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Worksheets
SOLUTIONSSOLUTIONS
Class
Term 1 (April to September)
10
MathematicsMathematics
(v)
Unit-I : Number System
1. Real Numbers
Summative Assessment
Ø Worksheets 1 - 11 3 - 14
Formative Assessment
Ø Worksheet 12 14 - 14
Unit-II : Algebra
2. Polynomials
Summative Assessment
Ø Worksheets 12 - 23 15 - 29
Formative Assessment
Ø Worksheet 24 29 - 30
3. Pair of Linear Equations in two Variables
Summative Assessment
Ø Worksheets 25 - 37 31 - 50
Formative Assessment
Ø Worksheet 38 50 - 51
Unit-III : Geometry
4. Triangles
Summative Assessment
Ø Worksheets 39 - 52 52 - 71
Formative Assessment
Ø Worksheet 53 72 - 72
Unit-IV : Trigonometry
5. Introduction to Trigonometry and Trigonometric Identities
Summative Assessment
Ø Worksheets 54 - 76 73 - 101
Formative Assessment
Ø Worksheet 77 101 - 102
Unit-V : Statistics
6. Statistics
Summative Assessment
Ø Worksheets 78 - 95 103 - 128
Formative Assessment
Ø Worksheet 96 129 - 129
CONTENTS
P-3S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-1
SECTION A1. We know that
a × b = L.C.M. × H.C.F.= 200 × 5 = 1000 1
2. Since = ±16 4So it is a rational number. 1
SECTION B3. Given H.C.F. = 18
L.C.M. = ?First term = 306
Second term = 1314 1we know that
First × Second term = L.C.M. × H.C.F.306 × 1314 = L.C.M. × 18
L.C.M. = ×306 131418 = 22388 1
SECTION C4. First we will find out HCF (1530, 1365)
1530 = 1365 × 1 + 1651365 = 165 × 8 + 45165 = 45 × 3 + 3045 = 30 × 1 + 1530 = 15 × 2 + 0
Now HCF (1530, 1365) = 15Now lets find out HCF (1305, 15)
1305 = 15 × 87 + 0∴ HCF (1530, 1365, 1305) = 15 1
SECTION D5. (a) Maximum number of parallel rows of each class = HCF of 104 and 96 1
Now, 104 = 2 × 2 × 2 × 1396 = 2 × 2 × 2 × 12 1
Hence HCF (104, 96) = 8
Real Numbers
1CHAPTERTerm-I
Summative Assessment
Unit I
Number System
P-4 1–MRETT I C S X--M A T E M AH
(b) No. of the students of class X in 1 row = 104
8 = 13
No. of students of class IX in 1 row = 968 = 12 1
(c) To minimise the tendency of the students to copy and to teach them value of honesty. 1
SUMMATIVE ASSESSMENT WORKSHEET-2
SECTION A
1. 145 = 29 × 5Hence, number of prime factors of 145 is 2. 1
2. p = a3b2
q = ab3c2
Hence HCF of p and q = ab2. 1
3. Decimal expansion of 3 223
2 5 will be terminating 1
SECTION B
4.
(
(
(
(
1656 4025 2
–3312
713 1656 2
–1426
230 713 3
– 690
23 230 10
–230
0
Hence HFC (1656, 4025) = 23. 2
SECTION C
5. z = 2 × 17 = 34 ½y = 2 × 34 = 68 ½x = 2 × 68 = 136 ½
Yes, value of x can be found without finding value of y or z as x = 2 × 2 × 2 × 17, which are primefactors of x. 1½
SECTION D
6. Time required by 3 children to complete a card together = LCM of 10, 16, 20Now 10 = 2 × 5
16 = 2 × 2 × 2 × 2 120 = 2 × 2 × 5 1
So required LCM = 2 × 5 × 2 × 2 × 2 1These children possess following values :Caring respect for elders, creative and helpful. 1
P-5S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-3
SECTION A
1. A rational number can be expressed as a terminating decimal of the denominator has factors 2 or5 only. 1
2. We know thata × b = HCF × LCM1800 = 12 × LCM
LCM = 1800
12 = 150 1
SECTION B
3. If the number 4n, were to end with the digit zero, then it should be divisible by 5. 1But 4n = (22)n = 22n
⇒ Only prime in the factorization of 4n is 2.So by fundamental theorem of Arithmetic, there are no other primes in the factorisation of 4n.½⇒ 4n can never end with the digit zero. ½
SECTION C
4. Let us assume on the contrary that −4 3 2 is rational. Then, there exist co-prime possible integersa and b such that
−4 3 2 = ab 1
⇒ −4ab = 3 2
⇒−4
3b a
b = 2 1
⇒ 2 is rational [∵ a, b are integers −4
3b a
b is a rotational number]
This, contradicts the fact that 2 is irrational.
So, our assumption is wrong. Hence −4 3 2 is an irrational number. 1
SECTION D
5. Let the number of columns be x.x is the largest number, which should divide both 104 and 96
104 = 96 × 1 + 8 196 = 8 × 12 + 0 1
∴ HCF of 104 and 96 is 8 1Hence, 8 columns are required. 1
SUMMATIVE ASSESSMENT WORKSHEET-4
SECTION A
1.38 = 3
32
= ××
3
3 33 52 5
= 337510
= 0·375 1
2.×3 4
64232 5
= ×
×4 46243 22 5
= 412486
10Hence decimal expansion of the rational number will terminate after 4 places of decimal. 1
P-6 1–MRETT I C S X--M A T E M AH
SECTION B
3.3
255 867765
1
2102 255
2042
51 102102
0∴ Required HCF = 51. 1
SECTION C
4. Let 5 be a rational number.
∴ 5 = ab
, (a, b are co-prime integers and b ≠ 0.)
a = b 5
a2 = 5b2
⇒ 5 is a factor of a2 1⇒ 5 is a factor of aLet a = 5c, (c is some integer)⇒ 25c2 = 5b2
⇒ 5c2 = b2
⇒ 5 is a factor of b2
5 is a factor of b∴ 5 is a common factor of a, b
But this contradicts the fact that a, b are co-primes.
∴ 5 is irrational
Let 2 – 5 be rational 1
∴ 2 – 5 = a
⇒ 2 – a = 5
2 – a is rational, so is 5 .
But 5 is not rational ⇒ contradiction
∴ 2 – 5 is irrational. 1
SECTION D5. Let 2 be a rational number.
∴ 2 = ab
, (a, b are co-prime integers and b ≠ 0)
a = 2 b
Squaring, a2 = 2b2 1⇒ 2 divides a2
⇒ 2 divides aSo we can write a = 2c for some integer c, substitute for a, 2b2 = 4c2, b2 = 2c2
This means 2 divides b2, so 2 divides b.∴ a and b have at least ‘2’ as a common factor. 1But this contradicts, that a, b have no common factors other than 1.
P-7S O L U T I O N S
∴ Our supposition is wrong. Hence, 2 is irrational.
Let32
= a, where a is a rational
a 2 = 3
2 = 3a
1
3a
is rational but 2 is not rational.
∴ Our supposition is wrong.
∴ 32
is irrational. 1
SUMMATIVE ASSESSMENT WORKSHEET - 5
SECTION A
1. According to Euclid’s division Lemma for any positive interger and 3 there exist unique integers qand r such that a = 3q + r, where r must satisfy 0 ≤ r < 3. 1
2. 1 = 1 × 1, 2 = 1 × 2, 3 = 1 × 3, 4 = 2 × 2, 5 = 1 × 56 = 2 × 3, 7 = 1 × 7, 8 = 2 × 2 × 2, 9 = 3 × 3, 10 = 2 × 5.Hence the least number that, divisible by all the numbers from 1 to 10 i.e., L.C.M.
= 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520. 1
SECTION B
3. By Euclid’s division algorithma = bq + r
Take b = 4 ∴ r = 0, 1, 2, 3 1So, a = 4q, 4q + 1, 4q + 2, 4q + 3Clearly, a = 4q, 4q + 2 are even, as they are divisible by 2.But 4q + 1, 4q + 3 are odd, as they are not divisible by 2.∴ any positive odd integer is of the form (4q + 1) or (4q + 3). 1
SECTION C
4. (i) The number of rooms will be minimum if each room accomodates maximum number ofparticipants. Since in each room the same number of participants are to be seated and all of themmust be of the same subject. Therefore, the number of paricipants in each room must be the HCFof 60, 84 and 108. The prime factorisations of 60, 84 and 108 are as under
60 = 22 × 3 × 584 = 22 × 3 × 7
108 = 22 × 33
Hence, HCF = 22 × 3 = 12Therefore, in each room 12 participants can be seated. 1
∴ Number of rooms required = Total number of participants
12
= 60 84 108
12+ +
= 25212
= 21. 1
(ii) HCF of numbers. ½
(iii) Liberty and equality are the pay marks of democracy. ½
P-8 1–MRETT I C S X--M A T E M AH
SECTION D
5. Let a = 4q + r, 0 ≤ r < 4
⇒ a = 4q, 4q + 1, 4q + 2 or 4q + 3 1
Case I. a2 = (4q)2 = 16q2 = 4(4q2) = 4m, m = 4q2 ½
Case II. a2 = (4q + 1)2 = 16q2 + 8q + 1 = 4(4q2 + 2q) + 1
= 4m + 1, where m = 4q2 + 2q ½
Cases III. a2 = (4q + 2)2 = 16q2 + 16q + 4
= 4(4q2 + 4q + 1) = 4m, where m = 4q2 + 4q + 1 ½
Cases IV. a2 = (4q + 3)2 = 16q2 + 24q + 9
= 4(4q2 + 6q + 2) + 1
= 4m +1, where m = 4q2 + 6q + 2 ½
From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form 4m or 4m + 1.1
SUMMATIVE ASSESSMENT WORKSHEET-6
SECTION A
1. The smallest prime number = 2and the smallest composite number = 4.Hence required HCF (4, 2) = 2 1
2.35 has terminating decimal expansion. 1
3. If q is some integer, then any positive odd integer is of the form 6q + 1. 1
SECTION B
4.
½
½
½
∴ Composite number x = 6762. ½
SECTION C
5. Let x = 0·3178178178... 110000 x = 3178.178178...
10 x = 3.178178...Subtracting, 9990 x = 3175 1
x = 31759990
6351998
= . 1
6762
2 3381
3
7
7
1127
161
23
= 7 × 161
= 1617
2 × 3381 =
P-9S O L U T I O N S
SECTION D
6. Let 2 be a rational number
So 2 = ,ab
[∵ a, b are co-prime integers and b ≠ 0]
⇒ a = 2bSquaring a2 = 2b2 ½⇒ 2 divides a2
⇒ 2 divides aso we can write a = 2c for same integer c, substitute for a, 2b2 = 4c2 ⇒ b2 = 2c2. ½This means 2 divides b2, so 2 divides b.∴ a and b have atleast ‘2’ as a common factor but this contradicts, that a, b have no commonfactors other than 1.∴ Our supposition is wrong. ½ Hence 2 is irrational.
Let 5 3 2+ = pq [HCF (p, q) = 1]
2 = 5
3p q
q−
2 is irrational but 5
3p q
q−
is a rational. ½
Irrational ≠ rational ⇒ Contradiction.∴ 5 3 2+ is an irrational number. 1
SUMMATIVE ASSESSMENT WORKSHEET - 7
SECTION A
1.3 3
4 4 4 46 6 6 2 6 2
·1250 2 5 2 5 10
× ×= = =× ×
Hence decimal expansion of the rational number will terminate after 4 places of decimal. 1
2. The decimal expansion of 15 1013
2 5× is terminating. 1
SECTION B
3. 240 = 228 × 1 + 12 1228 = 12 × 19 + 0
⇒ HCF of 240 and 228 = 12. 1
SECTION C
4. (i) In order to arrange the books as required, we have to find the largest number that divides 96,240 and 336 exactly, clearly, such a number is their HCF.We have,
96 = 25 × 3240 = 24 × 3 × 5
and 336 = 24 × 3× 7∴ HCF of 96, 240 and 336 is 24 × 3 = 48
P-10 1–MRETT I C S X--M A T E M AH
So, there must be 48 books in each stack. 1
∴ Number of stacks of English books = 9648 = 2
Number of stacks of Hindi books = 24048 = 5
Number of stacks of Sociology books = 33648 = 7 1
(ii) HCF of numbers. ½(iii) Cleanliness has been discussed in this question, it is a good habit that leads to good health.
½
SECTION D
5. Let x be any positive integer, then it is of the form 3q or 3q + 1 or 3q + 2.Squaring, we get
(3q)2 = 9q2 = 3 × 3q2 = 3m, m = 2q2
(3q + 1) = 9q2 + 6q + 1 1= 3(3q2 + 2q) + 1= 3m + 1, m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4 1= 9q2 + 12q + 3 + 1= 3(3q2 + 4q + 1) + 1= 3m + 1, m = 3q2 + 4q + 1 1
⇒ Square of any positive integer is of the form 3m or 3m + 1 for some integer m. 1
SUMMATIVE ASSESSMENT WORKSHEET-8
SECTION A
1.3 3
3 3 3 3189 189 189 2 189 2
·125 5 5 2 (10)
× ×= = =
×
The decimal expansion of 189125 will terminate after 3 places of decimal. 1
2.93
1500 = 2 2 2 393 31
3 5 2 5 2 5=
× × × ×Hence, decimal representation is terminating. 1
3. HCF of 33 × 5 and 32 × 52 = 32 × 5 = 9 × 5 = 45. 1
SECTION B
4. 90 = 2 × 32 × 5144 = 24 × 32
HCF = 2 × 32 = 18 1LCM = 24 × 32 × 5 = 720. 1
SECTION C
5. Let n be any +ve integer, thenn = 3q + r, r = 0, 1, 2 ½n = 3q or 3q + 1 or 3q + 2 ½
P-11S O L U T I O N S
Case I.When n = 3q, which is divisible by 3 ½
n + 2 = 3q + 2, which is not divisible by 3n + 4 = 3q + 4, which is not divisible by 3
Case II.When n = 3q + 1, which is not divisible by 3 ½
n + 2 = 3q + 1 + 2 = 3q + 3, which is divisible by 3n + 4 = 3q + 1 + 4 = 3q + 5, which is not divisible by 3
Case III.When n = 3q + 2, which is not divisble by 3 ½
n + 2 = 3q + 2 + 2 = 3q + 4, which is not divisible by 3n + 4 = 3q + 2 + 4 = 3q + 6, which is divisible by 3 ½
Case I, II and III ⇒ One and only one out of n, n + 2 or n + 4 is divisible by 3.
SECTION D
6. The greatest number of cartons is the HCF of 144 and 90 1144 = 24 × 32
90 = 2 × 32 × 5 1HCF = 2 × 32 = 18 1
∴ The greatest number of cartons = 18. 1
SUMMATIVE ASSESSMENT WORKSHEET-9
SECTION A
1. The product of a non-zero rational number and an irrational number is always irrational. 12. The product of two irrational numbers is always a non zero number. 1
SECTION B
3.
a = 90093003
= 3 ½
b = 1001143
= 7 ½
Since 143 = 11 × 13, so c = 11 or 13 ½d = 13 or 11. ½
18018
2 9009
a
3
b
c
3003
1001
143
d
P-12 1–MRETT I C S X--M A T E M AH
SECTION C
3. (i) Required number of mintues is the LCM of 18 and 12.18 = 2 × 32
and 12 = 22 × 3 1∴ LCM of 18 and 12 = 22 × 32 = 36Hence, Ravish and Priya will meet again at the starting point after 36 minutes. 1(ii) LCM of numbers 1(iii) Healthy competition is necessary for personal dvelopment and progress. 1
SECTION D
5. Any positive integer is of the form 2q or 2q + 1, for some integer q.∴ When n = 2q
n2 – n = (2q)2 – 2q = 2q(2q – 1) 1= 2m, when m = q(2q – 1)
Which is divisible by 2.When n = 2q + 1 1
n2 – n = (2q + 1) (2q + 1 – 1)= 2q (2q + 1)= 2m, when m = q(2q + 1) 1
Which is divisible by 2.Hence, n2 – n is divisible by 2 for every positive integer n. 1
SUMMATIVE ASSESSMENT WORKSHEET-10
SECTION A
1. (C) It is clear that 0·102003000..........is non-terminating non-repeating decimal, so it cannot be
expressed in the form of ·pq 1
2. (C) If n is an odd number, then (n2 – 1) is divisible by 8. 1
SECTION B
3. No.15 does not divide 175. 1LCM is exactly divisible by their HCF.∴ Two numbers cannot have their HCF as 15 and LCM as 175. 1
SECTION C
4. By Euclid’s division algorithm,510 = 92 × 5 + 5092 = 50 × 1 + 4250 = 42 × 1 + 842 = 8 × 5 + 2 18 = 2 × 4 + 0.
HCF (510, 92) = 292 = 22 × 23
510 = 2 × 3 × 5 × 17LCM = 22 × 23 × 3 × 5 × 17 = 23460
HCF × LCM = 2 × 23460 = 46920 1Product of 2 numbers = 510 × 92
⇒ HCF × LCM = Product of two numbers. 1
P-13S O L U T I O N S
SECTION D
5. Let 3 5− is a rational number
∴ 3 5− = ,pq 0q ≠
5 = 3 – pq =
3q pq−
∴ 3q p
q−
is rational number. 1
But 5 is a irrational number. 1
Since irrational number cannot be equal to rational number.∴ Our assumption is wrong. 1
∴ 3 5− is an irrational number. 1
SUMMATIVE ASSESSMENT WORKSHEET-11
SECTION A
1. Since LCM of a and b = p therefore LCM of 3a and 2b = 3 × 2 × p = 6p. 1
2. 5 2 7+ − is an irrational number. 1
3.32
is the rational number between 2 and 3. 1
SECTION B
4. (7 × 13 × 11) + 11 = 11 (7 × 13 + 1)= 11 (91 + 1)= 11 × 92 = 11 × 2 × 2 × 23 1
which is a composite number (more than one prime factors)and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 = 3 (7 × 6 × 5 × 4 × 2 × 1 + 1)
= 3 × (1681)= 3 × 41 × 41 1
which is a composite number (more than one prime factors)
SECTION C
5. We have 117 = 65 × 1 + 5265 = 52 × 1 + 1352 = 13 × 4 + 0 1
Hence, HCF = 13∴ 65m – 117 = 13⇒ 65m = 117 + 13 = 130
∴ m = 13065
= 2 1
Now, 65 = 13 × 5117 = 32 × 13
LCM = 13 × 5 × 32 = 585 1
P-14 1–MRETT I C S X--M A T E M AH
SECTION D
6. n3 – n = n(n2 – 1) = n(n + 1) (n – 1) = (n – 1) n(n + 1)= product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.Let a, a + 1, a + 2 be any three consecutive integers. ½Case I. If a = 3q.
a(a + 1) (a + 2) = 3q(3q + 1) (3q + 2)= 3q (even number, say 2r) = 6qr,
(∵ Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer)which is divisible by 6. 1Case II. If a = 3q + 1.∴ a(a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)= 6 r (q + 1), 1
which is divisible by 6.Case III. If a = 3q + 2.∴ a(a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q= 6r (say), 1
which is divisible by 6.Hence, the product of three consecutive integers is divisible by 6. ½
FORMATIVE ASSESSMENT WORKSHEET-12
Objective Type Questions
1. (D) 2. (C) 3. (B) 4. (B) 5. (B)
Oral Questions
1. 3 2 3 3 2 21323 1323
·(6 35 ) 2 3 5 7
=× × × ×
In the denominator 3 and 7 prime factors are involved hence the number is non-terminating.2. Product of numbers = H.C.F. × L.C.M.
H.C.F. = 480 672
3360×
= 96.
3. H.C.F. of 616 and 32 will be the answer616 = 2 × 2 × 2 × 11 × 732 = 2 × 2 × 2 × 2 × 2
H.C.F. = 2 × 2 × 2 = 8
Fill in the blanks1. Algorithm2. Rational3. Lemma4. Every composite number can be expressed as a product of primes and this decomposition is unique,
apart from the order in which the prime factors occur.5. Non-terminating.6. 2 and 57. Unique8. Euclid’s Division Algorithm9. Product.
●●
P-15S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-13
SECTION A
1. From the graph it is clear that the curve cuts the x-axis at four places, so number of zeroes is 4. 1
2. At the x-axis y = 0, so point on the x-axis is given by 2x – 5 = 0 ⇒ x = 52
1
So, required point is 5
, 0 ·2
SECTION B3. Since a and b are the zeroes of the polynomial, then
α + β = – 2Coefficient ofCoefficient of
xx
⇒ α + β = –1
1−
= 1 ...(1) 1
Given α – β = 9 ...(2)From (1) and (2) α = 5, β = – 4
αβ = 2Constant
Coefficient of xAgain α β = – k
(5) (– 4) = – k
⇒ k = 20 1
SECTION C3. f(x) = 4x2 + 4x – 3
f12
FHG
IKJ = 4
14
412
3FHG
IKJ + F
HGIKJ −
= 1 + 2 – 3 = 0
f −FHGIKJ
32 = 4
94
432
3FHG
IKJ + −FHG
IKJ −
= 9 – 6 – 3 = 0
Polynomials
2CHAPTERUnit II
Algebra
P-16 1–MRETT I C S X--M A T E M AH
∴12
32
,− are zeroes of polynomial 4x2 + 4x – 3 1
Sum of zeroes = 12
32
− = – 1 = −44
= –coeff. of coeff. of 2
xx
1
Product of zeroes = 1 3 32 2 4
æ öæ ö -ç ÷ç ÷- = =ç ÷ç ÷è ø è øconstant term
coeff. of 2x
∴Relation between zero and coeff. of polynomial is verfied. 1
SECTION D
2
2 4 3 2
4 3 2
2 2
3 2
2
2
2 3 53 4 2 3 12 28 20
2 6 8
3 4 28
3 9 12
5 16 20
5 15 20
+ ++ − + − + −
+ −− − +
− − +
+ −
+ −
+ −− − +
x xx x x x x x
x x x
x x x
x x x
x x
x x
xHence quotient = 2x2 + 3x + 5; and remainder = x 1
Hence, x should be substract from f (x) to make it exactly divisible by x2 + 3x – 4. 1
SUMMATIVE ASSESSMENT WORKSHEET-14
SECTION A
1. 2x2 + 5x + 1Since α and β are the zeroes of the polynomial, Therefore
(α + β) + αβ = – 52
12
+
= − +5 1
2 =
−42
= – 2. 1
2. Given α = – 5 and β = 4So, Polynomial = (x – α) (x – β)
= (x + 5) (x – 4)= x2 – 4x + 5x – 20= x2 + x – 20. 1
SECTION B
3. Sum of zeroes = a + b = –coeff. of coeff. of 2
xx
= – a
⇒ 2 a + b = 0 1
Product of zeroes = ab =constant
coeff. of 2x = b
⇒ a = 1 then b = – 2 1
5.
