Download Number System, HCF and LCM

•
Head Office : B-32, Shivalik Main Road, Malviya Nagar, New Delhi-110017
•
Sales Office : B-48, Shivalik Main Road, Malviya Nagar, New Delhi-110017
Tel. : 011-26691021 / 26691713
Price : ` 260
Typeset by Disha DTP Team
DISHA PUBLICATION
ALL RIGHTS RESERVED
© Publisher
No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the
publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried
and made our best efforts to provide accurate up-to-date information in this book.
For further information about the books from DISHA,
Log on to www.dishapublication.com or email to [email protected]
CONTENTS
SECTION A : QUANTITATIVE APTITUDE
1.
Number System & Simplification
A-1
- 10
2.
Simplification
A-11
- 14
3.
Surds, Indices, Square Roots and Cube Roots
A-15
- 19
4.
Ratio, Proportion and Partnership
A-20
- 26
5.
Average, Problem on Ages
A-27
- 33
6.
Percentage, Profit & Loss
A-34
- 42
7.
Time, Work & Pipes, Cisterns
A-43
- 51
8.
Time, Speed and Distance
A-52
- 58
9.
Simple Interest & Compound Interest
A-59
- 64
10.
Elementary Mensuration
A-65
- 74
11.
Permutation, Combination & Probability
A-75
- 82
12.
Data Interpretation
A-83
- 88
SECTION B : REASONING
1.
Analogy
2.
Classification
3.
B-1
-6
B-7
- 11
Series
B-12
- 16
4.
Coding & Decoding
B-17
- 24
5.
Word Formation
B-25
- 29
6.
Blood Relation
B-30
- 35
7.
Direction, Clock and Calender
B-36
- 43
8.
Logical Venn Diagram
B-44
- 48
9.
Syllogism
B-49
- 60
10.
Non-Verbal Series
B-61
- 64
11.
Problem Solving
B-65
- 73
12.
Input-Output
B-74
- 84
13.
Coded Inequalities
B-85
- 92
14.
Statement & Assumptions
15.
Data Sufficiency
B-93
- 100
B-101
- 108
SECTION C : ENGLISH LANGUAGE
1.
English Grammar & Vocabulary
C-1
- 28
2.
Reading Comprehension
C-29
- 50
3.
Cloze Test
C-51
- 56
4.
Parajumbles
C-57
- 64
D-1
- 10
SECTION D : GENERAL AWARENESS
1.
Geography
2.
Indian Polity
D-11
- 20
3.
Economy
D-21
- 36
4.
Miscellaneous
D-37
- 46
5.
Current Banking
D-47
- 54
6.
Current Affairs
D-55
- 68
SECTION A : QUANTITATIVE APTITUDE
C
H
A
P
T
E
R
1
Number System,
HCF and LCM
The ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are called digits, which can
represent any number.
Real numbers: Real numbers comprise the full spectrum of
numbers. They can take on any form – fractions or whole numbers,
decimal points or no decimal points. The full range of real numbers
includes decimals that can go on forever and ever without end.
3
For Example: 8, 6, 2 + 3 , etc.
5
Natural numbers: A natural number is a number that comes
naturally. Natural Numbers are counting numbers from 1, 2, 3,
4, 5, ........
Whole numbers: Whole numbers are just all the natural numbers
plus zero.
For Example: 0, 1, 2, 3, 4, 5, and so on upto infinity.
Integers: Integers incorporate all the qualities of whole numbers
and their opposites (or additive inverses of the whole numbers).
Integers can be described as being positive and negative whole
numbers.
For Example: … –3, –2, –1, 0, 1, 2, 3, . . .
p
Rational numbers: All numbers of the form
where p and q are
q
integers (q ¹ 0) are called Rational numbers.
For Example: 4, 3 , 0, ....
4
Irrational numbers: Irrational numbers are the opposite of rational
numbers. An irrational number cannot be written as a fraction,
and decimal values for irrational numbers never end and do not
have a repeating pattern in them. 'pi' with its never-ending decimal
places, is irrational.
7, 5, 2 + 2, p,.....
Even numbers: An even number is one that can be divided evenly
by two leaving no remainder, such as 2, 4, 6, and 8.
Odd numbers: An odd number is one that does not divide evenly
by two, such as 1, 3, 5, and 7.
Prime numbers: A prime number is a number which can be divided
only by 1 and itself. The prime number has only two factors, 1
and itself.
For example: 2, 3, 5, 7, 11, 13, 17, .... are prime numbers.
Composite Number: A Composite Number is a number which can
be divided into a number of factors other than 1 and itself . Any
composite number has additional factors than 1 and itself.
For example: 4, 6, 8, 9, 10 .....
Co-primes or Relatively prime numbers: A pair of numbers not
having any common factors other than 1 or –1. (Or alternatively
their greatest common factor is 1 or –1)
For Example: 15 and 28 are co-prime, because the factors of 15
(1,3,5,15), and the factors of 28 (1,2,4,7,14,28) are not in common
(except for 1).
