a) 69°40'
(b) 59°40'
(c) 35°80'
(d) 159°40'
3. In the given figure, OP bisects BOC and OQ bisects AOC. Then POQ is equal to
(a) 90°
(b) 120°
(c) 60°
(d) 100°
4. In the given, AB || CD. Then X is equal to
(a) 290°
(b) 300°
(c) 280°
(d) 285°
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5. In the adjoining figure, AB || CD, t is the transversal, EG and FG are the bisectors of BEF and DFE
respectively, then EGF is equal to
(a) 90°
(b) 75°
(c) 80°
(d)110°
6. In the given figure, AB ||CD and AC || BD. If EAC = 40°, FDG = 55°, HAB = x; then the value of x is
(a) 95°
(b) 70°
(c) 35°
(d) 85°
7. Find the measure of an angle, if six times its complement is 12° less than twice its supplement
(a) 48°
(b) 96°
(c) 24°
(d) 58°
8. If two parallel lines are intersected by a transversal, then the bisectors of the two pairs of interior angles
enclose a
(a) Trapezium
(b) Rectangle
(c) Square
(d) none of these
9. In fig., AB || CD, a is equal to
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(a) 93°
(b) 103°
(c) 83°
(d) 97°
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10. The complement of an angle exceeds the angle by 60°. Then the angle is equal to
(a) 25°
(b) 30°
(c) 15°
(d) 35°
11. In a ∆ABQ, if 2∠A = 3∠B = 6∠C, Then ∠A is equal to
(a) 60°
(b) 30°
(c) 90°
(d) 120°
12.
The side AB and AC of ∆ABC have been produced to D and is respectively. The bisectors of
∠CBD and ∠BCE meet at O. If ∠A = 40°, then ∠BOC is equal to
(a) 60°
(b) 65°
(c) 75°
(d) 70°
13.
In the given figure, AM BC and AN is the bisector of ∠A. What is the measure of ∠MAN
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(a) 17.5°
(b) 15. 5°
(c) 20°
(d) 25°
14.
H In the given figure, AB || DC, find the value of x
(a) x = 8
(b) x = 9
(c) x = 8 or 9
(d) x =10
15.
The areas of two similar s are 81 cm 2 and 144 cm 2 . If the largest side of the smaller is 27
cm, then the largest side of the larger is:
(a) 24 cm
(b) 48 cm
(c) 36 cm
(d) None of these
16.
A ladder 1 5 m long reaches a window which is 9 m above the ground on one side of street.
Keeping its foot at the same point, the ladder is turned to the other side of the street to reach 12 m
high. What is the width of the street
(a) 3 1 m
(b) 12 m
(c) 30 m
(d) 21 m
17.
A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower
casts the shadow 40 m long on the ground. Find the height of the tower.
(a) 600 m
(b) 160 m
(c) 60 m
(d) 52 m
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18. In an equilateral ABC, if AD BC, then
(a) 3AB2 = 2AD2
(b) 2AB2 = 3AD2
(c) 3AB2 = 4AD2
(d) 4AB2 = 3AD2
19.
In the given figure, ABD = CDB = PQB = 90°. Then
(a)
1 1 1
x y z
(b)
1 1 1
x z y
(b)
1 1 1
x y z
(b)
1 1 1
y x z
20. The area of two similar s are 121 cm 2 and 81 cm 2 respectively. What is the ratio of their
corresponding heights (altitudes)
(a) 11/9
(b) 22/9
(c)11/18
(d)None of these
21. ABC is a in which AB = AC and D is a point on AC such that BC2 = AC× CD. Then
(a) BD = DC
(b) BD = BC
(c) BD = AB
(d) BD = AD
22. In ABC, the median BE intersects AC at E, if BG = 6 cm, where G is the centroid, then BE is equal
to
(a) 7 cm
(b) 9 cm
(c) 8 cm
(d) 10 cm
23.
In the given figure, In a ABC, B = C. I f AM is the bisector of BAC and AN BC, then MAN is
equal to
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1
(B C)
2
1
(b) (C B)
2
1
(d) (B C)
2
(c) B C
24. In the given figure, the side BC of a ABC is produced on both sides. Then 1 + 2 is equal to
(a) A 1800
(c)
(b) 1800 A
1
(A 1800 )
2
(d) A 900
25. In fig, AB = AC, D is a point on AC and E on AB such that AD = ED = EC = BC. Then A : B
(a) 1 : 2
(b) 2 : 1
(c) 3 :1
(d) 1 : 3
Answers
1.(c)
2. (b)
3. (a)
4. (d)
5. (a)
6. (d)
7. (a)
8. (b)
9. (a)
10. (c)
11. (c)
12. (d)
13. (a)
14. (c)
15. (c)
16. (d)
17. (c)
18. (c)
19. (c)
20. (a)
21. (b)
22. (b)
23. (d)
24. (a)
25. (d)
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Chapter: 9 QUADRILATERALS AND PARALLELOGRAMS
Quadrilateral: A plane figure bounded by four line segments AB, EC, CD and DA is called a quadrilateral,
written as quad. ABCD or ABCD
Various types of quadrilaterals:
Parallelogram: A quadrilateral in which opposite sides are parallel is called parallelogram, written as || gm
Rectangle: A parallelogram each of whose angles is 90°, is called a rectangle, written as rect. ABCD.
Square: A rectangle having all sides equal is called a square.
Rhombus: A quadrilateral having all sides equal is called a rhombus
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Trapezium: A quadrilateral in which two opposite sides are parallel and two opposite sides are non-parallel, is
called a trapezium
Kite: A quadrilateral in which pairs of adjacent sides are equal is known as kite.
Key Results to Remember:
The sum of all the four angles of a quadrilateral is 3600.
In a parallelogram
(i) opposite sides are equal.
(ii) Opposite angles are equal.
(iii) Each diagonal bisectors the parallelogram
(iv) the diagonal bisect each other.
A quadrilateral is ||gm
(i) if both pairs of opposite sides are equal
or (ii) if both pairs of opposite angles are equal
or (iii) if the diagonals bisect each other.
or (iv) if a pair of opposite sides are equal and parallel.
The diagonals of a rectangle are equal.
The diagonals of a ||gm are equal, it is a rectangle.
Diagonals of a rhombus are perpendicular to each other.
Diagonals of a square are equal and perpendicular to each other.
The figure formed by joining the mid-points of the pairs of consecutive sides of a quadrilateral is a ||gm.
The quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rectangle.
The quadrilateral formed by joining the mid-points of the consecutive sides of a rhombus is a rectangle.
If the diagonals of a quadrilateral are perpendicular to each other, then the quadrilateral formed by
joining the mid-points of its sides, is a rectangle.
The quadrilateral formed by joining the mid-points of the sides of a square, is also a square.
POLYGONS
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Polygon: A closed plane figure bounded by line segments is called a polygon.
The line segments are called its sides and the points of intersection of consecutive sides are called its
vertices. An angle formed by two consecutive sides of a polygon is called an interior angels or simply an
angle of the polygon.
A polygon is named according to the number of sides, it has.
In general, a polygon of n sides is called n–gon. Thus, a polygon having 18 sides is called 18–gon.
Diagonal of a polygon: Line segment joining any two non–consecutive vertices of a polygon is called its diagonal.
Convex polygon: If all the (interior) angles of a polygon are less than 180°, it is called a convex polygon. In the
– figure given below, ABCDEF is a convex polygon. In fact, it is a convex hexagon.
(In other words, a polygon is a convex polygon if the line segment joining any two points inside it lies
completely inside the polygon).
Concave Polygon: If one or more of the (interior) angles of a polygon is greater than 180° i.e. reflex, it is
called
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concave (or re–entrant) polygon In the figure given below, ABCDEFG is a concave polygon. In fact, it is a
concave heptagon.
Exterior angle of convex polygon: If we produce a side of polygon, the angle it makes with the next side
is called an exterior angle. In the diagram given below, ABCDE is a pentagon. Its side AB has been produced
to P then CBP is an exterior angle.
Note: Corresponding to each interior angle, there is an exterior angle. Also, as an exterior angle and its adjacent
interior angle make a straight line, we have an exterior angle + adjacent interior angle = 180°
Regular polygon: A polygon is called regular polygon if all its sides have equal length and all its angles
have equal size.
Thus, in a regular polygon
(i) All sides are equal in length.
(ii) All interior angles are equal in size.
(iii) All exterior angles are equal size.
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Note: All regular polygons are convex.
KEY RESULTS TO REMEMBER:
(a) If there is a polygon of n sides (n > 3), we can cut it into (n 3), triangles with a common vertex, and
so the sum of the interior angles of a polygon of n sides would be
2n 4
(b) The there is a regular polygon of n sides (n 3), then its each interior angle is equal to
90
n
0
360
(c) Each exterior angle of a regular polygon of n sides is equal to
n
The sum of all the exterior angles formed by producing the sides of a convex polygon in the same
order is equal to four right angles.
EXPLANATION: If in a convex polygon P1 P2 P3P4 P5 all the sides are produced in order, forming exterior
angles 1, 2, 3, 4, and 5 then 1 2 3 4 5 4 right angles.
3600
.
If each exterior angle of a regular polygon is x°, then the number of sides in the polygon
x
Note: Greater the number of sides in a regular polygon, greater is the value of its each interior angle and
smaller is the value of each exterior angle.
n(n 1)
n.
If a polygon has n sides, then the number of diagonals of the polygon
2
CIRCLE AND TANGENTS
Circle: A circle is a set of all those points in a plane, each one of which is at given constant distance from a
given fixed point in the plane.
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The fixed point is called the centre and the given constant distance is called the radius of the circle. A circle
with centre O and radius r is usually denoted by C (O, r)
Tangent: A line meeting a circle in only one point is called a tangent to the circle. The point at which the
tangent line meets the circle is called the point of contact.
Secant: A line which intersects a circle in two distinct points is called a secant line.
KEY RESULTS TO REMEMBER:
The perpendicular from the centre of a circle to a chord bisects die chord.
EXPLANATION: If ON AB, then AN = NB.
Note: The converse of above theorem is true and can be stated as.
The line joining the centre of a circle to the midpoint of a chord Is perpendicular to the chord.
EXPLANATION: If AM = MB, then OM AB.
Cor. The perpendicular bisectors of two chords of a circle intersect at its centre.
Equal chords of a circle subtend equal angles at the center.
EXPLANATION: If AB = CD, then 1 = 2
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(Converse of above Theorem): If the angles subtended by two chords at the centre of a circle are equal
then the chords are equal.
EXPLANATION: If 1 = 2, then AB = CD.
Equal chords of a circle are equidistant from the centre.
EXPLANATION: If the chords AB and CD of a circle are equal and if OX⊥ AB and OY ⊥ CD then OX= OY.
(Converse of above Theorem) Chords equidistant from the center of the circle are equal.
EXPLANATION: If OX AB OY CD and OX = OF, then chords AB = CD
In equal circles (or in the some circle), equal chords cut of equal arcs.
EXPLANATION: If the chords AB = CD, then arc AB = are CD.
In equal circles (or in the same circle) if two arcs subtend equal angles at the centre (or at the
circumference), the arcs are equal.
EXPLANATION: If BOA= XOY, then arc AB = arc XY or if BPA= XQY, then arc AB = arc XY.
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The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point
on the remaining part of the circle.
(The theorem is popularly known as degree Measure Theorem)
EXPLANATION: A circle, centre O, with AOB at the centre, ACB at the circumference, standing on the
same arc AB, then AOB = 2 ACB
Angles in the same segment of a circle are equal.
EXPLANATION: A circle, centre O, ACB and ADB are angles at the circumference, standing on the same
arc, then
ACB = ADB
(Angles in same arc)
Or
(Angles in same segment)
The angle in a semicircle is a right angle
EXPLANATION: In the figure given below ACB – 90°
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(Converse of above theorem) The circle drawn with hypotenuse of a right triangle as diameter passes
through its opposite vertex.
EXPLANATION: The circle drawn with the hypotenuse AB of a right triangle ACB as diameter passes
through its opposite vertex.
If APB AQB and if P, Q are on the same side of AB, then A, B, Q, P are concyclic i.e. lie on the same
circle.
The sum of the either pair of the opposite angles of a cyclic quadrilateral is 1800.
EXPLANATION: If ABCD is a cyclic quadrilateral, then A + C = B + D = 180°
(Converse of above theorem): If the two angles of a pair of opposite angles of a quadrilateral are
supplementary then the quadrilateral is cyclic.
If a side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite
angle.
EXPLANATION: If the side AB of a cyclic quadrilateral ABCD is produced then 1 = 2.
THEOREMS ON TANGENTS
A tangent at any point of a circle is perpendicular to the radius through the point of contact.