P-17S O L U T I O N S
SECTION C
4. (i) Now p(x) = 2x3 – 11x2 + 17x – 6p(2) = 2(2)3 – 11(2)2 + 17(2) – 6
= 16 – 44 + 34 – 6= 50 – 50 = 0 1
Hence 2 is the zero of p(x).(ii) Again p(3) = 2 (3)3 – 11(3)2 + 17(3) – 6
= 54 – 99 + 51 – 6= 105 – 105 = 0 1
Hence, 3 is the zero of p(x).
(iii) Again p
12 = 2
312
– 11
212
+ 17
12 – 6
= 14
– 114
+ 172
– 6
= 0 1
Hence, 12
is also the zero of p(x).
SECTION D
5. Let p(x) = x3 – 10x2 + 31x – 30
Since 2 and 3 are zeroes of p(x).
So (x – 2) and (x – 3) are the factors of given polynomial. 1
i.e., (x – 2) (x – 3) = x2 – 5x + 6 is a factor x – 5
x2 – 5x + 6) x3 – 10x2 + 31x – 30x3 – 5x2 + 6x
– + –
– 5x2 + 25x – 30– 5x2 + 25x – 30+ – +
0
Hence other zero is (x – 5) i.e., x = 5. 1
SUMMATIVE ASSESSMENT WORKSHEET-15
SECTION A
1. The maximum number of zeroes that a polynomial of degree 3 can have is 3. 1
2. If 1 is the zero of the polynomial x2 + kx – 5, then
x2 + kx – 5 = 0
⇒ (1)2 + k(1) – (5) = 0
⇒ 1 + k – 5 = 0
⇒ k – 4 = 0
⇒ k = 4. 1
1
P-18 1–MRETT I C S X--M A T E M AH
SECTION B
3. Let p(x) = 3 x2 – 8x + 4 3
= 3 x2 – 6x – 2x + 4 3 1
= 3 x (x – 2 3 ) – 2 (x – 2 3 )
= ( 3 x – 2) (x – 2 3 ) 1
SECTION C
4. p(x) = x2 + 7x + 9
q(x) = x + 2
r(x) = – 1
g(x) = ?
p(x) = g(x)q(x) + r(x) 1
x2 + 7x + 9 = g(x) (x + 2) – 1
g(x) = x x
x
2 7 102
+ ++
1
= ( )( )
( )x x
x+ +
+2 5
2g(x) = x + 5. 1
SECTION D
5. p(x) = 2x2 + 5x + k
Sum of zeroes = –coeff. of
coeff. of 2x
x
= α + β = −52
1
Product of zeroes = constant
coeff. of 2x
= αβ = k2
According to question,
α2 + β2 + αβ = 214
1
⇒ (α + β)2 – 2αβ + αβ = 214
[∵ (α + β)2 = α2 + β2 + 2αβ]
⇒−F
HGIKJ
52
2
– k2
= 214
⇒254
214
− = k2
1
∴ 1 = k2
⇒ k = 2
Hence, k = 2 1
P-19S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-16
SECTION A
1. Given α = 3 and β = – 3So, Polynomial = (x – α) (x – β)
= (x – 3 ) (x + 3 )= x2 – 3. 1
2. If 3 is the one zero of the polynomial 2x2 + kx – 15, then2x2 + kx – 15 = 0
2(3)2 + k(3) – 15 = 018 + 3k – 15 = 0
3k + 3 = 0⇒ k + 1 = 0 ⇒ k = – 1Now, Polynomial = 2x2 – x – 15⇒ = 2x2 – 6x + 5x – 15⇒ = 2x(x – 3) + 5(x – 3)⇒ = (2x + 5) (x – 3)
Hence, other zero is –52
. 1
SECTION B
3. ∵ f (x) = x2 – 2x= x(x – 2) 1= x = 0 or x = 2
Hence, zeroes are 0 and 2. 1
SECTION C
4. According to question,
Sum of zeroes =218
Product of zeroes =5
161
So, Quadratic polynomial = x2 – (Sum of zeroes) x + Product of zeroes
= x2 – 218
FHG
IKJ x +
516
=116
(16x2 – 42x + 5) 1
= (16x2 – 42x + 5) 116
. 1
SECTION D
5. Given the polynomialf(x) = ax2 – 5x + c
Let the zeroes of f (x) are α and β, then according to question
Sum of zeroes, (α + β) = Product of zeroes, (αβ) = 10
Now, α + β = – coeff. of coeff. of 2
xx
= –−F
HGIKJ
5a 1
P-20 1–MRETT I C S X--M A T E M AH
⇒ 10 =+5a
1
⇒ a =12
and αβ =constant
coeff. of 2x =
ca
1
⇒ 10 = 2c⇒ c = 5
Hence, a =12
and c = 5. 1
SUMMATIVE ASSESSMENT WORKSHEET-17
SECTION A
1. (A) Let f (x) = ax2 + bx + c
Sum of zeros = α + β = coeff. of coeff. of 2
xx
= –ba
But one zero is 0
i.e., α + 0 = – ba
Hence other zero is –ba
· 1
2. (A) If a is added in the polynomial, thenp(x) = x2 – 5x + 4 + a
Given 3 is the zero∴ p(3) = 0⇒ (3)2 – 5(3) + 4 + a = 0⇒ – 6 + 4 + a = 0⇒ a = 2 1
SECTION B
3. Since, – 1 is a zero of polynomialp(x) = kx2 – 4x + k, then p(–1) = 0 1
∴ k(–1)2 – 4(–1) + k = 0⇒ k + 4 + k = 0⇒ 2k + 4 = 0⇒ 2k = – 4⇒ k = – 2Hence, k = – 2. 1
SECTION C
4. If (2x + 3) is a factor then – 32
is a zero of the polynomial 2x3 + 9x2 – x – b, then
2 23 3 32 9
2 2 2b − + − + −
= 0 1
27 81 64 4 4
b− + + − = 0
⇒ b = 604
= 15 1
Hence, b = 15. 1
P-21S O L U T I O N S
SECTION D
5. Since 53
and – 53
are two zeroes of f (x).
Therefore,
5 53 3
x x
− + =
2 25 1(3 5)
3 3x x − = −
is a factor of f (x)
Also, (3x2 – 5) is a factor of f (x)Let us now divide f (x) by (3x2 – 5)We have,
2
2 4 3 2
4 3 2
2 2
3 2
2
2
2 13 5 3 6 2 10 5
3 0 5
6 3 10 5
6 0 10
3 5
3 5
0
+ +− + − − −
+ −− − +
+ − −
+ −
−
−− +
x xx x x x x
x x x
x x x
x x x
x
x
By division alorithm, we have3x4 + 6x3 – 2x2 – 10x – 5 = (3x2 – 5) (x2 + 2x + 1)
= 2( 3 5) ( 3 5) ( 1)x x x+ − +
Hence, the zeros of f (x) are – 5 5
, ,3 3
– 1 and – 1
SUMMATIVE ASSESSMENT WORKSHEET-18
SECTION A
1. Let f (x) = x2 – 5x – 6
Let other zero be k, then 6 + k = – 5
1
æ ö-ç ÷ç ÷è ø = 5 1
∴ k = 5 – 6 = – 1 2. p(x) = ax2 + bx + c
Let zeroes of polynomial are α and − α, then
α + (− α) = – ba
⇒ 0 = – ba
⇒ b = 0 1
SECTION B
3. p(x) = ax2 + bx + cLet α and
1α
be the zeroes of p(x), then
Product of zeroes, α × 1α
=ca
1
So, required condition is c = a. 1
P-22 1–MRETT I C S X--M A T E M AH
SECTION C
4. Let p(x) = 5x3 + 8x – 4= 5x2 + 10x – 2x – 4= 5x(x + 2) – 2(x + 2)= (x + 2)(5x + 2) 1
Hence, zeroes of the quadratic polynomial 5x2 + 8 x – 4 are – 2 and 2
·5
Verification :
Sum of zeroes = – 2 + 25
= −85
Product of zeroes = (– 2) ×
25 =
−45
Again sum of zeroes = 2
coeff. of coeff. of
xx
= −85
1
Product of zeroes = 2
constantcoeff. of x
= −45
1
Thus relationship is verified.
SECTION D
5. 3x3 + 4x2 + 5x – 13 = (3x + 10) g(x) + (16x – 43) 1
⇒3 4 11 30
3 10
3 2x x xx
+ − ++
= g(x) 1
3 2 2
3 2
2
2
3 10 3 4 11 30 ( 2 3
3 10(–) (–)
– 6 11
– 6 – 20 ( ) ( ) 9 30 9 30 (–) (–) 0
x x x x x x
x x
x x
x x
xx
+ + − + − +
+
−
+ +++
1
Hence, g(x) = x2 – 2x + 3. 1
SUMMATIVE ASSESSMENT WORKSHEET-19
SECTION A
1. 4x2 – 12x + 9 = 04x2 – 6x – 6x + 9 = 0
⇒ 2x(2x – 3) – 3(2x – 3) = 0⇒ (2x – 3)(2x – 3) = 0
⇒ x = 32
and 32
Hence, zeroes of the polynomial are 32
, 32
P-23S O L U T I O N S
2. p(x) = 3x2 – kx + 6
Sum of the zeroes = 3 = – 2coefficient ofcoefficient of
xx
⇒ 3 = – (– )k
3⇒ k = 9.
SECTION B
3. Quadratic polynomial = ( 15) ( 15)x x− +
= 2 2( 15)x −= x2 – 15.
SECTION C
4.
2 3 2
3 2
2
2
2 23 2 1 6 2 4 3
6 – 4 2–
6 6 3
6 – 4 2 – – 2 1
xx x x x x
x x x
x x
x x
x
+− + + − +
++ −
− +
++ −
+
Quotient = 2x + 2; Remainder is – 2x + 1 1p(x) = g(x) q(x) + r(x)
= (3x2 – 2x + 1) (2x + 2) + (– 2x + 1)= 6x3 – 4x2 + 2x + 6x2 – 4x + 2 – 2x + 1 1= 6x3 + 2x2 – 4x + 3. Verified. 1
SECTION D
5. x2 – 2 5 x – 15
x – 5 ) x3 – 3 5 x2 – 5x + 15 5 2
x3 – 5 x2
– +
– 2 5 x2 – 5x
– 2 5 x2 + 10x + – – 15x + 15 5
– 15x + 15 5 + –
0 2
Now, x2 – 2 5 x – 15 = x2 – 3 5 x + 5 x – 15= x (x – 3 5 ) + 5 (x – 3 5 )= (x + 5 ) (x – 3 5 )
All the zeroes are 5 , – 5 , 3 5 . 2
P-24 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-20
SECTION A
1. f(x) = x2 – 7x – 8
Let other zero be k, then – 1 + k = – –71
FHG
IKJ = 7
k = 8 12. From the graph it is clear that the curve y = f (x) cuts the x-axis at two places between – 2 and 2.
∴ Required number of zeroes = 2. 1
SECTION B
3. Let, f (x) = 2x2 – 5x – 3Let the zeroes of polynomial are α and β, then
Sum of zeroes α + β =52
, product of zeroes αβ = – 32
According to question, zeroes of x2 + px + q are 2α and 2β
Sum of zeroes = –coeff. of coeff. of 2
xx
= − p1
= 2α + 2β = 2(α + β) = 2 × 52
= 5 ⇒ p = – 5 1
Product of zeroes = constant
coeff. of 2x =
q1
= 2α × 2β = 4αβ = 4 −FHGIKJ
32
= – 6
∴ p = – 5 and q = – 6. 1
SECTION C
4. p(x) = 3x2 – 4x – 7 and α and β are its zeroes.
Sum of zeroes α + β = 4 43 3
− − = 1
Product of zeroes αβ = 73
−
For new polynomial
Sum 1 1
+α β =
α + βαβ =
4 / 3 47 / 3 7
=−− 1
product 1 1
×α β =
1αβ =
1 37 / 3 7
=−−
The required polynomial = x2 – (sum of zeroes)x + product of zeroes
= 2 4 37 7
x x − − + −
= 21(7 4 3)
7x x+ −
= 2 1(7 4 3)
7x x+ − ½
P-25S O L U T I O N S
SECTION D
5. x4 – 4x + (8 – k)
x2 – 2x + k) x4 – 6x3 + 16x2 – 25x + 10 x4 – 2x3 + kx2
– + –
– 4x3 + (16 – k)x2 – 25x + 10 – 4x3 + 8x2 – 4kx 1
+ – +
(8 – k)x2 – (25 – 4k)x + 10 (8 – k)x2 – (16 – 2k)x + (8k – k2)
– + –
(2k – 9)x + (10 – 8k + k2) 2Given, remainder = x + a
⇒ 2k – 9 = 1 ⇒ k = 102
= 5
and a = 10 – 8k + k2 = 10 – 40 + 25 = – 5. 1
SUMMATIVE ASSESSMENT WORKSHEET-21
SECTION A
1. From the graph it is clear that curve cut the x-axis at three places. So number of zeroes is 3. 12. Let f (x) = ax2 + bx + c
Now product of zeroes = constant
coeff. of 2x =
ca
Sum of zeroes = – coeff. of coeff. of 2
xx
= – ba
1
SECTION B
3. Since α and β are the zeroes of the polynomialf(x) = x2 – px + q 1
∴ α + β = p, αβ = q
Now1 1+α β =
pq
α + β =αβ 1
SECTION C
4. Let α and β are the zeroes of the polynomial, then as per questionβ = 7α
∴ α + 7α = 8α = – −FHGIKJ
83
½
⇒ α = 13
½
and α × 7α = 2 1
3k +
⇒ 7α2 = 2 1
3k +
P-26 1–MRETT I C S X--M A T E M AH
⇒ 7 13
2FHG
IKJ =
2 13
k +1
⇒ 7 × 19
= 2 1
3k +
⇒73
– 1 = 2k
⇒23
= k. 1
SECTION D
5. Polynomial g(x) = (x + 2 ) (x – 2 ) = x2 – 2
x2 – 2 ) x4 – 5x3 + 2x2 + 10x – 8 ( x2 – 5x + 4
x4 – 2x2
– +
– 5x3 + 4x2 + 10x
– 5x3 + 10x + –
4x2 – 8 2 4x2 – 8 (–) (+)
0
x2 – 5x + 4 = (x – 4) (x – 1) 1Other zeroes are 4 and 1. 1
SUMMATIVE ASSESSMENT WORKSHEET-22
SECTION A
1. Given, p (x) = x2 – 5x + 6
Sum of zeroes = α + β = – −F
HGIKJ
51
= 5
Product of zeros = αβ = 61
= 6
Now α+ β – 3αβ = 5 – 3(6)
= 5 – 18 = – 13. 1
2. Required polynomial = ( 2) ( 2 2)x x− −
= 2 2 2 2 ( 2) (2 2)x x x− − +
= 2 3 2 4.x x− + 1
P-27S O L U T I O N S
SECTION B
3. f (x) = 2x2 – 7x + 3
Sum of roots = p + q = –coeff. of coeff. of 2
xx
= – −F
HGIKJ =7
272 ½
Product of roots = pq = constant
coeff. of 2x =
32
½
We know that(p + q)2 = p2 + q2 + 2pq
⇒ p2 + q2 = (p + q)2 – 2pq ½
= 72
3494
31
374
2FHG
IKJ − = − = . ½
SECTION C
4. 22 2 1x x+ -
2 4 3 24 3 2 8 14 2x x x x x ax b+ - + - + +
4 3 28 6 4x x x+ - – – +
3 2
3 2
8 2
8 6 4
x x ax
x x x
+ +
+ - – – +
24 ( 4)x a x b- + + + 24 3 2x x- - + + + –
( 7) 2a x b+ + -
For exact division, remainder is zero, then(a + 7) x + b – 2 = 0 2
a + 7 = 0, b – 2 = 0a = – 7, b = 2. 1
SECTION D
5.
2
2 4 3 22 2 1
4 3 2 8 14 2 8 12x x
x x x x x x+ −
+ − + − + −
8x4 + 6x3 – 4x2
– – +
8x3 + 2x2 + 8x8x3 + 6x2 – 4x– – +
– 4x2 + 12x – 12 1– 4x2 – 3x + 2 1+ + –
15x – 14 1
Subtract 15x – 14 or add – 15x + 14 for exact division. 1
P-28 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-23
SECTION A
1. Let 3, 3α = β = − be the given zeros and γ be the third zero. Then,
α + β + γ = – 41
−
3 3− + γ = 4 ⇒ x = 4.Hence third zero is 4. 1
2. If α, β and γ are the zeros of cubic polynomials f(x)Then f (x) = k{x3 – (α + β + γ) x2 + (αβ + βγ + γα) x – αβγ}Here, α + β + γ = 2, αβ + βγ + γα = – 7 and αβγ = – 14∴ f(x) = k(x3 – 2x2 – 7x + 14)where k is any non-zero real number. 1
SECTION B
3. Let p(x) = 3x2 + 11x – 4
= 3x2 + 12x – x – 4
⇒ = 3x(x + 4) – 1(x + 4) ½
⇒ = (3x – 1) (x + 4)
So, zeroes are : m = 13
and n = – 4. ½
Now,mn
nm
+ =
134
413
FHG
IKJ
−+ −FHG
IKJ
½
= 112− – 12 = –
14512
. ½
SECTION C
4. p(x) = 6y2 – 7y + 2
α + β = – 76
− =
76
αβ = 26
1 1+α β =
7 / 6 72 / 6 2
α + β = =αβ 1
1 1×α β =
1αβ = 3 1
The required polynomial = 2 73
2y y− +
= 2[2y2 – 7y + 6] 1
P-29S O L U T I O N S
5. Since α and β are the zeroes of polynomial 3x2 + 2x + 1.
Hence, α + β = −23
1
and αβ = 13
1
Now for the new polynomial,
Sum of the zeroes = 11
11
−+
+ −+
αα
ββ
= ( ) ( )
( )( )1 1
1 1− + − + + − −
+ +α β αβ α β αβ
α β
= 2 2
1
2 23
1 23
13
−+ + +
=−
− +
αβα β αβ
Sum of zeroes =
4323
= 2 1
Product of zeroes = 11 1
−+
FHG
IKJ +FHG
IKJ
αα
ββ
1 –
= ( )( )( )( )1 11 1
− −+ +
α βα β
= 11
11
− − ++ + +
= − + ++ + +
α β αβα β αβ
α β αβα β αβ
( )( )
Product of zeroes =
1 23
13
1 23
13
6323
3+ +
− += =
1
Hence, Required polynomial = x2 – (Sum of zeroes)x + Product of zeroes
= x2 – 2x + 3. 1
FORMATIVE ASSESSMENT WORKSHEET-24
Objective Type Questions
1. (?) 2. (D) 3. (A) 4. (C) 5. (D) 6. (B) 1 × 6
Fill in the blanks
1.
1. 1 3
2. 1 2
3. 3 4, 4, 2
4. 3 4, 2, 2
5. 0 —
−
− −
− −
S. No. No. of zeros Zeroes
1 × 5
Quiz
1. P(x) = 3x3 – 2x2 + 6x – 5P(2) = 3(2)3 – 2(2)2 + 6(2) – 5
= 3 × 8 – 2 × 4 + 12 – 5= 24 – 8 + 12 – 5= 36 – 13
P(2) = 23 1
P-30 1–MRETT I C S X--M A T E M AH
2. α + β = b
a−
= – 2
αβ = ca =
31
= 3
Hence the quadratic polynomial = x2 – (α + β) x + αβ= x2 – 2x + 3 1
3.1 1n m
+
11 11
·4 4
bb bm n aa
cmn a c ca
−− − −+ = = × = = =
− 1
4. For the polynomial t2 – 4t + 3
p + q = – ba = –
( 4)1
− = 4 ....(i)
pq = ca =
31
= 3
(p – q) = 2( ) 4p q pq+ −
= 2(4) 4 3− ×
= 16 12−
= 4 2= ±Hence p – q = + 2By (i) + (ii) 2p = 6 or 2p = 2Hence p = 3 or 1and q = 1 or 3
1 1 142
3pq
p q+ − + = 0
1 141 2 3 1
3 3i i+ − + = 0
1 141 6
3 3+ − + = 0
= 1 3 18 14
3+ − +
= 18 18
3−
= 0 1
●●
P-31S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-25
SECTION A
1. For a unique solution1
2
aa ≠
1
2
bb
i.e.,32
≠ 2
k−
⇒ – 4 ≠ ki.e., all real numbrs except – 4. 1
2. x = 6, y = 1. 1
SECTION B3. 3x – 2y = 4 ....(1)
2x + y = 5 ....(2)Putting x = 2 and y = 1 in equation (1), we have
L.H.S. = 3 × 2 – 2 × 1 = 4 = R.H.S. ½Putting x = 2 and y = 1 in equation (2), we have
L.H.S. = 2 × 2 + 1 × 1 = 5 = R.H.S. ½Thus, x = 2 and y = 1 satisfy both the equations of the given system.Hence x = 2, y = 1 is solution of the given system. 1
SECTION C4. Let the man can finish the work in x days and the boy can finish the same work in y days.
Work done by one man in one day = 1x
Now, work done by one boy in one day = 1y
According to question,
+2 7x y
=14
...(1)
+4 4x y
=13
...(2)
Pair of Linear Equationsin Two Variables
3CHAPTER
P-32 1–MRETT I C S X--M A T E M AH
Let 1x
= a and 1y = b, then
2a + 7b =14
...(3)
4a + 7b =13
...(4) 1
Multiply eqn. (3) by 2 and substract from it eqn. (4)
4a + 14b =12
4a + 4b =13
– – –
10b =16
1
=160
=1y
⇒ y = 60 days.