Twin Primes: A pair of prime numbers that differ by 2
(successive odd numbers that are both Prime numbers).
For Example: (3,5), (5,7), (11,13), ...
Place value : Place value is a positional system of notation in
which the position of a number with respect to a point determines
its value. In the decimal system, the value of the digits is based
on the number ten.
Each position in a decimal number has a value that is a power of
10. A decimal point separates the non-negative powers of 10,
(10)0=1, (10)1=10, (10)2=100, (10)3 =1000, etc.) on the left from the
negative powers of 10, (10)–1 =
1
1
1
, (10)–2 =
, (10)–3 =
,
10
100
1000
etc.) on the right.
Face value : The face value of a number is the value of the number
without regard to where it is in another number. So 4 7 always has
a face value of 7. However the place value includes the position
of the number in another number. So in the number 4,732, the 7
has a place value of 700, but has a face value of just 7.
1 : Place and face values of the digits in the number
495, 784:
For Example:
Number
Digit
Place value
Face value
495,784
4
9
400000
90000
4
9
5
7
5000
700
5
7
8
4
80
4
8
4
FRACTIONS
A fraction is known as a rational number and written in the form
p
where p and q are integers and q ¹ 0. The lower number
q
'q' is known as denominator and the upper number 'p' is known
as numerator.
of
A-2
Number System, HCF and LCM
TYPE OF FRACTIONS :
2 : Find three rational numbers between 3 and 5.
Proper Fraction: The fraction in which numerator is less than the
denominator is called a proper fraction.
For Example:
3+ 5 8
= =4
2
2
2nd rational number (i.e., between 3 and 4)
Sol. 1st rational number =
2 5 10
, ,
etc.
3 6 11
3+ 4 7
=
2
2
3rd rational number (i.e., between 4 and 5)
=
Improper fraction : The fraction in which numerator is greater
than the denominator is called improper fraction.
For Example :
4+5 9
= .
2
2
Both rational and irrational numbers can be represented in
number line. Thus real numbers is the set of the union of
rational and irrational numbers.
R =Q ÈQ'
Every real number is either rational or irrational.
3 6 8
. , , etc
2 5 7
=
Mixed fraction : Mixed fraction is a composite of a fraction and a
whole number.
1 3 6
For example: 2 , 3 ,5 etc.
2 4 7
Complex fraction: A complex fraction is that fraction in which
numerator or denominator or both are fractions.
Equivalent fractions/Equal fractions : Fractions with same
value.
2 2 3
For Example: 3 , 5 , 7 , etc.
4 6 5
7 6
For example :
Like fractions: Fractions with same denominators.
Decimal fraction: The fraction whose denominator is 10 or its
higher power, is called a decimal fraction.
For Example:
2 4 6 8 æ 2ö
, , , ç= ÷ .
3 6 9 12 è 3 ø
2 3 9 11
, , ,
7 7 7 7
Unlike fractions : Fractions with different denominators.
For example :
7 11 12
,
,
10 100 1000
2 4 9 9
, , ,
5 7 8 2
Simple fraction : Numerator and denominator are integers.
For example :
Continued fraction: Fractions which contain addition or
subtraction of fractions or a series of fractions generally in
denominator (sometimes in numerator also) are called continued
fractions.
3
2
and .
7
5
Vulgar fraction : Denominators are not the power of 10.
For example :
Formulae to Remember
n(n + 1)
2
v
Sum of first n natural numbers =
v
Sum of first n even numbers = n(n + 1)
v
Sum of first n odd numbers = n2
For example :
3 : Write 2.73 as a fraction.
Sol. 2.73 =
Composite Numbers : It is a natural number that has atleast one
divisor different from unity and itself.
Every composite number can be factorised into its prime factors.
For Example : 24 = 2 × 2 × 2 × 3. Hence, 24 is a composite number.
The smallest composite number is 4.
Whole Numbers : The natural numbers along with zero (0), form
the system of whole numbers.
It is denoted by W.
There is no largest whole number and
The smallest whole number is 0.
The number line : The number line is a straight line between
negative infinity on the left to positive infinity on the right.
-4 -3 -2 -1 0 1 2 3 4
3 9 5
.
, ,
7 2 193
273
100
4 : Express
Sol.
2
as a decimal fraction.
5
2 2´ 2 4
=
=
5 5 ´ 2 10
5 : After doing 3/5 of the Biology homework on
Monday night, Sanjay did 1/3 of the remaining homework on
Tuesday night. What fraction of the original homework would
Sanjay have to do on Wednesday night to complete the Biology
assignment ?
(a) 1/15
(b) 2/15
(c) 4/15
(d) 2/5
Number System, HCF and LCM
Sol. (c) Remaining homework on Monday night
=1–
A-3
3 2
=
5 5
Work done on Tuesday night
=
1
2 2
of =
3
5 15
Remaining homework to complete the biology
assignment =
2 2 6-2 4
=
=
5 15
15
15
6:
(a) Write 21.3751 upto two places of decimal.