EXPLANATION: If AB is a tangent at a point P to a circle C (O, r) then PO AB
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(converse of above theorem) A line drawn through the end of a radius and perpendicular to it, is a
tangent to the circle.
The lengths of two tangents drawn from an external point to a circle are equal.
EXPLANATION: If two tangents AP and AQ are drawn from a point A to a circle C (O, r), then AP =AQ
If two chords AB and CD intersect internally (ii) or externally (i) at a point P then PA PB PC PD
If PAB is a secant to a circle intersecting the circle at A and B is a tangent segment PA×PB =PT2 (refer the
figure below)
(Popularly known as Tangent-Secant theorem)
Alternate Segment Theorem:
In the figure below, if BAC is the tangent at A to a circle and if AD is any chord, then
DAC APD and PAB PDA (Angles in alternate segment)
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Note: The converse of the above theorem is true.
If two circles touch each other internally or externally, the point of contact lies on the line joining their
centers.
EXPLANATION: If two circles with centre O1and O2 which touch each other internally (i) or externally (ii),
at a point A then the point A lies on the line O1 O2, i.e. three points A, O1 and O2 are collinear.
SOME USEFUL RESULTS
There is one and only one circle passing through three non-collinear points.
Two circles are congruent if and only if they have equal radii.
Of any two chords of a circle, the one which is greater is nearer to the centre.
Of any two chords of a circle, the one which is nearer to the centre is greater.
If two circles intersect in two points, then the line through the centers is perpendicular bisector of the
common chord.
Angle in a major segment of a circle is acute and angle in a minor segment is obtuse.
If two tangents are drawn to a circle from an external point then.
(i) They subtend equal angles at the centre.
(ii) They are equally inclined to the segment, joining the centre to that point.
EXPLANATION: In a circle C (O, r), A is a point outside it and AP and AQ are the tangents drawn to the circle
Then, 1 = 2 and 3 = 4
If a circle touches all the four sides of a quadrilateral then the sum of opposite pair of sides are equal.
EXPLANATION: If ABCD is a circumscribed quadrilateral, then
Then, AB + CD = AD + EC
If two chords AB and AC of a circle are equal. Then the bisector of BAC passes through the centre O of
the circle
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The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
EXPLANATION: If ABCD is a cyclic quadrilateral in which AP, BP, CR and DR are the bisectors of
A, B, C and D respectively then
quadrilateral PQRS is also cyclic.
A cyclic trapezium is isosceles and its diagonals are equal.
EXPLANATION: If ABC cyclic trapezium such that AB || DC, then AD = BC and AC = BD
If two opposite sides of a cyclic quadrilateral are equal then the other two sides are parallel.
EXPLANATION: A cyclic quadrilateral ABCD in which AD = BC
Then, AB||CD
An isosceles trapezium is always cyclic.
EXPLANATION: A trapezium ABCD in which AB||CD and AD= BC
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Then ABCD is a cyclic trapezium.
Any four vertices of a regular pentagon are concyclic (lie on the same circle).
SAMPLE QUESTIONS
In each of the following questions a number of possible answers are given, out of which one
answer is correct. Find out the correct answer.
1. The diagonals of a rectangle ABCD meet at O. If BOC = 44°, then OAD is equal to
(a) 90°
(b) 60°
(c) 100°
(d) 68°
2. PQRS is a square. The SRP is equal to
(a) 45°
(b) 90°
(c) 100°
(d) 60°
3. ABCD is a rhombus with ABC = 56°, then ACD is equal to
(a) 90°
(b) 60°
(c) 56°
(d) 62°
4. ABCD is a parallelogram and X, Y are the .mid points of sides AB and CD respectively. Then
quadrilateral AXCY is a
(a) parallelogram
(b) rhombus
(c) square
(d) rectangle
5. X, Y is the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined
intersecting in P; CX and BY are joined intersecting in Q. Then PXQY is a
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(c) parallelogram
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(b) rhombus
(d) square
6. P is the mid-point of side AB to a parallelogram ABCD. A line through B parallel to PD meets DC at Q and
AD produced at R. Then BR is equal to
(a) BQ
(c) 2 BQ
(b)1/2
(d) None of these
7. ABCD is a trapezium in which AB|| CD. M and N are the mid-points of AD and BC respectively. If AB =
12 cm an MN = 14 cm. Find CD.
(a) 2 cm
(c) 12 cm
(b) 5 cm
(d) 16 cm
8. PQRS is a parallelogram. PX and QY are, respectively, the perpendicular form P and Q to SR and SR
Produced. The PX is equal to
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(a) QY
(b) 2QY
1
( c ) QY
(d) XR
2
9. In a ABC, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm
and AB = 30 cm. The perimeter of the quad. ARPQ is
(a) 91 cm
(b) 60 cm
(c) 51cm
(d) 70 cm
10. ABCD is a parallelogram. P is a point on AD such that AP = (1/3) AD and Q is a point on BC such that CQ =
(1/3) BC. Then AQCP is a
(a) parallelogram
(b) Rhombus
(c) rectangle
(d) square
11. The angles of a quadrilateral are respectively 100°, 98° and 92°. The fourth angle is equal to
(a) 90°
(b) 95°
(c) 70°
(d) 75°
12. The measure of each angle of a regular hexagon is
(a) 110°
(b) 130°
(c) 115°
(d) 120°
13. The interior angle of a regular polygon is 108°. The number of sides of the polygon is
(a) 6
(b) 7
(c) 8
(d) 5
14. The number of diagonals in a hexagon is
(a) 9
(b) 8
(c) 10
(d) 7
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15. If the number of diagonals of a polygon is 27, then the number of sides is
(a) 10
(b) 9
(c) 11
(d) 6
16. One angle of a pentagon is 140°. If the remaining angles are in the ratio 1 : 2 : 3 : 4 . The size of the
greatest angle is
(a) 150°
(b) 180°
(c) 160°
(d) 170°
17. The exterior angle of a regular polygon is one-third of its interior angle. The number of the sides of the
polygon is
(a) 9
(b) 8
(c) 10
(d) 12
18. The ratio of the measure of an angle of a regular octagon to the measure of its exterior angle is
(a) 3 : 1
(b) 2 : 1
(c) 1 : 3
(d) 1 : 2
19. ABCDE is a regular pentagon. Diagonal AD divides CDE in to parts, then the ratio of is equal
ADE
is
ADC
equal to
(a) 3 : 1
(b) 1 : 4
(c) 1 : 3
(d) 1 : 2
20. The difference between an exterior angle of ( n –1)sided regular polygon and an exterior angle of ( n +2)
sided regular polygon is 6°, then the value of n is
(a) 14
(b) 15
(c) 13
(d) 12
21. The radius of a circle is 13 cm and the length of one its chords is 10 cm. What is the distance of the chord
from the centre
(a) 10 cm
(b) 15 cm
(c) 12 cm
(d) 9 cm
22. In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A, B
and C. If ∠ADC = 130°. Find ∠CAB.
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(a) 40°
(b) 50°
(c) 30°
(d) 130°
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23. A cyclic quadrilateral whose opposite angles are equal, is a:
(a) Parallelogram but not a rhombus.
(b) Rhombus
(c) Rectangle
(d) Square but not a rectangle
24. In the given figure, TP and TQ are tangents to the circle. If ∠PAQ = 70°, what is ∠ PTQ?
(a) 30°
(b) 45°
(c) 60°
(d) 40°
25. In the given figure, PA and PB are tangents from a point P to a circle such that PA = 8 cm and ∠APB = 60°.
What is the length of the chord AB?
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(a) 8 cm
(b) 10 cm
(c) 6 cm
(d) 12 cm
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26. In the given figure O is the centre of the circle and PT is a tangent at T. If PC = 3 cm, PT = 6 cm, calculate
the radius of the circle.
(a) 9 cm
(b) 4.5 cm
(c) 8 cm
(d) 12 cm
27. In the given figure, ABCD is a quadrilateral in which ∠O = 90. A circle C(0, r) touches the sides AB, BC, CD
and DA at P, Q, R, S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find the value of r.
(a) 14 cm
(b) 15 cm
(c) 10 cm
(d) 16 cm
28. In the given figure, ABC is an isosceles ∆ in which AB = AC. A circle through B touches AC at its mid-point
D and intersects AB at P. Then which of the following is correct
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3
(a) AP AB
4
(b) AP
2.5
AB
3
4
(c) AP AB
5
1
(d) AP AB
4
29. ABC is a right angled ∆ with BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ∆ABC. The
radius of the circle is
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 1 cm
30. In the given figure, O is the centre of the circle. Then X + Y is equal to
Z
2
(a) 2 Z
(b)
(c) Z
(d) None
ANSWERS
1. (D)
9.(c)
17.( B)
25. (a)
2. (a)
10.(a)
18. (A)
26. (b)
3. (d)
11.(C)
19. (A)
27. (a)
4. (a)
12. (d)
20.( c)
28. (d)
5. (c)
13. (d)
21. (c)
29. (c)
6. (c)
14. (a)
22. (a)
30.(c)
7. (D)
15. (b)
23. (c)
8. (a)
16.(c)
24. (c)
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Chapter: 10 COORDINATE GEOMETRY
INTRODUCTION
Geometry begins with a point and straight line. Until now, we have studied geometry without any use of algebra.
In 1637, Descartes used algebra in the study of geometrical relationships. Thus, a new type of geometry was
introduced which was given the name analytical geometry or coordinate geometry. Thus, coordinate
geometry is that branch of mathematics in which geometry is studied algebraically, i.e. geometrical figures are
studied with the help of equations.
SOME BASIC FORMULAE
1. Distance Formula Distance between two points P x1 , y 1 and Q x 2 ,y 2 is given by
PQ
2
x 2 x1
y2 y1
2
Illustration 1 : Find the distance between the pair of points A(2, 5) and B(− 3, 7).
Sol :
AB
2
3 2
2
7 5 25 4 29.
2. Section Formulae
(a) Formula for internal division The coordinates of the point R(x, y) which divides the join of two given
points P(x1, y1) and Q (x2, y2) internally in the ratio m1 : m2 are given by
m1 x 2 m2 x 1 m1 y 2 m2 y 1
,
.
m1 m2
m1 m2
(b) Formula for external division The coordinates of the point R(x, y) which divides the join of two given
points P(x1, y1) and Q (x2, y2)
externally in the ratio m1 : m2 are given by
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PR m1 x1 x2 y 1 y 2
,
.
QR m2 2
2
Illustration 2 : Find the coordinates of the point which divides:
(i) the join of (2, 3) and (5, −3) internally in the ratio 1 :2
(ii) the join of (2, 1) and (3, 5) externally in the ratio 2 :3
Solution
Let (x, y) be the coordinates of the point of division. Then,
1 5 2 2 5 4
x
3
12
3
1 3 23 3 6
y
1.
12
3
Coordinates of the point of division are (3, 1).
(ii) Let (x, y) be the coordinates of the point of division.
Then, x
2 3 3 2
66
0
1
23
2 5 31 10 3
y
7
23
1
Coordinates of the point of division are (10, − 7).
Illustration 3 : Find the coordinates of the midpoint of the joint of points P(2, −1) and Q(− 3, 4).
Solution
The coordinates of the mid-point are
23
1
x
2
2
1 4 3
y
.
2
2
1 3
Coordinates of the mid-point are , .
2 2
Note: If the point R is given and we are required to find the ratio in which R divides the line segment PQ,
it is convenient to take the ratio k : 1.
Then, the coordinates of R are
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kx2 x1 ky 2 y 1
k 1 , x 1 .
Illustration 4: In what ratio does the point (6, − 6) divide the join of (1, 4) and (9, −12)?
Solution
Let the point R (6, −6) divides the join of P(1, 4) and Q (9, −12) in the ration k : 1.
By section formula, the coordinates of R are
k 9 1 1 k 12 1 4
9k 1 12k 4
,
,
.
, i.e.
k 1
k 1
k 1
k 1
But the coordinates of R are given to be (6, − 6).
9k 1
12k 4
6 and
6
k 1
k 1
9k 1 6k 6 and 12k 4 6k 6
3k 5 and 6k 10
5
In either case, k = ve
3
5
R divides PQ internally in the ratio :1 i.e. 5 : 3.
3
3. Centroid of a Triangle The point of concurrence of the medians of a triangle is called the centroid of
triangle. It divides the median in the ratio 2 : 1.
The coordinates of the centroid of a triangle whose vertices are x1 , y 1 , x2 ,y 2 and x3 ,y 3 are given by
x1 x2 x3 y 1 y 2 y 3
,
.
3
3
Illustration 5 Find the centroid of the triangle whose angular points are (3, −5), (−7, 4) and (10, −2),
respectively.