Put b = 160
in equation (3), we get
a = 1
15
So1
15=
1x
⇒ x = 15 days. 1
SECTION D
5. x + 3y = 6 ...(1)
⇒ y =6
3− x
3 6 0
1 0 2
x
y ½
2x – 3y = 12 ...(2)
⇒ y =2 12
3x −
0 6 3
4 0 2
x
y − − ½
Putting the above points and drawing a linejoing them, we get the graphs of the equationsx + 3y = 6 and 2x – 3y = 12. 1
Clearly, the two lines intersect at point B (6,0).
Hence, x = 6 and y = 0 is the solution of thesystem. 1
Again ∆ OAB is the region bounded by the line2x – 3y = 12 and both the co-ordinate axes.
1
–3 –2 –1 1 2 3 4 5 6
(6, 0)
(3, 1)(0, 2)
–1–2
–3–4
y'
xx'
y
12
2 – 3 = 12
xy
xy
+ 3 = 6
(0, –4)
A
(3, –2)
0B
P-33S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-26
SECTION A
1. The lines represented by the equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel, if
aa
1
2=
bb
1
2 ≠
cc1
2· 1
2. The lines represented by the equations a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0 are coincident, if
aa
1
2=
bb
1
2 =
cc1
2· 1
SECTION B3. 2x – y = 2 ...(i)
y = 2x – 2 ...(ii)x + 3y = 15
Substituting the value of y form (i) in (ii), we getx + 6x – 6 = 15 1
7x = 21x = 3
From (1), y = 2 × 3 – 2 = 4∴ x = 3 and y = 4. 1
SECTION C
4. (2m – 1)x + 3y – 5 = 0 ...(1)On comparing with the eqn.
a1x + b1y + c1 = 0a1 = 2m – 1, b1 = 3, c1 = – 5 ½
3x + (n – 1)y – 2 = 0 ...(2)On comparing with the eqn.
a2x + b2y + c2 = 0a2 = 3, b2 = (n – 1), c2 = – 2 ½
For a pair of equations to have infinite number of solutions.
aa
1
2= b
b1
2 = c
c1
2
2 13
m −=
31
52n −
= 1
2(2m –1) = 15 and 5 (n–1) = 6
⇒ m =174
, n = 115
· 1
SECTION D
5. Let the speed of the boat in still water be ‘x’ km/hr and speed of the stream be ‘y’ km/hr. Speed ofthe boat during upstream be (x – y) km/hr and during downstream be (x + y) km/hr.
∴ 30 28x y x y−
++
= 7 ; 21 21x y x y−
++
= 5 1
P-34 1–MRETT I C S X--M A T E M AH
Let 1
x y− = a and 1
x y+ = b, we get
30a + 28b = 7 ...(1)21a + 21b = 5 ...(2)
Multiplying eqn. (2) by 21 and eqn. (2) by 28 and then subtracting, we get630a + 558b = 147 ...(3)588a + 558b = 140 ...(4)
– – –On subtracting, 42a = 7 1
⇒ a =742
16
=Putting this value of a in eqn, (1), we get
28b = 7 – 30a = 7 – 30 × 16
= 2 ⇒ b = 1
14
Now, a =1 1
6x y−= ⇒ x – y = 6 ...(5) 1
b =1
x y+ =
114
⇒ x + y = 14 ...(6)
Solving (5) and (6), we get x = 10, y = 4Hence, speed of the boat in still water = 10 km/hr and speed of the stream = 4 km/hr.
SUMMATIVE ASSESSMENT WORKSHEET-27
SECTION A
1. x – y = 2 ...(1)and x + y = 4 ...(2)Adding both the equations, 2x = 6 ⇒ x = 3Put the value of x in the eqn. (2), we get
3 + y = 4 ⇒ y = 1Hence, a = 3, b = 1. 1
2. ad ≠ bc ⇒ ac
≠ bd
. Hence, the pair of linear equations has unique solution. 1
SECTION B3. Yes, for justification we have For the equation
2x + 3y = 9a1 = 2, b1 = 3 and c1 = – 9 ½
and for the equation, 4x + 6y = 18a2 = 4, b2 = 6 and c2 = – 18 ½
Hereaa
1
2=
24
12
36
12
1
2
= = =, bb
and cc
1
2
12
= =–9–18 ½
From above it is clear thataa
1
2=
bb
cc
1
2
1
2= ½
Hence system is consistent and dependent.
SECTION C
4. Given equations are 2x + 3y = 7 and 2αx + (α + β) y = 28.We know that the condition for a pair of linear equations to be consistent and having infinite
number of solutions isaa
1
2=
1
2
bb =
1
2
cc 1
P-35S O L U T I O N S
⇒2
2α3 7
28a b+= =
I II III
From I and III,2
2α =728
½
⇒ α = 4 ½
From II and III,3
α β+=
728
α +β = 12β = 12 – αβ = 12 – 4
⇒ β = 8Hence α = 4, and β = 8. 1
SECTION D
5. 2x + 3y = 12 ⇒ y = 12 2
3− x
0 6 3
4 0 2
x
y
x – y = 1 ⇒ y = x – 1
0 1 3
1 0 2
x
y −
Putting the above points and drawing a line joining them, we get the graph of the equations2x + 3y = 12 and x – y = 1.
–5 –4 –3 –2 –1 1 2 3 4 5 6 7 8
y'
–1–2
–3–4
12
3
5
4
x' x
y
B
P(3, 2)
x – y = 1
(0, – 1)2 3 = 12x + y
A
(0, 4)
M
(1, 0) (6, 0)
1
Clearly, the two lines interesect at point P (3, 2).Hence, x = 3 and y = 2 is the solution of the system. 1
Area of shaded region = Area of ∆ PAB
=12
× base × height = 12
× AB × PM
=12
× 5 × 3
= 7.5 square unit. 1
P-36 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-28
SECTION A
1. The equation of one line
4x + 3y = 14.
We know that if two lines a1x + b1y + c = 0 and a2x + b2y + c2 = 0 are parallel, then
aa
1
2=
bb
1
2 ≠
cc
1
2
Hence, second parallel line is – 12x = 9y. 1
2. Given, 3x + 4y = 7 and 3x + 4y = 16
Here,aa
1
2=
bb
1
2 ≠
cc1
2
Hence, pair of linear equations has no solution. 1
SECTION B
3. 99x + 101y = 499 ...(1)101x + 99y = 501 ...(2)
Adding equation (1) and (2), we get200x + 200y = 1000
⇒ x + y = 5 ...(3)Subtracting equation (2) from equation (1) ½
– 2x + 2y = – 2⇒ x – y = 1 ...(4) ½Adding equations (3) and (4)
2x = 6 ⇒ x = 3 ½Put the value of x in equation (3), we get y = 2. ½
SECTION C
4. (i) Let the fixed charge of taxi be Rs. x per km and the running charge be ` y per km.According to the question,
x + 10y = 75 ...(1)x + 15y = 110 ...(2) 1
Subtracting equation (2) from equation (1), we get– 5y = – 35
⇒ y = 7Putting y = 7 in equation (1), we get x = 5 1∴ Total charges for travelling a distance of 25 km
= x + 25y= ` (5 + 25 × 7)= ` (5 + 175)= ` 180 1
(ii) Pair of linear equations in two variables. ½(iii) We should always be justified in our dealings. ½
P-37S O L U T I O N S
SECTION D5. Let the sum of the ages of the 2 children be x and the age of the father be y years.
∴ y = 2x i.e., 2x – y = 0 ...(1) 1and 20 + y = x + 40
x – y = – 20 ...(2) 1Subtracting (2) from (1), we get x = 20From (1), y = 2x = 2 × 20 = 40
y = 40 1Hence, the age of the father = 40 years. 1
SUMMATIVE ASSESSMENT WORKSHEET-29
SECTION A
1. – 3x + 4y = 7 ...(1)
92
x – 6y + 212
= 0
⇒ –3x + 4y = 7 ...(2)
From eqns. (1) and (2),aa
1
2=
bb
1
2 =
cc1
21
Hence, pair of linear equations has infinite number of solutions.2. Put x = 2 and y = 3 in 2x – 3y + a = 0 and 2x + 3y – b + 2 = 0, we get
4 – 9 + a = 0 ⇒ a = 54 + 9 + 2 – b = 0 ⇒ b = 15
From above it is clear that 3a = b. 1
SECTION B
3. Given equations are :4x + py + 8 = 0 ...(1)2x + 2y + 2 = 0 ...(2)
The condition of unique solution,aa
1
2≠
bb
1
21
Hence,42
≠p2
or 21
≠ p2
∴ p ≠ 4. 1
SECTION C
4. (i) Let the monthly rent of the house be ` x and the mess charges per head per month be ` y.According to the given conditions,
x + 2y = 3900 ...(i)x + 5y = 7500 ...(ii) ½
Subtracting equation (ii) from equation (i), we get– 3y = – 3600
⇒ y =3600
3 = 1200 1
Putting this value of y in eqn. (i), we getx + 2400 = 3900
⇒ x = 3900 – 2400 = 1500Hence, monthly rent = ` 1500 1
P-38 1–MRETT I C S X--M A T E M AH
(ii) Mess charge per head per month = ` 1200 ½(iii) Pair of linear equations in two variables. ½(iv) Monitering is always good to control our of extreme habits. ½
SECTION D
5. 2(3x – y) = 5xy ...(1)
2(x + 3y) = 5xy ...(2)
Divide eqns. (1) and (2) by xy,
6 2y x
− = 5 ...(3)
2 6y x
+ = 5 ...(4) 1
Putting 1y = a and
1x
= b, then equations (3) and (4) become
6a – 2b = 5 ...(5)2a + 6b = 5 ....(6) 1
Multiplying eqn. (5) by 3 and then adding with eqn. (6), we get20a = 20 ⇒ a = 1
Putting this value of a in eqn. (5), we get
b =12
1
Now1y = a = 1 ⇒ y = 1
and1x
= b = 12
⇒ x = 2. 1
SUMMATIVE ASSESSMENT WORKSHEET-30
SECTION A
1. From the options it is clear that x = 2, y = 3 is a solution of the linear equation 2x + 3y –13 = 0.2. at y-axis x = 0, then
0 – y = 8 ⇒ y = – 8Hence, point of interrection is are (0, –8). 1
SECTION B
3. Pair of equationskx – 4y = 3
6x – 12y = 9Condition for infinite solutions :
aa
1
2=
bb
cc
1
2
1
2= 1
k6
=−
−=
412
39
⇒ k = 2. 1
P-39S O L U T I O N S
SECTION C
4. x – 5y = 6 ⇒ x = 6 + 5y
0 1 2
6 1 4
y
x
− −− 1
2x – 10y = 12 ⇒ x = 5y + 6
0 1 2
6 1 4
y
x
− −− 1
–5 –4 –3 –2 –1 1 2 3 4 5 6 7
x y – 5 = 6
2 – 10 = 12x y
–1
–2–3
y'
xx'
y
(6, 0)
(1, –1)
(–4, –2)
0
1
Since the lines are co-incident, so the system of linear equations is cosistent with infinite solutions.1
SECTION D
5. 2x + 3y = 12 ⇒ 12 2
3− x
x – y = 1 ⇒ x – 1
0 6 3
4 0 2
x
y 0 1 3
1 0 2
x
y − 1
Putting the above points and drawing a line joining them, we get the graphs of equations 2x + 3y= 12 and x – y – 1 = 0.
–3 –2 –1 1 2 3 4 5 6
–1–2
–3
y'
xx'
y
0B
1
2
3
42 + 3 = 12
xy
(3, 2)
A
C
x y – – 1 = 0
(0, 4)
(1, 0)(0, –1)
(6, 0)
1
Clearly, the two lines interesct at point (3, 2), Hence, x = 3 and y = 2 is the required solution. 1∆ ABC is the region between the two lines represented by the above equations and the x-axis. 1
P-40 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-31
SECTION A
1. For coincident lines
1
2
aa =
1 1
2 2
b cb c
=
⇒24
=k8
714
=
⇒ k = 4. 12. x + y = 0
x + y + 7 = 0
∵ 1
2
aa =
1 1
2 2
b cb c
¹
Hence, pair of equations has no solution. 1
SECTION B
3. The condition for no solution,
aa
bb
cc
1
2
1
2
1
2= ≠ ⇒
kk123 1
2= ≠ ½
When kk123= , we get k2 = 36 1
i.e., k = ± 6Since k ≠ 6, so k = – 6. ½
SECTION C
4. 7x – 4y = 49 ...(1)On comparing with the equation
a1x + b1y = 0a1 = 7, b1 = – 4, c1 = 49
Again, 5x – 6y = 57 ...(2)On comparing with the equation
a2x + b2y = c2
a2 = 5, b2 = – 6, c2 = 57
Since,aa
1
2=
75
and bb
1
2 =
46
1 1
2 2
a ba b
¹ 1
So, system has unique solution.Multiply eqn. (1) by 5 and multiply eqn. (2) by 7 and subtract,
35x – 20y = 24535x – 42y = 399 1– + –
22y = –154⇒ y = –7Put the value of y in eqn. (1)
5x – 6(–7) = 57 ⇒ 5x = 57 – 42 = 15⇒ x = 3Hence, x = 3 and y = –7. 1
P-41S O L U T I O N S
SECTION D
5.5
11
2x y−+
−= 2 ...(1)
61
32x y−
−−
= 1 ...(2) 1
Let1
1x −= p,
12y −
= q
Then the equations (1) and (2) will be5p + q = 2 ...(3)
6p – 3q = 1 ...(4) 1Multiplying eqn. (3) by 3 and then adding in eqn. (4), we get
21p = 7
⇒ p =721
13
= 1
Putting this value of p in eqn. (3), we get
q = 2 – 5p = 2 – 5 × 13
= 13
Now p =1
113x −
= ⇒ x –1 = 3
∴ x = 4
and q =1
2y− = 13
⇒ y – 2 = 3
y = 5 1∴ Solution is x = 4 , y = 5.
SUMMATIVE ASSESSMENT WORKSHEET-32
SECTION A
1. Since, line intersect y-axis, then x = 0
⇒ 3(0) – 2y = 6
⇒ y = – 3
Hence, required point is (0,– 3). 1
2. 4x – 5y = 5
kx + 3y = 3
For the condition of inconsistent
1
2
aa =
1 1
2 2
b cb c
¹
i.e.,4k
=−
≠5
353
⇒4k
=−53
⇒ k =−125
1
P-42 1–MRETT I C S X--M A T E M AH
SECTION B
3. From Fig., x + y = 22 ...(i)x – y = 16 ...(ii) ½
Adding (i) and (ii), we get 2x = 38or x = 19Put the value of x in equation (i), we get ½
19 + y = 22or y = 22 – 19 = 3 ½Hence, x = 19 and y = 3. ½
SECTION C
4. Let 1y
= a, the given equations become
4x + 6a = 15 ...(1)6x – 8a = 14 ...(2) ½
Multiply eqn (1) by 4 and eqn. (2) by 3 and adding16x + 24a = 6018x – 24a = 42
On adding, 34x = 102
⇒ x =10234
= 3 ½
Put the value of x in eqn. (1)4(3) + 6a = 15
6a = 15 – 12 = 3
a =36
= 12
∵ a =1y =
12
⇒ y = 2 1
Hence x = 3 and y = 2
Again y = px – 2 ⇒ 2 = p(3) –2 ⇒ 3p = 4 ⇒ p = 43
· 1
SECTION D
5. Let the speed of the boat be x km/hr and speed of the stream be y km/hr.∴ Upstream speed = x – y km/hrand Downstream speed = x + y km/hrAccording to the question,
32 36x y x y−
++ = 7
and40 48
x y x y−+
+ = 9
Let1
x y− = A, 1
x y+ = B, we get
32A + 36B = 7and 40A + 48B = 9
Solving these equations, we get A =18
, B = 12
1
P-43S O L U T I O N S
Hence A =18
= 1
x y−⇒ x – y = 8 ...(1) ½
and B =1
121=+x y
⇒ x + y = 12 ...(2) ½Adding equations (1) and (2), we get 2x = 20⇒ x = 10 1Putting this value of x in eqn.(1), we get
y = x – 8 = 10 – 8 = 2Hence, the speed of the boat = 10 km/hr and speed of the stream = 2 km/hr. 1
SUMMATIVE ASSESSMENT WORKSHEET-33
SECTION A
1. 2x + y = 76x – py = 21
For the condition of infinite number of solutions
aa
1
2=
bb
cc
1
2
1
2=
⇒26
=1 7
21−=
p
⇒13
=1
− p⇒ p = – 3. 1
2. If a pair of linear equations is consistent, then the lines represented by these equations will beintersecting (or) coincident. 1
SECTION B
3. Condition for unique solution,aa
1
2≠
bb
1
2k
12≠
3k
1
k2 ≠ 36k ≠ ± 6. 1
SECTION C
4. (i) Suppose the person invested ` x at the rate of 12% simple interst and ` y at the rate of 10%simple interest, then
yearly interest =12100
10100
x y+
∴12100
10100
x y+ = 130
⇒ 12x + 10y = 13000⇒ 6x + 5y = 6500 ..(1) ½If the invested amounts are interchanged, then yearly interest increases by ` 4.∴ 10x + 12y = 13400⇒ 5x + 6y = 6700 ...(2) ½
P-44 1–MRETT I C S X--M A T E M AH
Subtracting eqn. (2) from eqn. (1), we getx – y = – 200 ...(3)
Adding equations (1) and (2), we get11x + 11y = 13200
⇒ x + y = 1200 ...(4)Adding equations (3) and (4), we get
2x = 1000⇒ x = 500 1Putting x = 500 in equation (3), we get y = 700Thus, the person invested ` 500 at the rate of 12% per year and ` 700 at the rate of 10% per year.(ii) Pair of linear equations in two variables. ½(iii) Honesty is the best policy. ½
SECTION D
5. Let length = x and breadth = y
Then according to first condition,
(x – 5) (y + 3) = xy – 9
⇒ 3x – 5y = 6 ...(1) 1
According to second condition,
(x +3) (y +2) = xy + 67
⇒ 2x + 3y = 61 ...(2) 1
Multiplying eqn. (1) by 3 and eqn. (2) by 5 and then adding, we get
9x – 15y = 18
Aadding, 10x + 15y = 305
19x = 323 ⇒ x = 32319
= 17 1
Putting this value of x in eqn. (1), we get
3(17) – 5y = 6 ⇒ 5y = 51 – 6 ⇒ y = 9 1
Hence, perimeter = 2(x + y) = 2(17 + 9) = 52 units.
SUMMATIVE ASSESSMENT WORKSHEET-34
SECTION A
1. (B) First line is 10x – 8y –7 = 0
For parallel lines,aa
1
2=
bb
1
2≠
cc1
2So, second line is 15x –12y –7 = 0. 1
2. The area of the triangle formed by line x ya b
+ = 1 with the coordinate axis = ab. 1
SECTION B
3. Suppose my age is x years and my son’s age is y years. Thenx = 3y ...(1) ½
Five years later, my age will be (x + 5) years and my son’s age will be (y + 5) years.
∴ x + 5 =52
(y + 5)
⇒ 2x – 5y – 15 = 0 ...(2)
P-45S O L U T I O N S
Put x = 3y in equation (2)6y – 5y – 15 = 0 ⇒ y = 15
Putting y = 15 in equation (1)x = 45 1
Hence, my present age is 45 years and my son’s present age is 15 years.
