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Mathematics

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SARASWATI HOUSE PVT. LTD.(An ISO 9001:2008 Company)

EDUCATIONAL PUBLISHERSNew Delhi-110002

PULLOUT WORKSHEETS

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Surender VermaM.Sc. (Mathematics), B.Ed.

Delhi Public School,Dwarka, New Delhi

By

data

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CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

1. Real Numbers ............................................................................................... 7-15

� Worksheets (1 to 6) ��Assessment Sheets (1 and 2)��Chapter Test

2. Polynomials ................................................................................................ 16-28


3. Pair of Linear Equations in Two Variables ...................................... 29-50

� Worksheets (18 to 29) ��Assessment Sheets (5 and 6) ��Chapter Test

4. Triangles ...................................................................................................... 51-72


5. Introduction to Trigonometry ............................................................... 73-87


6. Statistics ...................................................................................................... 88-97


PRACTICE PAPERS (1 to 5) ......................................................................... 99-132

Solutions toSolutions toSolutions toSolutions toSolutions toPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETS[Summative Assessments][Summative Assessments][Summative Assessments][Summative Assessments][Summative Assessments]

[FIRST TERM]

7LAER REBMUN S

WORKSHEET–1

1. (B)3

4 3 3125 5 1

= = = 0.0625. 162 5 16 × 5

Clearly, the decimal form of 4 3125

.2 5 termi-

nates after four places.

2. (C) We know that the factors of a primeare 1 and the prime itself only.Therefore, the common factor of p and qwill be 1 only. Hence, HCF (p, q) = 1.

3. (A)As prime factors of 1005 are:1005 = 5 × 3 × 67.∴ 7 is not a prime factor of 1005.

4. Hint: 24 192= = 0.192

125 1000.

5. LCM =First number × Second number

HCF

=96 × 404

4= 24 × 404 = 9696.

6. (i) 660; (ii) 330Hint: Going in opposite direction to thefactor tree, we obtain2 × 165 = 330 (ii) and 2 × 330 = 660 (i).

7. HCF = 3; LCM = 420Hint: 12 = 22 × 3; 15 = 3 × 5; 21 = 3 × 7.

8. (i) Terminating

Hint: 1 3543 543

=250 2 × 5

.

(ii) Non-terminating repeating.

Hint: 2 19 1 1

= =108 12 2 × 3

.

9. Hint: Let 5 – 2 3 = ab

; b ≠ 0

⇒ 3 = 5 –2b a

b

As RHS of this equation is rational, butLHS is an irrational so a contradiction.

10. Let a be any odd positive integer and b= 4. By Euclid’s lemma there exist integersq and r such that

a = 4q + r, 0 ≤ r < 4... a = 4q or 4q + 1 or 4q + 2 or 4q + 3.Therefore, for a to be odd, we have to take

a = 4q +1 or 4q +3.

11. The maximum capacity of a bag will bethe HCF of 490, 588 and 882. Let us findout the required HCF by prime factorisa-tion method.

490 = 2 × 5 × 72

588 = 22 × 3 × 72

882 = 2 × 32 × 72

∴ HCF = 2 × 72 = 98

Thus, the maximum capacity of a bag is98 kg.

WORKSHEET–2

1. (A) HCF (p, q) = 1 ⇒ p and q are coprime.

If p and q are coprime with q ≠ 0 and pq

is

a rational number, then q has only 2 and 5as prime factors.i.e., q = 2m × 5n where, m and n are non-negative integers.

2. (B) Going to opposite direction to thefactor tree, we obtain

3 × 7 = 21 (ii) and 2 × 21 = 42 (i).

3. (A)Required number = 23×1449

161

=1449

7 = 207.

1Chapter

REAL NUMBERS

8 AM T H E M A T C SI X–

4. 2 = 1.414... and 3 = 1.732...

Therefore, we can take 1.5 = 32

as 2 < 32

< 3 .

5. Hint: As 12576 > 4052... 12576 = 4052 × 3 + 420Further 4052 = 420 × 9 + 272Further 420 = 272 × 1 + 148Further 272 = 148 × 1 + 124Further 148 = 124 × 1 + 24Further 124 = 24 × 5 + 4Further 24 = 4 × 6 + 0.In the last equation, remainder is zero.Hence, the required HCF = 4.

6. First given number is composite as5 × 3 × 11 + 11 = 11 (15 + 1) = 11 × 16

= 11 × 2 × 8But second given number is prime as5 × 7 + 7 × 3 + 3 = 35 + 21 + 3 = 59.