Sol :
The coordinates of centroid are
3 7 10 5 4 2
,
2, 1 .
3
3
4. Incentre of a Triangle Incentre of a triangle is the point of concurrence of the internal bisectors of the
angles of a triangle.
The coordinates of the incentre of a triangle whose vertices are x1 , y 1 , x2 ,y 2 and x3 ,y 3 are given by
ax1 bx 2 cx3 ay 1 by 2 cy 3
,
.
abc
a bc
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Illustration 5: Find the coordinates of in centre of a triangle having vertices A (0, 0), B (20, 15) and C
(−36, 15).
Sol :
We have,
2
2
a BC
20 36 15 15
b AC
36
c AB
20
2
15 39
2
15 25.
56
2
2
Coordinate of in centre are
ax1 bx2 cx3 56.0 39.20 25. 36
1.
a bc
56 39 25
ay by 2 cy 3 56.0 39.15 25.15
y 1
8
abc
56 39 25
Thus, I 1, 8 .
x
5. Area of a Triangle The area of a triangle whose vertices are
1
x1 , y1 , x2 ,y 2 and x3 ,y3 is given by A [x1 y 2 y 3 x2 y3 y 1 x3 y1 y 2 ]
2
Condition of Collinearity of Three Points
The points A x1 , y 1 , B x 2 , y 2 and C x3 , y 3 will be collinear (i.e. will lie on a straight line) if the area of the
triangle, assumed to be formed by joining them is zero.
SHORT-CUT METHOD FOR FINDING THE AREA
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1. Write the coordinates of the vertices taken in order in two columns. At the end, repeat the coordinates of
the first vertex.
2. Mark the arrow-heads as indicated. Each arrow-head shows the product,
3. The sign of the product remains the same for downward arrows while it changes for an upward arrow.
4. Divide the result by 2.
1
5. Thus, [ x1 y 2 x 2 y 1 x 2 y 3 x3 y 2 x 3 y 1 x1 y 3 ].
2
Illustration 6: Find the area of a triangle whose vertices are (4, 4) (3, − 2) and (−3, 16).
Solution
Required area
1
8 12 48 6 12 64
2
1
1
54 54 27 sq units.
2
2
Illustration 8 Show that the three points (−1, −1), (2, 3) and (8, 11) lie on a line.
Sol:
The area of the triangle whose vertices are (−1, −1), (2, 3) and (8, 11) is
1
1
3 2 22 24 8 11 0 0
2
2
since the area of the triangle is zero, the given points are collinear.
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Slope or Gradient of a Line
The tangent of the angle which a line makes with the positive direction of x-axis is called the slope or the
gradient of the line. It is generally denoted by m. If a line makes an angle with x-axis, then its slope
= tan , i.e., m = tan .
Note: 1. If a line is parallel to x-axis, m = tan 0 = 0.
2. If a line is parallel to y-axis, m = tan 90° = .
Illustration 7 : Find the slope of a line whose inclination with x-axis is 30°.
Solution
1
. Slope of a Line Joining Two Given Points
3
The slope of the line joining two points x1 , y 1 and x 2 , y 2 is
Slope , m tan 300
y 2 y 1 Difference of ordinates
x 2 x 1 Difference of abscissae
Illustration 8 : Find the slope of the line passing through the points (2, 3) and (4, 9).
Solution
y y1 9 3
3.
Slope of the line = 2
x2 x1 4 2
Parallel and Perpendicular Lines
(a) Two lines are parallel if and only if their slopes m1, m2 are equal, i.e. if m1 = m2.
(b) Two lines are perpendicular if and only if their slopes m1, m2 satisfy the condition m1 m2 = − 1.
Illustration 9: Show that the line joining (2, −3) and (−5, 1) is
(a) parallel to the line joining (7, −1) and (0, 3),
(b) perpendicular to the line joining (4, 5) and (0, −2).
Sol:
Let l1 be the line joining the points (2, −3) (−5, 1).
1 3
4
Slope of l1
.
5 2
7
m
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Let l2 be the line joining the points (7, − 1) and (0, 3).
3 1
4
Slope of l2
.
0 7
7
Slope of l1 slope of l2 (each 4 / 7).
Lines l1 and l2 are parallel.
(b) Let l3 be the line joining the points (4, 5) and (0, −2).
2 5 7 7
Slope of l3
.
0 4 4 4
4 7
Slope of l1 slope of l3 1,
7 4
the lines l1 and l3 are perpendicular.
Locus
When a point moves so that it always satisfies a given condition or conditions, the path traced out by it is
called its locus under these conditions.
Illustration 10: Let O be a given point in the plane of the paper and let a point P move on the paper so that
its distance from O is constant and is equal to a. All the positions of the moving point must lie on a circle
whose centre is O and radius is a. This circle is, therefore, the locus of P when it moves under the conditions
that its distance from O is equal to a constant a.
Shortcut Method to Find the Locus
1. Take a point on the locus and suppose that its coordinates are (x, y).
2. Apply the given conditions(s) to (x, y) and simplify the algebraic equation so formed.
3. The simplified equation is the required equation of the locus.
Illustration 11: A point moves so that its distance from (3, 0) is twice its distance from (−3, 0). Find the
equation of its locus.
Sol:
Let P(x, y) be any point of the locus. And, A (3, 0) and B (−3, 0) be the given points.
By the given conditions, PA = 2 PB.
2
2
x 3 y 0
2
2
x 3 y 0
Squaring both sides,
x2 y 2 6x 9 4 x2 y 2 6x 9
2
2
2
or 3x 3y 30x 27 0
or x2 y 2 10x 9 0.
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Sample Question
1. The distance between the points (7, 9) and (3, −7) is
(b) 4 17
(a) 4 15
(c) 17 4
(d) 17 5
2. The distance between (cos , sin ) and (sin , − cos ) is
(a) 3
(b) 2
(c) 1
(d) 0
3. The distance Between the points (4, p) and (1, 0) is 5 then p =
(a) ± 4
(b) 4
(c) −4
(d) 0
4. The distance between the points (a sin 60°, 0) and (0, a sin 30°) is
(a) a / 2
(b) a 2
(c) a / 3
(d) None of these.
5. The distance between the two points is 5. One of them is (3, 2) and the ordinate of the second is − 1,
then its x-coordinate is
(a) 7, −1
(b) −7, 1
(c) −7, −1
(d) 7, 1
6. A line is of length 10 and one end is (2, − 3). If the abscissa of the other end is 10 then its ordinate is
(a) 3 or 9
(b) −3 or −9
(c) 3 or − 9
(d) − 3 or 9
7. The nearest point from origin is
(a) (2, −3)
(b) (5, 0)
(c) (2, −1)
(d) (1, 3)
8. The vertices of a triangle are A (2, 2), 5 (− 4, − 4), C(5, − 8). Then, the length of the median through C is
(a) 65
(b) 117
(c) 85
(d) 113
9. P(3, 4), g(7, 7) are collinear with the point R where PR = 10. Then, R =
(a) (5, 2)
(b) (−5, 2)
(c) (−5, −2)
(d) (5, −2)
10. The third vertex of an equilateral triangle whose two vertices are (2, 4), (2, 6) is
(a) ( 3 , 5)
(b) (2 3,5)
(c) (2 3, 5)
(d) (2, 5)
11. Three points (0, 0), (3, 3 ), (3, ) form an equilateral triangle. Then, =
(a) 2
(b) −3
(c) − 4
(d) none of these.
12. The perimeter of a triangle formed by (0, 0), (1, 0), (0, 1) is
(a) 1 2
(b) 2 1
(c) 3
(d) 2 2
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13.
ABC is an isosceles triangle with B (1, 3) and C (− 2, 7) then A =
(a) (5/6, 6)
(b) (6, 5/6)
(c) (7, 1/8)
(d) None of these.
14. The area of the triangle formed by (a, a), (a + 1, a + 1), (a + 2, a) is
(a) a3
(b) 2a
(c) 1
(d) 2
15. The ratio in which (− 3, 4) divides the line joining (1, 2) and (7, −1) is
(a) 2 : − 5
(b) 5 : 2
(c) 1 : − 5
(d) 1 : 5
16. Mid-points of the sides AB and AC of ABC are (3, 5) and (−3, −3) respectively, then the length of
BC =
(a) 10
(b) 15
(c) 20
(d) 30
17. The point (t, 2t), (− 2, 6) and (3, 1) are collinear then t =
(a) 3/4
(b) 4/3
(c) 3
(d) 4
18. The base vertices of a right angled isosceles triangle are (2, 4) and (4, 2) then its third vertex is
(a) (1, 1) or (2, 2)
(b) (2, 2) or (4, 4)
(c) (1, 10 or (3, 3)
(d) (2, 2) or (3, 3)
19. The points (1, −1), (2, −1), (4, −-3) are the mid points of the sides of a triangle. Then its centroid is
(a) (7, − 5)
(b) (1/3, −1)
(c) (− 7, 5)
(d) (7/3, − 5/3)
20. The points k, 2 2k , 1 k,2k and 4 k,6 2k are collinear then k =
(a) − 1 or 1/2
(c) −1 or 1
(b) −1/2 or 1
(d) None of these.
Answers
1. (b)
2. (b)
3. (a)
4. (d)
5. (a)
6. (c)
7. (c)
8. (c)
9.(c)
10. (c)
11. (d)
12.. (d)
13. (a)
14. (c)
15. (a)
16. (c)
17. (b)
18. (b)
19. (d)
20. (a)
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Chapter: 11 RATIO AND PROPORTION
RATIO
A ratio is a comparison of two quantities by division. It is a relation that one quantity bears to another with
respect to magnitude. In other words, ratio means what part one quantity is of another. The quantities may be
of same kind or different kinds. For example, when we consider the ratio of the weight 45 kg of a bag of rice to
the weight 29 kg of a bag of suga’ we are considering the quantities of same kind but when we talk of allotting 2
cricket bats to 5 sportsmen, we are considering quantities
of different kinds. Normally, we consider the ratio between quantities of the same kind.
a
If a and b are two numbers, the ratio of a to b is
or a b and is denoted by a : b. The two quantities that
b
are being compared are called terms. The first is called antecedent and the second term is called consequent.
3
For example, the ratio 3 : 5 represents with antecedent 3 and consequent 5.
5
Note:
1. A ratio is a number, so to find the ratio of two quantities, they must be expressed in the same units.
2. A ratio does not change if both of its terms are multiplied or divided by the same number. Thus,
2 4 6
etc.
3 6 9
TYPES OF RATIOS
1. Duplicate Ratio The ratio of the squares of two numbers is called the duplicate ratio of the two numbers.
32
9
3
For example, 2 or
is called the duplicate ratio of .
4
16
4
2. Triplicate Ratio The ratio of the cubes of two numbers is called the triplicate ratio of the two
numbers.
33
27
3
For example, 3 or
is triplicate ratio of .
4
4
64
3. Sub-duplicate Ratio The ratio of the square roots of two numbers is called the sub-duplicate ratio of
3
9
.
two numbers. For example, is the sub-duplicate ratio of
4
16
4. Sub-triplicate Ratio The ratio of the cube roots of two numbers is called the sub-triplicate ratio of two
numbers.
2
8
For example, is the sub-triplicate ratio of
.
3
27
5. Inverse Ratio of Reciprocal Ratio If the antecedent and consequent of a ratio interchange their places, the
new ratio is called the inverse ratio of the first.
1 1
Thus, if a : b be the given ratio, then : or b : a is its inverse ratio.
a b
3
5
For example, is the inverse ratio of .
5
3
6. Compound Ratio The ratio of the product of the antecedents to that of the consequents of two or more
given ratios is called the compound ratio. Thus, if a : b and c : d are two given ratios, then ac : bd is the
compound ratio of the given ratios.
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for example, if
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3 4
5
3 4 5
3
, and be the given ratios, then their compound ratio is
, that is, .
4 5
7
4 5 7
7
PROPORTION
The equality of two ratios is called proportion.
a c
If , then a, b, c and d are said to be in proportion and we write a : b : : c : d. This is read as “a is to be as c is
b d
to d”.
3 6
For example, since , we write 3 : 4 : : 6 : 8 and say 3, 4, 6 and 8 are in proportion.
4 8
a
c
Each term of the ratio and is called a proportional a, b, c and d are respectively the first, second, third and
b
d
fourth proportionals.
Here, a, d are known as extremes and b, c are known as means.
1. If four quantities are in proportion, then
Product of Means = Product of Extremes
For example, in the proportion a : b : : C : d, we have bc = ad.
From this relation we see that if any three of the four quantities are given, the fourth can be determined.