(b) Write 3.27645 upto three places of decimal.
Sol. (a) 21.3751 = 21.38
(b) 3.27645 = 3.276
DECIMALS
1.
2.
The decimal expansion of a rational number is either
terminating or non-terminating recurring. More over , a
number whose decimal expansion is terminating or nonterminating recurring is rational.
The decimal expansion of an irrational number is nonterminating non recurring. Moreover, a number whose
decimal expansion is non-terminating non recurring is
irrational.
For example :
2 = 1.41421356237309504880...
p = 3.1415926535897932384626433...
7 : Find an irrational number between
Sol. We find by dividing,
1
2
and .
7
7
1
2
= 0.142857 and = 0.285714 .
7
7
DIVISIBILITY RULES
Divisibility by 2 : A number is divisible by 2 if its unit’s digit is
even or 0.
Divisibility by 3 : A number is divisible by 3 if the sum of its digits
are divisible by 3.
Divisibility by 4 : A number is divisible by 4 if the last 2 digits are
divisible by 4, or if the last two digits are 0’s.
Divisibility by 5 : A number is divisible by 5 if its unit’s digit is
5 or 0.
Divisibility by 6 : A number is divisible by 6 if it is simultaneously
divisible by 2 and 3.
Divisiblity by 7 : A number is divisible by 7 if unit’s place digit is
multiplied by 2 and subtracted from the remaining digits and the
number obtained is divisible by 7.
Divisibility by 8 : A number is divisible by 8 if the last 3 digits of
the number are divisible by 8, or if the last three digits of a number
are zeros.
Divisibility by 9 : A number is divisible by 9 if the sum of its digits
is divisible by 9.
Divisibility by 10 : A number is divisible by 10 if its unit’s digit is 0.
Divisibility by 11 : A number is divisible by 11 if the sum of digits
at odd and even places are equal or differ by a number divisible
by 11.
Divisibility by 12 : A number is divisible by 12 if the number is
divisible by both 4 and 3.
Divisibility by 13 : A number is divisible by 13 if its unit’s place
digit is multiplied by 4 and added to the remaining digits and the
number obtained is divisible by 13.
Divisibility by 14 : A number is divisible by 14 if the number is
divisible by both 2 and 7.
Divisibility by 15 : A number is divisible by 15 if the number is
divisible by both 3 and 5.
8 : Is 473312 divisible by 7?
Sol. 47331 – 2 × 2 = 47327
4732 – 2 × 7 = 4718
471 – 2 × 8 = 455
45 – 2 × 5 = 35
35 is divisible by 7, therefore, 473312 is divisible by 7.
9 : What is the value of M and N respectively if
M39048458N is divisible by 8 and 11, where M and N are single
digit integers?
(a) 7, 4
(b) 8, 6
(c) 6, 4
(d) 3, 2
Sol. (c) A number is divisible by 8 if the number formed by the
last three digits is divisible by 8.
i.e., 58N is divisible by 8.
Clearly,N = 4
Again, a number is divisible by 11 if the difference
between the sum of digits at even places and sum of
digits at the odd places is either 0 or is divisible by 11.
i.e. (M + 9 + 4 + 4 + 8) - (3 + 0 + 8 + 5 + N)
= M + 25 - (16 + N)
= M - N + 9 must be zero or it must be divisible by 11
i.e. M - N = 2
Þ M = 2+4 = 6
Hence, M = 6, N = 4
DIVISIONALGORITHM :
Dividend = (Divisor × Quotient) + Remainder
where, Dividend = The number which is being divided
Divisor = The number which performs the division process
Quotient = Greatest possible integer as a result of division
Remainder = Rest part of dividend which cannot be further divided
by the divisor.
A-4
Number System, HCF and LCM
10 : A certain number when divided by 899 leaves
the remainder 63. Find the remainder when the same number is
divided by 29.
(a) 5
(b) 4
(c) 1
(d) Cannotbe determined
Sol. (a) Number = 899Q + 63, where Q is quotient
= 31 × 29 Q + (58 + 5) = 29 [ 31Q + 2] + 5
\ Remainder = 5
Important Algebraic formulae
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a – b) (a + b) = a2 – b2
(a + b)2 + (a – b)2 = 2(a2 + b2)
(a + b)2 – (a – b)2 = 4ab
(a + b)3 = a3 + b3 + 3ab (a + b)
= a3 + 3a2b + 3ab2 + b3
7. (a – b)3 = a3 – 3a2b + 3 ab2 – b3
= a3 – b3 – 3ab (a – b)
3
3
8. a + b = (a + b) (a2 – ab + b2)
9. a3 – b3 = (a – b) (a2 + ab + b2)
10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
11. a3 + b3 + c3 = 3abc, if a + b + c = 0
1.
2.
3.
4.
5.
6.
