Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Study material of Math for class X
Document related concepts
Transcript
KENDRIYA VIDYALAYA CHENANI
dsUnzh; fo|ky;,puSuh
STUDY MATERIAL OF MATH FOR CLASS X
REAL NUMBER (GROUP NUMBER 1)
LEVEL WISE QUESTIONS AND SOLUTIONS
LEVEL- 1 (1 mark Each)
1. Find HCF of 128 and 216.
Solution 1:
Two numbers are : 216, 128
216 > 128
By using Euclid division lemma :
216 = 128 x 1 +88 ( R≠0 )
128 = 88 x 1 + 40 ( R ≠0 )
88 = 40 x 2 + 8
(R ≠ 0)
40 = 8 x 5 + 0
Here R = 0,
Divisor is 8, so HCF = 8.
1Mark
2. Given HCF(306,657) = 9. Find LCM(306,657)
Solution 2:
HCF = 9, Ist number = 306, IInd number = 657
We know that :
HCFxLCM = Products of two number
9 x LCM = 306 X 657
306x657
LCM =
9
LCM = 22338.
Mark
1
3. Prove that 3+2 5 is an irrational number.
Solution 3: Let us suppose that 3+2 5 is rational so, we can find co prime a and
b
where a and b are integers and b≠0.
a
b
So ,3+2 5 =
2 5 =
5 =
a
-3
b
a 3b
2b
Since a and b both are integers
a 3b
is rational number, so,
2b
5 is also rational number
But this contradicts the facts that
5 is irrational.
our supposition is wrong.
Hence 3+2 5 is rational.
Mark
4. Explain why 7×6×5×4×3×2×1+5 is a composite number.
Solution 4:
we have
7x6x5x4x3x2x1+5
5(7x6x1x4x3x2x1+1)
5(1008+1)
5(1009)
5045
The of factors of 5045 are 1,5,1009,5045.
It has more than two factors, so it is a composite number
1Mark
LEVEL -2 (2 marks)
1. Check whether 6n can end with the digit 0. For any natural number n.
Solution 1. : Let us suppose that 6n ends with the digits 0 for some n N
So,6n is divisible by 5
1
But prime factor of 6 are 2 and 3
prime factor of 6n are ( 2x3)n
It is clear that in prime factorization of 6n there is no place of 5.
So our supposition is wrong . Hence there exists no any natural number n for which 6n
ends with digit zero.
2 Marks
2. Find the LCM and HCF of 91 and 26, also verify that HCF×LCM= product of
two
Solution: Two number are 91 and 26
91 = 7x13
26 = 2x13
HCF = 13
LCM = 13x2x7 = 182
1
Mark
Verification
HCFXLCM = Products of two number
13x182 = 91x26
2366 2366
1Mark
3. The HCF of two number is 4 and their LCM is 9696,if one number is 96,find the
other number.
Solution:
HCF = 4
LCM = 9696, Ist number = 96,
We know that HCFXLCM = Ist x IInd
1Mark
4x 9696 = 96 x IInd
4x9696
= IIndIInd number = 404
96
1 Mark
4
.Find the LCM and HCF of 12 , 15 and 21 by prime factorization method.
Solution4 : numbers are 12, 15, 21
12 = 2x2x3
15 = 3x5
21 = 3x7
1Mark
So, HCF = 3
LCM = 3X2X2X5X7 = 420.
1
Mark
LEVEL -3 (3 marks)
1. Prove that
2 is irrational number.
Solution 1: Let us suppose that
p
q
2 is rational number, so it can be put in the form of
q≠0, and p and q are co prime
2=
p
q
squaring both side
2=
p2
q2
q2 =
p2
here p2 is divisible by 2, so p is also divisible by 2.
2
1
Mark
Let p = 2r
putting this value in above relation
4r 2
q =
2
2
so, r2 =
q2
, here q2 is divisible by 2, so q is also divisible by 2
2
from these two p and q both divisible by 2 but it is suppose p and q are co prime
which is contradiction so,
Mark
2 is irrational.
1
3
Show that the square of any positive integer is of the form 3m or 3m+1,for some
integer m.
Solution 3: Let a be any positive integer then it is of the form 3q,3q+1or 3q+2
If
a = 3q
squaring both sides,
a2 =9q2
= 3(3q2)
= 3m,
If
where m = 3q2
a = 3q+1
1Mark
squaring both sides,
a2 = (3q+1)2
= 9q2+1+2x3qx1
= 3(3q2+2q) +1
4
4
= 3m+1,
where m = (3q2+2q)
Where m is also an integer,
1 Mark
Hence , square of any positive integer is either of the form 3m or 3m+1 for some
integer m.
Find the LCM and HCF of 510 and 92 and verify the thatLCMxHCF = Product of
two no.
Solution 4. Two number are 510 and 92
510 = 2x3x5x17
92 = 2x2x23
HCF = 2
1
Mark
LCM = 2X2X3X5X17X23 = 23460
VERIFICATION :
LCMXHCF =23460X2 = 46920
Products of two numbers = 510x92 = 46920
Hence verified .
Mark
LEVEL -4 (4 marks)
1:- Show that 3√2 is irrational.
Sol. 1 :- Let 3√2 is rational.
1
Therefore x = 3√2
=> x/3 = √2
Marks
2
=> x/3 is rational and √2 is also rational.
But this contradicts the fact that √2 is irrational.
=> 3√2 is irrational.
Marks
2
2 :- Explain why ( 7 x 11 x 13) + 13 is a composite number.
Sol.2 :- ( 7 X 11 X 13) +13
Marks
2
= 13 x ( 7 x 11 + 1 ) x 1
So , it is product of more than 2 factors. Hence it is a composite number.
Marks
2
3 : Find the HCF of 616 and 32 using Euclid’s division algorithm.
Sol. 3:-
Using Euclid’s division lemma,
a = bq + r, . 0 ≤ r < b
Step 1 : 616 = 32 x 19 + 8
2 Marks
Step 2 : 32 = 8 x 4 + 0
HCF of 616 and 32 is 8.
Marks
2
4 .Use Euclid‟s division algoritha to find the largest number which divides 957 and
1280 leaving remainder 5 in each case.
Solution 4: 957-5=952 and 1280-5= 1275,are completely divisible by required number.
Now find the HCF by Euclid division lemma,
1275 > 952 by apply division lemma
1275 = 952 x 1 + 323 ( since R≠ 0 )
Marks
2
952 = 323 x 2 + 306 ( since R≠ 0 )
323 = 306 x 1 + 17
( since R ≠ 0)
306 = 17 x 18 + 0
here R = 0
Divisor in the last step is 17
HCF of 1275 and 952 is 17.
Hence required number is 17.
2 Marks
HOTS
REAL NUMBERS
2 MARKS QUESTIONS
1. Find the largest positive integer that will divide 122,150 and 115 leaving
remainder 5, 7 and 11 respectively.
Solution 1: 122-5 = 117 is exactly divisible by the required number
150-7 = 143,
115-11 = 104
So, required number is the HCF of 117,143,104
1 Mark
117 = 3x3x13
143 = 11x13
104 = 2x2x2x13
HCF ( 117,143,104 ) = 13
Hence, required number is 13.
1 Mark
2. Use Euclid‟s division algorithm to find the largest number which divides 957 and
1280 leaving remainder 5 in each case.
Solution 2: 957-5=952 and 1280-5= 1275,are completely divisible by required number.
Now find the HCF by Euclid division lemma,
1275 > 952 by apply division lemma
1275 = 952 x 1 + 323 ( since R≠ 0 )
952 = 323 x 2 + 306 ( since R≠ 0 )
323 = 306 x 1 + 17
306 = 17 x 18 + 0
1 Mark
( since R ≠ 0)
here R = 0
Divisor in the last step is 17
HCF of 1275 and 952 is 17.
Hence required number is 17.
1 Mark
3 MARKS QUESTIONS
1
In a school there are two sections of class 10th.there are 40 students in 1st section
and 48 students in second section. Determine the minimum number of books
required for their class library so that they can be distributed equally among
students of both sections.
Solution 1: Required number of books = LCM ( 40,48 )
40 = 2x2x2x5
48 = 2x2x2x2x3
LCM (40,48) = 2x2x2x2x3x5 = 240
Hence, required number of books are 240
2.
1 Mark
1 Mark
1 Marks
In a morning walk Nirmaljeet, Puneet, Rajiv step off together, their steps
measuring 240 cm , 90 cm, 120cm respectively. What is the minimum distance
each should walk so that one can cover the distance in complete steps?
Solution 2: 240 = 2x2x2x2x3x5
1 MARK
90 = 2x3x3x5
120 = 2x2x2x3x5
1 Mark
LCM = 2x2x2x2x3x3x5 = 720
Hence required distance 720 cm.
1 Mark
3. The sets of mathematics, physics and physical education books have to be stacked
in such a way that all the books are stored topic wise. The number of mathematics
, physics and physical education books are 14 , 18 and 22. Determine the number
of stacks of each books provided books are of the same thickness.
Solution 3: Firstly , to arrange the books as according to condition,
Find HCF of 14, 18 and 22.
14 = 2x7
18 = 2x3x3
22 = 2x11
1 Mark
HCF = 2
So ,there are only 2 books in each stack.
1 Mark
Number of stack of Mathematics books =
Number of stack of Physics books =
14
=7
2
18
=9
2
Number of stack of Physical Education books =
22
= 11.
2
1 Mark
Topic: REAL NUMBERS
Basic Concepts:
Natural Numbers:- All counting numbers 1,2,3,4…are called natural numbers. If N is
the set of natural numbers, then N = {1, 2, 3, 4,…}
1,2,3,4,… Natural Numbers
Whole Numbers:- The natural numbers including zero form the set of whole numbers. It
is denoted by W. Thus W = {0,1,2,3,4,…}
0
1,2,3,4,… Natural Numbers
Integers:- The natural numbers, their negatives and zero form the set of integers. It is
denoted by Z. Thus Z = {0,± 1, ±2, ±3, ±4,…}
Note:- Z is the first letter of the German word „Zahlen‟ means „to count‟ and „Zahl‟
means „number‟. Integers are also denoted by I.
Integers
0
1,2,3,4,… Natural Numbers
Classification of Natural Numbers
I) Natural numbers are divided into three groups:i)Unit:-1 is the only unit in natural numbers.
ii)Prime:-A natural number “a” is said to be Prime if it is divisible by 1 and „”a” itself.
For example 2,3,5,7,11,13,17,19,23,29,31,… are all prime numbers.
Observations:-The set of primes is infinite.
2 is the least and only even prime number.
iii)Composite:-A natural number “a “is said to be Composite if it has atleast one more
divisor except 1 and „a‟ itself.
For example 4,6,8,9,10,12,14,…are all composite numbers.
Natural Numbers
2,3,5,7,11,… Prime
1 Unit
Note:-In set of integers, Units are ±1 and an integer „a‟ is said to be prime if it is divisible
by
±1 and ±a.
II) Twin Primes:-A pair of prime numbers is said to be twin primes if they differ by 2.
For example 3,5; 5,7; 11,13; 17,19 are twin primes while as 7,11; 13,17 are not twin
primes.
III) Co-prime:-Two numbers are said to be co-prime or relatively prime if their
HCF/GCD is 1.
For example 4,9 are co-prime as HCF(4,9) = 1.
IV) Perfect number:-A number is said to be perfect if it is equal to the sum of its factors
other than itself.
OR
A number is said to be perfect if the sum of its all factors equals twice the number.
For example 6 is perfect number as sum of its factors 1,2,3 (except 6)=6
or 1+2+3+6=12=2(6)
28 is next perfect number, 496 is third perfect number.
Rational numbers:-A number in the form of
where
are integers;
and
(
)
i,e
are coprimes is called a rational number. Rational numbers are denoted
by Q (Q comes from word Quotient).
Thus Q = { ⁄
,
and (
)
Note:-The word „rational‟ is derived from the word „ratio‟
Rational Numbers
½, 3/4,22/7,31/37,…
Integers
Whole Numbers
0
}
Natural Numbers
1,2,3,…
Decimal Representation of Rational Numbers
Property I: Each rational number can be represented by a recurring or terminating
decimal fraction.
For example = 0.25; = 0.375;
= 0.33…= 0. ̅ ; = 0.66…= 0. ̅ ;
= 1.6875 are terminating decimals.
= 0.3636…= 0.̅̅̅̅;
= 3.142857142857…= 3.̅̅̅̅̅̅̅̅̅̅are all recurring decimals.
Property II: (Converse of Property I) Each decimal fraction (terminating or recurring)
represents a rational number.
For example i) 0.25 = 25/100 = , a rational number
ii) 0. ̅
Let x = 0. ̅ = 0.33…
(1)
Multiplying (1) by 10 both sides, we get
10x = 3.33…
(2)
Subtract (1) from (2), we get
9x = 3 => x = => x = , a rational number
iii) 0.̅̅̅̅
Let x = 0.̅̅̅̅ = 0.3636…
(1)
Multiplying (1) by 100 both sides, we get
100x = 36.3636…
(2)
(2) – (1) gives
99x = 36 => x =
=> x =
, a rational number
Note: A terminating decimal fraction is a particular case of a recurring decimal fraction.
For example 0.46 = 0.45999… = 0.45 ̅
Classification of Decimals
Decimals
Terminating Decimals
Non Terminating Decimals
(Rational Numbers)
(Infinite Decimals)
Non Terminating
Non Terminating non
Repeating Decimals
Repeating Decimals
(Irrational Numbers)
Pure Recurring Decimal
Mixed Recurring Decimal
(Rational Numbers)
(Rational Numbers)
Irrational Numbers:-A number that cannot be expressed in the form of
where
.
Or
A number that cannot be expressed in decimal form either in terminating or in repeating
decimals is known as Irrational numbers. Irrational numbers are non terminating and non
repeating decimals.
For example√ ,√ , √ , √ , , e…are Irrational numbers.
The set of irrational numbers is denoted by S.
Real numbers:-Rational numbers and Irrational numbers taken together are known as
real numbers. Thus every real number is either a rational number or an irrational number.
The set of real numbers is denoted by R.
Real numbers
Rational and Irrational numbers
𝜋
Integers
0. ̅
0.̅̅̅̅
e
√
Whole numbers
√
1,2,3,… Natural numbers
0
.
-1,-2,-3,…
ALGORITHM:-Algorithm is a series of well-defined steps which gives a procedure for
solving a type of problem.
LEMMA:-A lemma is a proven statement used for proving another statement.
EUCLID’S DIVISION LEMMA: Given positive integers a and b there exist unique
integers q and r satisfying a=bq+r, o≤r<b.
EUCLID’S DIVISION ALGOROTHM: It is a method to find HCF of two given
Positive integers. Euclid‟s division algorithm is used to find HCF of two positive
integers say m and n where m>n:
Step1: Find whole numbers q and r such that m=nq+r, o≤r<n.
Step2: If r=0 then n is the HCF of m and n. If r≠0 apply the division
lemma to n and r.
Step3: Continue the process till the remainder is 0. The divisor at
this stage will be the required HCF.
Note: Let a and b be positive integers. If a=bq+r, o≤r<b.
Then HCF(a,b)=HCF(b,r).
FUNDAMENTAL THEOREM OF ARITHMETIC:
Every composite number can be expressed as a product ofprimes, and this
factorization is unique except for the order inwhich they occur.
Every positive integer different from 1 can be expressed as product of non-negative
power of 2 and an odd number.
TERMINATNG DECIMAL REPRESENTATION
Let x=p/q be a rational number, such that the prime factorization of q is of the form
2 x5nwhere m and n are non-negative integers. Then x has a terminating decimal
expansion which terminates after k places of decimal, where k is the larger of m and n.
m
NON-TERMINATING DECIMAL REPRESENTATION
Let x=p/q be a rational number, such that the prime factorization of q is not of the form
2 x5n where m and n are non-negative integers. Then x has a non- terminating
repeating decimal expansion.
m
POLYNOMIALS
Basic concepts and Tips
Definitions of polynomials: an algebraic expression of the form
P(x) = a0xn + a1 xn-1 +…..an-1x + an where a0, a1 , a2…. an are the real number where a0≠ 0 ;
n is non –negative integer is called a polynomial in x .
OR
Polynomial is an algebraic expression in one variable with exponent of variable as whole
number only.e.g. x+1, x2 – x +3 , 3x3 + 5x2 – 3x +7
Concepts
•
Degree of polynomials- the highest power of x in p (x) is called the degree of the
polynomials.
•
•
The polynomial of the degree 1 is called linear polynomial
•
The polynomial of the degree 2 is called quadratic polynomial
•
The polynomial of the degree 3 is called cubic polynomial
•
The degree of Constant polynomial is 0.
•
The degree of zero polynomial is not defind.
Geometrical meaning of zero of polynomial –
•
Zero of linear polynomial is x-coordinate of the point of intersection of
straight line graph of linear polynomial with the x-axis.
•
Zeroes of Quadratic polynomial are the x-coordinates of the points of
intersection of graph with the x-axis. The graph of any Quadratic
polynomial is in the shape of parabola. Quadratic polynomial can have at
most two zeros.
•
Zeroes of Cubic polynomial are the x-coordinates of the points of
intersection of graph with the x-axis. Cubic polynomials can have at most
three zeroes.
•
•
Zeroes of any polynomial of degree n, are the x-coordinates of the points
of intersection of graph of polynomial with the x-axis. The zeroes of
polynomial of degree „n „are at most „n‟.
General form of polynomial –
•
ax + b ; a ≠ 0 is the linear polynomial
•
ax2 + bx + c; a≠ 0 a , b , c, are real number is a quadratic polynomials
•
ax3+ bx2 + cx+ d ; a≠ 0 , a , b , c , d , are the real number is a cubic
polynomials
•
value of polynomials – if p(x) is a polynomials in x, and if k is any real number
obtained by replacing x by k in p(x), is called the value of p(x) at x = k and
denoted by p(k)
•
zeroes of the polynomial- a real number k is said to be a zero of the polynomial
p(x) if p(k) = 0. In general , p(x) =ax + b
•
p (k)= a k + b
•
p (k) = 0
•
p (k) = 0
•
ak+b=0
•
k=-
it is clear that , zeroes of the linear polynomials is related to its coefficients.
•
If α and β are the zeroes of the quadratic polynomial p(x) = ax 2 + bx + c ; a ≠ 0
then
Sum of the zeroes = α +β == Product of zeroes = α β =
=
=
7.
Division algorithm for polynomial – If p (x) and g(x) are any two
polynomials with
g(x) ≠ 0 then we can find polynomial g(x) and r(x) such that;
p(x) = q(x) × g(x) + r(x)
Where r(x) = 0 or degree of r(x) < degree of g(x) . Here p(x) as dividend, g(x) as
divisor,
q (x) as quotient and r(x) as remainder
8. Forming a polynomial when its zeroes are given•
Let α ,β be the zeroes of a quadratic polynomial then its equation is
P(x) = k {x2 - (α + β) x + α β }
Where k is any constant.
Tips
•
Graph of linear polynomial ax + b is a straight line .if it intersects x axis at
point (x1 , 0) thus x1 is the only zero of the liner polynomial
•
Graph of quadratic polynomial ax2+ bx +c is an upward parabola or down
ward parabola according as a>0 or a<0
•
If two graph of ax2+ bx + c touch the x axis at point (x1 , 0) and (x2 , 0) then
x1 and x2 are two equal zeroes of ax2 +bx +c
•
If two graph of ax2 + bx +c intersects the x axis at two point (x1,0) and
(x2,0) thus x1 and x2 are the zeroes of quadratic polynomial ax2+ bx + c
•
The two zeroes of the quadratic polynomial , ax2 + bx + c has atmost 2
zeoes.
•
If the graph of ax2 + bx + c, neither intersecting nor touch the x-axis the
polynomial ax2 + bx + c, does not have any zero.
•
Polynomials f (x) divided by polynomial g (x) gives the quotient g(x) and
the remainder r(x) we express it as f(x) = g(x) × q(x) + r(x).
•
If f(x) is divided by a linear polynomial g(x) then the remainder is either 0
or a polynomial of degree is a constant real number.
•
If f(x) = g(x) × q(x) then to find the zeroes of f(x) we will find that the
zeroes of g(x) and q(x) ,all the zeroes of g(x) and q(x) are the zeroes of
f(x).
Formulaes & Key Points:
1. Degree of Linear polynomial is 1.
2. Degree of Quadratic polynomial is 2.
3. Degree of Cubic polynomial is 3.
4. Degree of Constant polynomial is 0.
5. Degree of Zero polynomial is not defind.
6. Zero polynomial has infinite many zeroes.
7. Constant polynomial has no zeroe.
8. Linear polynomial has one zeroes.
9. Linear polynomial has one zeroes.
10. Quadratic polynomial has at most two zeroes.
11. Cubic polynomial has three zeroes.
12. nth degree polynomial has at most „n‟ zereos.