2. Fourth proportional If a : b : : c : x, x is called the fourth proportional of a, b, c.
a c
b c
We have, or, x
.
b x
a
b c
Thus, fourth proportional of a, b, c is
.
a
Illustration 1 Find a fourth proportional to the numbers 2, 5, 4.
Sol:
2 4
Let x be the fourth proportional then 2 : 5 : : 4 : x or .
5 x
5 4
x
10 .
2
3. Third Proportional If a : b : : b : x, x is called the third proportional of a, b.
a b
b2
We have, or x
.
b x
a
b2
Thus, third proportional of a, b is
.
a
Illustration 2 Find a third proportional to the numbers 2.5, 1.5.
Sol :
Let x be the third proportional, then
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2.5 : 1.5 : : 1.5 : x or
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2.5 1.5
.
1.5
x
1.51.5
0.9 .
2.5
4. Mean Proportional If a : x : x : : x : b , x is called the mean or second proportional of a, b.
a x
We have, or x 2 ab or x ab .
x b
Mean proportional of a and b is ab .
We also say that a, x, b are in continued proportion.
Illustration 3 Find the mean proportional between 48 and 12.
Sol :
Let x be the mean proportional. Then,
48 x
or, x2 576 or,x 24 .
48 : x : : x : 12 or ,
x 12
a c
5. If , then
b d
ab cd
(Componendo)
(i)
b
d
ab cd
(ii)
(Dividendo)
b
d
ab cd
(iii)
(Componendo and Dividendo)
ab cd
a a c a c
(iv)
.
b bd bd
p
Illustration 4 The sum of two numbers is c and their quotient is . Find the numbers.
q
Sol :
Let the numbers be x, y.
Given : x + y + c
……(1)
x p
……(2)
and,
y q
x
p
x
p
[Using (1)]
x y pq
c pq
pc
x
.
pq
SOME USEFUL SHORT-CUT METHODS
1. (a) If two numbers are in the ratio of a : b and the sum of these numbers is x, then these numbers will
ax
bx
and
be
, respectively. Or
ab
ab
If in a mixture of x litres, two liquids A and B are in the ratio of a : b, then the quantities of liquids A and
ax
bx
litres and
B in the mixture will be
litres, respectively.
ab
ab
x
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(b) If three numbers are in the ratio of a : b : c and the sum of these numbers is x, then these numbers
ax
bx
cx
and
will be
,
respectively.
abc abc
abc
Illustration 5 Two numbers are in the ratio of 4 : 5 and the sum of these numbers is 27. Find the two numbers.
Sol :
Here, a = 4, b = 5 and x = 27.
ax
4 27
12
The first number =
a b 45
bx
5 27
15 .
and, the second number =
a b 4 5
Illustration 6 Three numbers are in the ratio of 3 : 4 : 8 and the sum of these numbers is 975. Find the three
numbers.
Sol :
Here, a = 3, b = 4, c = 8 and x = 975.
ax
3 975
195 .
The first number =
a b c 3 4 8
bx
4 975
260 .
The second number =
a b c 3 4 8
cx
8 975
520 .
and, the third number =
a b c 3 4 8
2. If two numbers are in the ratio of a : b and difference between these numbers is x, then these numbers will
be
ax
bx
(a)
and
, respectively (where a > b)
ab
ab
ax
bx
and
, respectively (where a < b).
(b)
ba
ba
Illustration 7 Two numbers are in the ratio of 4 : 5. If the difference between these numbers is 24, then find
the numbers.
Sol :
Here, a = 4, b = 5 and x = 24.
ax
4 24
bx 5 24
120 .
= 96 and, the second number =
The first number =
b a 5 4
ba 5 4
3. (a) If a : b = n1 : d1 and b : c = n 2 : d2, then a : b : c = n1 n2 : d1 n2 : d1 d2 .
(b) If a : b = n 1 : d1, b : c = n2 : d2 and c : d = n3 : d3, then
a : b : c : d = n1 n2 n3 : d1 n2 n3 : d1 d2 n3 : d1 d2 d3 .
Illustration 8 If A : B = 3 : 4 and B : C = 8 : 9, find A : B : C.
Sol :
Here, n1 = 3, n2 = 8, d1 = 4 and d2 = 9.
a : b : c n1 n2 : d1 n2 : d1 n2
= 3 8 : 4 8 : 4 9
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= 24 : 32 : 36 or, 6 : 8 : 9.
Illustration 9 If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D.
Sol :
Here, n1 = 2, n2 = 4, n3 = 6, d1 = 3, d2 = 5 and d3 = 7.
A : B :C : D n1 n2 n3 : d1 n2 n3 : d1 d2 : n3 : d1 d2 n3
= 2 4 6 : 3 4 6 : 3 5 6 : 3 5 7
= 48 : 72 : 90 : 105 or, 16 : 24 : 30 : 35.
Thus, A : D = 16 : 35.
4. (a) The ratio between two numbers is a : b. If x is added to each to these numbers, the ratio becomes
c : d. The two numbers are given us:
ax c d
bx c d
and
.
ad bc
ad bc
(b) The ratio between two numbers is a : b. If x is subtracted from each of these numbers, the ratio
becomes c : d.
The two numbers are given as :
ax d c
bx d c
.
and
ad bc
ad bc
Illustration 10 Given two numbers which are in the ratio of 3 : 4. If 8 is added to each of them, their ratio is
changed to 5 : 6. Find the two numbers.
Sol :
We have ,
a : b = 3 : 4, c : d = 5 : 6 and x = 8.
ax c d
The first number =
ad bc
3 8 5 6
12
3 6 4 5
and, the second number =
bx c d
ad bc
=
4 8 5 6
3 6 4 5
16 .
5. (a) If the ratio of two numbers is a : b, then the numbers that should be added to each of the numbers
in order to make this ratio c : d is given be
ad bc
.
c d
(b) If the ratio of two numbers is a : b, then the number that should be subtracted from each of the numbers in
bc ad
.
order to make this ratio c : d is given by
c d
Illustration 11 Find the number that must be subtracted from the terms of the ratio 5 : 6 to make it equal to 2 :
3.
Sol :
We have a : b = 5 : 6 and c : d = 2 : 3.
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The required number =
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bc ad
c d
6 2 5 3
3.
23
Illustration 12 Find the number that must be added to the terms of the ratio 11 : 29 to make it equal to 11 : 20.
Sol :
We have, a : b = 11 : 29 and to make it equal to 11 : 20.
ad bc
The required number =
c d
11 20 29 11
11.
=
11 20
6. There are four numbers a, b, c and d.
(i) The number that should be subtracted from each of these numbers so that the remaining numbers
ad bc
may be proportional is given by
.
a d b c
=
(ii) The number that should be added to each of these numbers so that the new numbers may be
proportional is given by
bc ad
.
a b b c
Illustration 13 Find the number subtracted from each of the numbers 54, 71, 75 99 leaves the remainders
which are proportional.
Sol :
We have, a = 54, b = 71, c = 75 and d = 99.
ad bc
The required number =
.
a d b c
54 99 7175
3.
54 99 71 75
7. The incomes of two persons are in the ratio of a : b and their expenditures are in the ratio of c : d. If
the saving of each person be Rs. S, then their incomes are given by
aS d c
bS d c
Rs.
and Rs.
. and their expenditures are given by
ad bc
ad bc
cS b a
dS b a
Rs.
.
and Rs.
ad bc
ad bc
Illustration 14 Annual income of A and B is in the ratio of 5 : 4 and their annual expenses bear a ratio of 4 : 3. If
each of them saves Rs. 500 at the end of the year, then find their annual income.
Sol :
We have, a : b = 5 : 4, c : d = 4 : 3 and S = 500.
aS d c
Annual income of A =
ad bc
5 500 3 4
=
5 3 4 4
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= Rs. 2500.
and annual income of B =
=
bS d c
ad bc
4 500 3 4
5 3 4 4
= Rs. 2000.
Illustration 15 The incomes of Mohan and Sohan are in the ratio 7 : 2 and their expenditures are in the ratio 4
: 1. If each saves Rs. 1000, find their expenditures.
Sol :
We have, a : b = 7 : 2, c : d = 4 : 1 and S = 1000.
cS b a 4 1000 2 7
A's expenditure
ad bc
7 1 2 4
= Rs. 20000
and, B’s expenditure =
dS b a
ad bc
1 1000 2 7
7 1 2 4
= Rs. 5000.
8. (a) If in a mixture or x litres of two liquids A and B, the ratio of liquids A and B is a : b, then the
quantity of liquid B to be added in order to make this ratio.
x ad bc
.
c : d is
c a b
(b) In a mixture of two liquids A and B, the ratio of liquids A and B is a : b. If on adding x litres of liquid B to the
ax
mixture, the ratio of A to B becomes a : c, then in the beginning the quantity of liquid A in the mixture was
cb
bx
litres and that of liquid B was
litres.
cb
Illustration 16 729 ml of a mixture contains milk and water in the ratio 7 : 2. How much more water is to be
added to get a new mixture containing milk and water in the ratio of 7 : 3.
Sol :
Here, x = 729, a : b = 7 : 2 and c : d = 7 : 3.
The quantity of water to be added
x ad bc 729 7 3 2 7
=
81 ml.
c a b
7 7 2
Illustration 17 A mixture contains alcohol and water in the ratio of 6 : 1. On adding 8 litres of water , the ratio
of alcohol to water becomes 6 : 5. Find the quantity of water in the mixture.
Sol :
We have, a : b = 6 : 1, a : c = 6 : 5 and x = 8.
The quantity of water in the mixture
bx
1 8
2 litres.
=
c b 51
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9. When two ingredients A and B of quantities q 1 and q 2 and cost price/unit c1 and c2 are mixed to get a
mixture c having cost price. Unit cm, then
q
c c
(a) 1 2 m and
q2 cm c1
c q c q
(b) cm 1 1 2 2 .
q1 q2
Illustration 18 In what ratio the two kinds of tea must be mixed together into one at Rs. 9 per kg and another
at Rs. 15 per kg, so that mixture may cost Rs. 10.2 per kg?
Sol :
We have, c 1 = 9, c2 = 15, cm = 10.2
q
c c
15 10.2 4.8 4
1 2 m
.
q2 cm c1
10.2 9 1.2 1
Thus, the two kinds of tea are mixed in the ratio 4 : 1.
Illustration 19 In a mixture of two types of oils O1 and O 1, the ratio O1 : O2 is 3 : 2. If the cost of oil O1 is Rs. 4
per litre and that of O2 is Rs. 9 per litre, then find the cost/litre of the resulting mixture.
Sol :
We have, q1 = q2 = 2, c1 = 4 and c2 = 9.
The cost of resulting mixture
c q c q
4 3 9 2 30
Rs. 6.
= 1 1 2 2
q1 q2
3 2
5
10.(a) If a mixture contains two ingredients A and B in the ratio a : b, then
a
b
100% and percentage of B in the mixture =
100% .
percentage of A in the mixture =
ab
ab
(b) If two mixtures M1 and M2 contain ingredients A and B in the ratios a : b and c : d respectively, then a third
mixture M3 obtained by mixing M1 and M2 in the ratio x : y will contain
cy
a x c y
ax
a b c d
ab cd
100% ingredient A, and 100%
x
y
x
y
dy
bx
or, a b c d 100% ingredient B.
xy
Illustration 20 If a mixture contains water and alcohol in the ratio 2 : 3, what is the percentage quantity of
water in the mixture?
Sol :
Here, a = 2, b = 3.
percentage quantity of water in the mixture
a
2
100
100%
=
ab
23
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2
200
100% =
or 40% .
5
5
SURDS AND INDICES
an is called a surd if n is a fraction and an is called an index if n is an integer. a is called the base.
SOME USEFUL FORMULAE
1.
2.
a m a n a m n
a m a n a m n
3.
a a
n m
m
m
a mn
m
5.
a n bn
b a
m
n
a b a m bn
6.
a
7.
n
a n b n ab , where ‘n’ is a +ve integer and ‘a’, ‘b’ are rational numbers
n
a na
, where ‘n’ is a +ve integer and ‘a’, ‘b’ are rational numbers
b
b
4.
n
8.
n
n
= a, where ‘n’ is a +ve integer and ‘a’ a + ve rational number
9.
m n
a mn a
10.
n m
a
k
m
n m
a , where ‘m’, ‘n’ are +ve integers and ‘a’ is a +ve rational number
n a k mn ak m , where ‘m’, ‘n’, ‘k’ are +ve integers and ‘a’ is a +ve rational number
11.
a a a
12.
a b ab
13. ( a b)2 = a + b + 2 ab
14. ( a b)2 a b 2 ab
15. a b c d a c and b d .
16.
17.