Cyclicity
Cyclicity of a number is used mainly for the calculation of
unit digits.
1.
Cyclicity of 1.
In 1n, unit digit will always be 1.
2.
Cyclicity of 1.
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
28 = 256
After every fourth interval 2, 4, 8, 6 are repeated
So cycle of 2 is 2, 4, 8, 6.
11 : Find unit digit of 254.
Sol. Here unit digit will repeat as 2, 4, 8, 6 after every four interval
till 52 next 53 will be 2 and 54 will be 4. So unit digit of 254 =
4.
12 : Find unit digit of 2323.
Sol. Here, 2, 4, 8, 6 will repeat after every four interval till 320 next
digit will be 2, 4, 8 , so unit digit of 2323 will be 8.
13 : Find unit digit of 133133.
Sol. Cycle of 3 is 3, 9, 7, 1 which repeats after every fourth interval
will 133132, so next unit digit will be 3.
14 : Find unit digit of 96363 ×7373.
Sol. Unit digit of 96363 = 7
Unit digit of 7373 = 3
So unit digit of 96363 × 7373 = 7 × 3 = 21.
i.e. 1.
15 : Find unit digit of 469 × 65.
Sol. Unit digit of 469 = 4
Unit digit of 65 = 6
So unit digit of 469 = 65 = 4 × 6 = 24 i.e. 4
Number System, HCF and LCM
A-5
E X ERCISE
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
The difference between a two-digit number and the number
obtained by interchanging the digits of the number is 18.
The sum of the digits of the number is 12. What is the
product of the digits of the two digits number ?
(a) 35
(b) 27
(c) 32
(d) Cannot be determined
There are 15 dozen candles in a box. If there are 39 such
boxes. How many candles are there in all the boxes together?
(a) 7020
(b) 6660
(c) 6552
(d) 3510
The product of two consecutive odd numbers is 19043.
Which is the smaller number?
(a) 137
(b) 131
(c) 133
(d) 129
3
of a number is 250 more than 40% of the same number..
5
What is the number?
(a) 1250
(b) 1180
(c) 1200
(d) 1220
7
1 2 5 6
If the fractions ] ] ]
] and
are arranged in
9
2 3 9 13
ascending order of their values, which one will be the
fourth ?
2
6
(a)
(b)
3
13
5
7
(c)
(d)
9
9
How many numbers, between 1 and 300 are divisible by 3
and 5 together?
(a) 16
(b) 18
(c) 20
(d) 100
What is the remainder when 496 is divided by 6?
(a) 0
(b) 2
(c) 3
(d) 4
The number 311311311311311311311 is :
(a) divisible by 3 but not by 11
(b) divisible by 11 but not by 3
(c) divisible by both 3 and 11
(d) neither divisible by 3 nor by 11
The divisor is 25 times the quotient and 5 times the
remainder. If the quotient is 16, the dividend is :
(a) 6400
(b) 6480
(c) 400
(d) 480
2 + 2 + 2 + ..... is equal to
2
(a)
(c) 2
(b) 2 2
(d) 3
11.
1
If 2 = x +
1+
3+
(a)
, then the value of x is :
1
1
4
18
17
(b)
13
(d)
17
The simplified value of
(c)
12.
21
17
12
17
1 öæ
1 ö
æ 1 ö æ 1 ö æ 1 ö ...... æ
çè1 – ÷ø çè1 - ÷ø çè 1– ÷ø
çè 1– ÷ø çè1–
÷ is
3
4
5
99
100 ø
(a)
2
99
1
50
The simplification of
(c)
13.
(b)
1
25
(d)
1
100
3.36 – 2.05 + 1.33 equals :
(a) 2.60
14.
15.
16.
17.
18.
19.
(b)
2.61
(c) 2.64
(d) 2.64
The sum of all the natural numbers from 51 to 100 is
(a) 5050
(b) 4275
(c) 4025
(d) 3775
If * is an operation such that x * y = 3x + 2y, then 2 * 3 + 3 * 4
is equal to :
(a) 18
(b) 29
(c) 32
(d) 38
Find the sum of all odd numbers between 100 and 175.
(a) 5069
(b) 5059
(c) 5049
(d) 5039
Product of two co-prime numbers is 117. Their L.C.M. should
be:
(a) 1
(b) 117
(c) equal to their H.C.F.
(d) cannot be calculated
The HCF and LCM of two numbers are 44 and 264
respectively. If the first number is devided by 2, the quotient
is 44. What is the other number?
(a) 108
(b) 44
(c) 124
(d) 132
The traffic lights at three different road crossings change
after every 48 seconds, 72 seconds and 108 seconds
respectively. If they all change simultaneously at 8 : 20 hours,
then at what time will they again change simultaneously ?
(a) 8 : 20 : 08 hrs
(b) 8 : 24 : 10 hrs
(c) 8 : 27 : 12 hrs
(d) 8 : 30 : 15 hrs
A-6
20.
21.
22.
23.