13. Sum of zeroes of quadratic polynomial =
14. Product of zeroes =
=-
=
15. Sum of zeroes of cubic polynomial =
=-
16. Sum of product of zeroes taken two at a time for cubic polynomail =
=
17. Product of zeroes of cubic polynomial =
HOTS POLYNOMIALS
2 MARKS QUESTIONS
= -
2
1. If α,β are the zeros of the polynomial 2x – 4x + 5 find the value of
2
2
a) α + β
2.
b) (α - β)
Ans:
(Ans: a) -1 , b) –6)
2
p (x) = 2 x – 4 x + 5
(
)
⁄
4/2 = 2
mark
c
5
mark
2
2
2
2
α + β = (α + β) – 2 α β
Substitute then we gets,
2
2
α + β = -1
2
2
(α - β) =(α + β) - 4 α β
2
Substitute, we get (α - β) = - 6
αβ=
a
4
(1/2mark)
(1/2mark)
3
2
2. On dividing the polynomial 4x - 5x - 39x - 46x – 2 by the polynomial g(x)
2
the quotient is x - 3x – 5 and the remainder is -5x + 8.Find the polynomial g(x).
2
(Ans:4 x +7x+2)
Ans: p(x) = g (x) q (x) + r (x)
(By Division Algorithm)
g(x) = p(x) − r(x)
q(x)
4
3
2
let p(x) = 4x – 5x – 39x – 46x – 2
mark
mark
2
q(x) = x – 3x – 5 and r (x) = -5x + 8
4
3
2
now p(x) – r(x) = 4x – 5x – 39x – 41x – 10
( )
g(x) =
( )
;
( )
–
g(x) =
mark
–
–
–
–
–
;
2
g(x) = 4x + 7x + 2
mark
3. If the squared difference of the zeros of the quadratic polynomial x2 + px + 45 is
equal to 144 , find the value of p.
(Ans: 18).
Ans: Let two zeros are α and β where α > β
According given condition
2
(α - β) = 144
2
mark
Let p(x) = x + px + 45
α + β =-b/a
αβ =
c/a = 45a
mark
2
now (α - β) = 144
2
(α + β) – 4 αβ = 144
2
(-p) – 4 (45) = 144
Solving this we get p = 18
(1 mark)
3 MARKS QUESTIONS
4
3
2
1. Find the value for K for which x + 10x + 25x + 15x + K exactly divisible by x + 7.
4
4
(Ans : K= - 91)
2
Ans: Let P(x) = x + 10x + 25x + 15x + K and g(x) = x + 7
Since P(x) exactly divisible by g(x)
∴ r (x) = 0
3
(1/2 mark)
2
x 3x 4x −13
now
x+7
∴ K + 91 = 0
4
3
4
3
2
x 10x 25x 15x K
x 7x
------------3
2
3x + 25 x
3
2
3x + 21x
------------------2
4x + 15 x
2
4x + 28x
------------------13x + K
- 13x - 91
---------------K + 91
------------
(2 mark)
K= -91
(1/2)
4
3
2
2. If two zeros of the polynomial f(x) = x - 6x - 26x + 138x – 35 are 2 √3.Find the
other zeros.
(Ans:7, -5)
Ans: Let the two zeros are 2 + 3 and 2 Sum of Zeros
=2+ 3+ 2=4
Product of Zeros = ( 2+ 3 )(2 =4–3
3
3)
3
=1
2
2
Quadratic polynomial is x – (sum) x + Product = x – 4x + 1
(1mark)
2
x – 2x – 35
2
x – 4x + 1 x 4 − 6 x 3 − 26 x 2 138 x − 35
x 4 − 4x 3 x2
----------------3
2
-2x – 27x + 138x
3
2
- 2x + 8x – 2x
----------------------2
-35x + 140x – 35
2
-35x + 140x – 35
-----------------------0
------------------------
(1marks)
2
∴ x – 2x – 35 = 0
(x – 7)(x + 5) = 0
x = 7, -5
(1/2 mark)
other two Zeros are 7 and -5
3. If α and β are the zeros of a Quadratic polynomial such that α + β = 24, α - β = 8.
2
Find a Quadratic polynomial having α and β as its zeros. (Ans: k(x – 24x + 128))
Ans: α+β = 24
α-β=8
----------2α = 32
α = 32/2 = 16, ∴
α = 16
1mark
Work the same way to α+β = 24
So, β = 8
½ mark
2
Q.P is x – (sum) x + product
2
= x – (16+8) x + 16 x 8
Solve this,
2
it is k (x – 24x + 128)
1½ mark
2
4. If α & ß are the zeroes of the polynomial 2x – 4x + 5, then find the value of
2
2
2
2
2
3
3
a. α + ß
b. 1/ α + 1/ ß c. (α - ß) d. 1/α + 1/ß e. α + ß
4
(Ans:-1, 4 ,-6, − ,-7)
5
25
2
Ans: Let p(x) = 2x – 4x +5
α+β =
− b = 4/2 = 2
a
c
αβ =
2
=5
a 2
(1/2 mark)
2
2
a) α +β = (α+β) - 2αβ
2 2
Substitute to get = α +β = -1
1
b)
(1/2 mark)
αβ
1
a+ β =
αβ
(1/2 mark)
1
a +
substitute , then we get
2
2
1
4
β
= 5
c) (α-β) = (α+β) - 4 αβ
(1/2 mark)
Therefore we get,
1
d)
α
1
α2
2
(α-β) = - 6
2
β
+ β2 =
(αβ 2
−½
2
= 5
2
2
∴
1
α
3
+ 1
β2
2
3
=
2
−4
25
2
e) α +β = (α+β)(α +β -αβ)
Substitute this,
3
3
to get, α +β = -7
(1/2 mark)
(1/2 mark)
3
2
5. If the ratios of the polynomial ax +3bx +3cx+d are in AP, Prove that
3
2
2b - 3abc+a d=0
3
2
Ans: Let p(x) = ax + 3bx + 3cx + d and α , β , ϒ are their three Zeros
but zero are in AP
let α = m – n , β = m, r = m + n
m – n+ m+ m + n = − b
a
Substitute this sum, to get m=
(1/2 mark)
(1/2mark)
Now taking two zeros as sum αβ +β r +αr =
(m-n)m + m(m+n) + (m + n)(m – n) =
=
Solve this problem , then we get
(1mark)
Product of roots = αβϒ =
(m-n)m(m+n) =
[(( ) )
=
On simplifying we get
3
2
2b – 3abc + a d = 0
(1 mark)
EASY AND SCORING QUESTIONS FOR SLOW BLOOMERS
CHAPTER- POLYNOMIAL
Level 1 (1 mark)
1. The number of zeroes, the polynomial f(x) = (x – 3)2+ 1 can have is :
(a) 0 (b) 1 (c) 2 (d) 3
Ans: c
2. The graph of the polynomial p(x) cuts the x-axis 5 times and touches it 3 times. The number of
zeroes of p(x) is : (a) 5 (b) 3 (c) 8 (d) 2
Ans: c
3. If the zeroes of the quadratic polynomial x + (a + 1)x+ b are 2 and –3, then :
(a) a = –7, b = –1 (b) a = 5, b = –1
(c) a = 2, b = –6 (d) a = 0, b = –6
2
Ans :d
4 .The zeroes of the quadratic polynomial x + 89x + 720 are :
(a) both are negative
(b) both are positive
(c) one is positive and one is negative
(d) both are equal
2
Ans:a
Level 2 (2marks)
Q.5 If α and β are zeros of the Polynomial 3x2+5x+2, Find the value of
2
Ans: 3x
+5x+2
α+β=
αβ =
+ =
=
Q.6 Find the zeros of the quadratic polynomial 6x2 – 7x – 3 and verify the relationship
between the zeros and the coefficients.
Ans: P(x) = 6x² - 7x – 3
= 6x² - 9x + 2x – 3
=3x (2x-3) +1(2x-3)
=(2x – 3)(3x+ 1)
x = 3/2 , x = -1/3
Now sum of zeroes = 3/2 – 1/3 = 7/6
Also sum of zeroes = -b/a = - (-7)/6 = 7/6
Product . of zeroes = 3/2 x -1/3 = -1/2
also product . of zeroes = c/a = -3/6 = -1/2
Q.7 Write the zeroes of the polynomial x2– x – 6.
Ans: x2– x – 6
x2-3x+2x-6 = x(x-3)+2(x-3)
= (x-3)(x+2), now zeroes of x2– x – 6 are x-3 =0 and x+2 =0
or x= 3 , x = -2
(1Marks)
(1 Marks)
Q8 Find a quadratic polynomial with sum of zeroes = 1/4 and product of zeroes 1/4.
Ans:A quadratic polynomial with sum of zeroes=S and product of zeroes=P is
= x²-Sx+p
= x²-x/4+1/4
=
Therefore, quadratic polynomial whose S= ¼, P=1/4 is 4x2 –x+1
Level 3 (3 marks)
Q.9. Find the zeroes of quadratic -2x-8 and verify the relationship between the zeroes and
their co-efficient.
Ans: .
We have f(x) = x2 – 2x – 8
= x2 – 4x + 2 x – 8
= x (x –4) + 2 (x – 4)
= (x -4)(x +2)
Zeroes of f(x) is f(x) = 0
(x + 2) and
(x – 4)=0
X+2=0
and x – 4 = 0
X = -2 and
x=4
Therefore Zeroes of f(x) is
Sum of zeroes =
And
And
= -2 + 4 = 2
(
=
Product of zeroes =
,
)
=2
= (-2)4 = -8
=
= -8
Q.10 Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeros are √5/3 and - √5/3
.
Ans:
Since √ and -√
are two zeroes of f(x)
√ )(
(
√ ) = x2-
is a factor of
3x2 – 5 is a factor of p(x)
3x+6x-2x-10x-5 = (x+√ ) (n-√ ) (n+1)(n+1)
( )
√ , -√ , -1, -1
Q.11Find the zeros of the polynomial 4√3x² + 5x - 2√3.
Ans: 4√3x2 + 5x - 2√
Product = 4√ x 2√ = 24
Sum = 5
We have F (x) = 4√ x2 + 8x – 3x - 2√
)-√ (√
F (x) = (√
F (x) = (√
)(
√ )
Zeroes of f[x] is given by
If F (x) = 0
(√
)(
√ )=0
(√
)
√
x=√
Hence Zeroes of f (x) is
=
)
√
x=
and
√
β=
√
Q.12 If m and n are the zeros of the polynomial 3x2+ 11x- 4, find the value of
Ans: Since m and n are the zeroes of 3x2 + 1 1x – 4
m+n = Now,
and mn = =
(
)
.
(
)
(
)
= -
Level 4 (4 marks)
13. If p and q are the zeroes of polynomial ax2 5x c , find the values of a and c,
if p+q = pq=10
Ans:
Given polynomial is
f (x) ax 2 5x c
sum of zeroes p+q =
5
a
Product of zeroes, pq =
Given, p q pq 10
5
1
10 a
a
2
Also,
c
a
(i)
c
10
a
c
10
1
2
2c = 10 c = 5
Hence, the values of a and c are
[ from Eq.(i)]
1
and 5.
2
14. If the sum of the squares of zeroes of the polynomial 6x 2+x+k is 25/36, find the value of k?
Ans: . a=6 , b=1 , c=k
α2+β2=25/36
α + β= -b/a = -1/6
αβ = c/a = k/6
Now, (α+β)2 = α2 + β2 + 2αβ
(-1/6)2 = 25/36 + 2xk/6
1/36 = 25/36 + 2xk/6
2k/6 = 1/36 – 25/36 = -24/36
k/3 = -24/36
k= = (
)x3 = -2
15. If α and β are two zeroes of the quadratic polynomial p(x) =2x 2 – 3x+7, find :a) 1/α + 1/β
b) α2+β2
Ans: 2x2-3x+7
a=2
b= -3 c= 7
α+β = - =
αβ = =
now, +
=
=
=
=
b) (α+β)2 = α2+ β2 + 2αβ
( )2= α2+ β2 + 2
α2+ β2 = – 7 = -
16. Find the value a for which polynomial x4 10 x3 25x2 15x a is exactly divisible by x+7
Ans: Let ( )
+25
g ( x) x 7
and
Since, p(x) is exactly divisible by g(x)
r(x) = 0
x3 3x 2 4 x 13
Now, x 7 x 4 10 x3 25 x 2 15 x a
x 4 7 x3
–
–
3
3x 25 x 2
3x3 21x 2
–
–
4 x2 15x
4 x 2 28x
–
–
– 13x + a
– 13x – 91
+
+
a+91
From Eq. (i)
a+91 = 0 a = -91
HOTS (LINEAR EQUATIONS IN TWO VARIABLES)
HOTS (2 MARKS)
Q 1.In a ΔABC, C = 3 B = 2 ( A + B). Find the three angles.
Solution:
Given that,
C = 3 B = 2( A + B)
3 B = 2( A + B)
3B = 2A + 2B
B = 2A
2 ∠ A − ∠ B = 0 ………………………………… (i)
We know that the sum of the measures of all angles of a triangle is 180°. Therefore,
A + B + C = 180°
A + B + 3 B = 180°
A + 4 B = 180° …………………………….. (ii)
Multiplying equation (i) by 4, we obtain
8 A − 4 ∠ = 0 …………………………….… (iii)
Adding equations (ii) and (iii), we obtain
9 A = 180°
A = 20°
From equation (ii), we obtain
20° + 4 B = 180°
4 B = 160°
B = 40°
C = 3 B
= 3 × 40° = 120°
Therefore, A, B, C are 20°, 40°, and 120° respectively.
Question 2:
ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.
Solution:
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Therefore, A + C = 180
4y + 20 − 4x = 180
− 4x + 4y = 160
x − y = − 40 ………………………(i)
Also, B + D = 180
3y − 5 − 7x + 5 = 180
− 7x + 3y = 180 …………………..(ii)
Multiplying equation (i) by 3, we obtain
3x − 3y = − 120 ……………………(iii)
Adding equations (ii) and (iii), we obtain
− 7x + 3x = 180 − 120
− 4x = 60
x = −15
By using equation (i), we obtain x − y = − 40
−15 − y = − 40
y = −15 + 40 = 25
A = 4y + 20 = 4(25) + 20 = 120°
B = 3y − 5 = 3(25) − 5 = 70°
C = − 4x = − 4(− 15) = 60°
D = − 7x + 5 = − 7(−15) + 5 = 110°
Q3.Solve the following pair of linear equations:
px + qy = p − q
qx − py = p + q
Solution:
px + qy = p − q ……………………..… (1)
qx − py = p + q ……………………..… (2)
Multiplying equation (1) by p and equation (2) by q, we obtain
p2x + pqy = p2 − pq …………………………… (3)
q2x − pqy = pq + q2 ………………………..… (4)
Adding equations (3) and (4), we obtain
p2x + q2 x = p2 + q2
(p2 + q2) x = p2 + q2
X=
=1, x=1
From equation (1), we obtain
p (1) + qy = p − q
qy = − q
y=−1
HOTS (3 MARKS)
Q1 For which values of a and b does the following pair of linear equations have an infinite
number of solutions?
2x + 3y =7
(a – b)x + (a + b)y = 3a +b –2
Solution:
2x + 3y -7 = 0
(a – b)x + (a + b)y - (3a +b –2) = 0
a1/a2 = 2/a-b = 1/2
b1/b2 = -7/a+b and
c1/c2 = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a1/a2 = b1/b2 = c1/c2
2/a-b = 7/3a+b-26a + 2b - 4 = 7a - 7b
a - 9b = -4 ... (i)
2/a-b = 3/a+b
2a + 2b = 3a - 3b
a - 5b = 0 ... (ii)
Subtracting equation (i) from (ii), we get
4b = 4
b=1
Putting this value in equation (ii), we get
a-5×1=0
a=5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many
solutions
2.Solve for x and y:
a(x+y)+ b(x-y)= a2-ab+b2
a(x+y) –b(x-y)=a2+b2+ab
Solution: a(x+y)+ b(x-y)= a2-ab+b2 --------------------- (1)
a(x+y) –b(x-y)=a2+b2+ab ---------------------(2)
Multiplying (i) by (a-b) and (2) by (a+b),we get
(a-b){ a(x+y)+ b(x-y)= a2-ab+b2}
(a+b) {a(x+y) –b(x-y)=a2+b2+ab}
(a2-b2)x+(a-b)2y= (a-b)(a2-ab+b2)----------------------------(3)
(a2-b2)x+(a+b)2y= (a+b)(a2+ab+b2)-------------------------(4)
Subtracting (4) from(3)
{(a-b)2-(a+b)2}y= a3-a2b+ab2-ba2+ab2-b3-b3-a3-a2b-ab2-ba2-ab2
-4aby= -4a2b-2b3
-4aby=-2b(2a2+b2)
y=
Put in equation (1),we have
(a+b)x= a2-ab+b2-(a-b)(
)
2a(a+b)x= 2a3- 2a2b+2ab2-2a3-ab2+2a2b+b3
2a(a+b)x= ab2+b3= b2(a+b)
x=
Q3. On selling a T,V at 5% gain and a fridge at 10% gain ,a shopkeeper gains Rs.2000. But if he
sells the T.V at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the transanction.Find
the actual price of the T.V and the fridge.
Solution: Let the actual price of the T.V and the fridge be x and y respectively.
ATQ,(
+
x-x) + (
y-y)= 2000
= 2000
x+2y =40000--------------------- (i)
And, (
-
x-x)- (y-
y)= 1500
= 1500
2x-y=30000 ----------------------- (ii)
Multiplying equation (ii) by 2,we get
4x-2y=60000 ---------------------------- (iii)
Adding equation (i) and (iii),we get
5x=100000
X=
= 20000
Substituting x=20000 in equation (i),we get
20000+ 2y= 4000
2y=40000-20000=20000
y=
= 10000
So,the solution of the given equations is x= 20000 and y=10000.
Q4.One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The
other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the
amount of their (respective) capital?
Solution: Let those friends were having Rs x and y with them.
Using the information given in the question, we obtain
x + 100 = 2(y − 100) x + 100 = 2y – 200
or x − 2y = −300 ……………….(i)
And,
6(x − 10) = (y + 10)
Or 6x − 60 = y + 10
Or 6x − y = 70 ………………………….(ii)
Multiplying equation (ii) by 2, we obtain
12x − 2y = 140 (iii)
Subtracting equation (i) from equation (iii), we obtain
11x = 140 + 300 11x = 440 x = 40
Using this in equation (i), we obtain
40 − 2y = −300
40 + 300 = 2y
2y = 340
y = 170
Therefore, those friends had Rs 40 and Rs 170 with them respectively.
Q5.The students of a class are made to stand in rows. If 3 students are extra in a row, there
would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the
number of students in the class.
Solution:
Let the number of rows be x and number of students in a row be y.
Total students of the class
= Number of rows × Number of students in a row
= xy
Using the information given in the question,
Condition 1
Total number of students = (x − 1) (y + 3)
Or xy = (x − 1) (y + 3) = xy − y + 3x − 3
Or 3x − y − 3 = 0
Or 3x − y = 3 ……………………………………. (i)
Condition 2
Total number of students = (x + 2) (y − 3)
Or xy = xy + 2y − 3x − 6
Or 3x − 2y = −6 ……………………………….(ii)
Subtracting equation (ii) from (i),
(3x − y) − (3x − 2y) = 3 − (−6)
− y + 2y = 3 + 6 y = 9
By using equation (i), we obtain
3x − 9 = 3
3x = 9 + 3 = 12
x=4
Number of rows = x = 4
Number of students in a row = y = 9
Number of total students in a class = xy = 4 × 9 = 36
TOPIC: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
BASIC CONCEPTS
LINEAR EQUATION: An equation of the form ax+by+c=0 or ax+by=d, where a,b,c,d are real
numbers, a2+b2 ≠0 and x,y are variables, is called linear equation in two variables.
METHODS OF SOLVING SIMULTANEOUS PAIR OF LINEAR EQUATIONS IN TWO
VARIABLES:
(i) Graphical Method
(ii) Algebraic Method
(i) Graphical Method:
(a) If
≠
, then graph will represent two intersecting lines. Point of intersection is the solution
of the system.
a1x+b1y=c1
a2x+b2y=c2
Intersecting lines represent unique solution.
(b) If
=
≠
, then graph will represent two parallel lines. System has no common solution.
a1x+b1y=c1
a2x+b2y=c2
Parallel lines represent no solution.
(c) If
=
=
, then graph will represent two coinciding lines. System has infinitely many
solutions.
a1x+b1y=c1
a2x+b2y=c2
Coincident lines represent many solutions
(ii) Algebraic Method:
Let the system of linear equation in two variables be a 1x+b1y=c1 and a2x+b2y=c2.
(a) If
≠
, then system has a unique solution and is known as consistent.
(b) If
=
≠
,then system has no solution and is known as inconsistent.
(c) If
=
=
,then system has infinitely many solutions and is known as dependent
consistent.
Examples
Pairs of lines
a1∕a2
b1∕b2 c1∕c2
Compare the Graphical
Algebraic
ratio
representatio interpretatio
Condition for
solvability
n
Intersecting
lines
x -2y = 0
3x + 4y-20=0
1∕3
-2/4
0∕-20
a1 b1
a 2 b2
2x+3y-9=0
4x+6y-18=0
2∕4
36
-9∕-18
a1 b1 c1 Coincident
a 2 b2 c 2 lines
x+2y-4=0
2x+4y-12=0
1∕2
2∕4
-4∕-12
a1 b1 c1 Parallel lines
a 2 b2 c 2
n
Exactly
unique
solution
Infinitely
many lines
No solution
System is consistent
System is
consistent(dependent)
System is inconsistent
GRAPHICAL METHOD:
Method of solving simultaneous linear equations in two variables:
STEP 1: Get three solutions of each of given linear equations.