7. No. Prime factors of 6n will be of type 2n × 3n.As it doesn't have 5 as a prime factor, so 6n

can't end with the digit 5.

8. Hint: Let a be any positive integer... a = 3q or 3q + 1 or 3q + 2... a2 = 9q2 = 3m; m = 3q2

or a2 = (3q +1)2 = 3m + 1, m = q (3q + 2)or a2 = (3q + 2)2 = 3m + 1, m = 3q2 + 4q + 1.

9. We represent 6, 72 and 120 in their primefactors.

6 = 2 × 372 = 23 × 32

120 = 23 × 3 × 5Now, HCF = 2 × 3 = 6And LCM = 23 × 32 × 5 = 360.

10. Hint: Let 2 5− = x, a rational number

⇒ 2 = x + 5Squaring both sides, we get

2 = x2 + 5 + 2x 5

⇒ 5 =2– – 32

xx

RHS of this last equation is rational, butLHS is an irrational which is a contradiction.

11. Length = 6 m 30 cm = 630 cmBreadth = 5 m 85 cm = 585 cmHeight = 3 m 60 cm = 360 cmThe required length of the tape will be theHighest Common Factor (HCF) of thenumbers 630, 585 and 360.Let us find out the HCF.

630 = 2 × 32 × 5 × 7585 = 32 × 5 × 13360 = 23 × 32 × 5

∴ HCF = 32 × 5 = 45Hence, the length of the tape will be 45 cm.

WORKSHEET–3

1. (C) 4 3 4 443 43 × 5 215

= = 2 × 5 (2 × 5) 10

= 0.0215

Hence, the number terminates after fourplaces of decimal.

2. (A) ( )( ) ( ) ( )2 22 3 2 3 2 3− + = −

= 2 – 3 = – 1.– 1 is a rational number.

3. (C) 128 = 27; 240 = 24 × 3 × 5.Now, HCF (128, 240) = 24 = 16.

4. Hint: First number =HCF × LCM

Second number= 232.

5. No.Hint: Prime factors of 15n will not be oftype 2n × 5n.

6. Rational number = 0.27Irrational number = 0.26010010001... .

7. (i) 145 29 8 232

= × = = 0.232625 125 8 1000

.

(ii) 7 125 875

× = = 0.087580 125 10000

.

8. Let us assume, to the contrary that 2 isrational. We can take integers a and b ≠ 0such that

3 = ab

, where a and b are coprime.

⇒ 3b2 = a2

⇒ a2 is divisible by 3⇒ a is divisible by 3 ...(i)We can write a = 3c for some integer c

9LAER REBMUN S

⇒ a2 = 9c2

⇒ 3b2 = 9c2 [∴a2 = 3b2]⇒ b2 = 3c2

⇒ b2 is divisible by 3⇒ b is divisible by 3 ...(ii)From (i) and (ii) we observe that a and bhave atleast 3 as a common factor. But thiscontradicts the fact that a and b are co-prime. This means that our assumption isnot correct.

Hence, 3 is an irrational number.

9. As: 1032 = 408 × 2 + 216 ...(i)408 = 216 × 1 + 192 ...(ii)216 = 192 × 1 + 24 ...(iii)192 = 24 × 8 + 0 ...(iv)

⇒ HCF = 24∴ From (iii)⇒ 24 = 216 – 192

= 216 – [408 – 216] {... Use (ii)}= 2 × 216 – 408= 2[1032 – 2 × 408] – 408

{... Use (i)}24 = 1032 × 2 – 5 × 408

⇒ m = 2.

10. Hint: Let x be any positive integer.Then it is of the form 3q or 3q + 1 or 3q + 2.If x = 3q, then

x3 = (3q)3 = 9m; m = 3q3

If x = 3q + 1, thenx3 = (3q + 1)3

= 9m + 1; m = q(3q2 + 3q + 1).If x = 3q + 2, then

x3 = (3q + 2)3

= 9m + 8; m = q (3q2 + 6q + 4).

11. The maximum number of columns mustbe the highest common factor (HCF) of616 and 32. Let us find out the HCF by themethod of Euclid's division lemma.Since 616 > 32, we apply division lemmato 616 and 32, to get

616 = 32 × 19 + 8Since the remainder 8 ≠ 0, we apply thedivision lemma to 32 and 8, to get

32 = 8 × 4 + 0

The remainder has now become zero, soour procedure stops. Since the divisor atthis stage is 8, the HCF of 616 and 32 is 8Hence, the maximum number of columnsis 8.