SECTION C
4. For (2p – 1)x + (p – 1)y – (2p + 1) = 0a1 = 2p – 1, b1 = p –1 and c1 = – (2p + 1) ½
and for 3x + y – 1 = 0a2 = 3, b2 = 1 and c2 = – 1 ½
For the condition of no solutionaa
1
2=
bb
1
2 ≠
cc1
22 1
3p −
=p p−
≠+1
12 1
11
By2 1
3p −
=p − 1
1⇒ 3p – 3 = 2p – 1⇒ 3p – 2p = 3 – 1
p = 2. 1
SECTION D
5. Let two digit number is 10x + y.According to question,
8(x + y) – 5 = 10x + y⇒ 2x – 7y + 5 = 0 ...(1)and 16(x – y) + 3 = 10x + y
6x – 17y + 3 = 0 ...(2) 1Solving eqns. (1) and (2) by cross-multiplication method, we get
x(–7)( ) – (–17)( )3 5 =
y( )( ) – ( )( )5 6 2 3 =
12 6( )(–7) – ( )(–7) 1
⇒x
− +21 85=
y30 6
134 42−
=− +
⇒x
64=
y24
18
=
⇒x8
=y3
= 1 1
Hence, x = 8, y = 3So required number = 10 × 8 + 3 = 83. 1
SUMMATIVE ASSESSMENT WORKSHEET-35
SECTION A
1. 3x + ky = 52x + y = 16
For unique solution,aa
1
2≠ b
b1
21
P-46 1–MRETT I C S X--M A T E M AH
⇒32
≠k1
⇒ k ≠32
SECTION B
2. For equation, 2x + 3y = 4a1 = 2, b1 = 3, c1 = – 4 ½
For equation, (k + 2) x + 6y = 3k + 2a2 = k + 2, b2 = 6, c2 = – (3k + 2) ½
For infinitely many solutionsaa
1
2=
bb
1
2 =
cc1
22
2k + =36
43 2
=+k
⇒ k = 2. 1
SECTION C
3. 141x + 93y = 189 ...(1)93x + 141y = 45 ...(2)
Adding equations (1) and (2), we get234x + 234y = 234
x + y = 1 ...(3)Subtracting equation (2) from equation (1), we get
48x – 48y = 144x – y = 3 ...(4) 1
Adding equations (3) and (4), we get2x = 4x = 2 1
Put the value of x in equation (3), we get2 + y = 1
y = 1 – 2 = – 1Hence, x = 2 and y = – 1 1
SECTION D
4. Let the cost of one pencil be ` x and the cost one chocolate be ` y.According to question,
2x + 3y = 11 ...(1)x + 2y = 7 ...(2)
Now, 2x + 3y = 11 ⇒ x =11 3
2− y
1 3 5
4 1 2
y
x − 1
and x + 2y = 7 ⇒ x = 7 – 2y
P-47S O L U T I O N S
0 1 3
7 5 1
y
x 1
–4 –3 –2 –1 1 2 3 4 5 6 7
y'
–1–2–3
12
3
5
4
x' x
y
6
7
8
0–5–6
–4–5–6–7
8
(1, 3)
(4, 1)(5, 1)
(7, 0)
2 + 3 = 11x y
x y + 2 = 7
(–2, 5)
1
Putting the above points and drawing the lines joining them, we get the graph of the aboveequations. Clearly, two lines interesect at point (1, 3).∴ Solution of eqns. (1) and (2) is x = 1 and y = 3∴ Cost of one pencil = ` 1 and cost of one chocolate = ` 3. 1
SUMMATIVE ASSESSMENT WORKSHEET-36
SECTION A
1. 4x –3y = 92x + ky = 11
For unique solution,aa
1
2=
bb
1
2≠
cc1
2
⇒42
=−3k
≠ 9
11
⇒ 2 =–3k
⇒ k =−32
· 1
SECTION B
2.x y+
+−1
21
3= 9
⇒ 3(x + 1) + 2(y –1) = 54⇒ 3x + 2y = 53 ...(1)
⇒x y−
++1
31
2= 8
P-48 1–MRETT I C S X--M A T E M AH
⇒ 2(x – 1) + 3 (y + 1) = 48⇒ 2x + 3y = 47 ... (2) ½Multiply eqn. (1) by (3), 9x + 6y = 53 × 3Multiply eqn. (2) by 2, 4x + 6y = 47 × 2
– – –
On subtracting 5x = 65
x =655
= 13 ½
Put the value of x in eqn. (2),
2(13) + 3y = 47
3y = 47 –26 = 21
y =213
= 7
Hence x = 13, y = 7. 1
SECTION C
3. ax + by =a b+
2or 2ax + 2by = a + b ...(i)and 3x + 5y = 4 ...(ii) 1Multiplying (i) by 5 and (ii) by 2b, we get
10ax + 10by = 5a + 5b6bx + 10by = 8b
– – –
On subtracting, x(10a – 6b ) = 5a – 3b 1
x =5 3
2 5 312
a ba b
−−
=( )
Putting x =12
in (1), we get y = 12
Hence x =12
and y = 12
· 1
SECTION D
4. Let the speed of bus be x km/hr and the speed of the train be y km/hr.
According to question,240 120
x y+ = 8
and120 240
x y+ = 7
Let 1x
= a, 1y
= b, then 1
240a + 120b = 8 ...(1)120a + 240b = 7 ...(2)
Apply [(1) × 2 – (2)], we get480a + 240b = 16120a + 240b = 7 ...(2)
– – –
On subtracting, 360a = 9
⇒ a =9
360140
= 1
P-49S O L U T I O N S
Putting this value of a in eqn.(1), we get
b =160
1
b =160
1=y ⇒ y = 60
a =140
1=
x ⇒ x = 40
Hence, speed of bus = 60 km/hr and speed of train = 40 km/hr. 1
SUMMATIVE ASSESSMENT WORKSHEET-37
SECTION A
1. The area of the triangle formed by the lines x = 3, y = 4 and x = y is = 12 sq unit. 1
SECTION B
2. 29x + 41y = 169 ...(1)41x + 29y = 181 ...(2)
Adding equations (1) and (2), we get70x + 70y = 350 1
x + y = 5 ...(3)Subtracting equation (1) from equation (2), we get
– 12x + 12y = – 12– x + y = – 1 ...(4)
Adding equations (3) and (4), we get2y = 4
⇒ y =42
= 2 ½
Putting this value of y in equation (3), we getx + 2 = 5 ⇒ x = 3
⇒ Hence, x = 3 and y = 2. ½
SECTION C
3. For x + 2y = 3a1 = 1, b1 = 2 , c1 = 3
For (k – 1) x + (k +1) y = k + 2a2 = k –1, b2 = k + 1, c2 = k + 2
For no solution,
aa
1
2=
bb
1
2 ≠
cc1
2
⇒1
1k − =2
1k + ≠ 3
2k + 1
I II III
From I and II,1
1k − =2
1k +⇒ k + 1 = 2k – 2 ⇒ k = 3
P-50 1–MRETT I C S X--M A T E M AH
From II and III,2
1k + ≠3
2k + ½
⇒ 2(k + 2) ≠ 3 (k + 1)⇒ 2k + 4 ≠ 3k + 3 ⇒ k ≠ 1 ½
From I and III,1
1k –≠
32k +
⇒ k + 2 ≠ 3k – 3
⇒ k ≠ –52
½
Hence, k = 3 but k ≠ 1 and k ≠ –52
· ½
SECTION D
4. Since BCDE and BECD with BC ⊥ CD, BCDE is a rectangle.∴ Opposite sides are equal.i.e., BE = CD ∴ x + y = 5 ...(i)and DE = BC = x – y 1Since perimeter of ABCDE is 21.∴ AB + BC + CD + DE + EA = 21
3 + x – y + x + y + x – y + 3 = 216 + 3x – y = 21
3x – y = 15 1Adding (i) and (ii), we get 4x = 20 ...(ii)
x = 5On putting the value of x in (i), we get
y = 0∴ x = 5 and y = 0. 1
FORMATIVE ASSESSMENT WORKSHEET - 38
Objective Type Questions
1. (C) 2. (B) 3. (B) 4. (B) 1 × 4 = 4
Word Problems
1. Let Leela’s present age = x and her daughter’s age = ySeven years ago :
y – 7 = 7 (x – 7)y – 7 = 7x – 49
y – 7x = – 42 ...(i) 1Three years later :
y + 3 = 3(x + 3)y + 3 = 3x + 9
y – 3x = 6 ...(ii) 12. Let one’s digit is y and ten’s digit is x.
According to the question :x + y = 9 ...(i) 1
number = 10x + yOn reversing digits number = 10y + xAgain according to the question,
9(10x + y) = 2(10y + x)90x + 9y = 20y + 2x
P-51S O L U T I O N S
90x – 2x + 9y – 20y = 088x – 11y = 0
8x – y = 0 ...(ii) 13. Let the rate of apple’s = Rs. x per kg
The rate of grapes = Rs. y per kgAccording to the question, 2x + y = 160 ...(i) 1After one month : 4x + 2y = 300. ...(ii) 1
Graph :
–3 –2 –1 0 1 2 3 4 5 6
4
3
2
1
0
–1
–2
–3
E
y
C
D
x' x
y'
A B
1. Points are A(2, 0) and B(4, 0). 22. Points are C(0, 4) and D(0, – 2). 23. Solution is (3, 1). 2
4. The area of triangle ABE = 12
× 2 × 1 = 1 unit2 2
5. The area of triangle CDE = 12
× 6 × 3 = 9 unit2 2
●●
P-52 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-39
SECTION A
1. We have
169121
=2
2
(26)(side of the smaller triangle)
⇒1311
=26
side of the smaller triangle
∴ side =11 26
13×
= 22 cm. 1
SECTION B
2. In triangles LMK and PNK, we have∠M = ∠N = 50º (Given) 1∠K = ∠K (Same)
∴ ∆LMK ~ ∆PNK (AA similarity)
∴LMPN
=KMKN
½
ax
=b c
c+
∴ x =bc
b c+ · ½
SECTION C
3. Proof : BA PQ ⇒ BR PQ and PR CA ⇒ PR CQA
R
Q
BCP D
Triangles
4CHAPTER
a
b c
P-53S O L U T I O N S
In ∆BRD, BR PQ
⇒BD
PD=
RD
QD(corr. of BPT) ....(1) 1
In ∆RPD, PR CQ
⇒RDQD =
PDCD (corr. of BPT) ....(2) 1
(1) and (2) ⇒ BD PD
PD CD= ⇒ PD2 = BD × CD
(12)2 = BD × CDBD × CD = 44 cm 1
SECTION D
4. Given : cos α =23
and OB = 3 cm
In ∆AOB cos α =23 =
AOAB
1
Let OA = 2x and AB = 3xIn ∆ AOB
AB2 = AO2 + OB2
(3x)2 = (2x)2 + (3)2
Þ 9x2 = 4x2 + 9 1Þ 5x2 = 9
x =95
= 35
Hence, OA = 2x = 235
= 65 cm
AB = 3x = 335
= 95
cm 1
So diagonal BD = 2 ´ OB = 2´3 = 6 cm ½and AC = 2´AO
= 2´65
=12
5 cm ½
SUMMATIVE ASSESSMENT WORKSHEET-40
SECTION A
1.ar ( DEF)ar ( ABC)
∆∆ =
22
2DE DE
ABAB =
⇒44
ar ( ABC)∆ =23
49
2FHG
IKJ =
⇒ ar (∆ABC) =44 9
4×
= 11 × 9 = 99 square unit. 1
A
D
B
CO
α
P-54 1–MRETT I C S X--M A T E M AH
SECTION B
2.
60°
P
70°
7 cmQ R
4.2
cm
3 3 cm
14 cm
8.4 cm6 3 cm
Z
YX
From given figures, PQ 4·2 1
;ZY 8·1 2
= = PR 3 2 1
;ZX 26 3
= = QR 7 1YX 14 2
= = ⇒ PQZY
PRZX
QRYX
= = 1
⇒ ∆PQR ~ ∆ZYX (SSS)∴ ∠X = ∠R 180º – (60º + 70º) = 50º 1
SECTION C
3. We have, PQR ~ PAB (∵ ∠P is common and PAPQ =
PBPR
) 1
⇒area PQRarea PAB
∆∆ =
PQPA
FHG
IKJ
2
32area PAB∆ =
43
2kk
FHG
IKJ 1
⇒ area ∆PAB = 18 cm2
∴ area of AQRB = area of ∆PQR – area of ∆PAB = 32 – 18 = 14 cm2 1
SECTION D
4. Let BD = DE = EC = x
BE = 2xBC = 3x
AE2 = AB2 + BE2 = AB2 + 4x2 ...(1) 1AC2 = AB2 + BC2 = AB2 + 9x2
AD2 = AB2 + BD2 = AB2 + x2 1Now, 8AE2 = 8AB2 + 32x2 [Multiply eqn. (1) by 8] ...(2)and 3AC2 + 5AD2 = 3(AB2 + 9x2) + 5 (AB2 + x2) 1
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2 ...(3)∴ 3AC2 + 5AD2 = 8AE2. [ From eqn. (2) & (3)] Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-41
SECTION A
1.Perimeter of ABCPerimeter of PQR
∆∆ =
length of AClength of PR
⇒2040
=AC8
⇒ AC =82
= 4 cm. 1
P
Q R
3k
A Bk
4k
A
B D E Cx x x
P-55S O L U T I O N S
SECTION B
2. Suppose the median AD intersects PQ at E.
Given, PQ || BC
⇒ ∠APE = ∠B and ∠AQE = ∠C
So in ∆APE and ∆ABD, ∠APE = ∠ABD
∠PAE = ∠BAD 1
∴ ∆APE ~ ∆ABD
⇒PEBD
=AEAD
...(1) 1
Similarly, ∆AQE ~ ∆ACD
QECD
=AEAD
...(2) ½
From eqns. (1) and (2), we have
PEBD
=QECD
⇒PEBD
=QEBD
, (as CD = BD)
⇒ PE = QEHence, AD bisects PQ. Proved. ½
SECTION C
3. Given :PSSQ
= PTTR
∠PST = ∠PRQTo prove : PQR is isosceles triangle.
Proof :PSSQ =
PTTR
By converse of B.P.T., we get ST || QR 1
∴ ∠PST = ∠PQR (Corresponding angles) 1
∴ ∠PST = ∠PRQ (Given)
∠PQR = ∠PRO
So, ∆PQR is isosceles triangle. Proved. 1
SECTION D
4. Given : ABC is a triangle in which DE || BC.
To prove :ADBD
=AECE
Construction : Draw DN ⊥ AE and EM ⊥ AD, Join BE and CD. 1
Proof : In ∆ADE, area (∆ADE) =12
× AE × DN ...(i)
In ∆DEC,
area (∆DCE) =12
× CE × DN ...(ii)
A
B C
P QE
D
P
Q R
S T
P-56 1–MRETT I C S X--M A T E M AH
By (i) / (ii) area ( ADE)area ( DEC)
∆∆
=
12
× ×
× ×
AE DN
12
CE DN
⇒area ( ADE)area ( DEC)
∆∆
=AECE
...(iii) ½
Now in ∆ADE, area (∆ADE) =12
× AD × EM ...(iv) ½
and in ∆DEB, area (∆DEB) =12
× EM × BD ...(v)
By (iv) / (v),area ( ADE)area ( DEB)
∆∆ =
12
AD EM
12
BD EM
× ×
× ×½
⇒area ( ADE)area ( DEB)
∆∆ =
ADBD
· ...(iv) ½
∆DEB and ∆DEC lies on the same base DE and between same parallel lines DE and BC.∴ area (∆DEB) = area (∆DEC)
From equation (iii),area ( ADE)area ( DEB)
∆∆
=AECE
...(vii)
From equations (vi) and (vii), we get
AECE
=ADBD
· Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-42
SECTION A
1. ∵ PQ || MN
So,KPPM
=KQQN
⇒KPPM
=KQ
KN KQ−
⇒4
13=
KQ20.4 – KQ
⇒ 4 × 20.4 – 4KQ = 13KQ
⇒ 17KQ = 4 ×20·4 ⇒ KQ = 20·4 4
17×
= 4·8 cm. 1
SECTION B
2. Since ABCD is a rhombus.So in ∆BOC
BC2 = OB2 + OC2
= 2BD
2 +
2AC2
1
BC2 = 2 2BD AC
4+
⇒ 4BC2 = AC2 + BD2 1
A
M N
D E
B C
A D
B C
O
P-57S O L U T I O N S
3. According to question,∠QPR = 90º
∴ QR2 = QP2 + PR2 1
∴ PR = 26 242 2−
= 100 = 10 cm
∠PKR = 90º
∴ PK = 2 210 8−
= 100 64−= 36 = 6 cm. 1
SECTION C
4. According to question, BD = CD = BC2
1
⇒ BC = 2BD
Using Pythagoras theorem in the right ∆ABC, we have
AC2 = AB2 + BC2
= AB2 + 4BD2 1
= (AB2 + BD2) + 3BD2
AC2 = AD2 + 3CD2. 1
SECTION D
5. Given: ABC is right angled at B and D is the mid-point of BC. ½
∴ BD = DC = 12
BC ½
A
B D C
In ∆ABD, AD2 = AB2 + BD2 (Pythagoras Theorem) ...(1) ½
In ∆ABC, AC2 = AB2 + BC2 (Pythagoras Theorem) ...(2) ½
From eqn. (1), AD2 = AB2 + BC2
2FHG
IKJ (D is the mid-point of BC) ½
⇒ 4AD2 = 4AB2 + BC2
⇒ BC2 = 4AD2 – 4AB2 ...(3) ½
Using this in (2), we get AC2 = AB2 + 4AD2 – 4AB2
AC2 = 4AD2 – 3AB2. Proved. 1
8 cm
CB
A
D
P-58 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-43
SECTION A
1. Given 2AB = DE and BC = 8 cm
∵ ∆ABC ~ ∆DEF
So,ABBC
=DEEF
⇒AB8
=2ABEF (1) (2)
⇒ EF = 2 × 8 = 16 cm. 1
SECTION B
2. Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.∴ ∆ABC ~ ∆PQR
⇒ABPQ =
BCQR =
ACPR =
3624
1
⇒ABPQ =
3624
⇒AB10 =
3624
⇒ AB =36 10
24×
= 15 cm. 1
3. Proof :
In quadrilateral ABCD,AOBO
=CODO
(Given)
⇒AOCO
=BODO
...(1)
In ∆ABD, EO || AB (Construction)
∴AEED
=BODO
(By BPT) ...(2) 1
From eqns. (1) and (2),AEED
=AOCO
In ∆ADC,AEED
=AOCO
⇒ EO || DC (Converse of BPT) 1
EO || AB (Construction)
∴ AB || DC
⇒ In quad. ABCD, AB || DC 1
⇒ ABCD is a trapezium. Proved.
A
B C
D
E F
A B
CD
OE
P-59S O L U T I O N S
SECTION C
1.APAB
=3 5
10 5..
= 13
⇒ AQAC
=39
= 13
1
In ∆ABC,APAB
=AQAC
and ∠A is common 1
⇒ ∆APQ ~ ∆ABC (SAS)
∴APAB
=PQBC
13
=4·5BC ⇒ BC = 13·5 cm. 1
SECTION D
4. In cyclic quadrilateral ABCD, we have∠A + ∠C = 180º
and ∠Β + ∠D = 180º 1∴ x + 7 + 3y + 23 = 180⇒ x + 3y = 150 ..(1) 1and y + 8 + 4x + 12 = 180⇒ 4x + y = 160 ...(2) 1Solving equations (1) and (2), we get
y = 40º∴ ∠ A = (x + 7)º = 37º
∠B = (y + 8)º = 48º∠C = (3y + 23)º = 143º∠D = (4x + 12)º = 132º 1
SUMMATIVE ASSESSMENT WORKSHEET-44
SECTION A
1. Given,Perimeter of ∆ABC = 32 cmPerimeter of ∆PQR = 48 cm
For similar triangles, we know thatABPQ =
ACPR
= BCQR =
Perimeter of ABCPerimeter of PQR
∆∆
⇒AC6
= 3248
⇒ AC =6 32
48×
= 4 cm.
A
B C
P
Q R
A
B
P
C
Q
3.5
7 6
3
4.5
P-60 1–MRETT I C S X--M A T E M AH
SECTION B
2. According to question, AO = 20 cm, BO = 12 cm, PB = 18 cmIn ∆AQO and ∆BPO,
∠AOP = ∠BOP (Vertically opposite angles)∠A = ∠B = 90º
∴ ∆AQO ~ ∆BPO 1
AQBP
=QOPO
= AOBO
AQ18
=2012
AQ = 30 cm. 1
SECTION C
3. Given : DB⊥BC, DE⊥AB and AC⊥BC.
To prove :BEDE
=ACBC
1
Proof : In ∆ABC, ∠1 + ∠2 = 90º (∠C = 90º) 1But ∠2 + ∠3 = 90º (Given)⇒ ∠1 = ∠3In ∆ABC and ∆BDE, ∠1 = ∠3 (Proved)
∠ACB = ∠DEB = 90º (Given)∴ ∆ABC ~ ∆BDE (AA) 1
⇒ACBC
=BEDE
·
SECTION D
4. Given : ∆ABC ~ ∆PQR,and ar ∆ABC = ar ∆PQR
A
B C
P
Q R
To prove : ∆ABC ≅ ∆PQRProof : ∆ABC ~ ∆PQR (Given)
⇒ar ( ABC)ar ( PQR)
∆∆ =
AB
PQ
BC
QR
CA
RP
2
2
2
2
2
2= = ..(1) 1
Also ar (∆ABC) = ar (∆PQR) (Given)
⇒ar ( ABC)ar ( PQR)
∆∆ = 1 1
From equation (1), we haveAB
PQ
2
2 =BC
QR
2
2 = CA
RP
2
2 =1
⇒ABPQ =
BCQR =
CARP
= 1 1
⇒ AB = PQBC = QRCA = RA
∴ By SSS congruency, ∆ABC ≅ ∆PQR. 1
P
BOA
Q
CB
A
D
E
1
23
P-61S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-45
SECTION A
1. For similar triangles,
area of triangle 1area of triangle 2 =
perimeter of 1perimeter of 2
∆∆
FHG
IKJ
2
=425
2FHG
IKJ =
16625
· 1
SECTION B
2. Since G is mid point of PQ,∴ PG = GQ
PGGQ = 1 1
According to question, GH || QR
∴PGGQ =
PHHR
(BPT)
1 =PHHR
⇒ PH = HR. 1Hence, H is mid-point of PR.
SECTION C
3. According to question, DE || AB
∴CDAD
=CEEB
(By BPT) ...(1) ½
Again since FE || DB,
∴CFFD
=CEEB
(By BPT) ...(2) ½
From equations (1) and (2),
CDAD
=CFFD
⇒ ADCD
FDCF
=
⇒ 1 + ADCD
=FDCF
+ 1 1
CD ADCD+
=FD FC
FC+
⇒ACCD
=CDFC
CD2 = AC.FC 1
P
Q R
G H
C
F
A B
ED
P-62 1–MRETT I C S X--M A T E M AH
SECTION D
4. Since, ∆ABC is equilateral, and AD || BC
∴ BD = DC = BC2
1
In right ∆ADB AD2 = AB2 – BD2
= AB2 + 2BC
4
= AB2 – 2AB
4
AD2 =34
AB2 = 4AD2 = 3AB2. (∵ AB = BC)
SUMMATIVE ASSESSMENT WORKSHEET-46
SECTION A
1. For similar triangles, we know that
area of ABCarea of DEF
∆∆ =
BC
EF
2
2
⇒80
area of DEF∆ =( )
( )
4
5
1625
2
2 =
⇒ area of ∆DEF =80 25
16×
= 125 cm2.
SECTION B
2. In ∆APB and ∆DPC, we have∠ A = ∠ D = 90°
and ∠ APB = ∠ DPC(vertically opposite angles) 1
Thus by AA criterion of similarity, we have∆APB ~ ∆DPC
⇒APDP
=PBPC
= AP × PC = PB × DP 1
SECTION C
3. To prove :ar ABCar DBC
( )( )∆∆ =
AODO
½
Construction : Draw AE ⊥BC and DF⊥BC.