STEP 2:Plot these points on graph in order to draw the lines representing these equations.
STEP 3: Get the solution of equations.
ALGEBRAIC METHOD:
Method of solving simultaneous linear equations in two variables using substitution.
STEP 1: From one equation find the value of one variable , (say y) in terms of other variable,
i.e., x.
STEP 2: Substitute the value of variable obtained in step 1, in the other equation to get an
equation in one variable.
STEP 3:Solve the equation obtained in step 2 to get the value of one variable.
STEP 4:Substitute the value of variable so obtained in any given equation to find the value of
other variable.
Method of solving simultaneous linear equations in two variables using Elimination method
by Equating the Coefficients:
STEP 1: Obtain the two equations.
Step 2: Multiply the equations so as to make the coefficients of one of the variables equal, which
is to be eliminated.
STEP 3: (i) If the coefficients of the variable to be eliminated are having same sign, then
subtract the equation obtained in step 2.
(ii) If the coefficients of the variable to be eliminated are having opposite sign, then add the
equation obtained in step 2.
STEP 4: Solve the equation obtained in step 3 to obtain the value of one variable.
STEP 5: Substitute the value of variable so obtained in any of the given equation to find the
value of other variable.
Solving system of linear equations by Method of comparison:
STEP 1: From each equation find the value of one variable in terms of other.
STEP 2: Equate them to get an equation in one variable and solve.
Method of Cross Multiplication for solving system of linear equations:
For system of linear equations a1x + b1y + c1= 0 and a2x + b2y + c2= 0 when,
solution is given by x =
≠
the unique
and y=
NOTE :- The following diagram helps in remembering the above solution .
x
b1
b2
Denominator for x
y
1
c1
a1
b1
c2
a2
b2
Denominator for y
Denominator for 1
Level – 1 ( 1 MARK EACH)
Fill in the blanks:
1. All ……………………triangles are similar.
(Equilateral)
2. If a line divides two sides of a triangle in the same ratio, the line is ..........................to the
third side.
(Parallel)
3. Sides of two similar triangles are in the ratio 4:9.The ratio of areas of these triangles is
…………….
(16:81)
4. Two triangle are said to be similar if................
(Corresponding sides are in
same ratio and corresponding angles are equal)
5. Traingle have ........diagonals.
(No)
6. The six elements of a triangle are its three angles and the ------------------- (three sides)
Level-2 ( 2 MARK EACH)
7. Find the length of x in the following cases:
In ABC
Ans.
BC2= AB2 + AC2
(by Pythagoras theorem)
x2 = 52 + 122
x2 = 25 + 144= 169
x = 169
x = 13
8. ABC is an isosceles triangle with AB = AC = 6 cm and BC = 8 cm. Find the length of the
altitude on BC and hence calculate the area.
Ans.
9. If ∆ ABC ~ ∆ DEF, such that AB = 1.2cm, and DE = 1.4cm. Find the ratio of areas of ∆
ABC and ∆ DEF.
Ans. Ratio of Areas of similar triangle is equal to ratio of square of corresponding sides.
arABC AB 2
Therefore
DE 2
arDEF
arABC 1.2
Or
arDEF 1.4
144
=
196
36
=
49
10. If the areas of two similar triangles are equal, prove that they are congruent.
2
Ans.Let us take two triangles ABC and PQR with equal areas.
Then, we have;
In this case;
Hence; the triangles are congruent.
Level – 3 ( 3 MARK EACH)
11. In ABC, AB = AC = x, BC = 5 cm and the area of the triangle ABC is 15 cm². Find x.
Ans.
x
x
B
C
D
Construction AD BC since ABC is an isosceles triangle therefore AD bisects BC i:e
BD = DC = 5/2
Area of ABC = ½ BC X AD
= ½ 5 X AD = ½ X 5 X
15 = ½ X 5 X
15 2
=
5
x 2 BD 2
x 2 BD 2
2
5
x2
2
Squaring both sides
25
36 = x2 4
25
x2 = 36 +
4
x = 6.5
Alternative method
Area of ABC = ½ BC X AD
15 = ½ 5 AD
Or AD = 30/5= 6
In right ABD
25
x2 = 36 +
(by Pythagoras theorem)
4
x= 6.5
12. If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC
~ Δ PQR, prove that
Solution:
Δ ABC ~ Δ PQR (Given)
Hence;
(A side and the median of one triangle are in same ratio as a corresponding side
and median of another triangle)
Proved
13. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the
same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Height of pole = AB = 6 m and its shadow = BC = 4 m
Height of tower = PQ = ?and its shadow = QR = 28 m
The angle of elevation of the sun will be same at a given time for both the triangles.
Hence; ΔABC ~ ΔPQR
This means;
Height of tower = 42 m
14. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that
CA2 = CB.CD.
Solution: In ΔBAC and ΔADC;
∠BAC = ∠ADC (given)
∠ACB = ∠DCA (Common angle)
Hence; ΔBAC ~ ΔADC
Hence;
(corresponding sides are in same ratio)
Or, CA x CA = CB x CD
Or, CA2 = CB x CD proved
Level – 4 ( 4 MARK EACH)
15. In the given figure, ABC and DBC are two triangles on the same base BC. If AD
intersects BC at O, show that
Solution: Let us draw altitudes AM and DN on BC; respectively from A and D
In ΔAMO and ΔDNO;
∠ AMO = ∠ DNO (Right angle)
∠ AOM = ∠ DON (Opposite angles)
Hence; ΔAMO ~ ΔDNO
Hence;
16. D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC.
Find the ratio of the areas of Δ DEF and Δ ABC.
Solution:
Since D, E and F are mid points of AB, BC and AC
Hence; ΔBAC ~ΔDFE
So,
So,
17. Prove that the area of an equilateral triangle described on one side of a square is
equal to half the area of the equilateral triangle described on one of its diagonals.
Solution: : Let us take a square with side ‘a’
Then the diagonal of square will be a 2
a√2
Area of equilateral triangle with side ‘a’
a
Area of equilateral triangle with side a√2
Ratio of two areas can be given as follows:
HOTS
Group-4
CHAPTER TRIANGLES
TWO MARKS
1. In figure PQ||MN and
KP
4
, KN = 20.4cm. Find KQ.
PM 13
K
Solution: In KMN ,
KP KQ
( PQ||MN given)
MP QN
KP
KQ
MP KN KQ
4
KQ
13 20.4 KQ
4(20.4 – KQ) = 13 KQ
81.6 – 4KQ = 13 KQ
KQ = 4.8cm
P
M
Q
N
2. In figure, DE||BC. If AD = x, DB = x-2, AE = x+2 and EC = x – 1, find the value of
„x‟.
Solution: In ABC ,
C
AD AE
(byThale‟s theorem)
DB EC
E
x
x2
x 2 x 1
x(x-1) = (x-2)(x+2)
A
2
B
D
2
x –x=x –4
x=4
3. AD is the bisector of A , if BD = 4 cm, DC = 3cm and AB = 6cm determine AC.
Solution: In ABC , AD is the bisector of A
A
BD AB
DC AC
4
6
3 AC
3X 6
=4.5 cm
AC=
4
B
A
C
D
A
THREE MARKS
4. ABC is a right-angled triangle, right angled at B. AD and CE are two medians drawn
from A and C respectively. If AC = 5 cm, and AD = (3 √ 5)/ 2 cm, find the length of CE.
Solution: In right angled triangle
A
Let BD = DC = x
AE = BE = y
In right angled triangle ABC
AB2 + BC2 = AC2
(2y)2 + (2x)2 = 25
E
4y2 + 4x2 = 25
….(i)
In right angled triangle ABD
AB2 + BD2 = AD2
B
D
C
2
3 5
4y + x =
2
45
4y2 + x2=
4
16y2 + 4x2= 45
……..(ii)
Subracting equation (i) from (ii)
12y2= 20
5
y2=
3
2
2
put this value in equation (i)
5
4 +4x2 = 25
3
20
4x2 = 25 3
55
x2 =
12
now in right triangle BEC
CE2 = BE2 + BC2
= y2 + (2x)2
=y2 + 4x2
5
55
= +4
3
12
= 20
CE = 2 5
5. In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥
AC and OF ⊥ AB. Show that
(a) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Solution: In ∆ AFO; AF2 = OA2 – OF2
In ∆ BDO; BD2 = OB2 – OD2
In ∆ CEO; CE2 = OC2 – OE2
Adding the above three equations, we get;
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 proved
(b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Solution: In ∆ AEO; AE2 = OA2 – OE2
In ∆ CDO; CD2 = OC2 – OD2
In ∆ BFO: BF2 = OB2 – OF2
Adding the above three equations, we get;
AE2 + CD2 + BF2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
From the previous solution, we also have;
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
Comparing the RHS of the above two equations, we get;
AF2 + BD2 + CE2 = AE2 + CD2 + BF2 proved
6. If ABC and DBC are two triangles on the same base BC and AD intersects BC
at O, show that
Solution: Let us draw altitudes AM and DN on BC; respectively from A and D
In ΔAMO and ΔDNO;
∠ AMO = ∠ DNO (Right angle)
∠ AOM = ∠ DON (Opposite angles)
Hence; ΔAMO ~ ΔDNO
Hence;
7. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution: In case of an equilateral triangle, an altitude will divide the triangle into two
congruent right triangles. In the right triangle thus formed, we have;
Hypotenuse = One of the sides of the equilateral triangle = 2a
Perpendicular = altitude of the equilateral triangle = p
Base = half of the side of the equilateral triangle = a
Using Pythagoras theorem, the perpendicular can be calculated as follows:
p2 = h2 – b2
Or, p2 = (2a)2 – a2
Or, p2 = 4a2 – a2 = 3a2
Or, p = a√3
8. Prove that the sum of the squares of the sides of a rhombus is equal to the
sum of the squares of its diagonals.
Solution: ABCD is a rhombus in which diagonals AC and BD intersect at point O.
To Prove: AB2 + BC2 + CD2 + AD2 = AC2 + BD2
In ∆ AOB; AB2 = AO2 + BO2
In ∆ BOC; BC2 = CO2 + BO2
In ∆ COD; CD2 = CO2 + DO2
In ∆ AOD; AD2 = DO2 + AO2
Adding the above four equations, we get;
AB2 + BC2 + CD2 + AD2
= AO2 + BO2 + CO2 + BO2 + CO2 + DO2 + DO2 + AO2
Or, AB2 + BC2 + CD2 + AD2 = 2(AO2 + BO2 + CO2 + DO2)
Or, AB2 + BC2 + CD2 + AD2 = 2(2AO2 + 2BO2)
(Because AO = CO and BO = DO)
Or, AB2 + BC2 + CD2 + AD2 = 4(AO2 + BO2) ………(1)
Now, let us take the sum of squares of diagonals;
AC2 + BD2 = (AO + CO)2 + (BO + DO)2
= (2AO)2 + (2BO)2
= 4AO2 + 4BO2 ……(2)
From equations (1) and (2), it is clear;
AB2 + BC2 + CD2 + AD2 = AC2 + BD2 proved
9. In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE
⊥ AC and OF ⊥ AB. Show that
(a) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Solution: In ∆ AFO; AF2 = OA2 – OF2
In ∆ BDO; BD2 = OB2 – OD2
In ∆ CEO; CE2 = OC2 – OE2
Adding the above three equations, we get;
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 proved
(b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Solution: In ∆ AEO; AE2 = OA2 – OE2
In ∆ CDO; CD2 = OC2 – OD2
In ∆ BFO: BF2 = OB2 – OF2
Adding the above three equations, we get;
AE2 + CD2 + BF2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
From the previous solution, we also have;
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
Comparing the RHS of the above two equations, we get;
AF2 + BD2 + CE2 = AE2 + CD2 + BF2 proved.
10. An aeroplane leaves and airport and flies due north at a speed of 1000 km per
hour. At the same time, another aeroplane leaves the same airport and flies
due west at a speed of 1200 km per hour. How far apart will be the two planes
after 1.5 hours?
Solution: Distance covered by the first plane in 1.5 hours = 1500 km
Distance covered by the second plane in 1.5 hours = 1800 km
The position of the two planes after 1.5 hour journey can be shown by a right triangle
and we need to find the hypotenuse to know the aerial distance between them.
Here; h = ? p = 1800 km and b = 1500 km
From Pythagoras theorem;
h2 = p2 + b2
Or, h2 = 18002 + 15002
= 3240000 + 2250000 = 5490000
Or, h = 300√61 km
CHAPTER TRIANGLES
KEY CONTENTS
1. TWO FIGURES ARE CALLED SIMILAR IF THEY HAVE SAME SHAPE , NOT
NECESSARILY THE SAME SIZE.
2. (AAA SIMILARITY) IF TWO TRIANGLES ARE EQUIANGULAR, THEN THE
TRIANGLES ARE SIMILAR
COR: (AA SIMILARITY): IF TWO ANGLES OF ONE TRIANGLE ARE
RESPECTIVELY EQUAL TO TWO ANGLES OF ANOTHER TRIANGLE, THEN THE
TWO TRIANGLES ARE SIMILAR
3. (SSS SIMILARITY) IF THE CORRESPONDING SIDES OF TWO TRIANGLES ARE
PROPORTIONAL THEN THEY ARE SIMILAR
4. (SAS SIMILARITY) IF IN TWO TRIANGLES, ONE PAIR OF CORRESPONDING
SIDES ARE PROPORTIONAL AND THE INCLUDED ANGLES ARE EQUAL,
THEN THE TRIANGLES ARE SIMILAR.
* THEOREM NO. 1: IF A LINE IS DRAWN PARRALLEL TO ONE SIDE OF A TRIANGLE
INTERSECTING THE OTHER TWO SIDES, THEN IT DIVIDES THE TWO SIDES IN THE
SAME RATIO. (BASIC PROPORTIONALITY THEOREM, BPT, THALES THEOREM).
* PROOF MAY BE ASKED.
* THEOREM NO 2: IF A LINE IS DRAWN INTERSECTING THE TWO SIDES OF A
TRIANGLE SUCH THAT IT DIVIDES THE TWO SIDES IN THE SAME RATIO, THEN
THE LINE IS PARALLEL TO THE THIRD SIDE. (CONVERSE OF BPT)
* PROOF MAY BE ASKED
THEOREM NO 3: THE INTERNAL BISECTOR OF AN ANGLE OF A TRIANGLE DIVIDES
THE OPPOSITE SIDE INTERNALLY IN THE RATIO OF THE SIDES CONTAINING THE
ANGLE.(ANGLE BISECTOR THEOREM)
THEOREM NO 4: IN A TRIANGLE ABC, IF D IS THE POINT ON BC SUCH THAT
BD/DC = AB/AC, PROVE THAT AD IS THE BISECTOR OF ANGLE A(CONVERSE)
THEOREM NO 5: THE EXTERNAL BISECTOR OF AN ANGLE OF A TRIANGLE DIVIDES
THE OPPOSITE SIDE EXTERNALLY IN THE RATIO OF THE SIDES CONTAINING THE
ANGLE.
THEOREM NO 6: THE LINE DRAWN FROM THE MID POINT OF ONE SIDE OF A
TRIANGLE PARALLEL TO THE OTHER SIDE BISECTS THE THIRD SIDE.
THEOREM NO 7:THE LINE JOINING THE MID POINTS OF TWO SIDES OF A TRIANGLE
IS PARALLEL TO THE THIRD SIDE.
THEOREM NO 8:PROVE THAT THE DIAGONALS OF A TRAPEZIUM DIVIDE EACH
OTHER PROPORTIONALLY.
THEOREM NO 9:IF THE DIAGONALS OF A QUADRILATERAL DIVIDE EACH OTHER
PROPORTIONALLY , THEN IT IS A TRAPEZIUM.
THEOREM NO 10:ANY LINE PARALLEL TO THE PARALLEL SIDES OF A TRAPEZIUM
DIVIDES THE NON PARALLEL SIDES PROPORTIONALLY.
THEOREM NO 11:If two triangles are equiangular, prove that:
i)
Ratio of the corresponding sides is the same as the ratio of the corresponding
medians.
ii)
Ratio of the corresponding sides is the same as the ratio of the corresponding angle
bisector segments.
iii)
Ratio of the corresponding sides is the same as the ratio of the corresponding
altitudes.
THEOREM NO 12:: IF ONE ANGLE OF A TRIANGLE IS EQUAL TO ONE ANGLE OF
ANOTHER TRIANGLE AND THE BISECTORS OF THESE EQUAL ANGLES DIVIDE THE
OPPOSITE SIDE IN THE SAME RATIO, PROVE THAT THE TRIANGLES ARE SIMILAR
THEOREM NO 13: IF TWO SIDES AND THE MEDIAN BISECTING ONE OF THESE SIDES
OF A TRIANGLE ARE RESPECTIVELY PROPORTIONAL TO THE TWO SIDES AND
THE CORRESPONDING MEDIAN OF ANOTHER TRIANGLE, THEN THE TRIANGLES
ARE SIMILAR.
THEOREM NO 14:IF TWO SIDES AND A MEDIAN BISECTING THE THIRD SIDE OF A
TRIANGLE ARE RESPECTIVELY PROPORTIONAL TO THE CORRESPONDING SIDES
AND THE MEDIAN OF ANOTHER TRIANGLE, THEN THE TWO TRIANGLES ARE
SIMILAR.
* THEOREM NO 15:THE RATIO OF THE AREAS OF TWO SIMILAR TRIANGLES IS
EQUAL TO THE RATIO OF THE SQUARES OF THE CORRESPONDING SIDES.
* PROOF MAY BE ASKED
THEOREM NO 16: THE AREAS OF TWO SIMILAR TRIANGLES ARE IN THE RATIO OF
I)
II)
III)
SQUARES OF THE CORRESPONDING ALTITUDES
SQUARES OF THE CORRESPONDING MEDIANS
SQUARES OF THE CORRESPONDING ANGLE BISECTORS.
THEOREM NO 17:IF THE AREAS OF TWO SIMILAR TRIANGLES ARE EQUAL THEN
THE TRIANGLES ARE CONGRUENT
5. * PYTHAGORAS THEOREM: IN A RIGHT ANGLED TRIANGLE, THE SQUARE OF
THE HYPOTENUSE IS EQUAL TO THE SUM OF THE SQUARES OF THE OTHER
TWO SIDES.
* PROOF MAY BE ASKED
* 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 TO THE FIRST SIDE IS A RIGHT ANGLE.
* PROOF MAY BE ASKED.
6. SOME IMPORTANT RESULTS ON PYTHAGORAS THEOREM:
1) ∆ ABC IS AN OBTUSE ANGLED TRIANGLE, OBTUSE ANGLED AT B. IF AD IS
PERPENDICULAR TO CB, PROVE THAT, AC2 = AB2 + BC2 +2 BC. BD
2) ∆ ABC IS AN ACUTE ANGLED TRIANGLE, ACUTE ANGLED AT B. IF AD IS
PERPENDICULAR TO CB, PROVE THAT, AC2 = AB2 + BC2 - 2 BC.
3) PROVE THAT IN ANY TRIANGLE, THE SUM OF THE SQUARES OF ANY TWO
SIDES IS EQUAL TO TWICE THE SQUARE OF HALF THE THIRD SIDE
TOGETHER WITH TWICE THE SQUARE OF THE MEDIAN WHICH BISECTS
THE THIRD SIDE. (APPOLONIUS HEOREM)
4) PROVE THAT THREE TIMES THE SUM OF THE SQUARES OF THE SIDES OF
THE TRIANGLE IS EQUAL TO FOUR TIMES THE SUM OF THE SQUARES OF
THE MEDIANS OF THE TRIANGLE.
INTRODUCTION TO TRIGONOMETRY
Level 1 (1 Mark)
1.What is the Value of Sin2 A + Cos2 A?
Sol.Sin2 A + Cos 2 A = 1
2.What is the Value of tan(90°- A) ?
Sol.cotA
3. If tanA = cot B,What is the Value of A+B ?
Sol.tanA = cot A
tanA = tanA (90°- B)
A = 90° - B
A + B = 90° Ans
4.What is the Value of sec 30° ?
Sol
√
Level 2 (2 Marks)
1. Evaluate cos 60° sin 30° + sin 60° + cos 30°
Solution:
cos 60° sin 30° + sin 60° + cos 30°
= 1/2 x 1/2 +√ 3/2 x √3/2
= 1/4 + 3/4 = 1+3/ 4 = 4/4 = 1 Ans
2. If sec2A (1 + sinA)(1-sinA) = k, find the value of k.
Sol:
sec2A (1- sin2A) = k
Sec2A (cos2A) = k
(1/cos2A) cos2A = k
k = 1 Ans
3. If sin A = 1/3, then find the value of (2 cot2A + 2).
Sol :
2 cot2 A +2 = 2 (cot2A + 1)
=
2 (cosec2 A)
=
2 (1/sin2 A)
= 2/(1/3)2 = 2/(1/9) = 2x9/1 = 18 Ans
4. 3
If tanA = cot B, prove that A + B = 90°
Sol:tanA = cot A
tanA = tanA (90°- B)
A = 90° - B
A + B = 90° Ans
Level 3 (3 Marks)
1.