1
a b
1
a b
( a b) a b
a b
a b ( a b) a b
If x = n(n + 1), then
(a)
x x x ... n and
(b)
x x x ... n 1 .
a b
ab
a b
ab
Illustration 1: Find the value of 243
0.8
0.4
243 .
Sol:
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243
0.8
= 243
243
0.4
0.4
= 35
2
5
243
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[ a m a n a mn ]
32 9 .
2/3
Illustration 2: Find the value of 27
64 4/3
Sol:
2/3
2/3
4/3
64 [ a
= 3 4 = 9 4 9 256 2304 .
27
64 4/3 33
3 4/3
2
m
b n a m bn ]
4
Illustration 3: Find the value of 3
2 2
4
Sol:
4
3
2 2
= 9
4
1
=
3
2 2
1
=
9
2 4
2 4
= 81
4
1
81
4
4
1 1
= 4 .
3 3
Illustration 4: Find the value of x if
Sol:
We have,
5
2x 7 3 0 5 2x 7 3
5
2x 7
5
5
2x 7 3 0 .
35
2x 7 243 [ ( n a )n a]
2x 250 or x 125 .
Illustration 5: Find the value of 5 64 5 512 .
Sol:
5
64 5 512
[ n a n b n ab
=
5
64 512
=
5
82 83 5 85 8.
[ n a n a ]
Illustration 6: Find the value of
Sol:
3 2
729 6 729
= 6 36 3 .
3 2
729 .
[ m n a mn a ]
[ n an a ]
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7 5
Illustration7: Find the value of
(217 )5
5 3
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.
5 3
(7 )
Sol:
Given expression
(21)7 n m p m n p
=
(a ) a
5
(7)5
21
3.
[ n a n a ]
=
7
7
Illustration 8: Simplify each of the following by rationalizing the denominator.
1
(i)
2 3
7 3 5 2
(ii)
48 18
Sol:
(i)
=
=
2
2
=
=
=
( 3)2
2 3
= 2 3.
4 3
(ii)
=
1
1
2 3
2 3 2 3 2 3
2 3
7 3 5 2
48 18
7 3 5 2
4 3 3 2
7 3 5 2
42 3 32 2
7 3 5 2
4 3 3 2
4 3 3 2
4 3 3 2
(7 3 5 2)(4 3 3 2)
(4 3 3 2)(4 3 3 2)
7 3 4 3 7 3 3 2 5 2 4 3 5 2 3 2
(4 3)2 (3 2)2
28 33 21 3 2 20 2 3 15 2 2
16 3 9 2
28 3 21 6 20 6 15 2 84 21 20 6 30 114 41 6
=
.
30
30
48 18
Illustration 9: If a and b are rational numbers, find the values of a and b in the following equation:
=
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3 2
3 2
Sol:
3 2
3 2
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ab 6 .
3 2
3 2
3 2
3 2
( 3 2)2
( 3)2 ( 2)2
3 2 2 3 2 52 6
32
1
= 52 6 .
3 2
a b 6 5 2 6 a b 6 .
3 2
On equating rational and irrational parts, we get
a = 5 and b = 2.
Illustration 10: Find the value of ( 72 72 72 ... ) ( 12 12 12 ... )
Sol:
Since 72 = 9 × 8, therefore, 72 + 72 72 ... 9
Also, since 12 = 4 × 3 therefore, 12 12 12 ... 3 .
Thus, the given expression =
9
3.
3
SAMPLE QUESTIONS
1. In the ratio 11 : 14, if the antecedent is 55, the consequent is
(a) 70
(b) 90
(c) 60
(d) None of these
2. The mean proportional between 64 and 81 is
(a) 48
(b) 68
(c) 72
(d) None of these
3. The mean proportional of 0.25 and 0.04 is
(a) 0.01
(b) 0.1
(c) 10 10
(d) None of these
4. The ratio of two numbers is 3: 4 and their sum is 420. The greater of the two numbers is
(a) 360
(b) 240
(c) 180
(d) None of these
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5. The ratio of boys and girls in a school is 9 : 5. If the total number of students in the school is 1050, then
the number of boys is
(a) 785
( b ) 890
(c) 675
(d) None of these
6. If A : B = 7 : 5 and B : C = 9 : 11, then A : B : C is equal to
(a) 55 : 45 : 63
(b) 63 : 45 : 55
(c) 45 : 63 : 55
(d) None of these
7. If A : B = 3/4, B : C = 4/5, C : D = 5/6, then A : D will be
(a) 2 : 3
(b) 4 : 3
(c) 1 : 2
(d) None of these
8. If 3A = 4B = 5C, then A : B : C is
(a) 16 : 20 : 18
(b) 15 : 20 : 16
(c) 20 : 15 : 12
(d) None of these
9. If 3A = 5B and 2B = 3C, then A : C is
(a) 5 : 2
(b) 7 : 2
(c) 3 : 2
(d) None of these
10. The ratio of present ages of Sita and Gita is 4 : 3. If, 4 years before, the ratio of their ages was 2 : 1, the
present age of Sita is
(a) 8 years
(b) 10 years
(c) 12 years
(d) None of these
11. Two numbers are in the ratio of 5 : 8. If 12 be added to each, they are in the ratio of 3 : 4. Find the sum of
two numbers.
(a) 43
(b) 39
(c) 47
(d) None of these
12. Two numbers are in the ratio of 5 : 7. If 25 be subtracted from each, they are in the ratio of 35: 59. Find
the difference of the two numbers,
(a) 48
(b) 52
(c) 24
(d) None of these
13. When x is added to each term of 7 : 13 the ratio becomes 2 : 3 . The value of x is
(a) 7
(b) 11
(c) 5
(d) None of these
14. Find the number which, when subtracted from the terms of the ratio 12 : 17 makes it equal to the ratio 2 :
3.
(a) 2
(b) 6
(c) 8
(d) None of these
15. In a mixture of 60 litres, the ratio of milk and water is 2 : 1. What amount of water must be added to
make the ratio of milk and water as 1 : 2?
(a) 75 litres
(b) 55 litres
(c) 60 litres
(d) None of these
16. A mixture contains alcohol and water in the ratio of 12 : 5. On adding 14 litres of water, the ratio
of alcohol to water becomes 4 : 3 . The quantity of alcohol in the mixture is
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(a) 18 litres
(c) 26 litres
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(b) 24 litres
(d) None of these
17. If an alloy contains copper and silver in the ratio 3 : 7, then the percentage quantity of silver in the alloy is
(a) 90%
(b) 70%
(c) 60%
(d) None of these
0
1/2
4/5
1
18. Simplify 64
32
64
1
3
(b) 17
(a) 17
8
8
7
7
(c) 11
(d) 17
8
8
1/2
19. The term r2 s
is approximately equal to r
to (85)1/2?
(a) 9.06
(c) 9.22
s
. Which of the following is the closest approximation
2r
(b) 9.34
(d) 9.28
1
2
1
3
1
4
20. The largest number in the sequence 1,2 ,3 ,4 is
1
(b) 22
(a) 1
(c) 3
1
3
(d) 4
1
4
ANSWERS
1.(a)
2. (c)
3. (b)
4. (b)
5. (c)
6. (b)
7. (c)
8. (c)
9. (a)
10. (a)
11. (b)
12. (c)
13. (c)
14. (a)
15. (c)
16 (c)
17. (b)
18. (a)
19.( c)
20. (c)
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Chapter: 12 PERMUTATIONS
Introduction
(1) The Factorial: Factorial notation: Let n be a positive integer. Then, the continued product of first n
natural numbers is called factorial n, to be denoted by n! or n .
Also, we define 0! = 1.
when n is negative or a fraction, n! is not defined.
Thus, n! = n(n –1)(n – 2)…..3.2.1.
(2) Exponent of Prime p in n!: Let p be a prime number and n be a positive integer. Then the last integer
n
n
amongst 1, 2, 3,….(n – 1), n which is divisible by p is p, where denotes the greatest integer less than
p
p
n
or equal to .
p
Definition of permutation
The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given
group of persons or objects with due regard being paid to the order of arrangement of selection are called
the (different) permutations.
For example: Three different things a, b and c are given, then different arrangements which can be made by
taking two things from three given things are ab, ac , bc, ba, ca, cb.
Therefore the number of permutations will be 6.
Number of permutations without repetition
(1) Arranging n objects, taken r at a time equivalent to filling r places from n things.
The number of ways of arranging =The number of ways of filling r places.
= n(n –1) (n – 2)……(n – r + 1)
n n – 1 n – 2 n – r 1 n – r !
n!
=
n Pr
n
1
!
n
r
!
(2) The number of arrangements of n different objects taken all a time = nPn = n!
n!
(i) n P0 1; n Pr n.n 1 Pr 1
n!
1
0 or r ! r N
(ii) 0! 1;
r !
Number of permutations with repetition
(1) The number of permutations(arrangements) of n different objects, taken r at a time, when each object
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may occur once, twice, thrice,…..upto r time in any arrangement =The number of ways of filling r places
where each place can be filled by any one of n objects.
The number of permutations = The number of ways of filling r places = (n)r.
(2) The number of arrangements that can be formed using n objects out of which p are identical (and of one
kind) q are identical (and of another kind)q are identical (and of another kind), r are identical(and of another
n!
kind) and the rest are distinct is
p!q!r!
Conditional permutations
(1) Number of permutations of n dissimilar things taken r at a time when p particular things always occur =
n–pC
r–p r!.
(2) Number of permutations of n dissimilar things taken r at a time when p particular thing never occur = n–
pC r!.
r
(3) The total number of permutations of n different things taken not more than r at a time, when each thing
may be repeated and number of times, is
n nr 1
n 1
(4) Number of permutations of n different things, taken all at a time, when m specified things always comes
together is
m! × (n – m + 1)!.
(5) Number of permutations of n different things, taken all at a time, when m specified things never come
together is
n! – m! × (n – m +1)!
(6) Let there be n objects of which m objects are alike of one kind, and the remaining (n – m) objects are
alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed
n!
from these objects is
.
m! n m !
The above theorem can be extended further i.e., if there are n objects, of which p1 are alike of one kind; p2 are
alike of another kind; p3 are alike of 3rd kind;.….; pr are alike of rth kind such that p1 + p2 +….. pr = n; then the
n!
number of permutations of these n objects is
p1 ! p2 ! ...... pr !
Circular permutations
In circular permutations, what really matters is the position of an object relative to the others. Thus, in
circular permutations, we fix the position of the one of the objects and then arrange the other objects in all
possible ways.
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There are two types of circular permutations:
(i) The circular permutations in which clockwise and the anticlockwise arrangements give rise to different
permutations, e.g. Seating arrangements of persons round a table.
(ii) The circular permutations in which clockwise and the anticlockwise arrangements give rise to same
permutations, e.g. arranging some beads to form a necklace.
Difference between clockwise and anti-clockwise arrangement: If anti-clockwise and clockwise order of
arrangement are not distinct e.g., arrangement of beads in a necklace, arrangement of flowers in garland etc.
then the number of circular permutations of n distinct items is
n 1 !
2
(i) Number of circular permutations of n different things taken r at a time, when clockwise and anticlockwise
r
Pr
.
r
(ii) Number of circular permutations of n different things taken r at a time, when clockwise and
r
P
anticlockwise orders are taken as different is r .
2r
Theorems on circular permutations
Theorem (i): The number of circular permutations of n different objects is (n–1)!
Theorem (ii): The number of ways in which n persons can be seated round a table is (n–1)!
Theorem (iii): The number of ways in which n different beads can be arranged to form a necklace,
orders are taken as different is
is
1
(n–1)!
2
COMBINATIONS
Definition
Each of the different groups or selections which can be formed by taking some or all of a number of objects,
irrespective of their arrangements, is called a combination.
n
Notation: The number of all combinations of n things, taken r at a time is denoted by C(n, r) or nCr or . nCr
r
is always a natural number.
Difference between a permutations and combination:
(i) In a combination only selection is made whereas in a permutation not only a selection is made but also an
arrangement in a definite order is considered.
(ii) Each combination corresponds to many permutations.
For example, the six permutations ABC, ACB, BCA, BAC, CBA and CAB correspond to the same combination
ABC.
Number of combinations without repetition
The number of combinations(selections of groups) that can be formed from n different objects taken r(0 r
n!
. Also nCr = nCn – r.
n) at a time is nCr =
r! n r !
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Let the total number of selections(or groups) = x. Each group contains r objects, which can be arranged in r!
ways. Hence the number of arrangement of r objects = x × (r!). But the number of arrangements = nPr.
n
x(r!) = nPr x =
Pr
n!
n Cr .
=
r! n r
r!
Number of combinations with repetition and all possible selections:
(1) The number of combinations of n distinct objects taken r at a time when any object may object may be
repeated any number of times.