What least number must be subtracted from 1936 so that
the remainder when divided by 9, 10, 15 will leave in each
case the same remainder 7?
(a) 29
(b) 39
(c) 49
(d) 59
Find the greatest number that will divide 55, 127 and 175 so
as to leave the same remainder in each case.
(a) 26
(b) 24
(c) 23
(d) 29
The H.C.F. of
2
3
(b)
2
81
(c)
160
3
(d)
160
81
The L.C.M. of
32.
(a)
25.
26.
27.
34.
The arrangement of the fractions
(b)
1
36
1 öæ
1 öæ 2 1
öæ 2 1
ö
æ
ç x + ÷ç x - ÷ ç x + 2 - 1 ÷ç x + 2 + 1÷ is equal to
x
x
è
øè
øè
x
x
øè
ø
(c) x8 +
28.
1
x6
1
x
8
2 5 1 3
, , , in ascending
9 8 3 4
order is
1
12
(d)
1365
455
Three numbers are in the ratio 1 : 2 : 3 and their H.C.F. is 12.
The numbers are :
(a) 4, 8, 12
(b) 5, 10, 15
(c) 10, 20, 30
(d) 12, 24, 36
The product of 2 number is 3024 and their LCM is 36.
Find their HCF.
(a) 81
(b) 85
(c) 86
(d) None of these
(a + b + c)2 – (a – b – c)2 = ?
(a) 4a(b + c)
(b) 2a(b + c)
(c) 3a(b + c)
(d) 4a(b – c)
(a) x6 +
5 2 5 3
, , ,
6 3 9 8
33.
2 3 4 9
, , , is :
3 5 7 13
(c)
24.
(b)
3 5 2 5
3 2 5 5
, , ,
, , ,
(d)
8 9 3 6
8 3 9 6
Martin has some marbles. He lost 14 and still had 6 left.
How many did he have to start with?
(a) 8
(b) 20
(c) 84
(d) None of these
(a)
(a) 36
3 5 2 5
, , ,
8 6 3 9
(c)
2 8 64
10
, ,
and
is :
3 9 81
27
(a)
31.
Number System, HCF and LCM
How many four digit numbers are there between 999 and
3000?
(a) 2001
(b) 2000
(c) 1999
(d) 1998
Arrangement of the fractions in ascending order is
(b) x8 +
(d) x6 –
35.
36.
37.
1
1
= 2 and x is real, then the value of x 17 + 19 is
x
x
(a) 1
(b) 0
(c) 2
(d) – 2
38.
39.
40.
If x +
41.
1
1
= 5, then x2 + 2 is :
x
x
29.
If x –
30.
(a) 5
(b) 25
(c) 27
(d) 23
Find the sum of the largest and least four-digit numbers.
(a) 9999
(b) 10000
(c) 10999
(d) 11999
3 5 1 2
, , ,
4 8 3 9
1 3 5 2
2 1 3 5
, , ,
, , ,
(d)
3 4 8 9
9 3 4 8
Which of the following statements is true ?
(a) 1 is not a prime number
(b) 1 is a prime number
(c) 1 is a composite number
(d) 2 is not a prime number.
The place value of '0' in 273045 is
(a) 1000
(b) 0
(c) 100
(d) 10
Find the sum of the AP : –37, –33, –29, ..........of first 12 terms
(a) –180
(b) 180
(c) 810
(d) –108
m
1
x6
(b)
(c)
1
x8
2 1 5 3
, , ,
9 3 8 4
42.
m
æ P ö æQö æRö
The value of ç ÷ ç ÷ ç ÷
èQø èRø è P ø
m
will be :
(a) P
(b) Q
(c) R
(d) 1
The place value of 1 in 0.0159 is –
(a) 1
(b) 1/10
(c) 1/100
(d) 1/1000
'm' and 'n' are two positive integers. If mn = 25, then what is
the value of nm ?
(a) 4
(b) 10
(c) 25
(d) 32
Sum of a number of two digits and the number obtained by
reversing the digits of the first number is 110. If the
difference of the digits is 4, then the number is
(a) 62
(b) 73
(c) 84
(d) 51
Divide 50 by half and add 20. From the same, subtract 35.
What do you get ?
(a) 10
(b) 85
(c) 15
(d) None of these
Number System, HCF and LCM
43. Sum of place values of 6 in 63606 is
(a) 6066
(b) 18
(c) 60606
(d) 6606
1
1
44. How many are in ?
8
2
(a) 2
(b) 16
(c) 8
(d) 4
45. The sum of all the factors of 100 is
(a) 223
(b) 115
(c) 216
(d) 217
46. The number of factors of 18 is
(a) 4
(b) 5
(c) 6
(d) 7
47. When 71777 is divided by 7, the remainder is
(a) 1
(b) 3
(c) 5
(d) 6
48. If 34*68 is to be divisible by 9, then the value of missing
digit, * should be
(a) 6
(b) 5
(c) 4
(d) 3
A-7
a b c 7a - 4b + 3c
= = =
, then the value of P is
3 4 7
P
(a) 25
(b) 26
(c) 27
(d) 28
49.