Sol:
If cot A = 7/8 then what is the value of (1+ cosA)(1- cosA) ∕ (1- sinA)(1 + sinA) ?
(1+cosA)(1-cosA) /(1-sinA)(1+sinA)
sin2A/cos2A = tan2A = 1/cot2
tan2A = 1/cot2 = 1/ (7/8)2 = (8/7)2 = 64/ 49 Ans
2.
Write the value of 2 cos2A +
Sol :
2 cos2A + 2/ 1 + cot2A
+ cot2A
2 cos2A + 2/ cosec2A
2 cos2A + 2 sin2A
2 (cos2A + sin2A)
2(1) = 2 Ans
3.Given that tan A =
, calculate sin A, cos A and sec A.
Sol :Let ABC be a triangle right angled at B.
As tan A =
Let BC = 12k,
AB = 5k.
Using Pythagoras Theorem,AC2 = CB2 + BA2
so,
(
4.√(
)
)
=
(12k)2 + (5k)2 = 169k2
AC
=
13k
sin A =
=
=
cos A =
=
=
andsec A=
=
= Sec A + tan A
Sol: L.H.S
√
x√
=
√(
)
√
=
=
+
=
=
R.H.S
Level 4 (4 Marks)
1
cos 2
sin 3
=1+sinθcosθ
1 tan sin cos
cos 2
sin 3
cos 3
sin 3
Sol: . L.H.S =
sin sin cos cos sin cos sin
1
cos
cos 3 sin 3 cos sin cos 2 sin 2 sin cos
cos sin
cos sin
=1+sinθcosθ
=
2.. Prove that
=
Sol: .
L.H.S
=
(
=
) (
=
=
R.H.S
1 cos
1 cos
2 cos ec .
1 cos
1 cos
3.: Prove that
Sol: L.H.S =
=
)
1 cos 1 cos
1 cos 1 cos
1 cos 1 cos
1 cos 1 cos
1 cos 2
1 cos
2
1 cos 2
1 cos
2
1 cos 1 cos
sin
sin
=
=
4.
If sec + tan
2
2 cos ec
sin
= p,prove that sin =
Sol: sec + tan
= p…………(i),
Sec2 – tan2 =1
⇒ (Sec – tan ) (Sec + tan )=1………….(ii)
Dividing ( ii) by (i) we get
(Sec – tan ) = …………………(iii)
Adding (i) and (ii) we get
Sec – tan +Sec + tan =
2sec =
…………………….(iv)
Similarly, 2 tan =
…………………(v)
Dividing (v) by (iv) we get
sin =
( using1-
)
HOTs
Chapter -Introduction to TRIGONOMETRY
Level 1 (2 Marks)
1.
Sol:
Prove that
sec2 A + cosec2A = sec2A . cosec2 A
LHS = sec2 A + cosec2A
=
1/cos2A + 1/ sin2A
=
sin2A + cos2A / cos2A . sin2A
=
1/cos2A x 1/sin2A = sec2A x cosec2A = RHS
2. Prove that ( sinθ + cosecθ )2 + (cosθ+secθ)2 = tan2θ + cot2θ + 7.
Ans.
L.H.S = sin2θ+cosec2θ+2sinθcosecθ+cos2θ+sec2θ+2cosθsecθ
= 1+1+cot2θ+1+tan2θ+2+2
= 7+tan2θ+cot2θ.
3. If cos (40° + x) = sin 30°, find the value of x, provided 40° + x is an acute angle.
Solution: Given that
cos (40° + x) = sin 30°
Now
RHS = sin 30° =
So,
cos (40° + x) =
We know that cos 60° = , therefore,
40 + x = 60°
or
x = 20°
Level 2 ( 3 Marks)
1 cos
1 cos
2 cos ec .
1 cos
1 cos
1. Prove that
Ans.
L.H.S
=
1 cos 1 cos
1 cos 1 cos
1 cos 1 cos
1 cos 1 cos
1 cos 2
1 cos
2
1 cos 2
1 cos
2
1 cos 1 cos
sin
sin
=
=
2
2 cos ec
sin
2. If cosec A = 2, find the value of cot A +
.
Solution:
Let us draw a right triangle ABC such that cosec A = 2
Let AC = 2k, BC = k, so that
AB
=√
= √(
)
=√
=√ k
Now, cot A =
=
sin A =
=
√
=√
( using 1-
)
cos A =
√
=
√
Putting these values in the given expression, we have
=√ +
cot A +
=√ +
√
√
√
=√ +
=√ +
[Rationalising the denominator]
√ =2
θ
3. Prove that :
θ
θ
θ
=
θ
θ
Ans.
LHS =
=
=
=
=
(
θ
θ
θ
θ
θ
θ) (
(
(
θ
θ
θ)
θ) (
θ
(
(
θ
(
θ
θ
4. Prove that: √
θ)
θ)
θ
θ
θ)
θ)
θ
θ
θ
θ
θ)
θ)(
θ
=
θ) (
(
θ
(
θ)
θ
θ)
(
θ
θ
θ
θ
= RHS
= Sec A + tan A.
θ)
Ans. L.H.S = √
=√
(
=√
)
=√
(
)(
)
(
)(
)
(
)
=
= sec A+ tan A = RHS
5.
Ans.
Show that ( sin6θ + cos6θ ) – 3 ( sin4θ + cos4θ ) + 1 = 0
(
)
LHS : 2(
=2[(
)
(
)
)
– 3(
)+1
3 sin cos sin cos
3sin cos 2 sin cos 1
2 sin 2 cos 2
2
2
3
2
2
2
2
2
2
2
[a3+b3 = (a+b)3 – 3ab(a+b)]
2 (1) 3 3 sin 2 cos 2 (1) 3 (1) 2 2 sin 2 cos 2 1
2 6 sin 2 cos 2 3 6 sin 2 cos 2 1 0
Chapter -Introduction to TRIGONOMETRY
KEY CONTENTS:
The word „trigonometry‟ is derived from the Greek words „tri‟ (meaning three) , „gon‟ (meaning
side) and „metron‟(meaning measure). In fact, trigonometry is the study of the relationships
between the sides and the angles of triangle.
C
TRIGONOMETRIC RATIOS:
The trigonometric ratios of the angle A in triangle ABC are defined as follow:
SinA = side opposite to angle A ∕ hypotenuse = BC/ AC
CosA = side adjacent to angle A / hypotenuse = AB/ AC
B
A
TanA = side opposite to angle A / side adjacent to angle A = BC/ AB
CosecA = 1/sin of angle A = hypotenuse/ side opposite to angle A = AC/BC
SecA = 1/ cos of angle A = hypotenuse/ side adjacent to angle A =AC/AB
CotA = 1/ tan of angle A = side adjacent to angle A/ side opposite to angle A = AB/BC
TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
Angle
0°
30°
45°
60°
90°
Sin A
0
½
1/√2
√3/2
1
Cos A
Tan A
1
0
√3/2
1/√3
1/√2
1
1/2
√3
0
Not defined
Cosec A
Not defined
2
√2
2/√3
1
Sec A
1
2/√3
√2
2
Not defined
Cot A
Not defined
√3
1
1/√3
0
TRIGONOMETRIC RATIOS OF COMPLEMENTRY ANGLES
Sin(90°- A) = cos A
Cos(90° – A) = sin A
Tan(90°- A) = cot A
Cot(90°- A) = tan A
Sec(90°- A) = Cosec A
Cosec(90°- A) = Sec A
TRIGONOMETRIC IDENTITIES
(a)
Cos2A + sin2A = 1
Sec2A – tan2A = 1
Cosec2 – cot2A = 1
(b)
(c)
TOPIC: STATISTICS
Important points and formulae pertaining to the chapter:
In this chapter we study about the techniques for finding mean, median and mode of
grouped data. We also learn about cumulative frequency graph of a frequency
distribution.
The three measures of central tendency are :
i. Mean
ii. Median
iii. Mode
MEAN
The mean for grouped data can be found by following methods:
(i)
DIRECT METHOD: x =
where xi (class mark)
,
and fi = corresponding frequency.
(ii) ASSUMED MEAN METHOD OR DEVIATION METHOD: x = a +
Where a = assumed mean and di =(xi-a)
(iii) THE STEP DEVIATION METHOD:
x = a +( ∑ fi ui / ∑ fi ) h
Where ui =
and h = class size
MODE
The mode for grouped data can be found by using the formula:
Mode= l+ (
)xh
Where
l = lower limit of modal class
h = class size
fi = Frequency of modal class
f0 = frequency of class preceding the modal class
f2 = Frequency of the class succeeding the modal class
,
Cumulative frequency: The cumulative frequency of a class is the frequency obtained
by adding the frequencies of all the classes preceding the given class.
MEDIAN
The median for grouped data is formed by using the formula:
Median= l + (
)
where l = Lower limit of median class
N= number of observations or ∑fi
cf = cumulative frequency of class preceding the median class
f = frequency of median class
h = class size
Cumulative frequencies curve is known as „ogive‟.
There are two types of ogives:
1. The “less than” type:
In this type, we start with the upper limits of the classes and go on corresponding
cumulative frequencies. When these frequencies are plotted against their
corresponding upper limits, we get a rising curve.
2. The “More than” type:
In this type, we start with the lower class limits of the classes and go on
Corresponding cumulative frequencies. When these frequencies are plotted against
their corresponding lower limits, we get a declining curve.
Ogives can be used to find the median of a grouped data. The median of grouped
data can be obtained graphically by plotting the Ogives of the less than type and more than type
and locate the point of intersection of both the Ogives. The x-coordinate of the point of
intersection of two Ogives gives the median of the grouped data.
RELATIONSHIP BETWEEN MEAN, MEDIAN AND MODE:
The relationship between mean, median and mode is depicted by Emperical Formula,
MODE = 3 MEDIAN- 2 MEAN
HIGH ORDER THINKING SKILLS (HOTS)
LEVEL 1 (2marks)
Q1. 6Marks obtained by 70 students are given below:
Marks
20
70
50
60
75
90
40
No. of
Students
8
12
18
6
9
5
12
Find the median.
(Ans:50)
Ans:
Marks
No . of
students
c.f
20
8
8
40
12
20
50
18
38
60
6
44
70
12
53
75
9
58
90
5
70
N = 70
N 70
35
2
2
The corresponding value of marks for 35 is 50
Q2. . Is the following statement correct: “median= mode+ (mean – mode)?” Justify your
answer.
Sol. Yes; as we know
Mode= 3 median – 2 mean
3 Median = mode + 2 mean
Median = mode + mean
= mode- mode + mean
= mode + ( mean – mode)
Q3. The mean of ‘n’ observation is x , if the first term is increased by 1, second by 2 and so on.
n 1
What will be the new mean.
(Ans: x +
)
2
Ans: Let the n observations be x1,x2 ,x3 … ….xn.
New observations are x1+1,x2+2 ,x3+3 … ….xn+n.
New mean =
(
)
(
) (
)
(
)
n(n 1)
(n 1)
2
X
The Mean of the new numbers is X +
.
n
2
Level 2 (3marks)
Q4 .
The mode of a distribution is 55 & the modal class is 45-60 and the frequency preceding the
modal class is 5 and the frequency after the modal class is 10.Find the frequency of the modal
class.
(Ans:15)
Ans:
mode = 55
Modal class = 45 – 60
Modal class preceding f1 = 5
After the modal class = f2 = 10
Mode = L +
f f1
xh
2 f f1 f 2
f 5
x 15
2 f 5 10
55 = 45 +
10 = (
f 5
) x 15
2 f 15
10
f 5
15 2 f 15
20 f - 150 = 15 f – 75 Or
f=
5 f = 75
75
= 15
5
Q5. The sum of deviations of a set of values x 1, x2, x3,…………xn, measured from
50 is -10 and the sum of deviations of the values from 46 is 70.
Find the value of n and the mean.
Ans:
(Ans:20,.49.5)
We have
n
( X i 50) = -10 and
i 1
n
(X
i 1
i
46) = 70
n
X
i 1
i
- 50n = -10
………… (1)
n
and
X
i 1
i
- 46 m = 70 …………..(2)
subtracting (2) from (1) , we get
- 4 n = - 80 we get n = 20
n
X
i 1
i
- 50 x 20 = -10
n
X
i 1
= 990
i
1 n
990
( Xi ) =
= 49.5
n i 1
20
Mean =
hence n = 20 and mean = 49.5
Q6
Prove that (xi - x ) = 0
n
Ans:
To prove
(X
i 1
We have, X =
i
X ) = 0 algebraic sum of deviation from mean is zero
1 n
( X i )
n i 1
n
X
nX =
i 1
i
n
Now,
(X
i 1
n
(X
i 1
i
i
X ) = (X1 - X ) + (X2 - X ) + ……… + ( Xn - X )
X ) = (X1 + X2 + ……… + Xn) - n X
n
(Xi X ) =
i 1
n
(X
i 1
i 1
X
i 1
i
- nX
i
X) =n X -n X
i
X)=0
n
(X
n
n
Hence,
(X
i 1
i
X)=0
Q7. Compute the median from the following data
Mid value
115
125
135
145
155
165
175
185
195
Frequency
6
25
48
72
116
60
38
22
3
(Ans:135.8)
Ans:
Here , we are given the mid values. So, we should first find the upper and lower
limits of the various classes. The difference between two consecutive values is h =
125 – 115 = 10
Lower limit of a class = Midvalue - h /2
Upper limit = Midvalue + h / 2
Calculate of Median
Mid – value
Class
Groups
FrequencyCumulative
frequency
115
110-120
6
6
125
120-130
25
31
135
130-140
48
79
145
140-150
72
151
155
150-160
116
267
165
160-170
60
327
175
170-180
38
365
185
180-190
22
387
195
190-200
3
390
N = fi = 390
We have,
N = 390 N / 2 = 390 / 2 = 195
The cumulative frequency first greater than N i.e. 195 is 267 and the corresponding class is
150 – 160, so, 150 – 160 is the median class.
L = 150, f = 116, h = 10, f = 151
Now,
n
f
2
Median = L +
xh
f
Median = 150 +
195 151
x 10 = 153.8
116
Q8. . The mean of the following frequency distribution is 50. But the frequencies f1 and f2 in class
20-40 and 60-80 are missing. Find the missing frequencies.
Class interval
Frequency
Sol.
Class interval
0-20
20-40
40-60
60-80
80-100
0-20
17
20-40
f1
Frequency (fi)
17
f1
32
f2
19
∑fi = 68+ f1 + f2
40-60
32
60-80
f2
Class mark (xi)
10
30
50
70
90
Now its given that total frequency is 120.
∑fi = 120
68+ f1 + f2 = 120
f1 + f2 = 120- 68
f1 + f2= 52
-------------------- (i)
Also mean of the data is 50 ( given)
Mean =
50 =
50 x 120 = 3480+30f1+70 f2
6000 – 3480= 30f1+70 f2
2520= 30f1+70 f2
30f1+70 f2 = 2520
10( 3 f1 + 7 f2) = 2520
3 f 1 + 7 f2 =
= 252 ---------------------- (ii)
Solving equation (i) and (ii), by elimination method,
Multiplying equation (i) by 3, we get
3f1 + 3f2= 156
Eqn (ii) is
3 f1 + 7 f2= 252
Now by subtracting these two equations, we get,
-4 f2= -96
80-100
19
fixi
170
30f1
1600
70 f2
1710
∑fixi = 3480+30f1+70 f2
Total
120
=> f2 = 24
Substitute the value of f2 in either of the equations (i) or (ii), we get, f1 = 28
CHAPTER : STATISTICS
LEVEL 1 (1 Mark)
Q1. Find the class mark of the class 10 –25
Ans:
Q2.Find the mean of first five natural numbers.
Ans :
Q3. If the mode of the distribution is 8 & mean is also 8 then find median
Ans: 3 median = mode + 2 mean
3 median = 8+ 2
Median =
8 = 24
=8.
Q4. Find the modal class of the following distribution
Class
frequency
0-6
7
6-12
5
12-18
10
18-24
12
24-30
6
Ans:Maximum frequency = 12
Modal class is 18-24.
LEVEL 2 (2 Marks)
Q5. Convert the following frequency distribution table into a less than type cumulative
frequency distribution table :
Marks
0–5
5 – 10
10 –15
15 – 20
20 –2 5
25– 30
No. of students.
Ans:
4
7
18
6
3
. less than type cumulative frequency table is
Marks
Less than 5
Less than 10
Less than 15
Less than 20
Less than 25
Less than 30
Q6.
12
f
4
7
12
18
6
3
c.f.
4
11
23
41
47
50
Find the mean of the following data
X1
10
15
20
25
30
F1
5
10
7
8
2
Ans:
x1
fi
fi xi
10
5
50
15
10
150
20
7
140
25
8
200
30
2
60
∑fi =32
Mean = ∑fi xi
∑fi
∑fi xi = 600
=
600
= 18.75
32
Q7. : If mean of the following data is 9, Find the value of K.
1 + 1 mark
x
y
3
4
6
K
12 15 9
1 6 4
Ans:
x
3
6
12
15
9
total
fx
f
y
4
xy
12
K
1
6
4
15+k
6k
12
90
36
150+6k
i
150 6 K
9
15 K
135 9 K 150 6 K
3K 15 K 5
x
i
i
Q8. Write a frequency distribution table for the following data:
Marks
Above 0
Above 10
Above 20
Above 30
Above 40
Above 50
No. of
students
30
28
21
15
10
0
Ans:
Marks
No. of students
0 - 10
2
10 - 20
7
20 - 30
6
30 - 40
5
40 - 50
10
Total
30
LEVEL 3 (3 Marks)
Q 9. The distribution below gives the weights of 30 students of a class. Find the median weight
of the students
Weight in
40-45
Kg
No. of
2
students
Ans:
45-50
50-55
55-60
60-65
65-70
70-75
3
8
6
6
3
2
Marks
Frequency
C.F
40 – 45
2
45 – 50
3
3+2 = 5
50 – 55
8
5+8= 13
55 – 60
6
13+6= 19
60 – 65
6
19+6= 25
65 – 70
3
25+3= 28
70 – 75
2
28+2= 30
N=30,
Median =
= 15,
l = 30,
n
2 cf
l +
f
= 50 +
h
x5
2
f= 3,
h= 5
= 50 +
x5
= 50 +
=
Q10. Find the mode of the given data
Family size 1-3
no. of
7
families
Ans- model class=3-5,
3-5
8
5-7
2
7-9
2
9-11
1
l=3, h= 2
f 1 =8, f 0 =7, f 2 =2
Mode = l + (f1`- f0)h/(2 f1 – f0 – f2 )
Solving mode = 3.286
Q11: Find x if mean of the following data is 62.8.
Class interval
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
5
8
x
12
7
8
Ans:
C.I.
0 -20
20-40
40-60
60-80
80-100
100-120
fx
f
i
f i 40 x
f i xi 2640 50 x
_
x
i
i
62.8 = (2640 + 50x)/(40+x)
xi
10
30
50
70
90
110
fi
5
8
x
12
7
8
fi X xi
50
240
50x
840
630
880
62.8(40+x) = 2640 +50x
2512+62.8x = 2640 +50x
62.8x – 50x = 2640 – 2512
12.8x = 128
X = 128/12.8 = 1280/128
x=10.
Q12. If the mean of the following distribution is 6 , find the value of p
X
f
2
3
4
2
6
3
10
1
P+5
2
Ans:
x
2
4
6
10
P+5
f
3
2
3
1
2
fx
6
8
18
10
2p+10
=mean
So
=6
2p=66-52
P=7
LEVEL
4 (4 Marks)
Q13. The median of the following data is 35 and the sum of all the frequencies is 170.
Class Interval Frequencies
0-10
10
10-20
20
20-30
30-40
40-50
50-60
60-70
f1
40
f2
25
15
Find f1 and f2, the missing frequencies.
Ans:
f1+f2+110=170
0-10
10
f1+f2=170-110 = 60 finding c.f
10-20
20-30
30
30+f1
30-40
70+f1
40-50
70+f1+f2
50-60
95+f1+f2
Median class = 30-40
l = 30, n= 170 ,cf = 30+f1 , h =10
n
cf
h
Median = l+ 2
f
35 = 30 + [170/2 – (30+f1)] X 10/ 40
35-30 = [85-(30+f1)]/4
5 X 4 = 85-30 –f1
20-85+30 = -f1
f1 = 35
therefore f2 = 60-35 = 25
Q14. The median of the following data is 525. Find the values of x and y, if the
Total frequency is 100.