= Coefficient of xr in (1 + x + x2 + …… + xr)n
= Coefficient of xr in (1 – x)– n = n + r – 1Cr
(2)The total number of ways in which it is possible to form groups by taking some or all of n things at a time
is nC1 + nC2 +…..+ nCn = 2n – 1.
(3) The total number of ways in which it is possible to make groups by taking some or all out of n=(n1 + n2
+…..) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is {(n1 + 1)(n2 + 1)…..} – 1
(4) The number of selections of r objects out of n identical objects is 1.
(5) Total number of selections of zero or more objects from n identical objects is n + 1.
(6) The number of selections taking at least one out of a1 + a2 + a3 +……+ an + k objects, where a1 are alike (of
nth kind) and k are distinct = [(a1 + 1)(a2 + 1)(a3 + 1)….. (an + 1)]2k –1.
Conditional combinations
(1) The number of ways in which r objects can be selected from n different objects if k particular objects are
(i) Always included = n – k Cr – k
(ii) Never included = n–kCr
(2) The number of combinations of n objects, of which p are identical, taken r at a time is
n–pC + n–pC
n–pC
n–pC if r p and n – pC + n–pC
n–pC
n–pC , if r > p.
r
r–1 +
r–2 +…..+
0,
r
r–1 +
r–2 +….. +
r–p
Division into groups
Case I:
(1) The number of ways in which n different things can be arranged into r different groups is n + r – 1Pn or n! n–
1C
r – 1 according as blank group are or are not admissible.
(2) The number of ways in which n different things can be distributed into r different group is rn –rC1(r –
1)n+rC2(r – 2)n–….. +(–1)n–1 nCr–1 or Coefficient of xn in n!(ex –1)r. Here blank groups are not allowed.
(3) Number of ways in which m × n different objects can be distributed equally among n persons (or
mn !n! mn !
numbered group) = (number of ways of dividing into groups) × (number of groups)! =
n
n
m! n! m!
Case II:
(1) The number of ways in which (m + n) different things can be divided into two groups which contain m
m n ! ,m n
and n things respectively is m+nCm.nCn =
m!n!
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Corollary: If m =n, then the groups are equal size. Division of these groups can be given by two types
Type I : If order of group is not important: The number of ways in which 2n different things can be
2n !
divided equally into two groups is
2
2! n!
Type II : If order of group is important: The number of ways in which 2n different things can be divided
equally into two distinct groups is
2n ! 2! 2!
2
2
2! n!
n!
(2) The number of ways in which (m + n + p) different things can be divided into three groups which contain
m n p ! ,m n p
m, n and p things respectively is m+n+pCm.n+pCn.pCp =
m!n!p!
Corollary: If m =n =p, then the groups are equal size Division of these groups can be given by two types
Type I: If order of group is not important: The number of ways in which 3p different things can be divided
3p ! .
equally into three groups is
3
3! p!
Type II: If order of group is important: The number of ways in which 2n different things can be divided
3p ! 3! 3p !
equally into two distinct groups is
3
3
3! p!
p!
(i) If order of group is not important: The number of ways in which mn different things can be divided
equally into m groups is
mn !
m
n! m!
(ii) If order of group is important: The number of ways in which mn different things can be divided equally
mn ! m! mn !
into m distinct groups is
m
m
n! m!
n!
Dearrangement
Any change in the given order of the things is called dearrangement.
If n things form an arrangement in a row, the number of ways in which they can be dearranged so that no
1 1 1
n 1
one of them occupies place is n! 1 .... 1 .
1! 2! 3!
n!
Some important results for geometrical problems
(1) Number of total different straight lines formed by joining the n points on a plane of which m (< n) are
collinear is nC2 – mC2 + 1.
(2) Number of total triangles formed by joining the n points on a plane of which m(<n) are collinear is nC3 –
mC .
3
(3) Number of diagonals in a polygon of n sides is nC2 – n.
(4) If m parallel lines in a plane are intersected by a family of other n parallel lines. Then total number of
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parallelograms so formed is mC2 × nC2 i.e.,
mn m 1 n 1
.
4
(5) Given n points on the circumference of a circle, then
(i) Number of straight lines = nC2
(ii) Number of triangles = mC3
(iii) Number of quadrilaterals = nC4.
(6) If n straight lines are drawn in the plane such that no two lines are parallel and no three lines are
concurrent. Then the number of part into which these lines divided the plane is = 1 + n.
n
(7) Number of rectangles of any size in a square of n × n is
r3 and number of squares of any size is
r 1
(8) In a rectangle of n × p (n < p) number of rectangles of any size is
n
r
2
.
r 1
np
n 1 p 1 and number of squares
4
n
of any size is
n 1 r p 1 r .
r 1
Multinomial theorem
Let x1, x2,…….,xm be integers. Then number of solutions to the equation x1 + x2 +…..+xm = n
Subject to the condition
a1 x1 b1, a2 x2 b2,……., am xm bm
………(ii)
is equal to the coefficient of xn in x a1 x a1 1 ....... x b1
x
a2
……..(i)
x a2 1 ..... x b2 ..... x am x am1 ....... x bm
……..(iii)
This is because the number of ways, in which sum of m integers in (i) equals n, is the same as the number of
times xn comes in (iii).
(1) Use of solution of linear equation and coefficient of a power in expansions to find the number of
ways of distribution: (i) The number of integral solutions of x1 + x2 + x3 + ……+xr = n where x1 0, x2 0,
……xr 0 is the same as the number of ways to distribute n identical things among r persons.
This is also equal to the coefficient of xn in the expansion of (x0 + x1 + x2 + x3 +……)r
1
= coefficient of in
1 x
= coefficient of xn in (1– x)–r
r
xn
r r 1 2
r r 1 r 2 ...... r n 1 n
= coefficient of xn in 1 rx
x .....
x ....
2!
n!
=
r r 1 r 2 ...... r n 1
n!
r n 1! n r 1 C
r 1
n! r 1!
(2) The number of integral solutions of x1 + x2 + x3 + ……+xr = n where x1 1, x2 1, ……xr 1 is the same as
the number of ways to distribute n identical things among r persons each getting at least 1. This is also equal
to the coefficient of xn in the expansion of (x1 + x2 + x3 +……)r
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In association with
r
x
= coefficient of xn in
1 x
= coefficient of xn in xr(1 – x)–r
r r 1 2
r r 1 r 2 ...... r n 1 n
= coefficient of xn in x r 1 rx
x .....
x ....
2!
n!
r r 1 2
r r 1 r 2 ...... r n 1 n
= coefficient of xn–r in 1 rx
x .....
x ....
2!
n!
=
r r 1 r 2 ...... r n r 1
n r !
r 1 r 2 ..... n 1
n r !
=
n 1 ! n1 C
r 1
n r ! r 1 !
Number of divisor
Let N= p11 .p22 .p33 .....pkk , where p1, p2, p3,…..pk are different primes and 1, 2, 3,…., k are natural number
then:
(1) The total number of divisors of N including 1 and N is = (1 +1) (2 +1)(3 +1)…..(k +1)
(2) The total number of divisors of N excluding 1 and N is = (1 +1) (2 +1)(3 +1)…..(k +1)–2
(3) The total number of divisors of N excluding 1 or N is = (1 +1) (2 +1)(3 +1)…..(k +1)–1.
(4) The sum of these divisors is p01 p11 p21 ..... p11
p
0
2
p21 p22 ..... p22 ....... p0k p1k p2k ..... pk2
(5) The number of ways in which N can be resolved as a product of two factor is
1
2 1 1 2 1 ..... k 1 ,If N is not a perfect square
1 1 1 ..... 1 1 ,If N is a perfect square
2
k
2 1
(6) The number of ways in which a composite number N can be resolved into two factors which are
relatively prime (or co prime) to each other is equal 2n –1 where n is the number of different factors
SAMPLE QUESTIONS
1. There are 6 candidates for 3 posts. In how many ways can the posts be filled?
(a) 120
(b) 130
(c) 100
(d) 110
2. From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In
how many ways can this be done?
(a) 1360
(b) 1260
(c) 1060
(d) 1160
3. There are 15 buses running between Delhi and Mumbai. In how many ways can a man go to
Mumbai and return by a different bus?
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(a) 280
(b) 310
(c) 240
(d) 210
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4. A teacher of a class wants to set one question from each of two exercises in a book. If there are 15 and 12
questions in the two exercises respectively, then in how many ways can the two questions be selected?
(a) 160
(b) 140
(c) 180
(d) 120
5. The students in a class are seated according to their marks in the previous examination. Once, it so
happens that four of the students got equal marks and therefore the same rank. To decide their seating
arrangement, the teacher wants to write down all possible arrangements one in each of separate bits of
paper in order to choose one of these by lots. How many bits of paper are required?
(a) 24
(b) 12
(c) 48
(d) 36
6. For a set of five true-or-false questions, no student has written all the correct answers, and no two
students have given the same sequence of answers. What is the maximum number of students in the class,
for this to be possible?
(a) 31
(b) 21
(c) 51
(d) 41
7. A code word is to consist of two English alphabets followed by two distinct numbers between 1 and 9. For
example, CA23 is a code word. How many such code words are there?
(a) 615800
(b) 46800
(c) 719500
(d) 410800
8. There are 6 multiple choice questions on an examination. How many sequences of answers are
possible, if the first three questions have 4 choices each and the next three have 5 each?
(a) 6000
(b) 5000
(c) 4000
(d) 8000
9. There are six multiple choice questions in an examination. How many sequences of answers are
possible, if the first two questions have 3 choices each, the next two have 4 choices each and the last two
have 5 choices each?
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(b) 3300
(c) 3600
(d) 3400
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10. Each section in the first year of plus two course has exactly 40 students. If there are 5 sections, in how
many ways can a set of 4 student representatives be selected, one from each section?
(a) 2560000
(b) 246500
(c) 2240000
(d) 2360000
11. There are 5 letters and 5 directed envelopes. Find the number of ways in which the letters can be put into
the envelopes so that all are not put in directed envelopes?
(a) 129
(b) 119
(c) 109
(d) 139
12. There horses H1, H2, H3 entered a field which has seven portions marked P1, P2, P3, P4, P5, P6 and P7. If no
two horses are allowed to enter the same portion of the field, in how many ways can the horses graze the
grass of the field?
(a) 195
(b) 205
(c) 185
(d) 210
13. How many different numbers of two digits can be formed with the digits 1, 2, 3, 4, 5, 6; no digits being
repeated?
(a) 40
(b) 30
(c) 35
(d) 45
14. How many three-digit odd numbers can be formed from the digits 1, 2, 3, 4, 5, 6 when
(i) repitition of digits is not allowed
(ii) repitition of digits is allowed?
(a) (i) 60, (ii) 108
(b) (i) 50, (ii) 98
(c) (i) 70, (ii) 118
(d) (i) 80, (ii) 128
15. How many two-digit odd numbers can be formed from the digits 1, 2, 3, 4 5 and 8, if repitition of digits
is allowed?
(a) 5
(b) 15
(c) 35
(d) 25
16. How many odd numbers less than 1000 can be formed using the digits 0, 2, 5, 7? (repetition of digits
is allowed).
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(a) 52
(b) 32
(c) 22
(d) 42
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17. How many 3-digit numbers each less than 600 can be formed from the digits 1, 2, 3, 4, 5 and 9, if
repetition of digits is allowed?
(a) 180
(b) 160
(c) 165
(d) 185
18. How many words (with or without meaning) of three distinct English alphabets are there?
(a) 15600
(b) 14650
(c) 12800
(d) 13700
19. How many numbers are there between 100 and 1000 in which all the digits are distinct?
(a) 548
(b) 648
(c) 748
(d) 756
20. How many integers between 1000 and 10000 have no digits other than 4, 5 or 6?
(a) 91
(b) 51
(c) 81
(d) 71
Answers
1. (a)
2. (c)
3. (d)
4.(c)
5.(a)
6. (a)
7.(b)
8. (d)
9.(c)
10.(a)
11.(b)
12.(d)
13.(b)
14.(a)
15. (b)
16.(b)
17.(a)
18 .(a)
19.(b)
20.(c)
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Section-B : MAT (Mental Aptitude Test )
Chapter: 1 Blood Relation
1.
S and R are brothers. T is daughter of S, U is the spouse of R and mother of Q. P is the daughter
of V, who is the spouse of T Who, is the cousin of Q?
(a) T
(b) V
(c) R
(d) P
2.
A man pointing to a photograph says, “The lady in the photograph is my nephew’s maternal
grandmother" How is the lady in the photograph related to the man’s sister who has no other sister.