Proof : ½
In ∆AOE and ∆DOF, ∠AOE = ∠DOF (Vertically opposite angles)∠AEO = ∠DFO = 90º (Construction)
⇒ ∆AOE ~ ∆DOF (AA)
∴AODO
=AEDF
...(1) 1
A
B D C
A
B
C
D
E
FO
A
B C
D
E F 4 cm 5 cm
A
B C
D
P
P-63S O L U T I O N S
Now,ar ABCar DBC
( )( )∆∆
=
12
BC AE
12
BC DF
× ×
× × =
AEDF
½
=AODO
[From equation (1)] ½
SECTION D
4. Proof : BC = 2 BD (AD is the median)
A
B D C
P
Q M R
and QR = 2QM (PM is the median) 1
Given,ABPQ =
ADPM
= BCQR
⇒ABPQ
=ADPM
= 2BD2QM
1
In triangles ABD and PQM,ABPQ
=ADPM
= BDQM
∴ ∆ABD ~ ∆PQM (SSS Similarty)⇒ ∠B = ∠Q, (By CPCT) 1
In ∆ABC and ∆PQR,ABPQ =
BCQR
and ∠B = ∠Q
∴ ∆ABC ~ ∆PQR. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-47
SECTION A
1. In the ∆ ABC,
DE || BC
∴ADDB
=AEEC
⇒x
x − 2=
xx
+−
21
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x2 – x = x2 – 4
⇒ x = 4. 1
A B
C
D
E
P-64 1–MRETT I C S X--M A T E M AH
SECTION B
2. Draw AC intersecting EF at G.In ∆CAB, GF || AB
⇒AGCG
=BFFC
...(i) 1
In ∆ADC, EG || DC
⇒AEED
=AGCG
...(ii)
From (i) and (ii), we haveAEED
=BFFC
· 1
SECTION C
3. In ∆ABC PQ || AB
∴BPPC
=AQQC (By BPT) ...(i) 1
Again in ∆BCD, PR || BD
⇒BPPC
=DRRC
(By BPT) ...(ii) 1
From (i) and (ii),AQQC =
DRRC
⇒ QR || AD (By converse of BPT) 1
SECTION D
4. ∵ ∆FEC ≅ ∆GBD⇒ EC = BD ...(1) 1It is given that ∠ 1 = ∠ 2⇒ AE = AD ...(2) 1
A
BF C
E1 2
G
D
3 4
From eqns. (1) and (2),
AEEC
=ADBD
⇒ DE || BC, (converse of BPT)⇒ ∠ 1 = ∠ 3 and ∠ 2 = ∠ 4 1Thus in ∆ADE and ∆ABC,
∠ A = ∠ A∠ 1 = ∠ 3∠ 2 = ∠ 4
So by AAA criterion of similarity, we have∆ADE ~ ∆ABC. Proved. 1
D C
BA
E F
B C
DA
R
P
Q
P-65S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-48
SECTION A
1. Since ST || QR
PSQS =
PTTR
⇒35 =
PTPR PT−
⇒35 =
PT28 PT−
⇒ 5PT = 84 – 3PT⇒ 8PT = 84
⇒ PT =848 = 10·5 cm 1
SECTION B
2. In ∆CAB, ∠A = ∠B (Given)∴ AC = CB (By isosceles triangle property) 1But, AD = BE (Given) ...(1)∴ AC – AD = CB – BE
CD = CE ...(2)
Dividing (2) by (1), we getCDAD
=CEBE
By converse of BPT, DE || AB. 1
SECTION C
3. In ∆ADB, AB2 = AD2 + BD2 (Pythagoras Theorem) ...(1)In ∆ADC, AC2 = AD2 + CD2 (Pythagoras Theorem) ...(2)Sultracting eqn. (2) from eqn. (1), we get
AB2 – AC2 = BD2 – CD2 1
=34
BC14
BC2 2F
HGIKJ − F
HGIKJ =
BC2
21
∴ 2(AB2 – AC2) = BC2
∴ 2(AB)2 = 2AC2 + BC2. 1
SECTION D
4. ∵ AB || PQ∴ ∠ABQ = ∠PQD (Corresponding angles) 1In ∆ADB and ∆PDQ, ∠ADB = ∠PDQ (Common)
∠ABQ = ∠PQDBy A.A. similarity, ∆ADB ~ ∆PDQ
∴DQDB
=PQAB
1
DQDB
=zx
...(i)
C
A B
D E
A
B CD3x x
A
P
C
B Q D
yz
x
P-66 1–MRETT I C S X--M A T E M AH
Similarly, ∆PBQ ~ ∆CBD
andBQDB
=zy ...(ii)
Adding (i) and (ii), we get zx
zy
+ = =DQ + BQDB
BDBD
1
⇒zx
zy
+ = 1
∴1 1x y
+ =1z · 1
SUMMATIVE ASSESSMENT WORKSHEET-49
SECTION A
1.ar ( ABC)ar ( PQR
ACPR
2
2
∆∆ )
= =ACPR
2FHG
IKJ
⇒81
169=
27·2PR
⇒27·2
PR
=29
13
⇒ 7·2PR =
913
⇒ PR =13 7·2
9×
= 10·4 cm. 1
SECTION B
2. ABCD is a square. ∆BCE is described on side BC is similar to ∆ACF described on diagonal AC.Since ABCD is a square.
So AB = BC = CD = DA and AC = 2 BCNow ∆BCE ~ ∆ACF
⇒Area ( BCE)Area ( ACF)
∆∆ =
2
2BCAC
1
⇒Area ( BCE)Area ( ACF)
∆∆ =
2
2BC 1
2( 2 BC)=
⇒ Area (∆BCE) = 12 Area (∆ACF) 1
SECTION C
3. Given : ∆ABC, right angled at A.BL and CM are medians.To prove : 4(BL2 + LM2) = 5BC2.Proof : In ∆ABL,
BL2 = AB2 + AL2 1
= AB2 + AC2
FHG
IKJ
2
(BL is median)
In ∆ACM, CM2 = AC2 + AM2
= AC2 + AB2
2FHG
IKJ , (CM is median) 1
BA
C
M
L
A B
C D
F
E
P-67S O L U T I O N S
BL2 + CM2 = AB2 + AC2 + AC
4AB
4
2 2+
4(BL2 + CM2) = 5AB2 + 5AC2
= 5(AB2 + AC2)
= 5BC2. Proved. 1
SECTION D
4. Given : In ∆ABC and ∆DEF, AP and DQ are medians, such that ABDE
=BCEF
= APDQ ...(i)
A
B CP
D
E FQ
To prove : ∆ABC ~ ∆DEF
Proof : From (1),ABDE
=12 BC
EF12
= APDQ 1
⇒ABDE
=BPEQ =
APDQ
⇒ ∆ABP ~ ∆DEQ [SSS similarity]⇒ ∠B = ∠E 1
In ∆ABC and ∆DEF,ABDE
=BCEF
1
and ∠B = ∠E, (By SAS criterion)∆ABC ~ ∆DEF. Proved 1
SUMMATIVE ASSESSMENT WORKSHEET-50
SECTION A
1.Perimeter of ABCPerimeter of PQR
∆∆ =
ABPQ
⇒6036 =
AB9
⇒ AB =60 9
36×
= 15 cm. 1
SECTION B
1. Construction : Join BD.Proof : AB || EF and AB | | CD∴ AB || CD || EF
In ∆ADB,AEED
=BGGD
(BPT)
In ∆BCD,BGGD
=BFFC
(BPT) 1
∴AEED
=BFFC
· 1A B
CD
GE F
P-68 1–MRETT I C S X--M A T E M AH
SECTION C
1. ∆AOB ~ ∆COD (A similarity)
By area theorem,ar ( AOB)ar ( COD)
∆∆ =
AB
DC
2
2 1
⇒ar ( AOB)ar ( COD)
∆∆ =
( )2 2CD
(CD)2 = 41
1
∴ ar (∆AOB) : ar (∆COD) = 4 : 1. 1
SECTION D
4. Given. The line segment XY is parallel to side AC of ∆ABC.Proof : In ∆BAC and ∆BXY, ∠B = ∠B (Common)
∠BAC = ∆BXY (Corresponding angles)∴ ∆BAC ~ ∆BXY (By AA similarity) 1
We have,ar ( BAC)ar ( BXY)
∆∆ =
BABX
2FHG
IKJ
2 ar ( BXY)ar ( BXY)× ∆
∆ =ABBX
FHG
IKJ
2
1
⇒ABBX
= 2
⇒BXAB
=12
⇒ 1 – BXAB
= 1 – 12
1
⇒AB BX
AB−
=2 1
2−
AXAB
=2 1
2−
· 1
SUMMATIVE ASSESSMENT WORKSHEET-51
SECTION A
1. We have
Area ( ABC)Area ( DEF)
∆∆ =
2
2BCEF
⇒9
16 =2
2(2·1)EF
⇒34
=2·1EF
⇒ EF =4 2·1
3×
= 2·8 cm 1
A B
CD
O
A
B CY
X
P-69S O L U T I O N S
SECTION B
2. Since, XY || QR
∴PXXQ =
PYYR
⇒12
=PY
PR PY− = 4
PR 4− 1
⇒ PR – 4 = 8 ⇒ PR = 12 cm
∴In right ∆PQR, QR2 = PR2 – PQ2
= 122 – 62 = 144 –36 = 108
QR = 6 3 cm 1
SECTION C
3. Given : In ∆ABC, AD⊥BC and AD2 = BD × CD.To prove : ABC is a right triangle.Proof : In right triangles ABD and ACD, applying Pythagoras theorem
AB2 = AD2 + BD2
and AC2 = AD2 + CD2 1∴ AB2 + AC2 = 2AD2 + BD2 + CD2
= 2(BD × CD) + BD2 +CD2,given AD2 = BD × CD 1
= (BD + CD)2 = BC2
i.e., AB2 + AC2 = BC2
⇒ ∆ABC is a right triangle. 1
SECTION D
4. In trapezium ABCD, AB || DC and DC = 2AB.
Also,BEEC
=43
In trapezium ABCD, EF || AB || CD
∴AFFD
BEBC
= =43
In ∆BGE and ∆BDC, ∠B = ∠B∠BEG = ∠BCD (Common corresponding angles)
∴ ∆BGE ~ ∆BDC (AA similarity) 1
⇒EGCD
=BEBC
...(1)
As,BEEC =
43 ⇒
ECBE
= 34
⇒ECBE
+ 1 =34
+ 1 1
⇒EC+ BE
BE=
74
RQ
P
X Y
4 cm
A
B D C
D C
EG
F
A B
P-70 1–MRETT I C S X--M A T E M AH
⇒BCBE
=74
⇒ BEBC
= 47
⇒EGCD
=47
⇒ EG =47
CD ...(i)
Similarly, ∆DGF ~ ∆DBA
⇒DFDA
=FGAB
1
⇒FGAB
=37
⇒ FG =37
AB
∵ AFFD
BEBD
ECBC
= =
⇒ =
L
N
MMMM
O
Q
PPPP
4737
...(ii)
Adding (i) and (ii), EG + FG =47
CD + 37
AB
⇒ EF =47
237
× +AB AB =87
AB+37
AB =117
ABb g∴ 7EF = 11AB. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-52
SECTION A
1.Area ( AOB)Area ( COD)
∆∆ =
41
⇒84
Area ( COD)∆ =41
Area (∆COD) =844
= 21 cm2. 1
SECTION B
2. Proof : In ∆POQ, AB || PQ, (Given)
∴OAAP
=OBBQ ...(i) (BPT) 1
In ∆OPR, AC || PR, (Given)
∴OAAP
=OCCR
...(ii) (BPT ) 1
From (i) and (ii), we getOBBQ =
OCCR
∴ BC || QR, (By converse of BPT) 1
A B
C D
O
P-71S O L U T I O N S
SECTION C
3. Since DE || BC.
By BPT, we have
ADDB
=AEEC
⇒DBAD
=ECAE
1
Adding 1 on both sides,
DBAD
+ 1 =ECAE
+ 1
DB ADAD+
=EC AE
AE+
1
ABAD
=ACAE
⇒ADAB
=AEAC
· 1
SECTION D
4. In ∆ABC, DP || BC
ADDB
=APPC
, (BPT) ...(1)
A
D
E
P
BQ C
Similarily, in ∆ABC EQ || AC
⇒BQQC =
BEEA
..(2) 1
From figure, EA = AD + DE (Given, BE = AD)= BE + ED 1= BD
Then (2) becomes,BQQC =
ADBD
...(3)
From (1) and (3), we get 1
APPC
=BQQC
∴ By converse of BPT, PQ || AB. 1
A
B
D
C
E
P-72 1–MRETT I C S X--M A T E M AH
FORMATIVE ASSESSMENT WORKSHEET-53
Objective Type Questions
1. (D) 2. (A) 3. (C) 4. (C) 5. (A)
Fill in the blanks
1. Same2. Parallel3. Similar4. Proportional5. SAS
Quiz
1. In given figure B'C'||BC. Find AB.
32
=AB3·6
AB =3·6 3
2×
AB = 5·42 cm2. In a right angled triangle, the square of hypotenuse is equal to the sum of the squares of the other
two sides.3. AB = AD + BD = 1·5 + 3 = 4·5 cm.
According to BPT,ADAB
=AEAC
or1·54·5 =
1AC
⇒13 =
1AC
⇒ AC = 3 cm.EC = AC – AE = 3 – 1 = 2 cm.
4. According to BPT,ADAB
=DBEC =
7·25·4 =
43
AD1·8 =
43
⇒ AD =1·8 4
3×
= 2·4 cm
5. No, because corresponding sides are not in proportion.
A
B C
2 3
4
P
Q R
6 5
4
●●
P-73S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET - 54
SECTION A
1. tan is not defined when θ is equal to 90º i.e., tan 90º = ∞. 1
2. cos 48º cos 42º – sin 48º sin 42º = cos (90º – 42º) cos (90º – 48º) – sin 48º sin 42º
= sin 42º sin 48º – sin 48º sin 42º = 0 1
SECTION B
3.sin 90°cos45° +
1cosec 30° =
11 / 2 +
12
1
= 2 + 12
=2 2+1
21
SECTION C
4. L.H.S. =1 cos1 cos
− θ+ θ =
1 cos 601 cos 60
− °+ °
=1212
11
−+
= 1 12 23 32 2
13
= = 1
Introduction toTrigonometry and
Trigonometric Identities
5CHAPTER
P-74 1–MRETT I C S X--M A T E M AH
R.H.S. =sin
1 cosθ
+ θ
=sin 60
1 cos 60°
+ °
=3 / 2
11+2
=3 / 2
3 / 21
=13
= L.H.S. 1
Hence relation is verified for q = 60°.
SECTION D
5. 4( )4 4sin 30°+cos 60° – 3( )2 2cos 45° – sin 90°
= 4
4 41 1+
2 2
é ùæ ö æ öê úç ÷ ç ÷ç ÷ ç ÷ê úè ø è øê úë û– 3 ( )
221
– 12
é ùæ öê úç ÷ç ÷ê úè øê úë û1
= 41 1
+16 16é ùê úê úë û
– 31
-12é ùê úê úë û
1
=1 3
+2 2
1
=42
= 2. 1
SUMMATIVE ASSESSMENT WORKSHEET-55
SECTION A
1. sin 38° sin 52° – cos 38° cos 52° = sin (90° – 52°) sin (90° – 38°) – cos 38° cos 52°= cos 52° cos 28° – cos 38° cos 52° = 0. 1
2. sin θ = cos θ
⇒sincos
θθ = 1
⇒ tan θ = 1 = tan 45°θ = 45°. 1
SECTION B
3. ∵ sin θ – cos θ =12
Squaring on both sides
(sin θ – cos) 2 =21
2æ öç ÷ç ÷è ø
P-75S O L U T I O N S
Þ sin2 θ + cos2 θ – 2 sin θ cos θ =14
1 – 2 sin θ cos θ =14
2 sin θ cos θ = 1 – 14
=34
1
Again (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ
= 1 + 34
= 74
Þ sin θ+ cos θ =74
= 7
21
SECTION C
4.2 2 2
2 25 cos 60° + 4 cos 30° – tan 45°
sin 30° + cos 60°=
222
2 2
1 35 4 (1)
2 21 12 2
+ − +
1
=
5 + 3 – 141 1+4 4
1
=
5 + 2412
=
1326 134 ·
1 4 22
= = 1
SECTION D
5. Given : sinq =2 2+
c
c d∵ cos2 q = 1 – sin2 q
= 1 –
2
2 2
c
c d
æ öç ÷ç ÷
+è ø1
= 1 – 2
2 2
cc d+
=2 2 2
2 2c +d c
c +d-
=2
2 2d
c +d1
Þ cos q = 2 2
d
c d+1
P-76 1–MRETT I C S X--M A T E M AH
Again, tanq =sincos
=2 2
2 2
/
/
c c d
d c d
+
+
=cd 1
SUMMATIVE ASSESSMENT WORKSHEET-56
SECTION A
1. sin 45º + cos 45º =12
12
22
2+ = = . 1
2. (sec A + tan A ) (1 – sinA) =1
cossin
AA
cos A+F
HGIKJ (1 – sin A)
=1 sin A
cos A+
(1 – sin A)
=1 2− sin
cosA
A
=coscos
cos2 AA
A= 1
SECTION B
3. L.H.S. =11
−+
coscos
AA
=11
11
−+
× −−
coscos
coscos
AA
AA
1
=2
2(1 cos A)(1 cos A)
−−
=2
2(1 cos A)
sin A−
=1 cos A
sin A−
=cos A1
sin A sin A−
= cosec A – cot A = R.H.S. Proved. 1
SECTION C
4. L.H.S. = cosec2q – tan2 (90° – q )
= 21
sin q – 2
2sin (90° )cos (90° )
-
-
P-77S O L U T I O N S
=12sin q
– 2
2sin (90° – )
sin
θ
θ1
=1
2sin q –
2
2cos
sin
q
q
=1 cos2
2sin
- q
q
= 2
2sinsin
θθ
1
= 1= sin2 θ + cos2 θ= sin2 θ + sin2 (90° – θ)= R.H.S. 1
SECTION D
Sol. cosec2 (90º – θ) = sec2 θsec2 θ – tan2 θ = 1 1
cos2 40º + cos2 50º = cos2 (90º – 50º) + cos2 50ºsin2 50º + cos2 50º = 1
tan2 30º =13
13
2FHG
IKJ = 1
sec2 52ºsin2 38º = sec2 52º.sin2 (90º – 52º) = sec2 52º.cos2 52º = 1
cosec2 70º – tan2 20º = cosec2 (90º – 20º) – tan2 20º = sec2 20º – tan2 20º = 1 1
∴ Given expression = 14
213
1
3 1−
× ×
( )
=14
29
9 836
136
− = =–
· 1
SUMMATIVE ASSESSMENT WORKSHEET-57
SECTION A
1. tan θ = 1,5 sin 4 cos5 sin 4 cos
q + qq - q =
5cos + 4 5 (1) + 4 9= =
5cos 4 5 (1) 4 1qq - - = 9. 1
2. sin2 A = 2 sin A⇒ sin A (sin A – 2) = 0Since, sin A ≠ 2, ∴ sin A = 0So, A = 0º. 1
SECTION B
3. 2 2 232 cosec 30 3 sin 60 tan 30
4° + ° − ° =
22 3 3 1
2(2) 32 4 3
+ − 1
=9 3
44 4
+ −
P-78 1–MRETT I C S X--M A T E M AH
=16 9 3
4+ −
=25 3
4−
1
SECTION C
4. L.H.S. =cos sin
cos sincos – sin
cos – sin
3 3 3 3θ θθ θ
θ θθ θ
++
+
=2 2(cos sin ) (cos sin sin cos )
(cos sin )θ + θ θ + θ − θ θ
θ + θ
+ 2 2(cos sin ) (cos sin sin cos )
(cos sin )θ − θ θ + θ + θ θ
θ − θ 1
= (1 – sin θ cos θ) + (1 + sin θ cos θ) 1= 2 – sin θ cos θ + sin θ cos θ 1= 2 = R.H.S. Proved
SECTION D
5. sin2 30° cos2 45° + 4 tan2 30° + 12
sin2 90° – 2 cos2 90° +124
=2 2 2
21 1 1 1 14 (1) 2(0)
2 2 242 3 × + + − +
1
= 1 1 4 1 1
+ + +4 2 3 2 24æ öç ÷ç ÷è ø
1
=1 4 1 1
+ + +8 3 24 2
=3+32+1+12
241
=4824
= 2. 1
SUMMATIVE ASSESSMENT WORKSHEET-58
SECTION A
1. sin2 60º – sin2 30º =
2 23 12 2
−
=3 14 4
− = 2 14 2
= · 1
2. x = a cos θ, y = b sin θNow, b2x2 + a2y2 – a2b2 = b2(a cos θ)2 + a2 (b sin θ)2 – a2b2
= a2b2 cos2 θ + a2b2 sin2 θ – a2b2
= a2b2 (sin2 θ + cos2 θ) – a2b2
= a2b2 – a2 b2 = 0. 1
P-79S O L U T I O N S
SECTION B
3. 2 cot2 A – 1 = 2(cosec2 A – 1) – 1
=22sin A
– 3
=2
2– 3
32
æ öç ÷ç ÷è ø
=8 1
33 3
-- = 1
SECTION C
4. According to question,
sin 3θ = cos (θ – 6º) 1
⇒ cos (90º – 3θ) = cos (θ – 6º)
⇒ 90º – 3θ = θ – 6º 1
⇒ 4θ = 90º + 6º = 96º
θ =964
º = 24º 1
SECTION D
5. tan θ =ABBC
= 15
In ∆ABC, AC2 = AB2 + BC2
= 1 + 5 = 6 1
⇒ AC = 6
(i)2 2
2 2cosec seccosec sec
θ − θθ + θ
=
22
22
6( 6)5
6( 6)5
−
+
=
665665
−
+
=2436
23
= · 1½
(ii) sin2 θ + cos2 θ =16
56
2 2FHG
IKJ +
FHG
IKJ
=16
56
+ = 1. ½
A
CB
16
5
θ
P-80 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-59
SECTION A
1. cosec θ – cot θ =14
⇒(cosec cot )(cosec cot )
(cosec cot )q - q q + q
q + q =14
⇒2 2cosec cot
cosec cotq - qq + q
=14
⇒2 21 cot cot
cosec cot+ q - q
q + q=
14
⇒ cosec θ + cot θ = 4. 1
2.11 11
2 2cot cosθ θ− =
2
2 2sin 1
11 –cos cos
θ θ θ
= 11 2
2sin 1
cos
θ − θ
= – 112
21 sin
cos
− θ θ
= – 11 cos
cos
2
2θθ
FHG
IKJ = – 11. 1
SECTION B
3. Here 3 sin θ – cos θ = 0 and 0º < θ < 90º
⇒ 3 sin θ = cos θ
⇒sincos
θθ =
13
1
⇒ tan θ =13
= tan 30º ∵ tansincos
θ θθ
=LNM
OQP
⇒ θ = 30º. 1
SECTION C
4. (i) Clearly, distance covered by the artist is equal to the lengthof the rope AC. Let AB be the vertical pole of height 12 m.It is given that ∠ACB = 30ºThus, in right-angled triangle ABC,
sin 30º =ABAC
⇒12
=12AC
⇒ AC = 24 m.Hence, the distance covered by the circus artist is 24 m. 2
12 mRope
30º
A
BC
P-81S O L U T I O N S
(ii) Trigonometric ratios of an acute angle of a right angled triangle. ½(iii) Single mindedness helps us to gain success in life. ½
SECTION D
5. L.H.S. =sec 1 sec 1sec 1 sec –1
q - q ++
q + q
=(sec 1) (sec 1)
(sec 1)(sec 1)q - + q +q + q - 1
=2
2sec 2sectansec 1
q q=
qq - 1
= 2 × 1 cos
cos sinq
´q q
= 2 × 1
sin q 1
= 2cosec θ= R.H.S.