Class interval
0-100
100-200
200-300
300-400
400-500
500-600
600-700
700-800
800-900
900-1000
Ans:
Frequency
2
5
X
12
17
20
Y
9
7
4
60-70
110+f1+f2
Class interval
0-100
100-200
200-300
300-400
400-500
500-600
600-700
700-800
800-900
900-1000
Frequency
2
5
X
12
17
20
Y
9
7
4
Cumulative frequency
2
7
7+x
19+x
36+x
56+x
56+x+y
65+x+y
72+x+y
76+x+y
It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24 (1)
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula : Median =l + [ (n/2 – cf) / f X h
we get
525 =500 +[50- 36-x] 100/20
i.e., 525 – 500 = (14 – x) × 5
i.e., 25 = 70 – 5x
i.e., 5x = 70 – 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e., y = 15
x=9 And y=15
Q15. . The mean of the following frequency table is 50. Find the missing frequencies
Class
Frequency
0-20
17
20-40
F1
40-60
32
60-80
F2
80-100
19
Total
120
Ans :
Class
fi
Xi
0-20
17
10
= -2
-34
20-40
F1
30
-f2
40-60
60-80
32
F2
50
70
= -1
0
=1
80-100
19
90
=2
38
F1 – F2 = 52 - (1)
Mean = a + h
=
50+20(
)
fiui
Ui =
0
F2
F1-F2 = 4 - (2 )
F1 = 28
F2 = 24
Q16 : The following distribution gives the daily income of 50 workers of a factory.
Daily income 100-120
120-140
140-160
160-180
180-200
(in Rs.)
Number of
12
14
8
6
10
workers
Convert the abovedata into a less than type Cumulative frequency distribution and draw its
ogive.
Ans:
Daily income(in Rs.)
Number f workers
(F1)
100-120
120-140
140-160
160-180
180-200
Total
12
14
8
6
10
€f1=n=50
Now , by drawing the points on the graph
i.e (120,12) ; (160,34) ; (180,40) ; (200,50).
We get graph of less than type Cumulative frequency.
Cumulative frequency less
than type
(x1)
12
12+14=26
26+8=34
34+6=40
40+10=50
LINEAR EQUATIONS IN TWO VARIABLES
FREQUENTLY ASKED QUESTIONS WITH SOLUTIONS
LEVEL-1( 1 MARK EACH)
Q1.Which of the following pairs of linear equations are consistent/inconsistent?
(i) x + y = 5, 2x + 2y = 10
Answer
(i) x + y = 5; 2x + 2y = 10
a1/a2 = 1/2
b1/b2 = 1/2 and
c1/c2 = 5/10 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of
possible solutions. Hence, the pair of linear equations is consistent.
Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat
and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically.
Solution: Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3x + 6y = 3900
Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2y = 1300
Q3.Find the number of solutions of the following pair of linear equations:
x + 2y -8=0
2x+ 4y=16
Solution: a1/a2 =1/2
b1/b2=1/2
c1/c2=1/2
Here a1/a2 = b1/b2= c1/c2
Therefore, the equations have infinite number of solutions.
Q4. Two lines are given to be parallel. The equation of one of the lines is 4x+ 3y=14. Find the
equation of the second line.
Solution: 8x+6y=28
LEVEL-2 ( 2 MARKS EACH)
Q1. Solve the following pair of linear equations by the substitution method.
x + y = 14 ; x – y = 4
Solution: x + y = 14 ... (i)
x – y = 4 ... (ii)
From equation (i), we get
x = 14 - y ... (iii)
Putting this value in equation (ii), we get
(14 - y) - y = 4
14 - 2y = 4
10 = 2y
y = 5 ... (iv)
Putting this in equation (iii), we get
x=9
x = 9 and y = 5
Q2. For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k –1)x + (k –1)y = 2k + 1
Solution:
3x + y -1 = 0
(2k –1)x + (k –1)y - (2k + 1) = 0
a1/a2 = 3/2k-1
b1/b2 = 1/k-1 and
c1/c2 = -1/-2k-1 = 1/2k+1
For no solutions,
a1/a2 = b1/b2 ≠ c1/c2
3/2k-1 = 1/k-1 ≠ 1/2k+1
3/2k-1 = 1/k-1
3k - 3 = 2k - 1
k=2
Hence, for k = 2, the given equation has no solution.
Q3. Solve the following pair of linear equations by the elimination method .
x + y =5 and 2x –3y = 4
Solution
x + y =5 and 2x –3y = 4
By elimination method
x + y =5 ... (i)
2x –3y = 4 ... (ii)
Multiplying equation (i) by (ii), we get
2x + 2y = 10 ... (iii)
2x –3y = 4 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 - (6/5) = 19/5
Hence, x = 19/5 and y = 6/5
Q4. On comparing the ratios a1/a2 , b1/b2 and c1/c2, find out whether the lines representing the
following pairs of linear equations intersect at a point, are parallel or coincident
5x – 4y + 8 = 0
7x + 6y – 9 = 0
Solution:Comparing these equation with
a1x + b1y + c1 = 0
a2x + b2y + c2= 0
We get
a1 = 5, b1 = -4, and c1 = 8
a2 =7, b2 = 6 and c2 = -9
a1/a2 = 5/7,
b1/b2 = -4/6 and
c1/c2 = 8/-9
Hence, a1/a2 ≠ b1/b2
Therefore, both are intersecting lines at one point.
LEVEL-3 ( 3 MARKS EACH)
Q1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more
than the number of boys, find the number of boys and girls who took part in the quiz. Find
the solutions graphically.
Solution
Let number of boys = x
Number of girls = y
Given that total number of student is 10 so that
x + y = 10
Subtract y both side we get
x = 10 – y
Putting y = 0 , 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0
x
10
5
y
0
5
Given that if the number of girls is 4 more than the number of boys
So that
y=x+4
Putting x = -4, 0, 4, and we get
y=-4+4=0
y=0+4=4
y=4+4=8
x
-4
0
4
y
0
4
8
Graphical representation
Therefore, number of boys = 3 and number of girls = 7.
Q2. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost
Rs 46. Find the cost of one pencil and that of one pen.
Solution:
Let cost of pencil = Rs x
Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 - 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = - 4
x = 10 – 7 × 15/5 = 10 – 21 = - 11
x
3
-4
-11
y
5
10
15
Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 - 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6
x
0
2
4
y
9.2
6.4
3.6
Graphical representation
Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.
Q3. Places A and B are 100 km apart on a highway. One car starts from A and another
from B at the same time. If the cars travel in the same direction at different speeds, they
meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds
of the two cars?
Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u - v) km/h
Respective speed of both cars while they are travelling in opposite directions i.e., travelling
towards each other = (u + v) km/h
According to the question,
5(u - v) = 100
u - v = 20 ... (i)
1(u + v) = 100 ... (ii)
Adding both the equations, we get
2u = 120
u = 60 km/h ... (iii)
Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
Q4.Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1
mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2
marks been deducted for each incorrect answer, then Yash would have scored 50 marks.
How many questions were there in the test?
Let the number of right answers and wrong answers be x and y respectively.
According to the question,
3x - y = 40 ... (i)
4x - 2y = 50
2x - y = 25 ... (ii)
Subtracting equation (ii) from equation (i), we get
x = 15 ... (iii)
Putting this value in equation (ii), we get
30 - y = 25
y=5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20
LEVEL-4 ( 4 MARKS EACH)
Q1.Draw the graphs of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0. Determine the
coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade
the triangular region.
Solution.
x-y+1=0
x=y-1
x
y
0
1
1
2
2
3
x
4
2
0
y
0
3
6
3x + 2y - 12 = 0
x = 12 - 2y/3
Graphical representation
From the figure, it can be observed that these lines are intersecting each other at point (2, 3)
and x-axis at ( - 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( - 1, 0), and (4,
0).
Q2. Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for
which y =mx +3.
Solution:
2x + 3y = 11 ... (i)
Subtracting 3y both side we get
2x = 11 – 3y … (ii)
Putting this value in equation second we get
2x – 4y = – 24 … (iii)
11- 3y – 4y = - 24
7y = - 24 – 11
-7y = - 35
y = - 35/-7
y=5
Putting this value in equation (iii) we get
2x = 11 – 3 × 5
2x = 11- 15
2x = - 4
Dividing by 2 we get
x=-2
Putting the value of x and y
y = mx + 3.
5 = -2m +3
2m = 3 – 5
m = -2/2
m = -1
Q3.The sum of the digits of a two-digit number is 9. Also, nine times this number is twice
the number obtained by reversing the order of the digits. Find the number.
Solution: Let the unit digit and tens digits of the number be x and y respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 ... (i)
9(10y + x) = 2(10x + y)
88y - 11x = 0
- x + 8y =0 ... (ii)
Adding equation (i) and (ii), we get
9y = 9
y = 1 ... (iii)
Putting the value in equation (i), we get
x=8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
.
Q4.Solve the following pairs of equations by reducing them to a pair of linear equations:
+
=4
Solution.
= -2
+
=4
-
= -2
Putting 1/x+y = p and 1/x-y = q in the given equations, we get:
10p + 2q = 4
10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/x-y = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y=2
Hence, x = 3 and y = 2
Minimum Package For SA-I
S .No
1
2
3
4
5
6
Name of the Chapter
Real Numbers
Polynomials
Pair of Linear
Equations in Two
Variables
Triangles
Introduction to
trigonometry
Statistics
Class-X
Contents
i. To find HCF by
Euclid‟s division lemma
& Prime Factorization.
ii. To Find LCM
iii)To Prove √x is an
irrational number.
i. To find the Zeroes &
Verify relation between
Zeroes & Co-efficient.
ii. Framing of
Polynomials with given
Zeroes.
i. To Find the solution by
Graphical method
ii. Find the value of
unknown in given system
of equations for a
particular condition
iii. Solution of equation
by any method
i)Proofs of all evaluative
theorems
ii)Simple Application of
theorems
i)Finding trigonometric
ratios
ii)Questions based on
Complementary angles
and specific angles
iii)Simple questions
based on identities
Remembering formula
for finding mean, mode
&median.
2015-2016
Expected Marks
7
6
10
8
12
17
Problems based on mean,
mode&median.
To find the ogive for
given data.
Total
60
Sample Paper -1 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
BLUE PRINT
S.No. Topic
1
2
3
4
5
Number
System
Algebra
Geometry
Trigonometry
Statistics
MCQ
SA I
SA II
/VSA
(2marks) (3marks)
(1mark)
1
LA
TOTAL
(4marks) (90)
2
11
1
1
1
1
4
3
2
2
2
11
23
17
22
17
90
2
1
2
1
6
2
2
3
2
10
Sample Paper -1 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
Time allowed: 3 hours
Max.Marks: 90
General Instructions:
All Questions are compulsory
The question paper consists of 31questions divided into four sections A,B,C,D
Section A comprises of 4 questions of 1 mark each, Section B
Comprises of 6 questions of 2 marks each, Section C comprises of 10 questions
of 3 marks each and Section D comprises of 11questions of 4 marks each.
Use of calculator is not permitted.
SECTION A (Q. No. 1 -4 each of 1mark)
1.
Find the zeros of polynomial X2 – 2X – 8 .
2.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the
foot of the ladder from base of the wall .
3.
4.
If sin A =
3
, calculate cos A .
4
Find the class mark of 15.5 – 20.5
SECTION B (Q. No. 5 -10 each of 2marks)
5.
If α and β are zeros of the Polynomial 3x2+5x+2, Find the value of
6.
For what value of k,will the following system of linear equations have infinitely many
solutions
2x +3 y =7
(a-b)x + (a+b)y = 3a+b-2
let ABC
DEF and their areas be respectively 64 cm2 and 121 cm2.If EF = 15.4 cm,
Find BC.
If sin4A = cos(A-200) , where 4A is an acute angle, find the value of A.
Convert the following frequency distribution table into a less than type cumulative
frequency distribution table:
Marks
No of Students
0-5
4
5-10
7
10-15
12
15-20
18
20-25
6
25-30
3
7.
8.
9.
10.
Evaluate cos48 – sin42
SECTION C (Q. No. 11 -20 each of 3 marks)
11.
12.
13.
14.
15.
16.
17.
Given HCF ( 306, 657 ) = 9. Find the LCM ( 306, 657 ).
Find the zeros of the quadratic polynomial 6x 2 – 7x – 3 and verify the relationship
between the zeros and the coefficients.
Solve: 3x – 5y =4
9x – 2y = 7 by using Elimination method.
If the areas of two similar triangles are equal then show that the triangles are congruent
ABC is an isosceles triangle right angled at C Prove that AB2 = 2AC2
Prove that
(sinA + cosecA)2 + (cosA + secA)2 = 7 + tan2A + cot2A
Without using t-tables evaluate the following
1
3cos68 cosec22 tan43 tan47 tan12 tan60 tan78
2
18.
19.
20.
If tan(A + B) =√3 and tan(A – B) =
1
, Find A and B
3
Find the median of the following distribution
Class
frequency
0-10
4
10-20
4
20-30
8
30-40
10
40-50
12
50-60
8
60-70
4
If the mean of the following distribution is 6 , find the value of p
X
f
2
3
4
2
6
3
10
1
P+5
2
SECTION D(Q. No. 21 -31 each of 4 marks)
21.
22
23.
24.
25.
Prove that √3is irrational.
Find all the zeros of the polynomial 2x 4 -3x3-3x2+6x-2 if two of its zeros are
√2 and - √2
8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys
can finish it in 14 days.Find the time taken by one man and by one boy alone to finish the
Work.What value is depicted ?
If tanA = 2 Evaluate secA sinA + tan 2A – cosecA
The marks obtained by 30 students of class X of certain school in Mathematics paper
Consisting of 100 marks are presented in the table below. Find the mode of this data
Class Interval
10-25
25-40
40-55
55-70
70-85
Number of Students
2
3
7
6
6
85-100
6
26.
In an equilateral triangle ABC , D is a point on side BC such that BD = 1/3 BC
Prove that 9 AD2 = 7 AB2
27
Prove that the ratios of the areas of two similar triangles is equal to the square of the
ratio of their corresponding sides.
28.
29.
Divide :
2t4 + 3t3 -2t2 – 9t -12 by t2 – 3
The median of the following data is 525 Find the values of x and y if the total
Frequency is 100
Class interval
Frequency
0-100
2
100-200
5
200-300
x
300-400
12
400-500
17
500-600
20
600-700
y
700-800
9
800-900
7
900-1000
4
30.
Prove that
–
31.
–
For a morning walk three persons step off together. There steps measure 80cm, 85cm, and
90cm respectively. What is the minimum distance each should walk so that they can cover
the distance in complete steps?
Sample Paper -1 (2014-15)
SUMMATIVE ASSESSMENT - 1
CLASS X
Marking Scheme
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity and maintain
uniformity. The answers given in the marking scheme are the best suggested answers.
2. Marking be done as per the instructions provided in the marking scheme. (It should not be
done according to one‟s own interpretation or any other consideration). Marking Scheme be
strictly adhered to and religiously followed.
3. Alternative methods be accepted. Proportional marks be awarded.
4. If a question is attempted twice and the candidate has not crossed any answer, only first
attempt be evaluated and „EXTRA‟ written with second attempt.
5. In case where no answers are given or answers are found wrong in this Marking Scheme,
correct answers may be found and used for evaluation purpose.
SECTION – A
1) The given polynomial is
P(x) =X2 – 2X – 8
Let P(x)=0
X2 – 2X – 8=0
X2 – 4X +2X – 8=0
X(X-4)+2(X-4)=0
(X-4)(X+2)=0
Either X-4=0 or X+2=0
X=4,-2
(By splitting middle term)
(1 marks)
2) AB2+BC2=AC2 (By PythagorasThm.)
A
82+BC2=102
64+BC2=100
BC2=100-64=36=62
B
C
BC=6m.
(1 marks)
3) Let ABC be any right triangle right angled at B.
3
BC
Now sinA = , but sinA =
4
AC
BC
3
=
=K where K is constant of proportionality
AC
4
BC = 3K, AC = 4K ,
By P.G.T
AC2 = AB2 +BC2
( 4K)2 =AB2 +( 3K)2
16K2 = AB2 + 9K2
AB =
cosA =
A
B
C
7K.
AB
7K
7
=
=
AC
4K
4
4. The class mark of 15.5 – 20.5=
(1 marks)
=18 (1)
SECTION – B
5. 3x2+5x+2
= b
α+β=
a
αβ
+ =
=c
=
a
=
6. Here a₁ = 2, b₁ = 3, c₁ = -7
a₂ = a-b, b₂ = a+b, c₂ = -( 3a+b-2 )
for infinitely many solutions,
(
marks)
(
marks)
(1 marks)
a1 b1
c
=
= 1 ,
a 2 b2 c 2
7
2
3
=
=
a b a b (3a b 2)
I
II
III
From I and III we have
a = 9b-4 …………(1)
from II and III
a-2b = 3 ………………(2)
by substituting the value of a from (1) in (2)
9b-4-2b = 3
b=1
Putting this value in (1) we get a = 5 .
1. Area of∆ ABC/area of ∆DEF =
64
=
121
64
=
121 ( )
BC = 11.2 cm.
( As ∆ABC ~ ∆DEF)
(1 marks)
(1 marks)
(1 marks)
(1)
8.Given that
Sin4A = cos(A-200)
Sin[900-(900- 4A)] = cos(A-200)
Cos(90o-4A) = cos(A-200)
(1 marks)
(90o-4A) = (A-200)
5A = 1100
A=220
(1 marks)
9. Less than type cumulative frequency table is
Marks
Less than 5
f
4
c.f.
4
Less than 10
Less than 15
Less than 20
Less than 25
Less than 30
7
12
18
6
3
11
23
41
47
50
(2 marks)
10. Cos 48° - sin 42°
Cos (90 – 42)° - sin 42°
Sin 42° - sin 42° = 0
(1 marks)
(1 marks)
SECTION- C
11.
HCF = 9, 1st number = 306, 2nd number = 657
We know that
HCFxLCM =Product of two numbers
9 x LCM
(1 marks)
= 306 x 657
306x657
9
LCM
=
LCM
= 22338.
(2 marks)
12. P(x) = 6x² - 7x – 3
6x² - 9x + 2x – 3
3x(2x-3) +1(2x-3)
(2x – 3)(3x+ 1)
x = 3/2 , x = -1/3
3 1 7 b
+(
)= =
2
3
6 a
1 c
3 1
product of zeroes = x(
)=
=
2
3
2 a
(1 marks)
Now sum of zeroes =
(2 marks)
13. 3x – 5y = 4 ……..(i)
9x – 2y = 7 ………(ii)
Multiply eq. (i) by 3
9x – 15y = 12 ……..(iii)
(1 marks)
Subtracting (ii) from (iii)
9x – 15y = 12
9x – 2y = 7
- +
-13y = 5
y=
5
13
(1 marks)
Putting value of y in (ii)
5
9x – 2(
)=7
13
9
x=
13
(1 marks)
14. Given that ∆ABC ~ ∆PQR
Area (∆ABC) = area (∆PQR)
OR area (∆ABC)/area(∆PQR) = 1
2
2
(1 marks)
2
AC
AB
BC
=
=
=1
2
2
PR 2
PQ
QR
AB = PQ, BC = QR, AC = PR
By SSS congruence condition
∆ABC ≈∆PQR
(2 marks)
15. ∆ABC is right angled at C.
AB² = AC² + BC² (By pyth. Th.)
AB²= AC² + AC² (AC = BC )
AB² = 2AC²
(1 marks)
(1 marks)
(1 marks)
16. LHS = (sin A + cosec A)² + (cos A + sec A)²
= sin²A + cosec²A + 2sinAcosecA + cos²A + sec²A + 2cosAsecA
=sin²A + cosec²A + 2 + cos²A + sec²A + 2
= sin²A + cos²A + cosec²A + sec²A + 4
=1 + 1+ cot²A + 1 + tan²A + 4
=7 + tan²A + cot²A = RHS
(1 marks)
(1 marks)
(1 marks)
17. 3 cos(90 – 22). Cosec22 –
1
tan43.Tan(90 – 43) tan12.tan(90 – 12).tan60(1)
2
1
3sin22.cosec22 – (tan43.cot43). (tan12.cot12). tan60
2
3x1–
3-
(1 marks)
1
x 1 x 1 x √3
2
3 6 3
=
2
2
(1 marks)
18. Tan(A + B) = √3
Tan (A + B) = tan60
A + B = 60 …………(i)
1
Tan(A – B) =
3
Tan(A – B) = tan 30
A – B = 30 ………(ii)
Adding (i) & (ii)
2A = 90
A = 45
Putting value of A in (i)
Putting the value of A in (i) we get
B = 15
(1 marks)
(1 marks)
45 + B = 60
(1 marks)
19. cumulative frequency table is
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
N= =25,f=10,cf=16,
And median class is 30-40
f
4
4
8
10
12
8
4
c.f.
4
8
16
26
38
46
50
Now use the formula to find the median.
20.
x
2
4
6
10
P+5
f
3
2
3
1
2
fx
6
8
18
10
2p+10
=mean
So
=6
2p=66-52
P=7
SECTION – D
21. Let us assume that √3 is rational.
Then, there exit co-prime positive integers a and b such that
√3 = a/b, b ≠0
a = √3b
squaring both sides
a² = 3b² ……..(i)
3divides a²
3 divides a
a = 3c, (where c is any integer)
a = 3c in (i)
9c² = 3b²
3c² = b²
It means 3 divide b² and 3 divides b.
3 is common factor of both a and b which is contradiction.