(a) Cousin
(b) Sister–in–law
(c) Mother
(d) Mother–in–law
3.
Pointing to Kapil, shilpa said, His mothers’ brother is the father of my son Ashish" How is
kapil related to shilpa
(a) Sister–in–law
(b) Nephew
(c) Niece
(d) Aunt
4.
A is the uncle of B, who is the daughter of C and C is the daughter–in–law of P. How is A
related to P?
(a) Brother
(b) Son
(c) Son–in–law
(d) Data inadequate
5.
Pointing towards a boy, Veena said, “He is the son of only son of my grandfather.” How is that
boy related to Veena?
(a) Uncle
(b) Brother
(c) Cousin
(d) Data inadequate
6.
Introducing Reena, Monika said, “She is the only daughter of my father’s only daughter.” How
is Monika related to Reena?
(a) Aunt
(b) Niece
(c) Cousin
(d) None of these
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Pointing to a man, a woman said, “His mother is the only daughter of my mother.” How is the
woman related to the mother?
(a) Mother
(b) Daughter
(c) Sister
(d) Grandmother
8.
If X is the brother of the son of Y’s son, how is X related to Y?
(a) Son
(b) Brother
(c) Cousin
(d) Grandson
9.
Pointing towards Rita, Nikhil said, “I am the only son of her mother’s son.” How is Rita related
to Nikhil?
(a) Aunt
(b) Niece
(c) Mother
(d) Cousin
10.
Pointing to a lady, a man said, “The son of her only brother is the brother of my wife.” How is
the lady related to the man?
(a) Mother’s sister
(b) Grandmother
(c) Mother-in-law
(d) Sister of father-in-law
11.
Pointing to Ketan, Namrata said, “He is the son of my father’s only son.” How is Ketan’s mother
related to Namrata ?
(a) Daughter
(b) Aunt
(c) Sister
(d) Sister-in-law
12.
Pointing to a man on the stage, Rashi said, “He is the brother of the daughter of the wife of my
husband.” How is the man on the stage related to Rashi ?
(a) Son
(b) Husband
(c) Cousin
(d) Nephew
13.
A woman introduces a man as the son of the brother of her mother. How is the man
related to the woman?
(a) Nephew
(b) Son
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(c) Cousin
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(d) Uncle
Introducing a man, a woman said, “He is the only son of my mother’s mother.” How is the
woman related to the man?
(a) Mother
(b) Aunt
(c) Sister
(d) Niece
15.
If X is brother of son of Y’s son, then how is X related to Y ?
(a) Brother
(b) Cousin
(c) Grandson
(d) Son
16.
Given that
1. A is the mother of B;
2. C is the son of A;
3. D is the brother of E;
4. E is the daughter of B.
The grandmother of D is
(a) A
(b) B
(c) C
(d) E
17.
A, B and C are sisters. D is the brother of E and E is the daughter of B. How is A related
to D?
(a) Sister
(b) Cousin
(c) Niece
(d) Aunt
18.
A and B are married couple. X and Y are brothers. X is the brother of A. How is Y
related to B?
(a) Brother-in-law
(b) Brother
(c) Cousin
(d) None of these
19.
Deepak has a brother Anil. Deepak is the son of Prem. Bimal is Prem’s father. In
terms of relationship, what is Anil of Bimal?
(a) Son
(b) Grandson
(c) Brother
(d) Grandfather
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B is the husband of P. Q is the only grandson of E, who is wife of D and mother-in-law
of P. How is B related to D?
(a) Nephew
(b) Cousin
(c) Son-in-law
(d) Son
Answers
1.(a)
8. (d)
15. (c)
2. (c)
9. (a)
16. (a)
3. (b)
10. (d)
17. (d)
4. (b)
11. (d)
18. (a)
5. (b)
12. (a)
19. (b)
6. (d)
13. (c)
20. (d)
7. (a)
14. (d)
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Chapter 2. Classification
1.
Choose out the odd one
(a) Lake
(b) Sea
(c) River
(d) Pool
2.
Choose out the odd one
(a) Arrow
(b) Axe
(c) Knife
(d) Dagger
3.
Choose out the odd one
(a) Sun
(b) Moon
(c) Star
(d) Universe
4.
Choose out the odd one
(a) House
(b) Cottage
(c) School
(d) Palace
5.
Choose out the odd one
(a) Tomato
(b) Cucumber
(c) Brinjal
(d) Carrot
6.
Choose out the odd one
(a) Brick
(b) Heart
(c) Bridge
(d) Spade
7.
Choose out the odd one
(a) Run
(b) Walk
(c) Think
(d) Jump
8.
Choose out the odd one
(a) Mumbai
(b) Cochin
(c) Kandla
(d) Mysore
9.
Choose out the odd one
(a) Tricycle
(b) Trident
(c) Trifle
(d) Tricolour
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Choose out the odd one
(a) Japan
(b) India
(c) Sri Lanka
(d) New Zealand
11.
Choose the pair in which the words are differently related
(a) Bouquet : Flowers
(b) Bunch : Grapes
(c) Furniture : Chair
(d) Album : Photos
12.
Choose the pair in which the words are differently related
(a) Waist : Belt
(b) Neck : Tie
(c) Wrist : Band
(d) Shoe : Laces
13.
Choose the pair in which the words are differently related
(a) Chaff: Wheat
(b) Grit: Pulses
(c) Grain : Crop
(d) Dregs : Wine
14.
Choose the pair in which the words are differently related
(a) Rice : Corn
(b) Tomato : Potato
(c) Student : Class
(d) Book : Library
15.
Choose the pair in which the words are differently related
(a) Ammeter : Current
(b) Hygrometer : Pressure
(c) Odometer : Speed
(d) Seismograph : Earthquakes
16.
Choose the pair in which the words are differently related
(a) Proteins : Marasmus
(b) Sodium ; Rickets
(c) Iodine : Goitre
(d) Iron Anemia
17.
Choose the pair in which the words are differently related
(a) Church : Monument
(b) Car : Bus
(c) Pond : Lake
(d) Pistol : Gun
18.
Choose the pair in which the words are differently related
(a) Sheep : Bleat
(b) Horse : Neigh
(c) Ass : Grunt
(d) Owl : Hoot
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Choose the pair in which the words are differently related
(a) Door : Bang
(b) Piano : Play
(c) Rain : Patter
(d) Drum : Beat
20.
Choose the pair in which the words are differently related
(a) Marble : Limestone
(b) Slate : Shale
(c) Quartzite : Sandstone
(d) Gneiss : Quartz
Answers
1.(c)
8. (d)
15. (b)
2. (a)
9. (c)
16. (b)
3. (d)
10. (b)
17. (a)
4. (c)
11. (c)
18. (c)
5. (d)
12. (d)
19. (b)
6. (a)
13. (c)
20. (d)
7. (c)
14. (b)
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Chapter 3. Direction Sense
From the positions in original figure, C and A move diagonally to opposite corners and then
one side each clockwise and anti–clockwise respectively. B and D move two sides each clockwise
and anti–clockwise respectively. Where is A now?
(a) At the south–west corner
(b) At the north–east corner
(c) At the south east corner
(d) At the south–west corner
2.
Divya journeys 10 km to east then 10 km to south–west. He . turns again and journeys 10 km
to North–West. Which direction is he in from the starting point?
(a) South
(b) North
(c) West
(d) East
3.
A child crawls 20 feet towards North, turns right and crawls 30 feet, turns right again and
crawls 35 feet. He turns left again and crawls 15 feet. He turns left again and crawls 15feet. Finally
he turns to his left to crawl another 15 feet. How far is he from his starting point and in which
direction?
(a) 45 feet North–East
(b) 30 feet East
(c) 30 feet West
(d) 15 feet West
4.
Saba was facing East. She walked 20 metres. Turning left she moved 15 metres and then
turning right moved 25 metres. Finally, she turned right and moved 15 metres more.
How far is she from her starting point.
(a) 25 metres
(b) 3 5 metres
(c) 50 metres
(d) 45 metres
5.
Jatin leaves his house and walks 12 km towards North. He turns right and walks another 12
km. He turns right again, walks 12 km more and turns left to walk 5 km. How far is he from his home
and in which direction ?
(a) 7 km East
(b) 10 km East
(c) 17 km East
(d) 24 km East
6.
Going 50 m to the south of her house, Radhika turns left and goes another 20 m. Then turning
to the North, she goes 30 m and then starts walking to her house. In which direction is she walking
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now ?
(a) North–west
(b) North
(c) South east
(d) East
7.
P, Q, R, S, T, U, V Ware sitting around a round table in the same order, for group discussion at
equal distance. Their position are clockwise. If V sits in the north, then what will be the position of S
?
(a)East
(b) South–east
(c) South
(d) South–west
8.
While facing East, Rohit turns to his left and walks 10metres, then he turns left and walks 10
meters. Now he turns 45° towards his right and goes straight to cover 25meters. In which direction
is he from his starting point?
(a) North–east
(b) North–west
(c) South–west
(d) South–east
9.
If all the directions are rotated, i.e., if North is changed toWest and East to North and so on,
then what will come in
place of North–West ?
(a) South–West
(b) North–East
(c) East–North
(d) East–West
10.
If a person is walking towards North, what direction should he follow so that he is walking
towards West ?
(a) right, right, left
(b) left, left, right
(c) left, right, left
(d) left, left, left
11.
Maya starts at point T, walks straight to point U which is 4 ft away. She turns left at 90° and
walks to W which is 4 ft away, turns 90° right and goes 3 ft to P, turns 90° right and walks 1 ft to Q,
turns left at 90° and goes to V, which is 1 ft away and once again turns 90° right and goes to R, 3 ft
away. What is the distance between T and R ?
(a) 4 ft
(b) 5 ft
(c) 7 ft
(d) 8 ft
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A villager went to meet his uncle in another–village situated 5 km away in the North–east
direction of his own village. From there he came to meet his father–in–law living in a village
situated 4 km in the south of his uncle’s village. How far away and in what direction is he now ?
(a) 3 km in the North
(b) 3 km in the East
(c) 4 km in the East
(d) 4 km in the West
13.
A person starts from a point A and travels 3 km eastwards to B and then turns left and
travels thrice that distance to reach C. He again turns left and travels five times the distance he
covered between A and B and reaches his destination D. The shortest distance between the starting
point and the destination is
(a) 12 km
(b) 15 km
(c) 16 km
(d) 18 km
14.
A girl leaves from her home. She first walks 30 metres in North–west direction and then 30
metres in South–west direction. Next, she walks 30 metres in South–east direction. Finally, she
turns towards her house. In which direction is she moving ?
(a) North–east
(b) North–west
(c) South–east
(d) South–west
15.
Sanjeev walks 10 meters towards the South. Turning to the left, he walks 20 metres and then
moves to his right. After moving a distance of 20 metres, he turns to the right and walks 20 metres.
Finally, he turns to the right and moves a distance of 10 metres. How far and in which direction is
he from the starting point?
(a) 10 metres North
(b) 20 metres South
(c) 20 metres North
(d) 10 metres South
16.
Ravi wants to go to the university. He starts from his home which is in the East and
comes to a crossing. The road to the left ends in a theatre, straight ahead is the hospital. In which
direction is the university?
(a) North
(b) South
(c) East
(d) West
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Of the six members of a panel sitting in a row, A is to the left of D, but on the right of E. C is
on the right of X, but is on the left of B who is to the left of F. Which two members are sitting
right in the middle?
(a) A and C
(b) C and B
(c) D and B
(d) D and C
18.
A, B, C and D are playing cards. A and B are partners. D faces towards North. If A faces
towards West, then who faces towards South?
(a) B
(b) C
(c) D
(d) Data inadequate
19.
P, Q, R and S are playing a game of carrom. P, R and S, Q are partners. S is to the right of R
who is facing west. Then, Q is facing
(a) North
(b) South
(c) East
(d) West
20.
The town of Paranda is located on Green Lake. The town of Akram is west of Paranda.
Tokhada is east of Akram but west of Paranda. Kakran is east of Bopri but west of Tokhada
and Akram. If they are all in the same district, which town is the farthest west ?
(a) Paranda
(b) Kakran
(c) Akram
(d) Bopri
Answers
1. (d)
8. (b)
15. (b)
2. (c)
9. (a)
16. (a)
3. (b)
10. (c)
17. (d)
4. (d)
11. (d)
18. (b)
5. (c)
12. (b)
19. (a)
6. (a)
13. (b
20. (d)
7. (d)
14. (a)
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Chapter 4. Coding Decoding
1.
Find the code of the work given in each of the following questions
In a certain code, PRODUCTIONS is written as QQPCVEUHPMT. How is ORIENTATION written in
that code?
