SOLUTIONS PULLOUT Worksheets · Class Term 1 (April to September) 10 MathematicsMathematics (v)...

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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


Ø Worksheets 12 - 23 15 - 29



3. Pair of Linear Equations in two Variables


Ø Worksheets 25 - 37 31 - 50



Unit-III : Geometry

4. Triangles


Ø Worksheets 39 - 52 52 - 71



Unit-IV : Trigonometry

5. Introduction to Trigonometry and Trigonometric Identities


Ø Worksheets 54 - 76 73 - 101



Unit-V : Statistics

6. Statistics


Ø Worksheets 78 - 95 103 - 128



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


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


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



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


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


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.


∴ 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


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


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 =


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


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


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


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


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


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


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


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


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


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


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.

●●



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


∴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


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.


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


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


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



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


⇒ 10 =+5a

1

⇒ a =12

and αβ =constant

coeff. of 2x =

ca

1

⇒ 10 = 2c⇒ c = 5

Hence, a =12

and c = 5. 1


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


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


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


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


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


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



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


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


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 +


⇒ 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


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


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



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


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



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


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

●●



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


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



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


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.


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


⇒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



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


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


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


(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


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


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



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


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.


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


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


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


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


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.


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)


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


SECTION A

1. 3x + ky = 52x + y = 16

For unique solution,aa

1

2≠ b

b1

21


⇒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


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


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


⇒ 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


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


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


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


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


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

●●



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


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


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


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


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


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


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


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


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



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


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


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


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



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


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)


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


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


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


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



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


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


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


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


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


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


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


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


⇒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


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


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




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

●●



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


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


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


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


Again, tanq =sincos

qq

=2 2

2 2

/

/

c c d

d c d

+

+

=cd 1


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° )

-

-

qq


=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


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

+ −


=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


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


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



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


(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.


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


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


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


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



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


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


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


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


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


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



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


=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


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


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



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


=(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


=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


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


=−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


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

−+

++−


=(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


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


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


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


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


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.


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

qq

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


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


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


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


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


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


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


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

++



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

++



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



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


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 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



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


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


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


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



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 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


⇒ 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


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


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


∴ 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


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)


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


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


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


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


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.



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


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


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


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


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)


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.


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)


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



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


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



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


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


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


∴ 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



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


(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


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


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


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


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



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.

●●

SOLUTIONS PULLOUT Worksheets · Class Term 1 (April to September) 10 MathematicsMathematics (v)...

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Transcript of SOLUTIONS PULLOUT Worksheets · Class Term 1 (April to September) 10 MathematicsMathematics (v)...