So, our assumption that √3 is rational is wrong
(1 marks)
(1 marks)
(1 marks)
(1 marks)
Hence √3 is irrational.
22. Given zeroes are √2 and - √2
So (x-√2 )(x+√2 )=x2-2 is a factor of the given polynomial
Now divide the given polynomial by x2-2 to get other two zeroes
From quotient 2x2-3x+1
2x2-2x-x+1
2x(x-1) -1(x-1)
(2x-1)(x-1)
Either 2x-1 =0, or x-1 =0
(2 marks)
1
X= or x= 1.
2
other two zeros are
1
,1.
2
(2 marks)
23. Let the time taken by one man = x days
And time taken by one boy = y days
ATQ
, + =
and + =
Now solve the equations to find x and y
24. Hypotenuse = √5
SecA = √5, sinA = 2/√5, cosecA = √5/2
(√5 x
2
25
2
5
) + (2)² -
5
2
+4 -
5
2
6
5 12 5
=
2
2
-
Class Interval
10-25
(1 marks)
(1 marks)
(1 marks)
(1 marks)
Number of Students
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
Since highest frequency is 7
Therefore modal class is 40-55
Mode = l + (
f1 f 0
)xh
2 f1 f 0 f 2
(2 marks)
= 40 + (
73
) x 15
14 3 6
= 40 + (
4
) x 15
5
= 40 + 12
= 52.
(2 marks)
26. In an equilateral triangle ABC, D is a point on side BC such that
BD= 1/3 BC
A
To Prove:- 9AD2=7AB2
Construction: Draw AP BC
Proof: In right triangle APB,
AB2 = AP2 +BP2
-(1)
(By Pythagoras theorem)
B
D
P
In right triangle APD,
AD2 = AP2 +DP2
-(2)
(By Pythagoras theorem)
From (2)
AP2 = AD2-DP2
-(3)
From (3) , putting the value of AP in (1) , we get
AB2 = AD2 – DP2 +BP2
= AD2- DP2 +(BC/2)2
In right triangle APB and APC,
Hyp. AB= Hyp. AC
AP = AP
(Common side)
Therefore ∆APB ≡∆ABC
(2 marks)
C
(RHS congruence criterion)
Therefore BP = CP
(CPCT)
Therefore, BP = CP = BC/2
AB2= AD2-DP2 +BC2/4
=AD2-(BP-BD)2+BC2/4
= AD2-(BP2+BD2-2BP.BD)+BC2/4
= AD2-BP2-BD2+2BP.BD+BC2/4
= AD2-(BC/2)2 – (BC/3)2+2(BC/2)(BC/3)+BC2/4
Because BP=BC/2 and BD=BC/3
= AD2-BC2/4 – BC2/9 +BC2/3 +BC2/4
= AD2 + 2/9 BC2
= AD2+2/9AB2
Because AB=BC
AB2(1-2/9)=AD2
7/9AB2 =AD2
7 AB2 = 9 AD2
(2 marks)
27) Given two Triangles ∆ABC and ∆DEF such that ∆ABC is similar to ∆DEF
To prove
ar∆ ABC
= AB2 = BC2= AC2
ar ∆DEF
DE2
Construction Draw AL BC
EF2
DF2
and DM EF
A
D
B
L
C
E
M
Proof: Since, similar triangles are equiangular and their corresponding sides are
proportional. Therefore,
∆ABC is similar to ∆DEF
∟A = ∟D, ∟B = ∟E, ∟C = ∟F
And AB
= BC
DE
=
AC
EF
-(1)
DF
Thus, in ∆ALB and ∆DME
∟ALB = ∟ DME
(each 90ᵒ)
∟B
(from eq.(1)
= ∟E
By AA similarity,∆ALB is similar ∆DME
AL
= AB
DM
-(2)
DE
From eq. (1) and (2), we get
AB
=
DE
BC
=
EF
AC
DF
=
AL
DM
Now,
ar(∆ABC)
ar (∆DEF)
= ½ (BC x AL)
½ (EF x DM)
[ because Area of triangle= ½ x Base x Altitude]
ar(∆ABC)
=
ar (∆DEF)
= BC
(BC x AL)
(EF x DM)
x
BC
-(3)
(2 marks)
F
EF
EF
ar(∆ABC)
=
-(4)
EF2
ar (∆DEF)
But
BC2
BC
=
EF
AB
= AC
DE
DF
(by similarity of ∆ABC and ∆DEF)
BC2
AB2
=
EF2
= AC2
DE2
-(5)
DF2
Therefore eq. (4) and (5), we get
ar∆ ABC
ar ∆DEF
28)
AB2
=
DE2
EF2
BC2
=
AC2
=
DF2
t² - 3 ) 2t⁴ + 3t³ - 2t² - 9t -12(2t² + 3t + 4
2t⁴-6t² (1)
- +
- 3t³ + 4t² - 9t -12
- 3t³
- 9t
+
(1)
4t² -12
4t² - 12
(1)
0
(1)
29).
Class
0-100
f
cf
2
2
(2 marks)
100-200
5
200-300
X
300-400
12
400-500
17
500-600
20
600-700
y
700-800
9
800-900
7
900-1000
4
f=20,cf=36+x,total of frequency given is 100
7
7+x
19+x
36+x
56+x
56+x+y
65+x+y
72+x+y
76+x+y
(2 marks)
And median class is 500-600
Use the median formula to find x and y
X=9 and y=15
(2 marks)
30) LHS=
–
=
–
Taking cos A common from numerator and denominator
(
)
(
)
(2 marks)
(
(
)
)
Dividing numerator and denominator by sinA, we get
–
=RHS
(2 marks)
31) Minimum distance = LCM of 80,85& 90
80 = 2 X 2 X 2 X 2 X 5
85 = 5 X 17
90= 2 X 3 X 3 X5
LCM= 2 X 2 X 2 X 2 X 3 X 3 X 5 X 17 = 12240
(1 marks)
(1 marks)
(1 marks)
(1 marks)
Sample Paper -2 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
Blue Print
S.No.
Topics
VSA
SA I
SA II
LA
TOTAL
1
Number system
2(1)=2
1(2)
1(3)
1(4)
5(11)
2
Algebra
1(2)
3(3)=9
3(4)=12
7(23)
3
Geometry
1(2)=2
2(3)=6
2(4)=8
6(17)
4
Trigonometry
2(2)=4
2(3)=6
3(4)=12
7(22)
5
Statistics
1(1)
1(2)=2
2(3)=6
2(4)=8
6(17)
Total
4
6(12)
10(30)
11(44)
31(90)
1(1)
Sample Paper -2 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
Time: 3hrs
Marks: 90
Max.
General Instruction:1. All questions are Compulsory.
2. The question paper consists of 31 questions divided into 4 sections, A,B,C and D. Section
– A comprises of 4 questions of 1 mark each. Section-B comprises of 6 questions of 2
marks each. Section C comprises of 10 questions of 3 marks each and Section- D
comprises of 11 questions of 4 marks each.
3. Question numbers 1 to 4 in Section are Very Short Answer type Questions to be
answered in one word or in one sentence or exact requirement of the question
4. Use of calculator is not permitted.
SECTION A
Questions 1 to 4 carry one mark each.
1. If HCF(120,225)=15, then find the LCM of 120 and 225.
2. Write the condition which should be satisfied by q so that rational number p/q has a terminating
decimal expansion.
3.In
ABC , AB 24cm, BC 10cm and AC 26cm.Is this a right triangle ? Give reason for your answer.
4.Write the relation connecting the measures of central tendencies.
SECTION B
Question 5to 10 carry two marks each.
5. Find H.C.F of 867, 255 using Euclid‟s division lemma.
6. Find the zeros of the polynomial 4√3x² + 5x - 2√3.
7.In figure PQ ||BC find QC
A
1.5cm
1.3cm
P
Q
3cm
B
C
8. If Sec4A = Cosec (A-20°) where 4A is an acute angle, find the value of A.
9. Simplify
1
1
sin
}
sin cos ec
10. Find the Mean of first five odd multiples of 5?
Section C
Question 11 to 20 carry three marks each.
11.Prove that √ is an irrational Number.
12. Find the zeroes of quadratic polynomial
zeroes and their
co-efficient.
-2x-8 and verify the relationship between the
13. For what value of k will the following system of linear equations has no solution?
3x+y=1
(2k-1)x+(k-1)y=2k+1
14. Evaluate: (Sin 470/Cos 430)2 +(Cos 430/Sin 470)2 -4Cos2450
15. A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8
is added to its
denominator. Find the fraction.
16. In fig ABC and AMP are two right triangles right angled at B and M respectively Prove
that
(i) ABC~ AMP
C
(ii)
=
M
√
A
17. Prove that
√
B
P
= Sec A + tan A
The distribution below gives the weights of 30 students of a class. Find the median weight of
the students
Weight in
Kg
No. of
students
40-45
45-50
50-55
55-60
60-65
65-70
70-75
2
3
8
6
6
3
2
19.In fig if AD⊥ BC prove that AB2+CD2 = BD2 +AC2
D
B
C
A
20. If the mean of the following distribution is 54.Find the value of p :
Class
frequency
0-20
7
20-40
p
40-60
10
60-80
9
80-100
13
Section D
(Q. No. 21 to Q. No. 31carry 4 marks each)
21. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeros are √5/3 and - √5/3
22. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of
their corresponding sides.
23. Draw the graphs of 2x+y=6 and 2x – y = 2.Shade the region bounded by these lines and xaxis. Find the area of the shaded region.
24. Prove that
=
25.The following distribution gives the daily income of 50 workers of a factory
Daily in come
Number of
workers
100-120
12
120-140
14
140-160
8
160-180
6
180-200
10
Convert the distribution above to a less than type cumulative frequency distribution and draw
itsOgive.
26. Without using trigonometric tables evaluate
(
) –
.
27. In a school students thought of planting trees in and around the school campus to reduce air
ands noise pollution. They planted two types of trees type A & type B . The total number of trees
planted are 25 and sum of type A and twice the number of type B trees is 40. Find the number of
each type of trees planted. What values can be imparted by planting trees.
28. Prove that
(SinØ+Cosec Ø )2 +(Cos Ø +SecØ)2 = 7+ Tan2Ø+ Cot2Ø
29. In fig ∆ ABC and ∆ DBC are two triangles on the same base BC. If AD intersects BC at O.
Show that
Area (∆ ABC)/Area (∆DBC) = AO/DO
A
D
B
C
30. The mean of the following frequency table is 50. Find the missing frequencies
Class
Frequency
0-20
17
20-40
F1
40-60
32
60-80
F2
80-100
19
Total
120
31. Prove that the square of any positive integer is of the form 3m or 3m+1 for some integer m.
Sample Paper -2 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
Marking Scheme
SECTION- A
Ques.1 LCM x HCF= Product of two numbers
LCM 15 120 25
120 225
1800
15
Ques2.q must be of the form 2n5m
LCM
Ques.3
Here, AB 2 (24) 2 576 , BC 2 (10) 2 100
and AC 2 (26) 2 676
So AC 2 AB 2 BC 2
Hence, the given ABC is a right triangle.
Ques4Mode= 3 Median-2 Mean
SECTION- B
Ques.5
867 = 255 x 3 + 102
[By using Euclid division lemma]
225 = 102 x 2 + 51
102 = 51x 2 + 0
Therefore HCF of 867 and 255 is 51
Ques.6
4√3x2 + 5x - 2√
Product = 4√ x 2√ = 24
(1marks)
Sum = 5
We have F (x) = 4√ x2 + 8x – 3x - 2√
)-√ (√
F (x) = (√
)
F (x) = (√
)(
√ )
Zeroes of f[x] is given by
If F (x) = 0
(√
)(
√ )=0
(√
)
√
x=√
x=
√
(1marks)
(1marks)
(1marks)
Hence Zeroes of f (x) is
=
√
and
√
β=
Ques.7Since PQ || BC
Therefore By using BasicProportionality Theorem
QC
=
=
(1marks)
=
(1marks)
2.6cm.
= cosec (A – 200)
Ques.8
= sec [ 900–(A – 200) ]
(1marks)
= sec (1100 – A)
4A =
1100 – A
(1marks)
A=
A=220
1
1
sin
sin
cos ec
(1marks)
= 1-sin sin =1- sin 2 = Cos 2
(1marks)
Ques. 9 sin
Ques.10
First five odd multiple of 5 are
5, 15, 25, 35, 45
Mean =
(1marks)
=
= 25
Section- C
Ques.11
(1marks)
Let √
b e a Rational no.
Therefore √
{where p, q are integers and q ≠o}
=
3=
3 q2 = p2
(1marks)
3 divides p2 3 divides p……1)
putting p= 3r
[from some integer ]
3q2 = (3r)2 =9r2
(1marks)
q2 = 3r2
3 divides q2 3 divides q
…..(2)
(1marks)
From eqn. 1 &2, 3 is a common factor of p & q which contradicts the fact that p & q are
coprime.So our assumption is wrong
√
We have f(x) = x2 – 2x – 8
Ques.12
= x2 – 4x + 2 x – 8
= x (x –4) + 2 (x – 4)
= (x -4)(x +2)
(1/2marks)
Zeroes of f(x) isf(x) = 0
(x – 4)=0
(x + 2)=0and
X+
X
2=0
= -2 and
and x – 4 = 0
x=4
(1/2marks)
Therefore Zeroes of f(x) is
,
Sum of zeroes =
And
Product of zeroes =
= -2 + 4 = 2
=
(
)
(1/2marks)
=2
= (-2)4 = -8
(1/2marks)
And
Ques.13
Here
=
= -8
a1 = 3b1= 1
c1 = 1
a2=(2k – 1)
b2 = (k – 1)
c2 = (2k + 1)
(1/2marks)
For no solution
(1/2marks)
(1/2marks)
&
3k - 3 = 2k - 1
,
2k + 1
k–1
3k - 2k = -1 +3
,
2k – k
-1 -1
(1/2marks)
K=2
,
k
Hence the given system of equations will have no solution if k =
2.(1marks)
Ques.14
{
2
(
)
+{
(
) 2
– 4 x ( )2
√
(1marks)
={
}2 + {
(1marks)
=2–2
=0
(1marks)
2
– 4( )2
√
=1+1–4x
Ques.15
Let the numerator be x and denominator be y, Fraction =
According to given condition
= And
(1marks)
3x – 3 =
3x – y = 3
- (1)
And
4x – y = 8
Subtracting eqn.
- (2)
(1 marks)
(1)from eqn. (2)
4x – y = 8
3x – y = 3
x=5
On putting the value of x in equation (1)
3x5–y=3
15 – y = 3
y = 12
Therefore fraction is =
(1marks)
Ques.16
(1)
In triangle ABC and AMP we have
0
∠
=∠
= 90 (each)
∠
∠ (common)
Therefore AA Criterion of similarity
ABC
AMP
(2)
(1marks)
(1marks)
(1marks)
Ques.17
1 sin A
1 sin A
1 sin A
1 sin A
L.H.S=
(1marks)
(1 sin A) 2 1 sin A
1
1 sin 2 A
CosA
CosA
SinA
CosA
R.H.S
SecA =tan
A
=
Ques.18
Marks
(1marks)
Frequency
(1 marks)
C.F
40 – 45
2
45 – 50
3
3+2 = 5
50 – 55
8
5+8= 13
55 – 60
6
13+6= 19
60 – 65
6
19+6= 25
65 – 70
3
25+3= 28
70 – 75
2
28+2= 30
2
(1marks)
N=30,
Median =
= 15,
l = 30,
n
cf
l +2
f
f= 3,
h= 5
h
(1marks)
= 50 +
x5
= 50 +
x5
= 50 +
=
(1marks)
In Δ
Ques.19
we have
AC2 = AD2 + CD2 (By Pythagoras theorem)-(1)
(1marks)
In Δ
AB2 = AD2 + BD2 (By Pythagoras theorem)–(2)
(1marks)
(2) - (1)
AB2 – AC2 = BD2 – CD2
AB2 – CD2 = BD2 +AC2
(1marks)
Ques.20
Hence proved.
Class
Mid value(xi)
fi ui =
0 – 20
10
7
-2
-14
20 – 40
30
p
-1
-p
40 – 60
50
10
0
0
60 – 80
70
09
1
9
80 – 100
90
13
2
6
(2marks)
= a+h(
)
54 = 50 + 20 (
P=11
(1marks)
fiui
)
SECTION - D
Since √ and -√
Ques.21
(
√ )(
are two zeroes of f(x)
√ ) = x2- is a factor of
(2marks)
3x2 – 5 is a factor of p(x)
3x+6x-2x-10x-5 = (x+√ ) (n-√ )(n+1)(n+1)
( )
√ , -√ , -1, -1
(2marks)
Ques22.Given two Triangles ∆ABC and ∆DEF such that ∆ABC is similar to ∆DEF
To prove
ABC
AB 2 BC 2 AC 2
DEF DE 2 EF 2 DF 2
A
D
Construction :Draw AL; BC and DM EF
B
L
C
E
M
Proof Since, similar triangles are equiangular and their corresponding sides are proportional.
Therefore,
∆ABC is similar to ∆DEF
∠A =∠ D, ∠B = ∠E, ∠C = ∠F
AB BC AC
DE EF DF
………………………………………………………………(I)
F
(2mark
Thus, in ∆ALB and ∆DME
∠ALB = ∠ DME
∠B
(each 90ᵒ)
= ∠E
(from eq.(1)
By AA similarity, ∆ALB is similar ∆DME
AL
AB
………………………………………………………………..(2)
DM DE
From eq. (1) and (2), we get
AB BC AC
AL
…………………………………………….(3)
DE EF DF DM
1
BC AL
arABC
2
arDEF 1
EF DM
2
(1marks)
[ because Area of triangle= ½ x Base x Altitude]
arABC
BC AL
arDEF EF DM
As
BC
AL
, fromeqn…………(3)
EF DM
arABC BC 2
………………………………………..(4)
arDEF EF 2
But
AB BC AC
…………………………………………..(5)
DE EF DF
(by similarity of ∆ABC and ∆DEF)
(1marks)
Therefore eq. (4) and (5), we get
arABC BC 2 AB 2
AC 2
arDEF EF 2 DE 2 DF 2
Ques.23
x
0
1
2
y
6
4
2
2x+y = 6
Y = 6 – 2x
2x-y = -2
x
0
1
2
y
2
3
6
y = 2x+ 2
(1 mark)
(2 mark)
Draw graph and then find the Area of shaded region
Ques.24
L.H.S
=
=
(
)
(1 mark)
(1 mark)
(tan sec ) (sec tan )(sec tan )
tan sec 1
mark)
=
tan sec
= sin 1
cos
=
(tan sec )(1 sec tan )
(2
tan sec 1
sin
1
cos cos
(1
mark)
=
Ques.25
Marks
R.H.S
No .of Students
Marks less than
C.F
100-120
12
120
12
120-140
14
140
26
140-160
8
160
34
160-180
6
180
40
180-200
10
200
50
(2 mark)
Cumulative frequency curve
(2 mark)
Ques.26
2
3 cos 430
cos 37 0 cos ec530
0
tan 5 0 tan 25 0 tan 45 0 tan 65 0 tan 85 0
sin 47
(1 mark)
2
3 cos(90 47 0 )
cos(90 0 530 ) cos ec530
=
sin 47 0
tan(90 0 85 0 ) tan 25 0 tan 45 0 tan(90 0 25 0 ) tan 85 0
2
3 cos(90 47 0
cos(90 0 530 ) cos ec530
sin 47 0
tan(90 0 85 0 ) tan 25 0 tan 45 0 tan(90 0 25 0 ) tan 85 0
(1 mark)
3 cos(90 47 0
sin 47 0
(1 mark)
2
sin 530 cos ec530
cot 85 0 tan 25 0 tan 45 0 cot 25 0 tan 85 0
1
3 sin 47
sin 530
0
1
1
sin 47
tan 25 0 1
tan 85 0
0
0
tan 25
tan 25
0
2
sin 530
3 1
2
1
1 1 1
{ tan 450 1}
=9-1
=8
(1 mark)
Ques.27
Let x ,y be the number of type A and type B trees
According to the question
x+ y=25…………………………………………………………(i)
x+2y =40………………………………………………………..(ii)
(1 mark)
Subtracting (ii) from (i)
Y=15
(1 mark)
Putting this value of y in eqn. (i)
X=10
(1 mark)
No. of type A trees = 10
No of type B trees=15
By involving students in such acts values like environmental consciousness and social
responsibilities are infused among them.
(1 mark)
Ques.28
L.H.S
sin Co sec 2 sin Co sec Cos Sec 2 2CosSec
2
2
2
Co sec2 1 cot 2
As we know that
(2 mark)
Sec 2 1 tan 2
1 1 Cot 2 2 1 tan 2 2 7 tan 2 Cot 2
R.H .S
Ques.29
(11/2 mark)
(2 mark)
Draw AL⊥BC and DM⊥BC
(Corresponding sides are proportional)
(
)
(
)
=
(11/2 mark)
(
)
(
)
=
(1 mark)
Ques.30
Class
fi
Xi
fiui
0-20
17
10
= -2
-34
20-40
F1
30
-f2
40-60
60-80
32
F2
50
70
= -1
0
=1
80-100
19
90
=2
38
Ui =
0
F2
(2 mark)
sin 2 cos 2 1
F1 – F2 = 52 - (1)
Mean = a + h
=
50+20(
)
F1-F2 = 4 - (2)
F1 = 28
(1 mark)
F2 = 24
(1 mark)
31.