WORKSHEET–4

1. (B) ( )( ) ( ) ( )2 26 5 6 5 6 5− + = −

= 6 – 5 = 1 = Rational number.2. (B)

Hint: Denominator is not in the exact formof 2m × 5n, where m, n are non-negativeintegers.

3. (C) 0 ≤ r < b.4. Hint: 107 = 4 × 26 + 3.5. Hint: 7 × 13 = (ii) and (ii) × 11 = (i).6. Let us represent each of the numbers 30,

72 and 432 as a product of primes. 30 = 2 × 3 × 5 72 = 23 × 32

432 = 24 × 33

Now, HCF = 2 × 3 = 6And LCM = 24 × 33 × 5 = 2160.

7. Here, 396 > 82.∴ 396 = 82 × 4 + 68Further 82 = 68 × 1 + 14Further 68 = 14 × 4 + 12Further 14 = 12 × 1 + 2Further 12 = 2 × 6 + 0In the last equation, the remainder is zeroand the divisor is 2.Hence, the required HCF = 2.

8. Hint: Let 3 + 2 5 = ab

; b ≠ 0

⇒ – 32

a bb

= 5 = Rational

Which is a contradiction as 5 is an irra-tional number.

Hence, 3 + 2 5 is an irrational number.9. (i) The given fraction can be written as

4 3 443 43 × 5

= = 0.02152 ·5 10

Hence, the given number terminates afterfour places of decimal.


(ii) The given fraction can be written as4

5 5 5359 2 × 359

=2 × 5 2 × 5

=5744

100000= 0.05744

Hence, the given number terminates afterfive places of decimal.

10. The required number of students will bethe highest common factor (HCF) of 312,260 and 156. Let us find out the HCF bythe method of prime factorisation.

312 = 23 × 3 × 13260 = 22 × 5 × 13156 = 22 × 3 × 13

∴ HCF = 22 × 13 = 52Number of buses required

= Total number of students

Number of students in one bus

= 312 + 260 + 156

= 1452

Thus, the maximum number of students ina bus and number of buses required are52 and 14 respectively.

11. Hint: Let x = any positive integerx = 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4Now take square of all form.

WORKSHEET–51. (C)

Hint: LCM of 18, 24, 30, 42 = 2520∴ Required number = 2520 + 1 = 2521.

2. (C) Let the quotient is m when n2 – 1 isdivided by 8.

∴ n2 – 1 = 8 × m

⇒ n2 – 1 = An even integer.

⇒ n2 = An even integer + 1 = Odd integer

∴ n = An odd integer.

3. (B)Hint: HCF (65, 117) = 13Now, 65m – 117 = 13.∴ m = 2 will satisfy this equation.

4. Prime factors of numbers 1 to 10 are:1 = 1; 2 = 1 × 2; 3 = 1 × 3; 4 = 1 × 22

5 = 1 × 5; 6 = 1 × 2 × 3; 7 = 1 × 7;8 = 1 × 23; 9 = 1 × 32; 10 = 1 × 2 × 5Now,LCM = 1 × 23 × 32 × 5 × 7

= 8 × 9 × 5 × 7 = 2520 is requirednumber.

5. Hint: 5 3

5 3

−+

= 2x – 15

⇒ 4 – 15 = 2x – 15

⇒ x = 2, which is a rational number.

6. Hint: Any odd positive integer will betype of 4q + 1 or 4q + 3∴ (4q + 1)2 = 16q2 + 8q + 1

= 8 (2q2 + q) + 1= 8n + 1

Also, (4q + 3)2 = 16q2 + 24q + 9= 8 (2q2 + 3q + 1) + 1= 8n + 1.

7. 35 cmHint: Find HCF.

8. Hint: Let 5 3 2− =ab

where a, b are integers and b ≠ 0Squaring on both sides,

5 18 6 10+ − =2

2ab

⇒2

223ab

− = 6 10

⇒−2 2

2

23

6

b a

b= 10 ... a contradiction.

9. (i) Terminating. (ii) Terminating.

10. The required number of burfis will be thehighest common factor of 420 and 130.Let us find out the HCF using Euclid'sdivision lemma.It is clear that 420 > 130. We apply Divisionlemma to 420 and 130, to get

420 = 130 × 3 + 30Since the remainder 30 ≠ 0, so we applyDivision lemma to 130 and 30, to get

130 = 30 × 4 + 10

11LAER REBMUN S

Again the remainder 10 ≠ 0, so we applyDivision lemma to 30 and 10, to get

30 = 10 × 3 + 0Now, the remainder is zero. So the HCF of420 and 130 is the divisor at the last stagethat is 10.Hence, the required number of burfis is 10.