SUMMATIVE ASSESSMENT WORKSHEET-60
SECTION A
9. tan θ + cot θ = 5On squaring both sides
tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 tan θ 1
tanθ = 25
⇒ tan2 θ + cot2 θ = 23. 12. sec 2A = cosec (A – 27º)
We know thatsec 2A = cosec (90º – 2A)
⇒ A – 27º = 90º – 2A
⇒ 3A = 117º
⇒ A =117
3º
= 39º. 1
SECTION B
3.6sin 23 sec79 3tan 48
cosec11 3cot 42 6cos 67°+ °+ °°+ °+ ° =
6cos (90° – 23)º + cosec (90° – 79)º+ 3cot (90° – 48)ºcosec 11º + 3cot 42º + 6co s 67º 1
=6cos 67º +cosec11º + 3cot 42ºcosec 11º + 3cot 42º + cos 67°
= 1. 1
P-82 1–MRETT I C S X--M A T E M AH
SECTION C
4. (i) Let AB be the tree broken at a point C such that the broken part CB takes the position CO andstrikes the ground at O. It is given that OA = 30 m and ∠AOC = 30º.Let AC = x and CB = y, then CO = y
In ∆OAC, we have tan 30º =ACOA
⇒13
=x
30
⇒ x =30
3 = 10 3 1
Again in ∆OAC, we have cos 30º =OAOC
⇒ 32
=30y
⇒ y =60
3 = 20 3
∴ Height of the tree = (x + y) = 10 3 + 20 3 = 30 3 1= 30 × 1·732 = 51.96 m
(ii) Trigonometric ratios of an acute angle of a right angled triangle. ½(iii) The problem of decreasing ratio of trees and land is discussed here. 1
SECTION D
5. 15 tan2 θ + 4 sec2 θ = 2315 tan2 θ + 4(tan2 θ + 1) = 23 115 tan2 θ + 4 tan2 θ + 4 = 23
19 tan2 θ = 19⇒ tan θ = 1 = tan 45º 1⇒ θ = 45ºNow, (sec θ + cosec θ)2 – sin2 θ = (sec 45º + cosec 45º)2 – sin2 45º
= 2 212
22
+ −FHG
IKJe j 1
= 2 212
2e j −
= 812
152
− = 1
SUMMATIVE ASSESSMENT WORKSHEET-61
SECTION A
1. Given :
sin θ =ab
From Fig. AB = b a2 2−
tan θ =ACAB
=−
a
b a2 2· 1
2. cos2 x + sin2 x = 1 1
y
x
A
C
O 30º30 m
y
B
B
C
Aθ
2 2b a−
ba
P-83S O L U T I O N S
SECTION B
3. Given : 5 cosec θ = 7
⇒ cosec θ =75
⇒ sin θ =57
1cosec
sin
θ = θ ∵ 1
sin θ + cos2 θ – 1 = sin θ – (1 – cos2 θ)
= sin θ – sin2 θ
=57
57
35 2549
1049
2− FHG
IKJ = =–
1
SECTION C
4. cos θ + sin θ = 2 cos θ
⇒ sin θ = cos θ ( 2 – 1)
⇒ sin θ =cos ( 2 1) ( 2 1)
( 2 1)
q - ++ 1
⇒ sin θ =cos ( )θ 2 1
2 1−
+
( 2 + 1) sin θ = cos θ 1
⇒ 2 sin q + sin θ = cos θ
⇒ cos θ – sin θ = 2 sin q . Proved. 1
SECTION D
5. sin A =15
, cos A = 1 sin A2−
∴ cos A = 115
2
−FHG
IKJ = 1
15
− = 25
1
sin B =110
, cos B = 1 2− sin B
∴ cos B = 1110
2
−FHG
IKJ = 1
110
− = 310
1
Now cos ( A + B) = cos Acos B – sin Asin B
=25
310
15
110
× ×–
=650
150
− = 550
= 5
2 212
= 1
cos (A + B) =12
= cos 45º
∴ A + B = 45º. 1
P-84 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-62
SECTION A1. ∵ sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = 45°
Also cos θ =12
Now, 2 tan θ + cos2 θ = 2(1) + 12
2FHG
IKJ
= 2 + 12
52
= · 1
2. cosec θ = 2
⇒ sin θ =12
= sin 30º
⇒ θ = 30ºNow cot θ = 3p
⇒ cot 30º = 3p⇒ 3 = 3p⇒ p = 1. 1
SECTION B
3.2 2 2 2 2
2 2 2 2sec (90º – )– cot 2cos 60º tan 28º tan 62º
–2(sin 25º +sin 65º ) 3(sec 43º – cot 47º )
q q
=2 2
2 2
2 2 2 2
1 12× × tan 28º×cot 28º(cosec – cot ) 2 2–2(sin 25º+ cos 25º ) 3[sec 43º – tan 43º ]
q q 1
=2
21 1× tan 28º×
1 2 tan 28º–2×(1) 3
½
=1 1
–2 6 =
13 ½
SECTION C
4. LHS =1 1 1 cosec A+cot A
cosec A–cot A sin A cosec A–cot A cosec A+cot A
æ öç ÷- = ´ç ÷è ø
– cosec A
=cos
cosec A + cot A
ec A – cot A2 2 – cosec A 1
=cosec A + cot A
1 – cosec A = cot A ½
Now RHS = 1 1
sin cosA ec A + cot A− = cosec A – 1
coscoscosec A + cot A
ec A – cot Aec A – cot A
× FHG
IKJ 1
= cosec A – 2 2
(cosec A–cot A)cosec A cot A-
= cosec A – (cosec A cot A)
1-
= cot A ½
∴ LHS = RHS Proved.
B
C
Aθ
1
1
2
P-85S O L U T I O N S
SECTION D
5. 2 sin (3x – 15°) = 3
sin (3x – 15)° =3
2sin (3x – 15)° = sin 60° 1
⇒ 3x – 15 = 60°⇒ 3x = 60 + 15 = 75°
⇒ x =753 = 25° 1
Now sin2 (2x + 10)° + tan2 (x + 5)° = sin2 (50 + 10)° + tan2 (25 + 5)°= sin2 60° + tan2 30°
=
2 23 12 3
+ 1
=3 14 3
+
=9 712+
= 13
·12
1
SUMMATIVE ASSESSMENT WORKSHEET-63
SECTION A
1. Given : sec θ + tan θ = 7
⇒(sec tan )(sec tan )
(sec tan )θ θ θ θ
θ θ+ −
−= 7
⇒sec tan
sec tan
2 2θ θθ θ
−−
= 7
⇒1
sec tanθ θ− = 7
⇒ sec θ – tan θ =17
· 1
2.2
21 sec A
cosec A 1−
− =1 1
1 1
2
2− ++ −( tan )
cot
A
A
=− =tan
cot
– tan
/ tan
2
2
2
21
A
A
A
A = – tan4 A 1
SECTION B
3. Given : 2 sin 2θ = 3
⇒ sin 2θ =3
2 = sin 60º 1
⇒ 2θ = 60º
Hence, cos 2θ = cos 60º = 12
· 1
P-86 1–MRETT I C S X--M A T E M AH
SECTION C4. (i) Let A be the kite and CA be the string attched to the kite such that its one end is tied to a point
C on the ground. The inclination of the string CA with the ground is 60º.
In ∆ABC, we are given that ∠C = 60º and perpendicular AB = 60 m.
∴ sin C =ABAC
⇒ sin 60º =ABAC
⇒3
2=
60AC
⇒ AC =120
3 = 40 3 m
Hence, the length of the string is 40 3 m. 2(ii) Trigonometric ratios of an acute angle of a right angled triangle. ½(iii) Playing makes children’s mind and body healthy. ½
SECTION D
5. Consider an equilateral triangle of sides ‘a’ unitsA
B C
a
30°
60° 60°
a
a D
∴ BD = CD = ·2a
. Also ∠BAD = ∠CAD = 30°
As ∠A = ∠B = ∠C = 60°. In the right ∆ADB, ∠ADB = 90°∴ AD2 = AB2 – BD2 (By Pythagoras Theorem) 1
⇒ AD2 = a2 – 2
4a
= 23
4a
1
AD =3
2a
tan 30° =BDAB
= / 2
3 / 2a
a = 1
·3 1
SUMMATIVE ASSESSMENT WORKSHEET-64
SECTION A1. The maximum value of sin θ is 1. 1
2. tan θ =78
( sin )( sin )( cos )( cos )1 11 1
+ −+ −
θ θθ θ =
1
1
2
2−−
sin
cos
θθ
= cos
sin
2
2θθ
= cot2 θ = 1
2tan θ
=1
7 8
64492( / )
= ·
60º
A
BC
60 m
P-87S O L U T I O N S
SECTION B
3. L.H.S. = – 1 + sin sin ( º )
cot ( º )A A
A90
90−
−
= – 1 + sin A cos A
tan A½
= – 1 + sin A cos A × cot A ½
= – 1 + sin Acos A × cos Asin A
½
= – 1 + cos2 A = – (1 – cos2 A) ½
= – sin2 A = R.H.S.
SECTION C
4. Let AB = x∵ AC – AB = 1⇒ AC = x + 1∵ AC2 = AB2 + BC2
∴ (x + 1)2 = x2 + (5)2
⇒ x2 + 2x + 1 = x2 + 25
⇒ 2x = 24 ⇒ x = 242
= 12 1
Hence, AB = 12, AC = 13
sin C =ABAC
1213
=
cos C =BCAC
513
= 1
Now11
++
sincos
CC
=1 12
13
1 513
25131813
2518
+
+= = · 1
SECTION D
5. L.H.S. =tan sintan sin
θ θθ θ
+−
=
sinsin
cossin
sincos
q + qqq - qq
1½
=
1sin 1
cos 1
sin –1cos
æ öç ÷q +ç ÷qè øæ öç ÷qç ÷qè ø
1½
=sec 1sec 1
q+q - = R.H.S. Proved. 1
C
A
B
x x + 1
5 cm
P-88 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-65
SECTION A
1. Given : 5 tan θ = 4
Now5 35 3
sin cossin cos
θ θθ θ
−+ =
5 3
5 3
5 35 3
sincossincos
tantan
θθθθ
θθ
FHG
IKJ −
FHG
IKJ +
= −+
(Divide Numerator & Denominato by cos θ)
=4 34 3
14 3
17
−+
=+
= · 1
2. Given : 3x = sec θ, and 3x
= tan θ
∴ (3x)2 + 3 2
xFHG
IKJ = sec2 θ – tan2 θ
⇒ 22
19 x
x − = 1
∴ x2 – 12x
=19
1
SECTION B
3. R.H.S. =2 2
2 2
1 (cosec cot ) –1
1 (cosec cot ) 1
- q + q=
+ q + q +p
p
=2 2
2 2
cosec cot 2cosec .cot –1
cosec cot 2cosec .cot 1
q + q + q qq + q + q q +
=2 2
2 2
1 cot cot 2cosec cot 1
cosec cosec 1 2cosec cot 1
+ q + q + q q -q + q - + q q +
1
=2cot (cot cosec )
2 cosec (cosec cot )q q + qq q + q
=cossin
θθ × sin θ = cos θ = LHS. Proved 1
SECTION C
4. L.H.S. = cos sincot
A1 – tan A
A A
+−1
=cos
sincos
sinAAA
Acos Asin A
1 1− FHG
IKJ
+− FHG
IKJ
1
=cos
cos sinsin
sin cos
2 2AA A
AA A−
+−
=cos
cos sin–
sincos sin
2 2AA A
AA A− −
1
P-89S O L U T I O N S
=cos sincos sin
2 2A AA A
−−
=(cos sin )(cos sin )
(cos sin )A A A A
A A− +
−= cos A + sin A = R.H.S. Proved. 1
SECTION D
5. L.H.S. =cosec A cosec A
cosec A–1 cosec A+1+ 1
=2 2cosec A+cosec A+cosec A–cosec A
(cosec A–1)(cosec A+1)
=2
2
2cosec A
cosec A–1
=2
2
2cosec A
cot A1
=2
2
2
2sin Acos Asin A
=2
2 2
2 sin Asin A cos A
´ 1
= 2
2cos A
= 2 sec2 A = R.H.S. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-66
SECTION A
1. Given : 5 tan θ = 12
tan θ =125
So13
3sinθ
=133
1213
FHG
IKJ = 4. 1
2.sin
cosθ
θ1 + =sin ( cos )
( cos )( cos )θ θ
θ θ1
1 1−
+ −
=sin ( cos )
cos
sin ( cos )
sin
θ θθ
θ θθ
1
1
12 2
−−
= − =
1 − cossin
θθ 1.
SECTION B
3. Given : 4 cos θ = 11sin θ
⇒ cos θ =114
sin θ
B
C
Aθ
12
5
13
P-90 1–MRETT I C S X--M A T E M AH
Now11 711 7
cos sincos sin
θ θθ θ
−+ =
1111 sin 7sin
411
11 sin 7sin4
´ q - q
´ q + q1
=sin
sin
θ
θ
1214
7
1214
7
−FHG
IKJ
+FHG
IKJ
= 121 28121 28
93149
–+
= ·
SECTION C
4. Let sec θ + tan θ = λ ...(1)
We know that
sec2 θ – tan2 θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1 1
⇒ λ(sec θ – tan θ) = 1
⇒ sec θ – tan θ =1λ ...(2)
Adding eqns. (1) and (2), we get
2sec θ = λ + 1λ
⇒ 2 xx
+FHG
IKJ
14
= λ + 1λ
⇒ 2x + 12x
= λ + 1λ 1
Comparing both sides, we get
λ = 2x or λ = 12x
⇒ sec θ + tan θ = 2x or 12x
· 1
SECTION D
5. Given : cosec θ = 5 Þ sin θ = 15
So, cos2 θ = 1 – sin2 θ = 1 – 15
45
=
∴ cos θ =25
1
(i) cot θ – cosec θ =θ − θθ
coscosec
sin
= −2 / 55
1 / 5
= 2 – 5 1½
(ii) sin2 θ + cos2 θ =15
25
15
45
2 2FHG
IKJ +
FHG
IKJ = + = 1. 1½
P-91S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-67
SECTION A
1. 22tan 30º
1 tan 30º+= 2
1 2 223 3 3
1 41 11 3 33
= = ++
=2 3
43× =
32
= sin 60º. 1
2. tan A B
2+F
HGIKJ = tan
C90º –
2æ öç ÷ç ÷è ø
∵ A2
B2
C2
+ + =LNM
OQP90º
= cot C2
· 1
SECTION B
3. In the ∆ABP, sin 30º =ABAP
⇒12
=50AP
⇒ AP = 100 cm ½
In the ∆ AQD, sin 30º =ADAQ ½
⇒12
=20AQ ⇒ AQ = 40 cm ½
Now, the length of (AP + AQ) = 100 + 40 = 140 cm.
SECTION C
4.2 2cos (45º+ )+cos (45º – )
tan (60º+ ) tan (30º – )q qq q
+ cosec (75º + θ) – sec (15º – θ)
= 2 2cos (45º + )+sin (90º – 45º + )
tan (60º + )cot (90º – 30º + )q qq q
+ cosec (75º + θ) – cosec (90º – 15º + θ) 1
= 2 2cos (45º + )+sin (45º+ )
tan (60º + ).cot (60º + )q qq q
+ cosec (75º + θ) – cosec (75º + θ) 1
= 11
= 1. 1
SECTION D
5. L.H.S. = (1 + cot A – cosec A) (1 + tan A + sec A)
= 11
11+ −F
HGIKJ + +FHG
IKJ
cossin sin
sincos cos
AA A
AA A 1
=sin cos
sinsin cos
cosA A
A
A AA
+ −FHG
IKJ
+ +FHG
IKJ
1 1
P-92 1–MRETT I C S X--M A T E M AH
=(sin cos )
sin cosA A
A A+ −2 21
1
=sin cos sin cos –
sin cos
2 2 2 1A A A AA A
+ +1
=1 2+ sin cos
sin cosA A – 1
A A= 2 = R.H.S. 1
SUMMATIVE ASSESSMENT WORKSHEET-68SECTION A
1. sin (x –20)º = cos (3x –10)º= sin [90º – (3x –10)º]
⇒ x – 20 = 90 – 3x + 10⇒ 4x = 90 + 10 + 20 = 120
⇒ x =120
4 = 30. 1
2. tan θ =64 28
=h
h =6 28
4×
= 42. 1
SECTION B3. L.H.S. = (cosec θ – cot θ)2
⇒21 cos
sin sinθ − θ θ
⇒21 cos
sin− θ
θ
=2
2(1 cos )1 cos
− θ− θ
=2(1 cos )
(1 cos )(1 cos )− θ
− θ + θ
=1 cos1 cos
− θ+ θ
= R.H.S.
SECTION C4. We have m2 = a2cos2 θ + 2absin θ cos θ + b2sin2 θ ...(i) 1
and n2 = a2sin2 θ – 2absin θ cos θ + b2cos2 θ ...(ii) 1Adding (i) and (ii), we get
m2 + n2 = a2(cos2 θ + sin2 θ) + b2(cos2 θ + sin2 θ)= a2 (1) + b2 (1)= a2 + b2 = RHS. Proved. 1
SECTION D
5. L.H.S. =tan
cotcot
tanθ
θθ
θ1 1−+
−
=
1tan tan
1 1 tan1tan
q q+- q-
q
C
A Bθ
6
4
C
A Bθ
h
28
P-93S O L U T I O N S
=2tan 1
tan 1 tan (1 tan )q +
q- q - q 1½
=3tan 1
(tan 1) tanq -
q - q 1
=2 2(tan 1)(tan tan 1) tan tan 1
(tan 1)(tan ) tanq - q + q + q + q +
=q - q q 1
= tan θ + 1 + cot θ = R.H.S. ½
SUMMATIVE ASSESSMENT WORKSHEET-69
SECTION A
1. tan A =8
15
Now cosec2 A –1 =12sin A
– 1 = 1 2
2− sin
sin
A
A
=cos2 A
sin A2 = cot2 A
=1 1
8 152 2tan ( / )A= =
22564
1
2. tan x = sin 45º cos 45º + sin 30º
=12
12
12
FHG
IKJFHG
IKJ +
=12
12
+ = 1 = tan 45º
⇒ x = 45º. 1
SECTION B
3. ∠C = 90º (Angle in a semi-circle)
∴ AB = ( ) ( )3 2 9 4 132 2+ = + = ½
tan A =BCAC
23
= ½
tan B =ACBC
32
= ½
∴ tan A tan B =23
32
. = 1. ½
SECTION C
4. cosec θ =1312
sin θ =1213
cos2 θ = 1 – sin2 θ
= 1 –
21213
1
√13 cm
P-94 1–MRETT I C S X--M A T E M AH
=−169 144
169 = 25
169
⇒ cos θ =5
13
Nowθ − θθ − θ
2 sin 3 cos4 sin 9 cos
=
× − × × − ×
12 52 313 1312 54 913 13
1
=−−
24 1548 45 =
93 = 3. 1
SECTION D
5. L.H.S. =tan sectan sec
θ θθ θ
+ −− +
11
=2 2tan sec (sec tan )
tan sec 1q + q - q - q
q - q +[∵ 1 = sec2 θ – tan2 θ]
=(tan sec ) [(sec tan ) (sec tan )]
tan sec 1θ + θ − θ + θ θ − θ
θ − θ + 1
= (tan θ + sec θ) [ (sec tan )]
tan sec1
1− −
− +θ θ
θ θ 1
= (tan θ + sec θ) [ sec tan ][ sec tan ]11
− +− +
θ θθ θ 1
=sincos cos
θθ θ
+1
=1 + sin
cosθ
θ = R.H.S. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-70
SECTION A
1. Maximum value of θ1
,sec i.e., cos θ = 1 1
SECTION B
2.cos 45° 1
+sec 30° sec 60° =
112 +
2 23
=12
32
12
× +
=6
412
+ 1
=6 24+
·
SECTION C
3. L.H.S. =sin cossin cos
sin cossin cos
θ θθ θ
θ θθ θ
−+
++−
P-95S O L U T I O N S
=(sin cos ) (sin cos )
sin cos
θ θ θ θθ θ
− + +−
2 2
2 2
=(sin cos ) – sin cos (sin cos ) sin cos
sin ( sin )
2 2 2 2
2 22 2
1
θ θ θ θ θ θ θ θθ θ
+ + + +− −
1
=1 1
12 2+
− +sin sinθ θ1
=2
2 12sin θ − = R.H.S. Proved. 1
4. sec º .sin º cos º .cos º (tan º .tan º .tan º)
(sin º sin º )
41 49 29 6123
20 60 70
3 31 592 2
+ −
+
ec
=2 2
2cosec (90º– 41º)sin 49º cos 29º.sec (90º– 61º) [tan 20º. 3 cot (90º– 70º)]
33[sin 31º cos (90º –59º )]
+ -
+1
=2 2
2cosec 49º.sin 49º cos 29º.sec 29º [tan 20º 3.cot 20º ]
33(sin 31º cos 31º )
+ -
+1
=1 1 2
3+ −
= 2 2
3−
= 0 1
SECTION D5. L.H.S. = 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ
= 2 sec2 θ – (sec2 θ)2 – 2 cosec2 θ + (cosec2 θ)2 1= 2 (1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2 1= 2 + 2 tan2 θ – 1 – 2 tan2 θ – tan4 θ – 2 – 2 cot2 θ + 1 + 2 cot2 θ + cot4 θ 1= cot4 θ – tan4 θ= R.H.S. 1
SUMMATIVE ASSESSMENT WORKSHEET-71
SECTION A
1. AB = (AC) BC)2 2− (
= ( ) ( )17 8 289 642 2− = −= 225 = 15
15sec A + 8cot A = 151715
FHG
IKJ + 8
158
FHG
IKJ
= 17 + 15 = 32. 1
2. sin2 A =12
tan2 45º
=12
× (1)2 = 12
2FHG
IKJ
⇒ sin A =12
= sin 45º
⇒ A = 45º 1
A
C
B
8
15
17
A
P-96 1–MRETT I C S X--M A T E M AH
SECTION B
3. cos (A – B) =3
2 = cos 30º ⇒ A – B = 30º ...(1) ½
sin (A + B) =3
2 = sin 60º ⇒ A + B = 60º ...(2) ½
Adding equations (1) and (2), 2A = 90º ⇒ A = 45º ½
From (2), B = 60º – A = 60º – 45º = 15º ½
SECTION C
4. we know that,sec (90º – θ) = cosec θ, tan (90º – θ) = cot θ, cot (90º – θ) = tan θ, cosec (90º – θ) = sec θ 1
Hence, cosecsin .sec (90º – ) tan tan (90º– )
–(90° – ) cos .cot (90° – ) cotq q q q
q q q q = sin .cos .tan
sec .cos .tancotcot
θ θ θθ θ θ
θθ
ec − 1
=sin
sintan
coscos .tan
θθ
θ
θθ θ
× ×
×−
1
11 = 1 – 1 = 0 1
SECTION D
5. sin θ = 34
⇒ cos θ = 7
4 and tan θ =
37
½+½
∴2 2
2
cosec cot2 cot
sec 1
q- q+ q
q - = 7
cos+ qx
⇒ 2
1 72
3tan+ ´
q=
7 74x
+ 1
⇒1 2 7
tan 3+
q=
7 74x
+
⇒7 2 7 7
3 3 4+ - =
7x
1
⇒4 7 7
4-
= 7x
⇒3 7
4=
7x
⇒ x = 43
· 1
SUMMATIVE ASSESSMENT WORKSHEET-72
SECTION A
1. cos2 17º– sin2 73º = cos2 17º – sin2 ( 90 – 17)º= cos2 17º – cos2 17º= 0. 1
2. A = 30º, sin 2A = sin 2 (30º) = sin 60º = 3
2· 1
P-97S O L U T I O N S
SECTION B
3.2 2 2
2 22 cos 60 3 sec 30 2 tan 45
sin 30 cos 45° + ° − °
° + °=
222
22
1 22 3 2(1)2 3
1 12 2
+ − +
1
=
2 4 24
1 14 2
+ −
+
=10
·3 1
SECTION C
4. We have sin (A + B – C ) =12
= sin 30º
∴ A + B – C = 30º ...(1) 1
and cos ( B + C – A ) =12
= cos 45º
∴ B + C – A = 45º ...(2) 1Adding eqs. (1) and (2), we get
2B = 75º⇒ B = 37.5ºNow subtracting eqn. (2) from eqn. (1), we get
2(A – C) = – 15º ⇒ A – C = – 7·5º ...(3)We know that, A + B + C = 180º 1
A + C = 142·5º ...(4)Adding eqns. (3) and (4), we get 2A = 135º
A = 67·5º⇒ C = 75ºHence, ∠A = 67·5º, ∠B = 37·5º, ∠C = 75º. 1
5. sec A =178 ⇒ cos A =
817
sin A = 21 cos A− = 28
117
− =
1517 1
L.H.S. =2
23 4 sin A4 cos A – 2
−
=
2253 4289
644 3289
− −
=867 900256 867
−− 1
=33611
−− =
33·
611
P-98 1–MRETT I C S X--M A T E M AH
R.H.S. =2
23 tan A
1 3 tan A−
−
=
2
2
1538151 38
− −
1
=192 22564 675
−− =
33611 1
∴ L.H.S. = R.H.S.