Let x be any positive integer and b=3.
According to Euclid's division lemma, we can say that
x=3q+r,0≤r<3
(1/2 mark)
Therefore, all possible values of x are:
x=3q,(3q+1) or (3q+2)
(1/2 mark)
Now lets square each one of them one by one.
(i)
(3q)2=9q2
(1mark)
Let m=3q2 be some integer, we get 9q2=3×3q2=3m
(ii) (3q+1)2=9q2+6q+1=3(3q2+2q)+1
Let m=3q2+2q be some integer, we get
(3q+1)2=3m+1
(1mark)
Unit/Topic
VSA
SA(I)
SA(II)
LA
Total
Number System
--
4(2)
3(1)
4(1)
11(4)
Algebra
1(1)
4(2)
6(3)
12(3)
23(8)
Geometry
1(1)
2(1)
6(2)
8(2)
17(6)
Real Numbers
(iii) (3q+2)2=9q2+4+12q=9q2+12q+3+1=3(3q2+4q+1)+1
Let m=(3q2+4q+1) be some integer, we get
(3q+2)2=3m+1
Hence, square of any positive integer is either of the form 3m or 3m+1 for some integer m.
(1mark)
Trigonometry
1(1)
---
9(3)
12(3)
22(70
Statistics
1(1)
2(1)
6(2)
8(2)
17(6)
Total
4(4)
12(6)
30(10)
44(11)
90(31)
SAMPLE PAPER –3
Summative Assessment I (2015-16)
Mathematics
Blue Print
CLASS : X
SAMPLE PAPER –3
Summative Assessment I (2015-16)
Mathematics
CLASS : X
Time: 3hrs
Max. Marks: 90
General Instruction:1. All questions are Compulsory.
2. The question paper consists of 31 questions divided into 4 sections, A,B,C and D. Section – A
comprises of 4 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each
,Section- C comprises of 10 questions of 3 marks each and Section – D comprises of 11 questions of
4 marks each.
3. Use of calculator is not permitted.
Section A (Q1 to Q4 carry one mark each)
Q.1)
How many solutions are there if the lines l1and l2 are parallel?
Q.2)
If ∆ ABC ~ ∆ DEF, BC = 3 cm EF =4 cm and area of ∆ ABC = 54 cm 2 then find the area of ∆ DEF .
Q 3) If Sin θ = 3/5 then find the value of Cos θ
Q.4)
Find the median of 30, 5, 2, 22, 14, 26, 10 .
Section B (Q5 to Q10 carry 2 marks each)
Q.5) Find the quadratic polynomial, the sum & the product of whose zeros are 3 & 2 respectively.
Q.6) . At a certain time in a deer park, the number of heads and the number of legs of deer and human
visitors were counted and it was found there were 39 heads & 132 legs.Find the number of deer and
human visitors in the park.
Q.7). A ladder 10m long reaches a window 8m above the ground. Find the distance of the foot of the
ladder
form base of the wall.
Q.8)
Show that Cos 38o Cos 52o – Sin 38 o Sin 52 o = 0.
Q.9).
Find the mean of the following data
xi
10
15
20
25
30
fi
5
10
7
8
2
Q.10)Evaluate
tan 65
cot 25
Section C(Q11 to Q20 carry 3 marks each)
Q.11)
The HCF of two numbers is 4 and their LCM is 9696. If one number is 96. Find the other number.
Q.12)
If α,β are the zeros of the polynomial 2x2 – 4x + 5 find the value of α2 + β 2
Q.13)
Find the value of x and y for the pair of linear equations
7(y + 3) ─ 2(x + 2) = 14,
4 (y ─ 2) + 3(x ─ 3) = 2
Q.14) In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 =
7AB2
Q.15) ABC is an Isosceles triangles right angled at C. prove that AB 2 = 2AC2
Q.16)
Prove the identity
cos ec cot 2 1 cos
1 cos
Q.17)
If tanA+sinA = m and tanA-sinA = n, show that m2-n2 = 4√
Q.18)
The following distribution given the daily income of 50 workers of a factory
Daily income in Rs
100-120
120-140
140-160
160-180
180-200
No of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution and draw
its ogive.
Q.19)
Q.20)
A survey conducted on 20 house holds in a locality by a group of students resulted in the
following frequency table for the number of family members in a house hold. Find the mode of
the data.
Family size
1-3
3-5
5-7
7-9
9-11
No of families
7
8
2
2
1
Given sec =
13
Calculate all other trigonometric ratios.
12
Section D(Q21 to 31 carry 4 marks each)
Q.21). Obtain all other zeroes of 3x4 + 6x 3 – 2x 2 -10x – 5 if two of its zeroes are
5
5
and
3
3
Q.22) A fraction becomes 9 if 2 is added to both the numerator and the denominator. If 3 is added both
the
11
numerator and denominator its becomes 5 . Find the fraction.
6
Q.23)
Solve the pair of equation by reducing them to a pair of linear equation
6x + 3y = 6xy
2x +4y = 5xy
Q.24) The ratio of the areas of two similar triangles is equal to the square of the ratio of their
corresponding sides.
Q.25)
Two poles of heights 6m and 11m stand on a plane ground. If the distance between the feet of
the poles is 12m. Find the distance between their tops.
Q.26) If A ,B and C are interior angles of a triangle ABC the show that Sin( B+C) = Cos A
2
Q.27)
Evaluate
5 cos2 60o + 4 Sec2 30o - tan 45o
Sin2 30o + Cos2 30o
Q.28)
If the median of the distribution given below is 28.5 find the value of x & y
C.I
0-10
10-20
20-30
30-40
40-50
50-60
Total
fi
5
X
20
15
y
5
60
Q.29)
The following tables give production field per hectare of wheat of 100 forms of a village
2
Production in
(kg)
50-55
55-60
60-65
65-70
70-75
75-80
No. of forms
2
8
12
24
38
16
Change the distribution to a more than type distribution and draw its ogive
Q.30) An army contingent of 616 members is to march behind an army band of 32 members in a
parade on
the occasion of republic day. The two groups are to march in the same number of column:a)
b)
c)
What is the maximum number of column in which they can march?
Which mathematical concept is used in the above problem?
What is the importance of an army for any country?
Use Euclid's division lemma to show that the square of any positive integer is either
of form 3m or 3m + 1 for some integer m.
Q.31)
SOLUTION SAMPLE PAPER –3 (2015-16) (SA I)
Mathematics
CLASS : X
Section A
Q.1)
No Solution
Q.2)
=>
ar ∆ ABC = (BC)2
ar ∆DEF
(EF)2
54
=
9
1 mark
ar ∆ DEF
16
ar∆ DEF= (54 x 16 )/9
Q.3)
Q.4)
Option (a) Sin θ = 3/5
BC = 4 ( by Pythagoras thm)
Cos θ = 4/5
=
96 cm2
3
1 mark
5
4
1 mark
Option (a) Arranging the data in increasing order
2, 5, 10, 14, 22, 26, 30
n = 7 (odd)
Median = (n + 1)
2
term
= 4th term =14
1 mark
Section B
Q.5) Sum of zeros= -3
Product of zeros = 2
1 mark
Quadratic polynomial in x= x2 –(sum of zeros) x+ product of zeros.
= x2 + 3x+ 2.
1mark
Q.6)
Let the no. of deers be x
And no. of humans be y
ATQ :
x + y = 39 ---- (1)
4 x + 2 y = 132 ----- (2)
1 mark
Multiply (1) by 4 and subtracting equation (2) from equation (1)
On solving, we get …
x = 27 and y= 12
1 mark
Q.7)
Let AB be the ladder, CB be distance of foot of the ladder from the wall then in right
angled ∆ABC
Therefore, By Pythagoras Theorem
A
(AB)2 = (AC)2 + (BC)2
1 mark
(10)2 = AC2 + 8 2
100 = 64 + AC2
AC2 = 36,
C
B
AC = 6
The foot of the ladder is at a distance of 6m from the base of the wall.
Q.8)
Cos 38o x Cos (90o- 52o) – Sin 38 o x Sin ( 90o- 52 o )
1mark
1mark
Cos 38o x Sin38o – sin 38o x Cos 38o
=0
1mark
Q.9)
xi
fi
fi xi
10
5
50
10
150
20
7
140
25
8
200
30
2
60
15
∑fi =32
∑fi xi =
600
1mark for correct table
Mean = ∑fi xi
=
∑fi
600
= 18.75
1 mark
32
Q.10) We know cot A= tan(90° - A)
So cot 25° = tan (90° - 25°) = tan 65°
1 mark
Therefore tan 65° / cot 25° = tan 65° / tan 65° = 1
1 mark
Section C
Q.11)
One number X second no = LCM X HCF
(1)
96 X 2nd no. = 9696 X 4
(1)
2nd no = 404
(1)
Q.12)
p (x) = 2 x2 – 4 x + 5
a=2, b= -4 & c=5
α+β=
2
αβ= =
1 mark
α2 + β2 = (α + β)2 – 2 α β
1 mark
2
2
2
Substitute then we get, α + β = 2 -2x = 4-5 = -1
Q.13) 7(y + 3) – 2 (x+ 2) = 14 --------- (1)
4(y– 2) + 3(x – 3) = 2 ----------(2)
From (1) 7y +21 – 2x – 4 = 14
On solving, we will get….
2x – 7y – 3 = 0 ------------- (3)
1 mark
From (2) 4y – 8 + 3x -9 =2
On solving, we will get….
3x + 4y – 19 =0 ----------------- (4)
1 mark
2x– 7y – 3 =0 ------------- (3)
3x + 4y – 19 =0------------- (4)
Multiplying equations 3 & 4 by 3 & 2 respectively ,we get
6x– 21y – 9 =0 ------------- (5)
6x + 8y – 38 =0------------- (6)
1 mark
Subtracting equation 5 from6,we get
-29y +29=0
-29y = -29
y=1
Putting y=1 in equation 3,we get
2x-10 =0
x =5
x = 5 and y = 1
Q 14
Answer
1 mark
1 mark (figure)
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
BE = EC = BC/2 = a/2
And, AE = a√3/2
Given that, BD = 1/3BC
BD = a/3
DE = BE - BD = a/2 - a/3 = a/6
Applying Pythagoras theorem in ΔADE, we get
AD2 = AE2 + DE2
9 AD2 = 7 AB2
1 mark
1 mark
Q.15)
ABC is a isosceles right triangle right angle at C
(AB)2 = (BC)2 + (AC)2
(AB)2 = (AC)2 + (AC)2
AB2 = 2AC2
1 mark
∙* BC = AC+
2 mark
Q.16)
LHS
(Cosec Ө - cot Ө) 2
=
( 1
-
Sin Ө
=
Cos Ө)2
Sin Ө
1mark
- Cos Ө)2
(1
Sin 2Ө
=
(1 - Cos Ө)( 1- Cos Ө)
Sin 2Ө
=
(1 - Cos Ө)( 1- Cos Ө)
1mark
1- Cos2 Ө
=
(1 - Cos Ө)( 1- Cos Ө)
(1 – Cos Ө) (1 + Cos Ө)
=
(1 – Cos Ө)
(1 + Cos Ө)
1 mark
= RHS
Q.17) Ans: tanA + SinA = m tanA – SinA = n.
m2-n2= (tanA + SinA)2-(tanA - SinA)2
= 4 tanA SinA
= √
√
= √
√
= tan 2 A tan 2 A. cos 2 A
=4 tan 2 A Sin 2 A
1 mark
1 mark
=4 (tan A sin A)(tan A sin A)
m2 – n2 = 4 mn
Q.18)
1 mark
Converting the given distribution to less than type cumulative frequency distribution, we get,
Daily income in RS
Cumulative frequency
Less than 120
12
Less than 140
26
Less than 160
34
Less than 180
40
Less than 200
50
1+2(for correct graph)
mark
Let us now plot the points corresponding to the ordered pairs (120, 12), ( 140, 26), (160, 34),
(180, 4),
( 200, 50) on a graph paper and join them by a free hand smooth curve
Q.19)
Here the maximum class frequency is 8 and the class corresponding to this frequency is 3- 5, so
the modal class is 3-5
1mark
Lower limit (l) of modal class = 3
Class size (h) = 2
Frequency (f1) = 8
Frequency (f0) =7
Frequency f2 = 2
Mode = l
f1 f 0
h
2 f1 f 0 f 2
1mark
Mode = 3.286
1 mark
Q.20) Given secѲ = 13/12
Finding all trigonometric ratios & value of third side.
Section D
Q.21)
So x
x2
Since two zeros are
5
5
and
3
3
5
5
and x
are the factors of the given polynomial
3
3
5
is a factor then
3
½ mark each
( 3 x2 – 5 ) is a also factor of the given polynomial
Applying the division algorithm to the given polynomial
1mark
x2 +2x +1
3x2 – 5
3x4 + 6x 3 – 2x 2 -10x – 5
3x4
– 5x 2
+
6x 3
+ 3x 2 - 10 x
6x 3
-10 x
-
+
3x 2 - 5
3x 2 - 5
-
+
0
3x4 + 6x 3 – 2x 2 -10x – 5
= (3x2 – 5) (x2 +2x +1)
= x2 +2x +1
=
(x +1)(x + 1)
So its other zeros are -1 and -1
Q.22)
2 marks for division
1mark for correct zeroes
let the fraction be = x
y
Then according to the given condition, we have
X+2 = 9
Y+2
1 mark
11
and x + 3 =
y+3
5
1mark
6
or 11 x + 22 = 9y + 18 and 6x +18 = 5y +15
11x – 9y = 18 -22 and 6x – 5y =15-18
11x – 9y = -4 --------------I
6x – 5y = -3 ---------------II
1 mark
Solve it by any algebraic method and get x = 7 and y = 9
1 mark
Fraction = 7
9
Q.23)
On dividing each one of the given equation by x y we get
3 +6 = 6 and
x y
taking 1 = u
4+2=5
½ mark
x y
and 1 = v
3u + 6v = 6---------------I
x
1 mark
4u + 2v = 5 -------------II
Apply any algebraic method and solve
1 mark
y
u = 1, v = ½
Put u = 1 in =n I we get
1 mark
3 x 1+6v = 6
6v = 6 – 3
v=½
now u = 1
1
=
1
x
x=1
u=½
1 mark
=> y= 2
Hence the given system of equation has one solution x = 1 and y = 2
Q.24)
Given , to prove ,construction
Proof
----------------1 ½ mark
.
----------------2 ½ mark
Q.25) Let AB = 11 m, CD = 6m
1 mark
A
be the two poles such that BD = 12 m
11
Draw CE ┴AB and join AC
E
C
CE = DB = 12m
AE =AB – BE = AB – CD
1 mark
6
=11 -6 = 5m
In rt ∆ACE, By Pythagoras theorem
B
1 mark
D
We have AC2 = CE2 + AE2
AC2 = (12)2 + (5)2
= 144+25
= 169
AC = 13m
1 mark
Hence the distance between the tops of the two poles is 13 m
Q.26)
Since A, B and C are the interior angles of a ∆ABC therefore
∠A + ∠B + ∠C = 180
Or
A+B+C
1 mark
= 90
2
So,
B + C = 90-A
2
Sin ( B+C)
2
= Sin (90-A)
2
Sin ( B+C)
2
= Cos A
2
Q.27)
1mark
1 mark
1 mark
2
Evaluate
5 cos2 60o + 4 Sec2 30o - tan 45o
Sin2 30o + Cos2 30o
=
5 ( ½ )2 + 4 (2/√3)2 - (1)2
( ½ )2 + (√3/2)2
=
5 x 1 + 4 x 4 -1
2 mark
4
3
1 +3
4
4
= 5 + 16 -1
4
3
1 mark
4
4
= 15+64 -12
1 mark
12
=
67
12
Q.28)
CI
F
Cf
0-10
5
5
10-20
x
5+x
20-30
20
25 + x
30-40
15
40+x
40-50
Y
40+x+y
50-60
Here n=60
5
45+x+y
1 mark
=> n = 30
2
Median = 28.5
l= 20 , h=10
cf = 5+x
f= 20
n
cf
median = l 2
f
h
1 mark
28.5 = 20 + 30-(5+x) x 10
20
28.5 = 20 + 25-x
2
57 = 40 + 25 –x
=> 57 -65 = -x
=> x = 8
Also 45 + x + y = 60
2 mark
y=7
Q.29)
Converting the given distribution to a more than type distribution, we get
Production field (Kg)
Cumulative frequency
More than or equal to 50
100
More than or equal to 55
100-2= 98
More than or equal to 60
98 – 8 = 90
More than or equal to 65
90 -12 = 78
More than or equal to 70
78 – 24 = 54
More than or equal to 75
54 – 38 = 16
Now draw the o give by plotting the points (50, 100) (55,98) (60, 90) (65,78) (70,54) and (75, 16)
1 mark
on the graph paper and join them by a free hand smooth curve.
3
mark
Q.30) Given integers are 32 & 616 clearly 616>32.
Therefore, applying Euclid’s division lemma to 616 & 32 we get
616 = 32x19+8
Since the remainder is 8 which is not equal to zero.
So, applying the division lemma again, we get
32 = 8x4+0
Therefore Remainder=0
2 marks
b)
Hence the maximum number of column in which they can march =8
1 mark
c)
Any importance of army
1 mark
Q.31) Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
1 mark
1 mark
Or,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
2 marks
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
Sample Paper -4 (2015-16)
SUMMATIVE ASSESSMENT - 1
CLASS X
MATHEMATICS
Unit
Number
System
Algebra
(i)Polynomials
(ii)Linear
equations in
two varriables
Geometry
(i)Triangles
Trigonometry
(i)Introduction
to
trigonometry
Statistics
VSA
(1)
1(1)
BLUE PRINT
SA-I
SA-II
(2)
(3)
---6(2)
LA
(4)
4(1)
Total
1(1)
4(2)
6(2)
12(3)
23(8)
1(1)
2(1)
6(2)
8(2)
17(6)
----
4(2)
6(2)
12(3)
22(7)
1(1)
4(4)
2(1)
12(6)
6(2)
30(10)
8(2)
44(11)
17(6)
90(31)
11(4)
Class-X-Maths
Sample Paper:4 (2015-16)
SA 1
Time: 3hrs
M.M: 90
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 31 questions divided into four sections A, B, C, and D.
Section – A comprises of 4 questions of 1 mark each, Section – B comprises of 6 questions
of 2 marks each, Section – C comprises of 10 questions of 3 marks each and Section – D
comprises of 11 questions of 4 marks each.
3. Question numbers 1 to 4 in Section – A are multiple choice questions where you are to
select one correct option out of the given four.
4. Use of calculator is not permitted.
5. An additional 15 minutes has been allotted to read this question paper only.
Questions:
Q1. Triangle ABC is similar to triangle DEF and their areas are64 cm 2 and 121
cm2 respectively. If EF= 15.4 cm, then find BC.
Q2. The graph y= p(x) is shown below. How many zeroes does the polynomial
p(x) have?
Q3.What is the relation between Mode, Median and Mean.
Q4.What type of number 7 x 11 x 13 + 13+13 x 2 is?
SECTION-B
Q5.In figure, If AD
BC, then prove that AB2 + CD2 = AC2 + BD2
Q6.Find LCM and HCF of 180 and220 by fundamental theorem of Arithmetic.
Q7.Solve the following pair of linear equations.
3x + 4y = 10 and 2x - 2y = 2
Q8.Prove thatcos4 +sin4 +2 sin² cos² = 1.
Q9. If the mean of 6,4,7,p and 10 is 8, find the value of p.
Q10.If x + 2 is a factor of x3 + ax2 + x - 14, find the value of a?
SECTION -C
Q11.Find the mode of the following data:
Marks
0 - 10
10 - 20
20 - 30
No. of students 3
12
32
Q12.In
30 - 40
20
40 - 50
6
ABC, if AD is the median, then show that AB2 + AC2 = 2[AD2 + BD2].
Q13.Prove that:
Q14.The sum of the numerator and denominator of a fraction is 8. If 3 is added to
both the numerator and the denominator the fraction becomes . Find the fraction.
.
Q15.Prove that
7 is irrational.
.
Q16Find the median of the following data.
Class Interval 0-20
20-40
40-60
Frequency
7
8
12
60-80
10
80-100
8
100-120
5
Q17.In figure,
.
Q18.Without using trigonometric tables evaluate:
Q19.Use Euclid's division lemma to show that the square of any positive integer is
either of the form 3m or 3m + 1 for some integer m.
Q 20.For what values of a and b does the following pairs of linear equations have
an infinite number of solutions:
2x + 3y = 7; a (x + y) -b (x - y) = 3a + b – 2
SECTION-D
Q21.Use Euclid's division lemma to show that the cube of any positive integer is
either of the form 9m or 9m 1 for some integer m.