If
50.
If
51.
a 1
a+b
= , then
is equal to
b 2
a -b
(a) –3
(b)
1
2
(c) 2
(d)
10
7
In number 97580, when the digits 7 and 5 as interchanged
its place, then the difference between the original and the
new number is
(a) 1800
(b) 1080
(c) 1008
(d) 1000
ANSW ER KEY
1
(a)
11
(b)
21
(b)
31
(c)
41
(b)
2
(a)
12
(c)
22
(b)
32
(c)
42
(a)
3
(a)
13
(d)
23
(a)
33
(b )
43
(c)
4
(a)
14
(d)
24
(d)
34
(a)
44
(d)
5
(a)
15
(b)
25
(d)
35
(a)
45
(d)
6
(c)
16
(a)
26
(a)
36
(b )
46
(a)
7
(d )
17
(b)
27
(d)
37
(a)
47
(d)
8
(d )
18
(d)
28
(c)
38
(d )
48
(a)
9
(b )
19
(c)
29
(c)
39
(c)
49
(b)
10
(c)
20
(b)
30
(c)
40
(d )
50
(a)
51
(a)
HINTS AND SOLUTIONS
1.
(a) Let the two-digit number be = 10x + y, where x > y
According to the question,
4.
3x 2 x
= 250
5
5
Þ x = 250 × 5 = 1250
Then
10x + y – 10y – x = 18
or, 9x – 9y = 18
or, 9(x – y) = 18
18
or, x – y =
=2
9
and, x + y = 12
From equations (i) and (ii)
2.
3.
...(i)
...(ii)
14
=7
2
From equation (i)
y= 7– 2 = 5
\ Required product = xy = 7 × 5 = 35
(a) Total number of candles = 15 × 12 × 39 = 7020
(a) Out of the given alternatives,
137 × 139 = 19043
\ Required smaller number = 137
2x = 14 Þ x =
(a) Let the number be x
5.
7
1 2 5 6
, , ,
and
9
2 3 9 13
LCM of their denominators is 234
(a) The given fractions are
117, 78, 26,18, 26
234
117, 2 ´ 78,5 ´ 26, 6 ´ 18, 7 ´ 26
234
117,156,130,108,182
234
On arranging the numerators in ascending order
108, 117, 130, 156, 182.
\ Ascending order of the fraction is
6 1 5 2 7
< < < <
13 2 9 3 9
\
A-8
6.
7.
8.
9.
Number System, HCF and LCM
(c) LCM of 3 and 5 = 15
300
\
= 20 numbers
15
(d) Let us divide the different powers of 4 by 6 and find
the remainder.
So remainder for 41 = 4, 42 = 4, 43 = 4, 44 = 4, 45 = 4,
46 = 4 and so on.
Hence remainder for any power of 4 will be 4 only.
(d) Sum of digits = 35 and so it is not divisible by 3.
Now, (Sum of digits at odd places)
– (Sum of digits at even places)
= (19 – 16) = 3, not divisible by 11.
So, the given number is neither divisible by 3 nor by 11.
(c)
13.
(d) 3.36 - 2.05 + 1.33
=3
= 2+
14.
x
25
15.
....(i)
16.
= Divisor × Quotient + remainder = x ´
x x
+
25 5
xæx ö
ç + 1÷
5è5 ø
16 ´ 25 ´ 405
= 6480
25
OR, Divisor = 25 × 16 = 400
=
Remainder =
11.
(c) Let x = 2 + 2 + 2 + .....
x2 = 2 +
2 + 2 + 2 + .....
x2 = 2 + x Þ x2 – x – 2 = 0
x2 – 2x + x – 2 = 0 Þ x (x – 2) + 1(x – 2) = 0
(x – 2) (x + 1) = 0
x = 2 or –1
(b) 2 = x +
1
1+
Þ2= x+
1
1
3+
4
Þ2= x+
100
= 50 odd numbers
2
Sum of first 50 odd numbers = (50)2 = 2500
174
= 87 odd numbers
2
Sum of first 87 odd numbers = (87)2 = 7569
\ Required sum = (7569 – 2500) = 5069
17. (b) H.C.F of co-prime numbers is 1.
So, L.C.M. = 117/1 = 117.
18. (d) The first number = 2 × 44 = 88
HCF ´ LCM 44 ´ 264
=
= 132
88
88
19.
\ Dividend
= 400 × 16 + 80 = 6480
Þ
Þ
Þ
Þ
(a) Upto 100 there are
\ The second number =
400
= 80
5
100 ´ 101 50 ´ 51
–
2
2
= 5050 – 1275 = 3775
(b) Here, x * y = 3x + 2y
\2*3+3*4
= (3 × 2 + 2 × 3) + (3 × 3 + 2 × 4)
= 12 + 17 = 29
Upto 174 there are
16 ´ 25 æ 25 ´ 16 ö
+ 1÷ [Putting the value of x]
=
ç
5 è 5
ø
\
n(n + 1)
\ 51 + 52 + ..... + 100
2
= (1 + 2 + ..... 100) – (1 + 2 + ... + 50) =
x
5
Given, quotient = 16
x
= 16
So,
25
\ x = 25 × 16
Dividend
10.