(a) PQJDOVBSJNO
(b) PQJDOUBUJPO
(c) PSJFOVBSJNO
(d) NSHFMVBSJNO
2.
Find the code of the work given in each of the following questions
If, in a code, MIND becomes KGLB and ARGUE becomes YPESC, then what will DIAGRAM be
in that code?
(a) BGYEPYK
(b) BGYPYEK
(c) GLPEYKB
(d) LKBGYPK
3.
Find the code of the work given in each of the following questions
In a certain code, BASIC is written as DDULE, How is LEADER written in that code ?
(a) NGCFGT
(b) NHCGGU
(c) OGDFHT
(d) OHDGHU
4.
In a certain language, SIGHT is written as FVTUG. How is REVEAL written in the same’
language?
(a) YNRIRE
(b) DQHQMX
(c) FSJSOZ
(d) ERIRNY
5.
If in a certain language, MIRACLE is coded as NKUEHRL, then how is GAMBLE coded in
that language?
(a) JDOCMF
(b) CLEMNK
(c) HCPFQK
(d) AELGMN
6.
Find the code of the work given in each of the following questions
If in a certain code, GLAMOUR is written as IJCNMWP and MISRULE is written as OGUSSNC,
then how will TOPICAL be written in that code ?
(a) VMRJECN
(b) VMRHACJ
(c) VMRJACJ
(d) VNRJABJ
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Find the code of the work given in each of the following questions
In a certain code, BELIEF is written as AFKKDI. How is SELDOM written in that code ?
(a) RDKCNL
(b) RFKENM
(c) RFKFNP
(d) TFKENP
8.
Find the code of the work given in each of the following questions
In a certain code language, THANKS is written as SKNTHA. How is STUPID written in that code
language ?
(a) DIPUTS
(b) DISPUT
(c) DIPUST
(d) None of these
9.
Find the code of the work given in each of the following questions
SPIDER is written as PSDIRE in a certain code, how would COMMON be written in that code ?
(a) OCOMMO
(b) OCMMNO
(c) OCMOMN
(d) OCMMON
10.
Find the code of the work given in each of the following questions
In a certain code, CALANDER is written as CLANAEDR, How is CIRCULAR written in that code ?
(a) ICCRLURA
(b) CRIUCLRA
(c) ICRCLUAR
(d) CRIUCALR
11.
If each of the letters in the English alphabet is assigned odd numerical value beginning A =
1, B = 3 and so on, what will be the total value of the letters of the word INDIAN ?
(a) 86
(b) 88
(c) 89
(d) 96
12.
Find the code of the work given in each of the following questions in a certain code, the
word DEAL is coded as 4–5–1–12. Following the same rule of coding, what should be the code for
the word LADY?
(a) 12–4–1–25
(b)12–1–4–25
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(c) 10 – 1 – 4 – 23
13.
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(d) 12 – 1 – 4 – 22
If DRIVER = 12, PEDESTRIAN = 20, ACCIDENT = 16, then CAR =?
(a) 3
(b) 6
(c) 8
(d) 10
14.
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If ‘rose’ is called ‘poppy’, ‘poppy’ is called ‘lily’, ‘lily’ is called ‘lotus’ and ‘lotus’ is called
‘gladiola’, which is the king of flowers?
(a) Rose
(b) Lotus
(c) Poppy
(d) Gladiola
15.
If ‘rat’ is called ‘dog’, ‘dog’ is called ‘mongoose’, ‘mongoose’ is called ‘lion’, ‘lion’ is called
‘snake’ and ‘snake’ is called ‘elephant’, which is reared as pet ?
(a) Rat
(b) Dog
(c) Mongoose
(d) Lion
16.
If ‘blue’ means ‘green’, ‘green’ means ‘white’, ‘white’ means ‘yellow’, ‘yellow’ means ‘black’,
‘black’ means ‘red’ and ‘red’ means ‘brown’, then what is the colour of milk ?
(a) Black
(b) Brown
(c) Blue
(d) Green
17.
If ‘paper’ is called ‘wood’, ‘wood’ is called ‘straw’, ‘straw’ is called ‘grass’, ‘grass’ is called
‘rubber’ and ‘rubber’ is called ‘cloth’, what is the furniture made up of ?
(a) Paper
(b) Wood
(c) Straw
(d) Grass
18.
If ‘man’ is called ‘girl’, ‘girl’ is called ‘woman’, ‘woman’ is called ‘boy’, ‘boy’ is called ‘butler’
and ‘butler’ is called ‘rogue’, who will serve in a restaurant ?
(a) Butler
(b) Girl
(c) Man
(d) Rogue
19.
Find the code of the work given in each of the following questions
In a certain coding system, ‘rbm std bro pus’ means ‘the cat is beautiful’, ‘tnh pus dim std’
means ‘the dog is brown’, ‘pus dim bro pus cus’ means ‘the dog has the cat’. What is the
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In association with
code for ‘has’?
(a) std
(b) dim
(c) bro
(d) cus
20.
Find the code of the work given in each of the following questionsIn a certain code
language, ‘put tir fin’ means ‘delicious juicy fruit’, ‘tie dip sig’ means ‘beautiful white
lily’ and ‘stg Ion fin’ means ‘lily and fruit’. Which of the following stands for ‘and’ in that
language ?
(a) Ion
(b) sig
(c) fin
(d) None of these
Answers
1. (a)
8. (d)
15. (c)
2. (a)
9. (b)
16. (d)
3. (b)
10. (d)
17. (c)
4. (d)
11. (d)
18. (d)
5. (c)
12. (b)
19. (d)
6. (c)
13. (b)
20. (a)
7. (c)
14. (d)
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Chapter 5. In complete Figures
1. To complete Pattern, which figure is placed in the blank space of fig. (X)
2. To complete Pattern, which figure is placed in the blank space of fig. (X)
3. To complete Pattern, which figure is placed in the blank space of fig. (X)
4. To complete Pattern, which figure is placed in the blank space of fig. (X)
5. To complete Pattern, which figure is placed in the blank space of fig. (X)
6. To complete Pattern, which figure is placed in the blank space of fig. (X)
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7. To complete Pattern, which figure is placed in the blank space of fig. (X)
8. To complete Pattern, which figure is placed in the blank space of fig. (X)
9. To complete Pattern, which figure is placed in the blank space of fig. (X)
10. To complete Pattern, which figure is placed in the blank space of fig. (X)
11. To complete Pattern, which figure is placed in the blank space of fig. (X)
12. To complete Pattern, which figure is placed in the blank space of fig. (X)
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13. To complete Pattern, which figure is placed in the blank space of fig. (X)
14. To complete Pattern, which figure is placed in the blank space of fig. (X)
15. To complete Pattern, which figure is placed in the blank space of fig. (X)
16. To complete Pattern, which figure is placed in the blank space of fig. (X)
17. To complete Pattern, which figure is placed in the blank space of fig. (X)
18. To complete Pattern, which figure is placed in the blank space of fig. (X)
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19. To complete Pattern, which figure is placed in the blank space of fig. (X)
20. To complete Pattern, which figure is placed in the blank space of fig. (X)
ANSWERS
1.(c)
2.(a)
3.(a)
4. (c)
5.(d)
6.(c)
7.(a)
8.(d)
9.(d)
10.(d)
11.(c)
12.(a)
13.(c)
14.(b)
15.(c)
16.(d)
17.(b)
18.(c)
19.(d)
20.(c)
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Chapter 6. Mirror Images
1. GEOGRAPHY
2. INFORMATIONS
3. REASONING
4. EFFECTIVE
5. MAGAZINE
6. ANS43Q1 2
7. DL 9 C G4 7 2 8
8. BR4AQ16HI
9. UTZFY6KH
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10. TA RAI N10 14 A
11. Which of the following collections of letters will look the same in the mirror ?
12. Choose the correct mirror-image of the Fig. (X)
13. Choose the correct mirror-image of the Fig. (X)
14. Choose the correct mirror-image of the Fig. (X)
15. Choose the correct mirror-image of the Fig. (X)
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16. Choose the correct mirror-image of the Fig. (X)
17. Choose the correct mirror-image of the Fig. (X)
18. Choose the correct mirror-image of the Fig. (X)
19. Choose the correct mirror-image of the Fig. (X)
20. Choose the correct mirror-image of the Fig. (X)
Answers
1.(a)
2.(c)
3.(b)
4.(a)
5.(d)
6.(b)
7.(c)
8.(a)
9.(d)
10.(d)
11.(d)
12.(d)
13.(c)
14.(d)
15.(d)
16.(b)
17.(c)
18.(d)
19.(c)
20.(b)
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Chapter 7. Paper Cutting
1. Choose a figure which would most closely resemble the unfolded form of fig.
2. Choose a figure which would most closely resemble the unfolded form of fig.
3. Choose a figure which would most closely resemble the unfolded form of fig.
4. Choose a figure which would most closely resemble the unfolded form of fig.
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5. Choose a figure which would most closely resemble the unfolded form of fig. (
6. Choose a figure which would most closely resemble the unfolded form of fig.
7. Choose a figure which would most closely resemble the unfolded form of fig.
8. Choose a figure which would most closely resemble the unfolded form of fig.
9. Choose a figure which would most closely resemble the unfolded form of fig.
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10. Choose a figure which would most closely resemble the unfolded form of fig.
11. Choose a figure which would most closely resemble the unfolded form of fig.
12. Choose a figure which would most closely resemble the unfolded form of fig.
13. Choose a figure which would most closely resemble the unfolded form of fig.
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14. Choose a figure which would most closely resemble the unfolded form of fig.
15. Choose a figure which would most closely resemble the unfolded form of fig.
16. Choose a figure which would most closely resemble the unfolded form of fig.
17. Choose a figure which would most closely resemble the unfolded form of fig.
18. Choose a figure which would most closely resemble the unfolded form of fig.
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19. Choose a figure which would most closely resemble the unfolded form of fig.
20. Choose a figure which would most closely resemble the unfolded form of fig.
ANSWERS
1.(c)
2.(a)
3.(b)
4.(a)
5.(b)
6. (c)
7.(b)
8.(d)
9.(b)
10.(d)
11.(a)
12.(b)
13.(c)
14.(c)
15.(d)
16.(c)
17.(d)
18.(c)
19.(c)
20.(d)
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Chapter: 8Analytical Reasoning
1. Find the minimum number of straight lines required to make the given figure.
(a) 9
(b) 11
(c) 15
(d) 16
2. Find the minimum number of straight lines required to make the given figu
(a) 13
(b) 15
(c) 17
(d) 19
3. Find the number of triangles in the given figure.
(a) 16
(b) 18
(c) 14
(d) 15
4. Find the number of triangles in the given figure.
(a) 12
(b) 18
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(c) 22
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(d) 26
5. Find the number of triangles in the given figure.
(a) 21
(b) 23
(c) 25
(d) 27
6. Find the number of triangles in the given figure.
(a) 12
(b) 13
(c) 14
(d) 15
7. Find the number of triangles in the given figure.
(a) 8
(b) 10
(c) 11
(d) 12
8. Find the number of triangles in the given figure.
(a) 18
(b) 20
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(c) 28
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(d) 34
9. Find the number of triangles in the given figure.
(a) 18
(b) 20
(c) 24
(d) 27
10. Find the number of triangles in the given figure.
(a) 11
(b) 13
(c) 15
(d) 17
11. Find the number of triangles in the given figure.
(a) 16
(b) 18
(c) 19
(d) 21
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12. Find the number of triangles in the given figure.
(a) 10
(b) 12
(c) 14
(d) 16
13. Find the number of triangles in the given figure.
(a) 20
(b) 24
(c) 28
(d) 32
14. Find the number of triangles in the given figure.
(a) 23
(b) 27
(c) 29
(d) 31
15. Find the number of triangles in the given figure.
(a) 28
(b) 32
(c) 36
(d) 40
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16. What is the number of triangles that can be formed whose vertices are the vertices of
an octagon but have only one side common with that of the octagon?
(a) 64
(b) 32
(c) 24
(d) 16
17. Count the number of squares in the given figure.
(a) 8
(b) 12
(c) 15
(d) 18
18. Count the number of squares in the given figure.
(a) 18
(b) 19
(c) 25
(d) 27
19. Count the number of squares in the given figure.
(a) 12
(b) 13
(c) 16
(d) 17
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20. Count the number of squares in the given figure.
(a) 11
(b) 21
(c) 24
(d) 26
ANSWERS
1.(b)
2.(a)
3.(b)
4.(b)
5.(d)
6.(d)
7.(b)
8.(c)
9.(c)
10.(c)
11.(d)
12.(c)
13.(c)
14.(c)
15.(c)
16.(b)
17.(c)
18.(d)
19.(d)
20.(c)
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