SUMMATIVE ASSESSMENT WORKSHEET-73
SECTION A
1. cos ( A + B) = cos (180º – C)= cos C = cos 90º = 0. 1
2. sin (45º + θ) – cos (45º – θ) = sin [90º – (45º + θ)] – cos (45º – θ)= cos (45º – θ) – cos (45º – θ)= 0. 1
SECTION B
3. 3 tan θ = 1
tan θ =13 = tan 30° 1
⇒ θ = 30°∴ sin2 θ – cos2 θ = sin2 30° – cos2 30°
=
221 32 2
−
=1 34 4
− = – 24
= – 1
·2
1
SECTION C
4. Given : xsin θ = y cos θ
⇒ x =ycossin
θθ ...(1) 1
and x sin3 θ + ycos3 θ = sin θcos θ ...(2)
Eliminating x from (1) and (2), we get
cossin
y . sin3 θ + ycos3 θ = sin θcos θ
⇒ y cos θ sin2 θ + ycos3 θ = sin θcos θy cos θ [sin2 θ + cos2 θ] = sin θ cos θ
⇒ y = sin θ ...(3) 1
P-99S O L U T I O N S
Substituting this value of y in (3), we get
x = cos θ ...(4) 1
∴ Squaring and adding (3) and (4), we get
x2 + y2 = cos2 θ + sin2 θ = 1. Proved.
SECTION D
5. L.H.S. =cos
tansin
sin cos
2 3
1θ
θθ
θ θ−+
−
=cos
sincos
sinsin cos
2 3
1
θθθ
θθ θ−
+−
1
=cos
cos sin–
sincos sin
3 3θθ θ
θθ θ− −
1
=cos sin
cos sin
3 3θ θθ θ
−−
1
=(cos sin )(cos sin sin cos )
(cos sin )θ θ θ θ θ θ
θ θ− + +
−
2 2
= 1 + sin θ cos θ = R.H.S. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-74
SECTION A
1. tan θ + cot θ = 2Squaring on both sides,
tan 2 θ + cot2 θ + 2tan θ cot θ = 4⇒ tan2 θ + cot2 θ = 4 – 2 = 2, as tan θ cot θ = 1. 1
2. tan 3A = cot (A – 26°)⇒ cot (90º – 3A) = cot (A – 26º)⇒ 90º – 3A = A – 26º
⇒ 4A = 90º + 26º = 116º
⇒ Α =116
4º
= 29º. 1
SECTION C
4. cos 50º = cos (90° – 40)º = sin 40º
cosec2 59º = cosec2 (90º – 31º) = sec2 31º 1
tan 78º = tan (90° – 12º) = cot 12º
Hence, 2 2
2
cos 50º 4(cosec 59º tan 31º ) 22sin 40º 33tan 45º
-+ - tan 12º tan 78º. sin 90º 1
=sin ºsin º
(sec º– tan º)
( )
402 40
4 31 31
3 1
23
2 2
2+×
− tan 12º cot 12º × 1
=12
43
23
76
+ − = 1
P-100 1–MRETT I C S X--M A T E M AH
SECTION D
5. Let cot θ = x, 3 cot2 θ – 4cot θ + 3 = 0 becomes
3 x2 – 4x + 3 = 0 1
or (x – 3 ) ( 3 x – 1) = 0
∴ x = 3 or 13
1
⇒ cot θ = 3 or cot θ = 13
∴ θ = 30º or θ = 60º
If θ = 30º, then cot2 30º + tan2 30º = ( )313
313
22
+FHG
IKJ = + =
103
1
If θ = 60º, then cot2 60º + tan2 60º = 13
313
32
2FHG
IKJ + = +( ) =
103
1
SUMMATIVE ASSESSMENT WORKSHEET-75
SECTION A
1. cos 2θ = sin 4θ⇒ sin (90º – 2θ) = sin 4θ⇒ 90º – 2θ = 4θ ⇒ 6θ = 90º
⇒ θ =906
º = 15º. 1
2. tan x = sin 45º cos 45º + cos 60º
=12
12
12
FHG
IKJFHG
IKJ +
=12
12
+ = 1 = tan 45º
⇒ x = 45º. 1
SECTION B
3. cos 68º + tan 76º = cos (90º –22º) + tan (90º – 14º) 1 = sin 22º + cot 14º, [∵ cos (90º – θ) = sin θ and tan (90º – θ) = cot θ ] 1
SECTION C
4. Given : cos θ + sin θ = p and sec θ + cosec θ = q∴ L.H.S. = q(p2 – 1) = (sec θ + cosec θ) [(cos θ + sin θ)2 – 1]
= (sec θ + cosec θ) (1 + 2 sin θ cos θ –1) ½
=1 1
cos sinθ θ+F
HGIKJ (2 sinθ cos θ) 1
=sin coscos sin
θ θθ θ+
× 2 sin θ cos θ
= 2(sin θ +cos θ) 1= 2p= R.H.S. Proved. ½
P-101S O L U T I O N S
SECTION D
5. Since, x2 = r2 sin2 A cos2 C
y2 = r2 sin2 A sin2 C
z2 = r2 cos2 A 1
x2 + y2 + z2 = r2 sin2 A cos2 C + r2 sin2 A sin2 C + r2 cos2 A 1
= r2 sin2 A (cos2 C + sin2 C) + r2 cos2 A
= r2 sin2 A + r2cos2 A 1
= r2 (sin2 A + cos2 A)
= r2. Proved. 1
SUMMATIVE ASSESSMENT WORKSHEET-76
SECTION A
1. ∵ sin 20° = sin (90° – 70°) = cos 70°3 sin2 20° – 2 tan2 45° + 3 sin2 70° = 3 cos2 70° + sin2 70°) – 2 × (1)2
= 3 × 1 – 2 = 1 12. (1 + tan2 θ) cos2 θ = sec2 θ cos2 θ
= 21
cos θ = cos2 θ = 1. 1
SECTION B
3. sin (36 +θ)º = cos (16 + θ)º⇒ cos [90º –(36 + θ)º] = cos (16 + θ)º 1⇒ 90 – 36 – θ = 16 + θ⇒ 2θ = 90 – 36 – 16 = 38
⇒ θ =382
= 19.
SECTION C
4. Given :(sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C)Multiply both the sides by (sec A – tan A) (sec B –tan B) (sec C – tan C)(sec A + tan A) (sec B + tan B) (sec C + tan C ) × (sec A – tan A) (sec B – tan B) (sec C – tan C)= (sec A – tan A)2 (sec B – tan B)2 (sec C –tan C)2 1⇒ (sec2 A – tan2 A) (sec2 B – tan2 B) (sec2 C – tan2 C)
= (sec A – tan A)2 (sec B – tan B)2 (sec C – tan C)2
⇒ 1 = [ (sec A – tan A) (sec B – tan B) (sec C – tan C)]2
⇒ (sec A – tan A) (sec B – tan B) (sec C – tan C) = ± 1 1Similarly, multiply both sides by (sec A + tan A) (sec B + tan B) (sec C + tan C), we get (sec A + tan A) (sec B +tan B) (sec C + tan C) = ± 1. Proved. 1
5. Given expression =tan .cos cot (90º – 50º )
sin tan 50ºq q +
q – [cos2 20º + cos2 (90º – 20º)] 1
=sin cos tan 50º
.cos sin tan 50º
q q+
q q – [cos2 20º + sin2 20º] 1
= 1 + 1 – 1 = 1. 1
P-102 1–MRETT I C S X--M A T E M AH
SECTION D
6.cos ºsin º
6525
= cos 65º cos 65º
sin (90º 65º ) cos 65º=
- = 1 1
tan ºcot º
2070
= tan(90º–70º ) cot 70º
cot 70º cot 70º= = 1
sin 90º = 1tan 5º tan 35º tan 60º tan 55º tan85º = tan (90º – 85º) tan (90º – 55º) tan 55º tan 60º.tan 85º. 1
= cot 85º.tan 85º.cot 55ºtan 55º. 3= 1 × 1 × 3 = 3 1
∴ Given expression = 1 – 1 – 1 + 3 = 3 – 1.
FORMATIVE ASSESSMENT WORKSHEET-77Objective Type Questions
1. (C) 2. (B) 3. (A) 4. (D) 5. (A) 6. (C) 7. (C)
Fill in the blanks
1. (I) 25° (II) 65 ° (III) 59° (IV) 51° (V) 55°(VI) 50° (VII) 1° (VIII) 10° (IX) 23° (X) 53°
Error’s Correction
2. (I)sin º sin º
cos º cos º
2 2
2 220 70
20 70
++
=sin ( º º ) sin º
cos ( º º ) cos º
2 2
2 290 70 70
90 70 70
− +− +
=cos º sin º
sin º cos º
2 2
2 270 70
70 70
++
= 11
= 1
(II)sin º sin º
cos º cos º
2 2
2 220 70
20 70
++
=sin ( º º ) sin º
cos ( º º ) cos º
2 2
2 290 70 70
90 70 70
− +− +
=cos º sin º
sin º cos º
2 2
2 270 70
70 70
++
= 11
= 1
(III) Correct (No Error)
(IV)sin º sin º
cos º cos º
2 2
2 220 70
20 70
++
=sin ( º º ) sin º
cos ( º º ) cos º
2 2
2 290 70 70
90 70 70
− +− +
cos º sin º
sin º cos º
2 2
2 270 70
70 70
++
=11
= 1
(V) Correct (No error)
Answers
1. sin C = 5
13
2. sin A = sin A′
⇒25
=B'C'20
⇒ B'C' =2 20
5×
= 83. TRUE / FALSE
(I) F, (II) F, (III) T, (IV) F, (V) T, (VI) F.■
A B
C
5 2 20
A' B'
C'
A
B C
5 cm12 cm
13 cm
P-103S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET - 78
SECTION A
1. Class mark of the class 10 – 25 =10 25
2352
+= = 17·5. 1
2. The abscissa of the point of intersection of the ‘‘less than type’’ and ‘‘more than type’’ cumulativefrequency curve of a grouped data is median. 1
SECTION B
3. Marks No. of students c.f.
0 – 10 5 5
10 – 20 15 20
20 – 30 30 50
30 – 40 8 58
40 – 50 2 60
N = 60
HereN2
=602
= 30
So, median class = 20 – 30l = 20, f = 30, c.f. = 20, h = 10
Median = l +
N2
c.f.æ öç ÷-ç ÷ç ÷ç ÷ç ÷è ø
f ´ h 1
= 20 + 30 20
30
æ ö-ç ÷ç ÷è ø ´ 10
= 20 + 10030
= 20 + 103
= 20 + 3.33 = 23.33. 1
Statistics
6CHAPTER
P-104 1–MRETT I C S X--M A T E M AH
SECTION C
4. Class marks (xi) fi fixi
27 4 108
32 14 448
37 22 814
42 16 672
47 6 282
52 5 260
57 3 171
Σf = 70 Σfx = 2755 2
∴ Mean =i i
i
f xf
åå ½
=275570
= 39·36. ½
SECTION D
5. Weight (in Kg) Cumulative Frequency
More than or equal to 0 120
More than or equal to 10 106
More than or equal to 20 89
More than or equal to 30 67
More than or equal to 40 41
More than or equal to 50 18 2
Plotting the points
120
110
100
90
80
70
60
50
40
30
20
10
(0, 120)
(10, 106)
(20, 89)
(30, 67)
(40, 41)
(50, 18)
0 10 20 30 40 50x
y
Cum
ulat
ive
freq
uen
cy
Scale-axis 2 cm = 10 units-axis 1 cm = 10 units
xy
More than ogive
Lower limits 60
2
P-105S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-79
SECTION A
1. The mean of observations 1 2, , ......, =nx x x x then the mean of x1 + a, x2 + a ...... xn + a = .+x a 1
SECTION B
2. Class Interval Frequency
0 – 50 8
50 – 100 15
100 – 150 32
150 – 200 26
200 – 250 12
250 – 300 7
Total 100 2
SECTION C
3. Modal class : 30 – 40,Here l = 30, f1 = 45, f2 = 12, f0 = 30, h = 10
Mode =1 0
1 0 22 −
+ × − −
f fl h
f f f 1
=45 30
30 1090 30 12
− + × − − 1
= 30 + 3·125 = 33·125 1
SECTION D
4. Class f c.f x fx
0 – 50 2 2 25 50
50 – 100 3 5 75 225
100 – 150 5 10 125 625
150 – 200 6 16 175 1050
200 – 250 5 21 225 1125
250 – 300 3 24 275 825
300 – 350 1 25 325 325
Total Σf = 25 Σfx = 4225 2
∴ Mean =4225
16925
Σ = =Σfxf
Mode = l + 1 0
1 0 22 − − −
f ff f f × h 1
From table modal class = 150 – 200so f1 = 6, f0 = 5, f2 = 5, l = 150, h = 50
P-106 1–MRETT I C S X--M A T E M AH
So Mode = 150 + (6 5)
12 5 5−
− − × 50
= 150 + 502
= 150 + 25 = 175 1
Now Median =13 mode +
23 mean
=1 2
(175) 1693 3
+ ×
= 58·33 + 112·66 = 170·83 1
SUMMATIVE ASSESSMENT WORKSHEET-80
SECTION A
1. Given frequency distribution = 8·1Σ fixi = 132 + K
Σ fi = 20
∵ 8·1 =ΣΣ
i i
i
f xf =
13220
+ K
⇒ K = 6 1
SECTION B
2. Modal class : 20 – 30Here l = 20, f1 = 40, f0 = 24, f2 = 36, h = 10 ½
Mode = lf f
f f fh+
−− −
×( )1 0
1 0 22½
= 20 + ( – )
– –40 24
80 24 36 × 10 ½
= 20 + 16 10
20×
= 28 ½
SECTION C
3. xi fi xifi
3 10 30
9 p 9p
15 4 60
21 7 147
27 q 27q
33 4 132
39 1 39
Total Σfi = 26 + p + q Σxifi = 408 + 9p + 27q
P-107S O L U T I O N S
Given, Σ fi = 40,⇒ 26 + p + q = 40⇒ p + q = 14 ...(i) ½
∴ Mean, x– =ΣΣx ffi i
i½
14·7 =408 9 27
40+ +p q
588 = 408 + 9p + 27q 1180 = 9p + 27q
p + 3q = 20 ...(2)Subtracting eq. (1) from eq. (2), we get
2q = 6⇒ q = 3 ½Putting this value of q in eq. (1), we get
p = 14 – q = 14 – 3 = 11 ½
SECTION D
4. Class Mid values fi fixi c.fixi
25 – 35 30 7 210 7
35 – 45 40 31 1240 38
45 – 55 50 33 1650 71
55 – 65 60 17 1020 88
65 – 75 70 11 770 99
75 – 85 80 1 80 100
Total 100 4970 2
Mean x =i i
i
f xf
ΣΣ =
4970100 = 49·7 1
From table N2
= 100
2 = 50
Hence median class = 45 – 45so, l = 45, c.f. = 38, f = 33, h = 10
Median =.
2n c f
l hf
− + ×
= 45 + 50 38
33−
× 10
= 45 + 3·64 = 48·64. 1
P-108 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-81
SECTION A
1. According to question,Mode – Mean = k (Median – Mean)
i.e., k =Mode – Mean
Median – Mean
=3Median – 2Mean – Mean
Median – Mean
=3(Median – Mean)(Median – Mean) = 3. 1
2. The median of give data is 20·5. 1
SECTION B
3. Marks Cumulative frequency
More than or equal to 0 38
More than or equal to 10 35
More than or equal to 20 27
More than or equal to 30 12
More than or equal to 40 5 2
SECTION C
4. Classes fi c.f.
0 – 100 15 15
100 – 200 17 32
200 – 300 f 32 + f
300 – 400 12 44 + f
400 – 500 9 53 + f
500 – 600 5 58 + f
600 – 700 2 60 + f
From table, n = 60 + f
⇒ n f2
602
=+
1
Since, median = 240∴ Median class is 200 – 300. ½
Median = l
nc f
fh+
−L
NMMM
O
QPPP ×2
. .½
⇒ 240 = 200 +
602
32100
+ −L
NMMM
O
QPPP ×
f
f½
P-109S O L U T I O N S
⇒ 40 =60 64
2+ −L
NMOQP
ff 100
⇒ 8f = 10 f – 40
⇒ 2f = 40
∴ f = 20. ½
SECTION D
5. Class Interval c.f. No. of students
0 – 10 7 7
10 – 20 21 14
20 – 30 34 13
30 – 40 46 12
40 – 50 66 20
50 – 60 77 11
60 – 70 92 15
70 – 80 100 8 2
From table, maximum frequency = 20. So modal class = 40 – 50
Now l = 40, f1 = 20, f2 = 11, f0 = 12, h = 10
Mode = l + 1 0
1 0 22
æ ö-ç ÷ç ÷- -è ø
f ff f f ´ h 1
= 40 + 20 12
40 12 11
æ ö-ç ÷ç ÷- -è ø ´ 10
= 40 + 8
17 ´10
= 40 + 8017
= 40 + 4·7 = 44·7. 1
SUMMATIVE ASSESSMENT WORKSHEET-82
SECTION A
1. The modal class of the distribution is 40 – 50. 1
SECTION B
2. The multiples of 5, according to the problem are :
5, 15, 25, 35, 45 ½
Mean =5 15 25 35 45
5+ + + +
1
=125
5 = 25. ½
P-110 1–MRETT I C S X--M A T E M AH
SECTION C
3. For mean,
Class Interval xi fi fixi
0 – 10 5 3 15
10 – 20 15 8 120
20 – 30 25 10 250
30 – 40 35 15 525
40 – 50 45 7 315
50 – 60 55 4 220
60 – 70 65 3 195
Σf = 50 Σfx = 1640 1
∵ Mean =i i
i
f xf
åå =
164050
= 32·8 ½
For mode,
Modal class = 30 – 40
and l = 30, f1 = 15, f2 = 7, f0 = 10, h = 10
Mode = l + 1 0
1 0 22-
- -f f
f f f ´ h
= 30 + 15 10
30 10 7-
- - ´ 10 1
= 30 + 5
13 ´ 10
= 30 + 5013
= 30 + 3·85
∴ Mode = 33·85. 1
SECTION D
4. Classes f c.f.
5 – 10 2 2
10 – 15 12 14
15 – 20 2 16
20 – 25 4 20
25 – 30 3 23
30 – 35 4 27
35 – 40 3 30 2
Total Σf = 30 = N
Since,N2
= 15
P-111S O L U T I O N S
∴ Median class is 15 – 20
Median = l + 2N . .c f
f
æ öç ÷-ç ÷ç ÷ç ÷ç ÷è ø
× h 1
From table, l = 15, N = 30, c.f. = 14, f = 2, h = 5
Median = 15 + 15 14
2−F
HGIKJ × 5
= 15 + 2·5 = 17·5. 1
SUMMATIVE ASSESSMENT WORKSHEET-83
SECTION A
1. According to formula for an even number of 2n terms.Median = Mean of nth term and (n + 1)th term 1
SECTION B2. Age Number of Patients
Less than 20 60Less than 30 102Less than 40 157Less than 50 227 2Less than 60 280Less than 70 300
SECTION C3.