64
= 2.64
99
(d) We have,
1 + 2 + 3 ..... + n
=
Remainder =
=
36
05 33
36
05
33
- 2 +1
= 3+
-2+1+
99
99 99
99
99
99
æ 36 5 33 ö
æ 36 - 5 + 33 ö
= (3 - 2 + 1) + ç - + ÷ = 2 + ç
÷ø
è 99 99 99 ø
è
99
(b) Let the divisor be x
According to the question
Quotient will be =
2 3 4
98 99
´ ´ ´ ..... ´ ´
3 4 5
99 100
2
1
=
=
100 50
12.
1
1+
1
12 + 1
4
1
1
x+
x+
13 + 4 Þ 2 =
4 Þ2=
17
1+
13
13
13
1
21
13
34 –13
13
=
=
Þ2= x+
Þx= 2–
17
17
17
17
(c) The traffic lights will again change at three different
road crossings simultaneously after the LCM of 48, 72
and 108
i.e., after every (432 sec) 7 minutes and 12 seconds, i.e.
the earliest at 8 : 27 : 12 hours.
20. (b) The LCM of 9, 10 and 15 = 90
On dividing 1936 by 90, the remainder = 46
But 7 is also a part of this remainder.
\ the required number = 46 – 7 = 39
21. (b) Let x be the remainder, then the numbers (55 – x),
(127 – x) and (175 – x) are exactly divisible by the
required number.
Now, we know that if two numbers are divisible by a
certain number, then their difference is also divisible
by the number. Hence the numbers (127 – x) – (55 – x),
(175 – x) – (127 – x) and (175 – x) – (55 – x) or, 72, 48 and
120 are divisible by the required number. HCF of 48, 72
and 120 = 24, therefore the required number = 24.
22.
(b) Required H.C.F. =
H.C.F.of 2,8, 64,10 2
= ×
L.C.M.of 3,9,81, 27 81
Number System, HCF and LCM
23.
24.
25.
(a) Required L.C.M =
L.C.M.of 2,3, 4,9 36
=
= 36.
H.C.F.of 3,5,7,13 1
(d) Let the required number be x, 2x and 3x. Then, their
H.C.F. = x. So, x = 12.
\ The numbers are 12, 24 and 36.
(d) Product of 2 numbers = Product of (LCM × HCF)
Þ 3024 = 36 × HCF Þ HCF =
26.
27.
A-9
33.
34.
(b) Let no. of marbles = x
According to question
x – 14 = 6
x = 6 + 14 = 20
(a) Converting fractions into decimals :
2
5
1
= 0.222; = 0.625; = 0.333;
9
8
3
3024
= 84
36
(a) (a + b + c)2 – (a – b – c)2
= (a + b + c + a – b – c) (a + b + c – a + b + c)
= 2a(2b + 2c) = 4a(b + c)
3
= 0.750
4
Ascending order is
36.
(b) 273045
1 öæ
1ö
æ
(d) ç x + ÷ç x - ÷
x
xø
è
øè
0 × 100
Þ 0.
æ 2 1 öæ 2 1
öæ 2 1
ö
ç x - 2 ÷ ç x + 2 - 1÷ç x + 2 + 1÷
x
øè
x
ø
è
x øè
37.
(a) Sum of A.P. =
Sum =
æ 2 1 ö éæ 4 1
öù
= ç x - 2 ÷ êç x + 4 + 2 –1÷ú
è
x ø ëè
x
øû
m
29.
31.
32.
(d)
39.
(c)
0.0154
40.
41.
1
=5
x
On squaring both sides,
1
– 2 = 25 Þ x2 +
1
= 27
x
x2
(c) Largest four digit no. = 9999
smallest four digit no. = 1000
sum = 9999 + 1000 = 10,999
(c) Required number = 2999 – 1000 = 1999
(c) Converting fractions into decimals.
2
m
Þ
3
5
2
= 0.375; = 0555; = 0.666 ;
8
9
3
5
= 0.833
6
3 5 2 5
\ 8 , 9 , 3 , 6 are in ascending order..
Pm
Q
m
´
Qm
R
m
´
Rm
Pm
–2
The place value of 1 is
(c) x –
x2 +
30.
38.