Q22.100 surnames were randomly picked up from a local telephone directory and
the frequency distribution of the number of letters in the English alphabets in the
surnames was obtained as follows:
No. of letters
1-4
4-7
7-10
10-13
13-16
16-19
No. of
6
30
40
16
4
4
surnames
What is the average length of a surname? Which average you will use? Justify.
Q23. Prove that: (cosec A - sin A) (sec A - cos A) (tan A + cot A) = 1
Q 24.State and prove basic proportionality theorem.
Q25.If tan + sin = m and tan - sin = n. Show that m2 - n2=
3
2
.
Q26.What must be subtracted from x - 6x - 15x + 80 so that the result is exactly
divisible by x2 + x - 12?
Q27.Solve graphically the pair of equations 2x + 3y = 11 and 2x - 4y = -24. Hence
find the value of coordinate of the vertices of the triangle so formed.
Q28.The mode of the following frequency distribution is 55. Find the value of f 1
and f2.
Class Interval 0 - 10
15 - 30
30 - 45
45 - 60 60 - 75 75 - 90 Total
Frequency
6
7
f1
15
10
f2
Q29. Prove that the ratio of areas of two similar triangles is equal to ratio of square
of their corresponding sides.
Q30. Prove that :
-
=
-
Q31*.Ankur and Vinay are friends, their ages differ by 2 years. Ankur‟s father
is1 times as old as Ankur and Vinayand Vinay‟sisteris half the age of Ankur and
Vinay.The ages of Ankur‟s father and Vinay‟sister differ by 30 . Find their ages.
If your best friend has got leg injury and has not able to attend the school for 10
days. What will you do?
51
Sample paper -4
Answer Key
SA1
Class X
SECTION-A
1]
(1Mark)
2] Zeroes of a polynomial are the x- coordinates of the points where its graph crosses or touches
the X- axis. Graph of y = p(x) intersects the X-axis at 4 points. Therefore, the polynomial
p(x) has 4 zeroes
(1Mark)
3] 3 Median = Mode + 2 Mean(1Mark)
4] Composite Number.(1Mark)
SECTION -B
5] In
In
ADC, AD2 = AC2 - CD2
………… (1)
ABD, AD2 = AB2 - DB2 ……… (2)
[ By Pythagoras theorem]
[ By Pythagoras theorem]
(1Mark)
From (1) and (2)
AB2 - BD2 = AC2 - CD2
AB2 + CD2 = AC2 + BD2
(1 Mark)
6]
180 = 22 32 5
Mark)
220 = 22 5
Mark)
(½
11
(½
LCM = 22 32 5 11 = 1980
(½ Mark)
HCF = 22 5 = 20
(½ Mark)
7] 3x + 4y = 10
……. (1)
x–y= 1
……….(2)
multiplying Eq (2) by 4 we get
3x + 4y= 10
……….(3)
4x - 4y = 4
[By adding (1) and (3)]
x=2
x-y=1
(1 Mark)
1/2
y=x-1=2-1=1
[ By putting x=2 in eq(2)]
(1 Mark)
The solution x = 2, y = 1
8] We know that sin2 + cos2 =1 Taking Square of both sides (sin2 + cos2 )2 =12
(1Mark)
(sin2 )2 + (cos2)2 + 2 sin2 .cos2)= 1
Therefore, sin4 + cos4+ 2sin2cos2=1
9] 6 4 7 p 10
8
5
27+p = 40
p = 13(1Mark)
(1Mark)
(1Mark)
10] Let p(x) = x3 + ax2 + x - 14
g(x) = x + 2
is a factor of p(x)
p(-2) = 0
so, p(-2)
(1Mark)
= -8 + 4a - 2 - 14 = 0
= 4a = -24
(1Mark)
= a = -6
SECTION _C
11. Modal Class = 20-30, l = 20, f0 = 12, f1 = 32, f2 = 20, h = 10 (1Mark)
Mode =
(1Mark)
=
= 20+6.25 = 26.25
(1Mark)
12]
Given:
Construction - draw AE
BC
(1Mark)
In right triangle AEB and AEC
AB2 + AC2 = AE2 + BE2 + EC2 + AE2
(1Mark)
= 2AE2 + (BD - ED)2 + (ED + DC)2
= 2AE2 + 2ED2 + BD2 + DC2
AB2 + AC2 = 2AE2 + 2ED2 + 2BD2
(½ Mark)
= 2 [AE2 + ED2] + 2BD2
= 2 (AD2 + BD2)
(½ Mark)
13]
LHS =
(1Mark)
(1Mark)
=
= 2 cosec A = RHS(1Mark)
14]
Let the fraction be:
According to the question
x+y=8
(1Mark)
(1Mark)
On solving we get x = 3, y = 5
The fraction is
(1Mark)
OR
Let the tens and units digits of the number be x and y respectively then
(1Mark)
7(10x + y) = 4 (10y + x)
70x + 7y = 40y + 4x
66x = 33y
(1Mark)
y = 2x
Also, y - x = 3
On solving, x = 3, y = 6
Number = 36. (1Mark)
15] Let
7 be a rational number
Let
7=
(p and q are coprime integers with q 0)
7q =p
7q2 = p2
7 divides p2
hence 7 divides p.
(i)
(1Mark)
Let p = 7c
7q2 = 49 c2
Or q2 = 7c2
7 divides q2(1Mark)
Hence 7 divides q
(ii)
From (i) and (ii) p and q have a common factor 7 which contradicts our assumption.
Hence
7 is irrational.
(1Mark)
16] Class Interval
Frequency
0 - 20
7
7
20 - 40
8
15
40 - 60
12
27
60 - 80
10
37
80 - 100
8
45
100 - 120
5
50
Total
50
Median class 40 - 60
= 40, f = 12 CF = 15 h = 20
Median =
+
(1Mark)
(1Mark)
= 40 +
= 40 +
= 40 +
= 56.7
(1Mark)
17] In
ABC and
BAC =
C=
DAC
ADC
(1Mark)
C
(1Mark)
(1Mark)
18]
(1Mark)
(1Mark)
19] p(x) = x3 - 3x2 + x + 2
g(x) = ?
q(x) = x - 2
r(x) = -2x + 4
p(x) = g(x) + r(x)
x3 - 3x2 + x + 2 = g(x) (x - 2) - 2x + 4
(1Mark)
(1Mark)
g(x) =
=
= x2 - x + 1
20] The system has infinitely many solutions:
(1Mark)
(½ Mark)
(1Mark)
Equating (1) and (2), we get a = 5b
Equating (2) and (3), we get 2a - 4b = 6
(1Mark)
On solving, we get b = 1 and a = 5.
(½ Mark)
SECTION -D
21] If a and b are any two positive integers. Then a = bq + r, 0
1, 2 Therefore, a = 3q or a = 3q + 1 or a = 3q + 2
If a3 = (3q)3 = 27q3 = 9(3q3) = 9m where m = 3q3
r <b Let b = 3 Therefore r = 0,
(1Mark)
(1Mark)
If a = 3q + 1
a3 = 27q3 +1+9q(3q+1) = 27q3+ 27q2+ 9q+1= 9(3q3+3q2+q) + 1 = 9m+1 where m =
3q3+3q2+q (1Mark)
if a = 3q + 2
a3 = 27q3 + 8+18q(3q+2) = 27q3+54q2+36q+ 8= 27q3+54q2+36q+9-1,9(3q3+6q2+4q+1)1=9m-1
where m=3q3+6q2+4q+1
(1Mark)
Therefore, the cube of any positive integer is either of the form 9m,9m + 1 or 9m- 1.
22] C.I
1-4
4-7
7-10
10-13
13-16
16-19
fi
6
30
40
16
4
4
xi
2.5
5.5
8.5
11.5
14.5
17.5
Average most suitable here is the Mode because we are interested in knowing the length of
surname for maximum no. of people
Since the maximum frequency is 40 and it lies in the class interval 7-10.(2 Mark)
Therefore, modal class = 7-10
?= 7, h=3, f0=30 , f1=40 , f2=16
Mode =
(1Mark)
= 7+
= 7 +.88 = 7.88 years(approx.)
(1Mark)
23]
LHS=(cosec A - sin A) (sec A - cos A) (tan A + cot A)
(1Mark)
=
(1Mark)
=
cos 2 A sin 2 A
1
sin A
cos A cos A.sin A
(1Mark)
=1(1Mark)
= RHS
24] If a line drawn parallel one side of a triangle intersect other two sides at distinct points then
it divides other two sides in same ratio.
Given:A triangle ABC in which a line parallel to side BC intersects other two sides
AB and AC at D and E respectively (see fig.)
To prove that
=
Construction: Let us join BE and CD and then draw DM
Mark)
AC and EN
AB.
(1½
Proof:
Now, area of ∆ADE = x base x height =
x AD x EN
Note that ∆BDE and ∆DEC are on the same base DE and between the same parallels BC and
DE.
So, ar(BDE) = ar(DEG)
Mark)
(1½
Therefore, from (1), (2) and (3), we have :
Hence proved.
(1Mark)
=
.
Q25.m2 - n2 = (m + n) (m - n)
[tanᶿ +sinᶿ - (tanᶿ -sin ᶿ)]
= [tan
= 2 tan
. 2 sin
= 4 tan
. sin
(1Mark)
(1Mark)
=4
(1Mark)
=4
= 4 √(sec2 .sin2 - sin2 )
(as sec2 .sin2 =
x sin2
=
= tan2 )
(tan2 - sin2 = ( tan +sin )(tan – sin )=mn)
=4
(1Mark)
=4
26] Let p(x) = x3 -6x2- 15x + 80
Let say that we subtracted ax + b so that it is exactly divisible by
x2 + x - 12
s(x) = x3- 6x2- 15x + 80 - (ax + b)
= x3 - 6x2 - (15 + a)x + (80 - b) (1Mark)
Dividend = Divisor x Quotient + Remainder
But remainder = 0
Dividend = Divisor x Quotient
s(x) = (x2 + x -12) x quotient
s(x) = x3 - 6x2- (15 + a)x + (80 - b)
(1Mark)
x (x2 + x - 12) - 7(x2 + x - 12)
= x2 + x2- 7x2- 12x - 7x + 84
= x2 - 6x2- 19x + 84 (1Mark)
Hence, x2 - 6x2 - 19x + 84 = x2 - 6x2 - (15 + a)x + (80 - b)
-15 - a = - 19
and
-a = -19+15
a = +4
80 - b = 84-b = 84 - 80
b = -4
Hence if in p(x) we subtracted 4x - 4 then it is exactly divisible by
x2 + x -12. (1Mark)
27] We have to solve the pair of equations graphically
2x + 3y = 11 … (1)
2x – 4y = -24 … (2)
For (1)
X
1
4
-2
y
3
1
5
(1Mark)
For (2)
X
-12
0
-10
y
0
6
1
(1Mark)
point of intersection x = -2, = 5
The triangle formed is shaded as
ABC coordinates are
A (-2,5) B (-12,0) C(5.5,0).(2Mark)
28]
Class Interval
Frequency
0 – 15
6
15 – 30
7
30 – 45
f1
45 – 60
15
60 – 75
10
75 – 90
f2
Total
51
Mode = 55 (Given)
Modal Class 45 - 60
(1Mark)
= 45, fo = 10 and f1 = 15
f2 = 10 h = 15
38 + f1 + f2 = 51
f1 + f2 = 51 - 38
f1 + f2 = 13
…(1)
55 = 45 +
(1Mark)
(1Mark)
10 =
200 - f1 = 225 - 15f1
5f1 = 25
f1 = 5
f1+f2=13
f2=13-5=8
The missing frequencies are 5 and 8.
(1Mark)
29]
Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of
their corresponding sides.
Given: ∆ABC ~ ∆PQR
To Prove:
AB 2 BC 2 AC 2
arABC
arPQR PQ 2 QR 2 PR 2
Construction: Draw AD?BC and PS?QR
Proof:
(
)
(
)
=
=
(1Mark)
(1Mark)
∆ADB ~ ∆PSQ (AA) Therefore,
=
But ∆ABC ~ ∆PQR
Therefore,
(1Mark)
(iii)
Therefore,
… (iv)
Therefore,
(1Mark)
From (iii)
30]
LHS =
(1Mark)
(1Mark)
RHS =
(1Mark)
= tan
(1Mark)
Hence, LHS = RHS
31. Let present ages of two friends Vinay and Ankur be x and y years respectively By
supposing x>y
x-y = 2….(1)
NowAnkur‟s father is 1 times as old as Vinay and Ankur
Ankur‟s father =
(x+y) .,…..(2)(1Mark)
Again Vinay‟s sister is half the age of Vinay and Ankur
Vinay‟s sisterage = (x+y) ………(3)(1Mark)
Nowdifference between theages of Ankur‟s father and Vinay‟s sister is 30
(x+y) - (x+y) =30
y+ x = 30 ………..(4)
adding (1) and (4)
2x = 32 or x=16
From (1) put x=16 we get y = 14(1Mark)
Hence ages of Ankur and Vinay are 14 and 16 years respectively.
Ans: Ring him up and inform all the work being done in school daily
(1Mark)
TERM -I
VALUE BASED QUESTIONS
Q1. Mohit and Sahil are driving on two roads represented by the equation 2x+3y= 7 and
4x + 6y =12.
They drive within the speed limit.
(a) Will they meet at some point? Justify your answer.
(b) What value is depicted by Mohit and Sahil?
Solution: (i) They will never meet as the equations have no solution.
(ii)They are following the traffic rules.
Q2.A test contains „true‟ or „false‟ questions. One mark is awarded for every correct answer
while ¼ mark is deducted for every wrong answer. A student knew answers to some of the
questions. Rest of the questions he attempted by cheating. He answered 120 questions and
got 90 marks. If answer to all questions he attempted by cheating were wrong, then how
many questions did he answer correctly? How the habit of cheating will effect his character
building.
Solution: 96 questions.
Self respect, trust, personal integrity, wastage of time, reputation
Q3.A man wished to donate some money to a group of poor people, he decided to give Rs.
120 to each person and found that he fell short of Rs. 60, when he wanted to give to all the
people present. He therefore, distributed Rs. 90 to each person and found that Rs 90 were left
over. How much money did he have and how many people were there? What value is
depicted by the man?
Solution. No of people= 5
Total money= 540
Values- Peace of mind, care, sympathy
Q4.A trust has Rs. 30,000. It is invested in two different types of bonds. The first bond pays
5% interest per annum which will be given to orphanage and second bond pays 7% interest
per annum which will be given to an NGO for cancer aid. Trust obtains an annual interest of
Rs 1800.
(a) How much amount is invested in each type of bond?
(b) Which value is depicted by trust?
Solution: Rs. 15000 in each bond.
Values depicted: Caring, showing concern
Q5. (i) Find the HCF of No of girls 861 and No of boys 1353 using Euclid‟s Division
Algorithm.
(ii) Deduce the LCM of 861 and 1353.
(iii) What do think about gender ratio nowdays?
Solution. (i) 123
(ii) 9471
(iii) The Gender ratio should be balance by saving girls.
Q6.(i) If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2±√ , find the other zeroes.
(ii) Deduce the other two zeroes are of opposite signs.
(iii) Meenu confesses that the sum of the absolute values of the other two zeros is greater
than their sum. Is she correct? If so, which value of Meenu is depicted by her confession?
(iv) Which mathematical concept has been covered in this problem?
Solution. (i) The other two zeroes of the given polynomial are 7 and -5
(ii) As 7 is positive and -5 is negative, so the other two zeroes are of opposite signs.
(iii) Meenu is correct.
The value „ wiseness „ is depicted by Meenu‟s confession.
(iv) The mathematical concept „polynomials‟ has been used in the solution.
Q7.(i) If secA + tan A = p, find the value of sec A- tanA
(ii) A student claims that the value of sinA is
.Is he correct? If so, which value is
depicted by his claim?
Solution.(i) secA- tan A=
(ii) sin A=
=
(
)
(
)
=
The student is correct.
So, the value „skilful‟ is depicted in his claim.
Q8.In figure, LM ||AB. If AL= x-3, AC=2X, BM=x-2 and BC= 2x+3, find the value of x.
(ii) Deduce the ratio
(iii) Deduce the ratio
(iv) Shalini asserts that the value of x is a perfect square. Is She correct? If so, which value of
Shalini is depicted by her assertion?
(v) Which mathematical concept has been covered in this problem?
C
L
A
Solution: (i) x= 9
(ii)
(iii)
M
B
(iv) She is correct. The value „ intelligence‟ is depicted by her assertion
(v) The mathematical concept „Triangle‟ has been covered in this problem.
Q9. Raj, on his birthday, distributed chocolates in an orphanage. He gave 2 chocolates to
each child and 20 chocolates to adults. Considering number of children as x and total number
of chocolates distributed is y, form the linear equation. If the total number of chocolates
distributed is 150, how many children are there in the orphanage. Explain the value depicted
by Raj in the question.
Solution: 2x+20=y ; y=150
Number of children= 65
Sympathetic/empathetic attitude of Raj towards orphan children is depicted here.
Q10. The percentage of salary that 10 households donate to an orphanage is given below: 5,
3, 10, 5, 2, 4, 7, 8, 1, 5 Find the mean, median and mode of the data. Also tell the values
depicted by the persons of these households.
Solution: Mean= 5
Median= 5
Mode= 5
These persons have shown their responsibility towards society. Also they have shown
sympathetic attitude towards orphans.
Q11 .A man hires a taxi to cover a certain distance. The fare is Rs 50 for first kilometre and
Rs 25 for subsequent kilometers. Taking total distance covered as x km and total fare as y:
( a). Write a linear equation for this.
(b). The man covers a distance of 10 km and gave Rs 300 to the driver. Driver said “It is not
the correct amount” and returned him the balance. Find the correct fare and the amount paid
back by the driver.
(c) Which value is depicted by the driver in the question?
Solution. (a) 25x-y+25=0
(b) Correct fare= Rs 275
(c) The values depicted by the driver in the question are honesty and truthfulness
Q12. (i) Find the value of p for which the arithmetic mean of the following distribution is 52:
Wages( in rupees)
Number of workers
10-20
5
20-30
3
30-40
4
40-50
P
50-60
2
60-70
6
70-80
13
(ii) Is p an even or an odd number?
(iii) Mithilesh thinks that the value p can be expressed as the sum of two prime numbers. Is
she correct? If so, which value of Mithilesh is depicted by her thought?
(iv) Which mathematical concept has been covered in this problem?
Solution: (i) p= 7
(ii) Odd
(iii) 7 can be expressed as the sum of two prime numbers.
The value „ Good Arithmetical Knowledge‟ is depicted by her thought.
(iv) The mathematical concept „statistics‟ has been covered in the solution
Q13.In a housing society, people decided to do rainwater harvesting. Rainwater is collected
in the underground tank at the rate of 30 cm3/sec. Taking volume of water collected in x
seconds as y cm3.
(a) Form a linear equation.
(b) Write it in standard form as ax + by + c = 0.
(c) Which values are promoted by the members of this society?
Solution: Rate at which rainwater is collected in the tank = 30 cm3/sec .
Time for which water is collected = x seconds.
Total amount of water collected = y cm3
(a) According to the given condition, linear equation formed is y = 30x
(b) The equation in standard form is 30x – y + 0 = 0
(c) Values promoted by the members of the society are environmental protection and
cooperation.
Q14.After every 6 months, price of petrol increases at the rate of Rs 4 per litre. Taking price
of petrol in December 2010 as x and present price of petrol as y,
(a) Form a linear equation showing the price of petrol in December 2014.
(b) Due to continuous rise in the price of petrol, people are more interesting in CNG whose
price is increasing at the rate of Rs 3 per litre in a year. Form a linear equation taking price of
CNG in December 2010 as „a‟ and in December 2014 as „b‟.
(c) Which value is depicted by using CNG over petrol?
Solution: (a) Price of petrol in December 2010 = x Price of petrol in December 2014 = y
Price of petrol increased in 1 year = 4 × 2 = Rs 8 Price of petrol increased in 4 years
(December 2010- December 2014) = 8 × 4 = Rs 32
Equation representing the price of petrol in December 2014 = y = x + 32
(b) Price of CNG in December 2010 = a Price of CNG in December 2014 =b Price of CNG
increased in 1 year = Rs 3 Price of CNG increased in 4 years (December 2010- December
2014) = 3 × 4 = Rs 12
Equation representing the price of CNG in December 2014 = b = a + 12
(c) The value depicted by using CNG over petrol is environmental protection.
Q15. In a seminar on the topic „Liberty and Equality‟ the number of participants in hindi,
social science and English are 60 , 84 and 108 respectively.
(i)
Find the minimum number of rooms required if in each room the same number of
participants are to be seated and all of them being from the same subject.
(ii)
Which mathematical concept has been used in this problem?
(iii) Which values are discussed in the above problem?
Solution: (i) No of rooms required= 21
(ii) HCF
(iii) Liberty and equality are the pay marks of democracy.
Q16. Some people of a society decorated their area with flags and tricolor ribbons on
republic day. The following data shows the number of persons in different age group who
participated in the decoration:
Age in
5-15
15-25
25-35
35-45
45-55
55-65
years
No. of
6
11
21
23
14
5
patients
(i) Find the Mode of above data.
(ii) What values do these persons possess?
Solution: (i) Mode= 36.8
(ii) National integrity, unity, beauty