Sol. : xi fi xifi
3 10 30
9 p 9p
15 4 60
21 7 147
27 q 27q
33 4 132
39 1 39
Total Σfi = 26 + p + q Σxifi = 408 + 9p + 27q
Given, Σfi = 40,⇒ 26 + p + q = 40⇒ p + q = 14 ...(1) ½
∴ Mean, x– =ΣΣx ffi i
i½
14.7 =408 9 27
40+ +p q
588 = 408 + 9p + 27q 1180 = 9p + 27q
p + 3q = 20 ...(2)
P-112 1–MRETT I C S X--M A T E M AH
Subtracting eq. (1) from eq. (2), we get2q = 6
⇒ q = 3 ½Putting this value of q in eq. (1), we get
p = 14 – q = 14 – 3 = 11. ½
SECTION D
4. Daily income (classes) No. of workers (c.f.)
less than 250 10less than 300 15less than 350 26less than 400 34less than 450 40less than 500 50 1
50
45
40
35
30
25
20
15
10
5
250 300 350 400 450 500Class limits
Cu
mul
ativ
e fr
eque
ncy Scale
-axis 1 cm = 50 units-axis 1 cm = 5 units
xy
y
x
2
From graph,N2
502
= = 25
Hence, Median daily income = ` 350. 1
SUMMATIVE ASSESSMENT WORKSHEET-84
SECTION A
1. The model class is 10 – 15, than f ≥ 8. 1
SECTION B
2. According to question mode = 24·5and mean = 29·75The relationship connecting measures of central tendencies is :We know that 3 Median = Mode + 2 Mean ½
3 Median = 24·5 + 2 × 29·75= 24·5 + 59·50 ½
3 Median = 84·0
∴ Median =843
= 28 1
P-113S O L U T I O N S
SECTION C
3. Class Class marks ui = x -ai
hfi fiui
x i
20 – 30 25 –2010
= –2 25 –50
30 – 40 35 –1010
= –1 40 –40
40 – 50 45 = a 0 = 0 42 0
50 – 60 551010
= 1 33 33
60 – 70 652010
= 2 10 20
=150ifå Σfiui = – 37 1
Mean = a + f ui ifi
åå ´ h 1
= 45 +37
150
æ ö-ç ÷ç ÷è ø ´10 = 42·5 approx. 1
SECTION D
4. Class xi (class mark) fi fixi
0 – 100 50 12 600
100 – 200 150 16 2400
200 – 300 250 6 1500
300 – 400 350 7 2450
400 – 500 450 9 4050
Total Σfi = 50 Σfixi = 11000 2
Mean =ΣΣx ffi i
i= 11000
50 = 220·00 1
∴ Average daily income = ` 220·00. 1
SUMMATIVE ASSESSMENT WORKSHEET-85
SECTION A1. Given mode – median = 24
Using formula mode = 3 median – 2 Meandifference between median and mean = a.
SECTION B2. The frequency table of the given data is as given below :
14 15 16 18 20 25
1 3 1 3 3 4 1i
i
x x
f 1
It is given that the mode of the given data is 25. So it must have the maximum frequency that ispossible only when xi = 25 hence x = 25. 1
P-114 1–MRETT I C S X--M A T E M AH
SECTION C
Sol. Classes Frequency Mid points fixi
fi xi
0 – 20 6 10 60
20 – 40 8 30 240
40 – 60 10 50 500
60 – 80 12 70 840
80 – 100 8 90 720
100 – 120 6 110 660
Total Σf = 50 Σfx = 3020 2
∴ Mean, x– =ΣΣ
i i
i
f xf =
302050
= 60.4 1
SECTION D
Sol. Class Interval Frequency Cumulative frequency
0 –100 2 2
100 – 200 5 7
200 – 300 x 7 + x
300 – 400 12 19 + x
400 – 500 17 36 + x
500 – 600 20 56 + x
600 – 700 y 56 + x + y
700 – 800 9 65 + x + y
800 – 900 7 72 + x + y
900 – 1000 4 76 + x + y
N = 100 2
Hence, 76 + x + y = 100⇒ x + y = 100 – 76 = 24 ....(1)Median = 525, so median class = 500 – 600
Now, median = l + . .
2n c f
f
-´h
525 = 500 +
100(36 )
220
é ù- +ê ú
ê úê úê úê úë û
x´100
25 = (50 – 36 – x) 5
Þ (14 – x) = 255
= 5
Þ x = 14 – 5 = 9 1Put the value of x in equation (1), we get
y = 24 – 9 = 15. 1
3.
4.
P-115S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-86
SECTION A1. The median of the data = 20·5. 1
2. The mean of first five prime numbers =2 3 5 7 11
5+ + + +
=285 = 5·6 1
SECTION B
3.
3 9 15 21 27
7 5 10 12 6
21 45 150 252 162
x
f f
fx fx
= 40
= 630
ΣΣ
Σ fx = 630, Σ f = 40 1
∴ Mean =63040 = 15·75. 1
SECTION C
4. Class xi (class marks) fi fixi Cum. Frequency
0 – 10 5 8 40 810 – 20 15 16 240 24
20 – 30 25 36 900 60
30 – 40 35 34 1190 94
40 – 50 45 6 270 100 1
Total Σfi = 100 Σfixi= 2640
∴ Mean =i i
i
f xf
åå =
2640100
= 26.4 1
Here, median class : 20 – 30
∴ Median = 20 + 50 24
36−
× 10
= 20 + 7·22 = 27·22 1
SECTION D
5. Class Interval Frequency Cumulative Frequency
0 – 10 5 510 – 20 f1 5 + f120 – 30 20 25 + f130 – 40 15 40 + f140 – 50 f2 40 + f1 + f250 – 60 5 45 + f1 + f2
Total N = 60 2
Hence, 45 + f1 + f2 = 60f1 + f2 = 15
and N = 60
P-116 1–MRETT I C S X--M A T E M AH
Median = 28.5, ∴ Median class is 20 – 30.
Median = l + .
2N
c f
f
− × h ...(1)
28.5 = 20 + 30 5
201– − f
× 10 1
⇒ 8.5 × 2 = 25 – f1
⇒ f1 = 25 – 17 = 8
From (1), f2 = 15 – f1 = 15 – 8 = 7
∴ f1 = 8 and f2 = 7. 1
SUMMATIVE ASSESSMENT WORKSHEET-87
SECTION A
1. 27.x = 1
2. Median = 1 13 1
observation2 2+ + =
th thn observation = 7th observation. 1
SECTION B
3. Heights No. of girls
120 and more 50
130 and more 48
140 and more 40
150 and more 28
160 and more 8 2
SECTION C
4. (i) No. of children (xi) No. of famlies (fi) fi xi
0 5 0
1 11 11
2 25 50
3 12 36
4 5 20
5 2 10
Total Σfi = 60 Σfixi = 127 1
Mean =f xfi i
i
∑
∑
=12760
= 2·12 approx. 1
(ii) Mean of ungrouped data. ½(iii) For progress, we should decrease population growth. ½
P-117S O L U T I O N S
SECTION D5.
Wages c.f.
More than 80 200
More than 100 180
More than 120 150
More than 140 130
More than 160 90 2 2
SUMMATIVE ASSESSMENT WORKSHEET-88
SECTION A1. From the table it is clear that the frequency is maximum for the class 30 – 40, so modal class is
30 – 40. 1
SECTION B
2. Modal class : 60 – 80Here, l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20 ½
Mode = 1 0
1 0 22f f
l hf f f
−+ ×
− − ½
=61 52
60 20122 52 38
−+ ×− − ½
=9
60 20 65·6232
+ × = ½
SECTION C
3. Class Interval Frequency Cumulative Frequency
135 – 140 4 4
140 – 145 7 11
145 – 150 18 29
150 – 155 11 40
155 – 160 6 46
160 – 165 5 51
Total N = 60 3
x
y
Scale-axis 2 cm = 10 units-axis 1 cm = 10 units
xy
Lower Lts
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
080 100 120 140 160
More than ogive
(80, 200)
(100, 180)
(120, 150)
(140, 130)
(160, 90)
P-118 1–MRETT I C S X--M A T E M AH
SECTION D
4. Let assumed mean, a = 649·5 and h = 100
Life time xi ui = x ah
i − fi fiui
400 – 499 449·5 – 2 24 – 28
500 – 599 549·5 – 1 47 – 47
600 – 699 649·5 0 39 0
700 – 799 749·5 1 42 42
800 – 899 849·5 2 34 68
900 – 999 949·5 3 14 42
Total Σfi = 200 Σfiui = 77 2
∴ Mean, x– = a + ΣΣf uf
hi i
i×
FHG
IKJ 1
= 649·5 + 77200
× 100
= 649·5 + 38.5= 688. 1
Average life time is 688 hours.
SUMMATIVE ASSESSMENT WORKSHEET-89
SECTION A
1. Median =13 mode +
23 mean
=1 2
(8) (8)3 3
+
=16 8
3+
=243 = 8. 1
2. The median of the data is 42·5.
SECTION B
3. Modal class : 5 – 7
Here l = 5, f1 = 80, f0 = 45, h = 2, f2 = 55 ½
Mode = l + f f
f f fh1 0
1 0 22−
− −× ½
= 5 + 80 45
160 45 55–
– – × 2
= 5 + 35 2
60×
= 6·17. 1
(in hrs)
P-119S O L U T I O N S
SECTION C
3. Let assumed mean, a = 35 and h = 10.
xi ui=x a
hi −
fi fiui
(Class Marks)
5 – 3 5 – 15
15 – 2 13 – 26
25 – 1 20 – 20
35 0 15 0
45 1 7 7
55 2 5 10
Total Σfi = 65 Σfiui = – 44 1½
∴ Mean, x– = a + ΣΣ
i i
i
f uf × h ½
= 35 + −4465
× 10
= 35 – 6·76
x– = 28·24. 1
SECTION D4. Draw the ‘less then’ ogive and ‘more than’ ogive simultaneously.
Less than c.f. More than c.f.
30 10 20 10040 18 30 9050 30 40 8260 54 50 7070 60 60 4680 85 70 4090 100 80 15 2
100
90
80
70
60
50
40
30
20
10
0 10 20 30 40 50x
y
Cum
ula
tive
freq
uen
cy
From The Graph Median = 58.360 70 80 90
(20, 100)
(30, 90)
(40, 82)
(50, 70)
(60, 46)
(70, 40)
(80, 15)
(90, 10)
(80, 85)
(70, 60)
(60, 54)
(50, 30)
(40, 18)
(30, 10)
More than ogive
Less than ogive
2
P-120 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-90
SECTION A
1. In the given formula l represents the lower limit of the class with highest frequency. 1
2. The require formula is 3 Median = Mode + 2 Mean 1
SECTION B
3. Class Cumulative frequency
More than 50 60
More than 60 48
More than 70 30
More than 80 20
More than 90 5 2
SECTION C
3. (i) Here class intervals are not in inclusive form. So, we first convert them in inclusive form bysubtracting 1/2 from the lower limit and adding 1/2 to the upper limit of each cases. where h isthe difference between the lower limit of a class and the upper limit of the preceding class. The
given frequency distribution in inclusive form is as follows :
Age (in years) No. of cases
4·5 – 14·5 6
14·5 – 24·5 11
24·5 – 34·5 21
34·5 – 44·5 23
44·5 – 54·5 14
54·5 – 64·5 5
we observe that the class 34·5 – 44·5 has the maximum frequency. So, it is the modal class suchthat
Here, l = 34.5, h = 10, f1 = 23, f0 = 21, f2 = 14 1
Now, Mode = lf f
f f fh+
–2 –
0
0
1
1 2–×
⇒ Mode = 34 5 10. +23 – 21
46 – 21 – 14×
= 34·5 + 2
11× 10
= 34·5 + 1·81
= 36·31 1
(ii) Mode of grouped data. ½(iii) If we practise habit of cleanliness we will be able to put disease at on arm’s length. ½
P-121S O L U T I O N S
SECTION D
4. We prepare cumulative frequency table :
Height (in cm) Frequency Height Less than Cumulative Frequency
140 – 143 3 143 3
143 – 146 9 146 12
146 – 149 26 149 38
149 – 152 31 152 69
152 – 155 45 155 114
155 – 158 64 155 178
158 – 161 78 161 256
161 – 164 85 164 341
164 – 167 96 167 437
167 – 170 72 170 509 2
Now we mark upper class limits on X-axis and cumulative frequency on Y-axis. We plot
(140, 0), (143, 3), (146, 12), (149, 38), (152, 69), (155, 114), (158, 178), (161, 256),
(164, 341), (167, 437), (170, 509)
These points are joined by line segments to obtain the cumulative frequency polygon as shown
in Figure
143 146 149 152 155 158 161 164 167 170
550
500
450
400
350
300
250
200
150
100
50
140
Y
X
Heights
Cuf
P-122 1–MRETT I C S X--M A T E M AH
SUMMATIVE ASSESSMENT WORKSHEET-91
SECTION A1. In the distribution, median class = 20 – 25
Hence, lower limit of median class = 20and modal class = 25 – 30So, lower limit of modal class = 25Sum of lower limits of median class and lower limit of modal class = 25 + 20 = 45. 1
SECTION B
2. xi fi xifi
1 1 13 2 65 1 57 5 359 6 5411 2 2213 3 39
Total 20 162 1
Mean =ΣΣ
i i
i
f xf ½
x– =16220
= 8·1
∴ Mean number of plants per house is 8·1.
SECTION C
3. xi fi fixi(Class marks)
10 12 12030 15 45050 32 160070 p 70p90 13 1170
Total Σfi = 72 + p Σfixi = 3340 + 70p 1
We know that Mean, x– =ΣΣf xfi i
i½
⇒ 53 =3340 70
72++
pp ½
⇒ 3340 + 70p = 53 (72 + p)⇒ 3340 + 70p = 3816 + 53p⇒ 70p – 53p = 3816 – 3340⇒ 17p = 476
p =47617
= 28. 1
P-123S O L U T I O N S
SECTION D
4. Frequency distribution given below :
Age (in years) Frequency Age less than Cumulative Frequency
0·5 – 9·5 5 9·5 5
9·5 – 19·5 15 19·5 20
19·5 – 29·5 20 29·5 40
29·5 – 39·5 23 39·5 63
39·5 – 49·5 17 49·5 80
49·5 – 59·5 11 59·5 91
59·5 – 69·5 9 69·5 100 2
Now we plot points (9·5, 5), (19·5, 20), (29·5, 40), (39·5, 63), (49·5, 80) (59·5, 91) and (69·5, 100) andjoin them to obtain the required ogive as shown in figure :
9.5 19.5 29.5 49.5 59.5 69.539.5
90
80
70
60
50
40
30
20
10
–0.5
SUMMATIVE ASSESSMENT WORKSHEET-92
SECTION A
1. In the given distribution,Median class = 160 – 165
So upper limit of Median class = 165and Modal class = 150 – 155So Lower limit of Modal class = 150Hence, sum of upper limit of Median class and lower limit of modal class
= 165 + 150 = 315. 1
SECTION B
2. Mode = l + 1 0
1 0 22f f
f f f-
- - × h 1
Here, modal class : 30 – 40,l = 30, f1 = 25, f0 = 20, f2 = 12, h = 10 ½
P-124 1–MRETT I C S X--M A T E M AH
∴ Mode = 30 + 25 20
50 20 12–
– –×10
= 30 + 5 10
18×
= 30 + 2·77 = 32·77 (Modal age). ½
SECTION C
3. Class Interval Frequency
0 – 10 8
10 – 20 12
20 – 30 25
30 – 40 13
40 – 50 12
Total 70 1
Here, modal class 20 – 30and l = 20, f1 = 25, f2 = 13, f0 = 12, h =10 ½
Mode = l + 1 0
1 0 22
æ ö-ç ÷ç ÷- -è ø
f ff f f × h ½
= 20 + 25 12
50 12 13−
− −FHG
IKJ × 10 ½
= 20 + 1325
× 10
= 20 + 5·2 = 25·2.
SECTION D
4. Height (in cm) Number of girls (fi) xi ui = ix - ah fiui
120 – 130 2 125 – 2 – 4
130 – 140 8 135 – 1 – 8
140 – 150 12 145 0 0
150 – 160 20 155 1 20
160 – 170 8 165 2 16
Total Σfi = 50 Σfiui = 24
Let assumed mean, a = 145 and h = 10
∴ Mean, x– = a + h × i i
i
f uf
ΣΣ 2
x– = 145 + 2450
× 10 1
= 145 + 4·8x– = 149·8. 1
P-125S O L U T I O N S
SUMMATIVE ASSESSMENT WORKSHEET-93
SECTION A
1. Time Frequency Cumulative frequency
0 – 10 8 8
10 – 20 10 18
20 – 30 12 30
30 – 40 22 52
40 – 50 30 82
50 – 60 18 100
NowN2
= 50
Hence, median class is 30 – 40. 1
SECTION B
2. Here, the modal class : 30 – 40l = 30, f1 = 25, f0 = 10, f2 = 12, h = 10 ½
Mode = l + 1 0
1 0 22f f
f f f−
− − × h ½
= 30 + 25 10
50 10 12−
− − × 10 ½
= 30 + 1528 × 10 = 35·35 ½
SECTION C
3. (i) Let the assumed mean, A = 1400 and h = 400.Calculation of Mean
Height (in m) x1 No. of Villages f1 D = xi –1400 ui = xi − 1400
400fiui
200 142 – 1200 – 3 – 426600 265 – 800 – 2 – 5301000 560 – 400 – 1 – 5601400 271 0 0 01800 89 400 1 892200 16 800 2 32
Total N= Σfi= 1343 Σfiui = –1395 1
We have A = 1400, h = 400, Σfiui = –1395 and N = 1343.
Mean = A + h f ui i1N ∑RST
UVW= 1400 + 400 ×
–13951343
FHG
IKJ
= 1400 – 415·49 = 984·51 1
P-126 1–MRETT I C S X--M A T E M AH
(ii) Mean by assumed mean method. ½(iii) Villages are much necessary to keep a balance between the ecological problems. ½
SECTION D
4. More than c.f.
0 100
10 90
20 72
30 32
40 12 1
2
From graph,N2
=100
2 = 50
Hence, Median = 25. 1
SUMMATIVE ASSESSMENT WORKSHEET-94
SECTION A
1. The number of athletes who completed the race in less than 14·4 seconds = 2 + 14 + 16 = 32. 1
SECTION B
2. From the cumulative frequency distribution
15 + x = 28
⇒ x = 28 – 15 = 13 1
and 43 + 18 = y
y = 61 1
Hence, x = 13 and y = 61.
SECTION C
3. Modal class : 201 – 202
Here l = 201, f1 = 26, f0 = 12, f2 = 6, h = 1 1
Mode = l + 1 0
1 0 22f f
f f f−
− − × h
Mode = 201 + 26 12
52 12 6−
− − × 1 1
= 201 + 1434 = 201·41. 1
100
90
80
70
60
50
40
30
20
10
0 10 20 30 40x
y
Class
P-127S O L U T I O N S
SECTION D
4. xi fi fixi(Class marks)
15 12 18045 21 94575 x 75x505 52 5460135 y 135y165 11 1815
Total Σfi = 150 Σfixi = 8400 + 75x + 135y 1
x + y = 54
∴ x– =ΣΣf xfi i
i
91 =8400 75 135
150+ +x y
13650 = 8400 + 75x + 135y
75x + 135y = 5250 ⇒ 5x + 9y = 350 ...(i) 1
From table, 96 + x + y = 150 ...(ii) 1
Solving eqs. (i) and (ii), we get x + y = 54
x = 34 and y = 20. 1
SUMMATIVE ASSESSMENT WORKSHEET-95
SECTION A
1. Class Frequency Cumulative frequency
20 – 40 10 10
40 – 60 12 22
60 – 80 20 42
80 – 100 22 64
∵ N2
= 32
Hence, Median class is 60 – 80. 1
SECTION B
2. Modal class : 40 – 60 1Also, l = 40, f1 = 28, f2 = 20, f0 = 16, h = 20 1
Mode = l + 1 0
1 0 22f f
f f f−
− − × h ½
= 40 + 28 16
56 16 20−
− − × 20
= 40 + 12 = 52. ½
P-128 1–MRETT I C S X--M A T E M AH
SECTION C
3. From the given table,12 + a = 25 ½
⇒ a = 25 – 12 = 13 ½25 + 10 = b
⇒ b= 35 ½⇒ b + c = 43 ½⇒ c = 43 – b
= 43 – 35 = 8and 48 + 2 = d⇒ d = 50 ½
SECTION D
4. Now we draw cumulative frequency distribution table by less than method.
Marks Number of students Marks less than CumulativeFrequency
0 – 10 6 10 6
10 – 20 25 20 31
20 – 30 48 30 79
30 – 40 72 40 151
40 – 50 116 50 267
50 – 60 63 60 330 2
Now we plot the points (0, 0), (10, 6), (20, 31), (30, 79), (40, 150), (50, 267), (60, 330) we getfollowing ogive :
10 20 30 50 6040
350
300
250
200
150
100
50
0
Y
X
P-129S O L U T I O N S
FORMATIVE ASSESSMENT WORKSHEET-96
Objective Type Questions 10 × 1 = 10
1. (B) 2. (C) 3. (A) 4. (C) 5. (B)6. (A) 7. (B) 8. (C) 9. (A) 10. (D)
Fill in the blanks 10 × 1 = 10
1. mode2. uniform3. mode4. mode + 2 mean5. median6. mode
7.2
a b+
8.i i
i
f xf
ΣΣ
9.i i
i
f dx
fΣ
=Σ
10. mode.
●●