=1+1=2
x19
m
æ P ö æQö æR ö
çç ÷÷ ç ÷ ç ÷
èQø èRø è P ø
10
1
(c) x + = 2
x
Þ
x2 – 2x + 1 = 0
Þ
(x – 1)2 = 0 Þ x = 1
1
12
[2 × – 37 + (12 – 1) 4]
2
= 6 (–74 + 11 × 4) = 6(–30) = – 180
æ 2 1 öæ 4 1
ö
1
= ç x - 2 ÷ç x + 4 + 1 ÷ = x6 – 6
è
x øè
x
ø
x
\ x17 +
n
(2a + (n–1)d)
2
where , n = no. of terms
a = first term
d = common difference
2
æ 2 1 ö éæ 2 1 ö ù
ê
x
x
+
-1ú
ç
÷
ç
÷
=
è
x2 ø êëè
x2 ø úû
28.
2 1 5 3
, , ,
9 3 8 4
1
.
100
(d) mn = 25
52 = 25
Here, m = 5, n = 2
value of n m = 25 = 32
(b) Let the two digit number be 10x + y
Reversing the digit, number become
10y + x.
sum = 10x + y + 10y + x
Þ 11x + 11y = 110
...(1)
Þ x + y = 10
x – y = 4 (Given)
...(2)
From equ. (1) & (2)
2x = 14 Þ x = 7
\ y=3
Hence the number is 73.
50
+ 20 – 35 = 45 – 35 = 10
2
42.
(a)
43.
(c) Place values of all 6 in
63606
60000
600
6
Sum = 60000 + 600 + 6 = 60606
=1
A-10
44.
(d) Let ‘x’
x ×
1
1
are in
8
2
48.
1 1
=
8 2
49.
1
×8=4
2
(d) Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
Sum of all the factors of 100
= 1+ 2 + 4 + 5 + 10 + 20 + 25 + 50 + 100
= 217
\x=
45.
46. (a)
(b)
a b c
= = =k
3 4 7
\ a = 3k, b = 4k, c = 7k
7a - 4b + 3c a
=
P
3
7 ´ 3k - 4 ´ 4k + 3 ´ 7k 3k
=
P
3
2 18
3 9
3
Number System, HCF and LCM
(a) According to divisibility rule of 9, sum of all the digits
should be divisible by 9.
\ 3 + 4 + * + 6 + 8 = 21+*
So, * = 6 Þ 21 + 6 = 27 is divisible by 9.
3
21k - 16k + 21k
=k
P
1
Þ 21 - 16 + 21 = P Þ P = 26
\ 18 = 2 × 3 × 3 × 1
Number of factors = 4
50.
51.
Remainder
1
b
2
1
1
3
b+b
+1
a+b 2
2
2
=
=
=
=–3
1
1
a-b 1
b-b
-1 2
2
2
47. (d) 7
71777 10253
7
17
14
37
35
27
21
6
(a) a =
(a) Original number = 97580
New number = 95780
Reqd. difference = 97580 – 95780 = 1800
C
H
A
P
T
E
R
2
Simplification
Simplification : Algebraic expressions contain alphabetic
symbols as well as numbers. The process to find the value of
given expression is called simplification.
Rule of Simplification : To solve the expression we use a rule
‘V-BODMAS’.
3 : What value should come in place of the question
mark (?) in the following question ?
3
5 2
of
of of 1680 = ?
8
7
5
Symbol
Full name
Sign of the symbol
(a) 150
(b) 180
V
B
O
D
M
A
S
Viniculam
Bracket
Of
Division
Multiplication
Addition
Subtraction
– (Bar)
( ), { }, [ ]
(as multiply)
÷
×
+
–
(c) 210
(d) 240
Before applying the ‘BODMAS’ rule, we simplify the expression
under the viniculam. After removing the brackets, we must the
operations in the given order.
1 : What value should come in place of the question
mark (?) in the following question ?
432×66 –1562 = ?
(a) 23450
(b) 24360
(c) 25890
(d) 26950
Sol. (d) ? = 432 × 66 – 1562 = 28512 – 1562 = 26950
2 : What value should come in place of the question
mark (?) in the following question ?
6
(a) 1
(c) 1
Sol. (b)
2
3
4
5
÷ 4 =?
1
3
(b) 1
12
19
(d) 1
7
18
5
8
Sol. (b)
2 5 3
? = 1680 ´ ´ ´ =180
5 7 8
4:
4´ 2 + 6
=?
5 ´ 16 - 2
(a) 5
(b)
16
35
1
5
(d)
7
39
(c)
Sol. (d) ? =
8+ 6
14
7
4´2+6
=
=
=
5 ´ 16 - 2 80 - 2 78 39
5 : The solution of
is
(a) 1
(c) 3
Sol. (a)
0.827 ´ 0.827 - 0.173 ´ 0.173
0.654
(b) 2
(d) 4
0.827 ´ 0.827 - 0.173 ´ 0.173
0.654
Þ
Þ
20 24 = 20 ´ 5 = 25 = 1 7
?=
¸
3 24 18
18
3
5
(0.827)2 - (0.173)2
0.654
(0.827 + 0.173) (0.827 - 0.173)
0.654
[\ a 2 - b2 = (a + b) (a - b)]
Þ
(1) (0.654)
=1
0.654