11. Let n = 3q, 3q + 1 or 3q + 2.Case I: If n = 3q, then

n = 3q divisible by 3n + 2 = 3q + 2 ⇒ Not divisible by 3n + 4 = 3q + 4 = 3(q + 1) + 1

⇒ Not divisible by 3.Case II: If n = 3q + 1 then only

n + 2 = 3q + 1 + 2 = 3q + 3= 3(q + 1) is divisible by 3.

and if n = 3q + 2 then onlyn + 4 = 3q + 6 = 3(q + 2)

is divisible by 3.

WORKSHEET–6

1. (C) 3825 = 52 × 32 × 17So, 11 is not a prime factor of 3825.

2. (C) As p and p + 1 are two consecutivenatural numbers, HCF = 1 andLCM = p (p + 1).

3. (A)

Hint: The given number is 51 17

or 1500 500

∴ Denominator = 500 = 22 × 53

Clearly, the denominator is exactly in theform 2m × 5n, where m and n are non-negative integers; so the given number isa terminating decimal expansion.

4. Hint: ............... 8 = 23; 9 = 32; 25 = 52

∴ HCF (8, 9, 25) = 1

LCM (8, 9, 25) = 1800.

5. Hint: HCF (210, 55) = 5∴ 210 × 5 + 55y = 5⇒ 55y = 5 – 1050

⇒ y = 104555

− = – 19.

6. Irrational

Hint: 2 32 3

−+

=3

x

⇒ 7 – 4 3 = 3

x

⇒ 7 3 – 12 = x = Irrational.

7. Rational Number = 0.55Irrational number = 0.5477477747... .

8. 15Hint: HCF (1380, 1455, 1620) = 15.

9. (i) 0.052. (ii) 5.8352.10. We know that any positive integer is either

of the form 3q, 3q + 1 or 3q + 2 for someinteger q.Now, three cases arise.Case I. When p = 3q,

p + 2 = 3q + 2 and p + 4 = 3q + 4Here, p = 3q is exactly divisible by 3

p + 2 = 3q + 2 leaves 2 as remain-der when it is divided by 3

p + 4 = 3q + 4 or 3 (q + 1) + 1 leaves1 as remainder when it isdivided by 3.

Case II. When p = 3q + 1, p + 2 = 3q + 3 and p + 4 = 3q + 5

Here, p = 3q + 1 leaves 1 as remainder when it is divided by 3

p + 2 = 3q + 3 or 3 (q + 1) is exactly divisible by 3

p + 4 = 3q + 5 or 3(q + 1) + 2 leaves 2 as remainder when it is divided by 3.

Case III. When p = 3q + 2, p + 2 = 3q + 4and p + 4 = 3q + 6Here, p = 3q + 2 leaves 2 as remainderwhen it is divided by 3.p + 2 = 3q + 4 or 3(q + 1) + 1 leaves 1 asremainder when it is divided by 3p + 4 = 3q + 6 or 3(q + 2) is exactly divisibleby 3.

Hence, in all the cases, one and one numberout of p, p + 2 and p + 4 is divisible by 3,where p is any positive integer.


ORAny positive odd integer is type of 2q + 1where q is a whole number.∴ (2q + 1)2 = 4q2 + 4q + 1 = 4q (q + 1) + 1

...(i)Now, q(q + 1) is either 0 or evenSo it is 2m, where m is a whole number.∴ from (i) ⇒ (2q + 1)2 = 8m + 1.

11. Since, height of each stack is the same,therefore, the number of books in each stackis equal to the HCF of 96, 240 and 336.Let us find their HCF

96 = 24 × 2 × 3240 = 24 × 3 × 5336 = 24 × 3 × 7

So, HCF = 24 × 3 = 48.Now, number of stacks of English books

=9648

= 2

Number of stacks of Hindi books

= 24048

= 5

Number of stacks of Mathematics books

= 33648

= 7.

ASSESSMENT SHEET–11. (D) The denominator of each fraction in the

options (A), (B) and (C) can be expressed inthe form 2n 5m, where m, n being wholenumbers.

2. (A) Let x be any positive integer then it isof the form 3q or 3q + 1 or 3q + 2. So, x2

canbe written in the form 3m or 3m + 1.

3. HCF × LCM = Product of the two numbers⇒ 40 × 252 × p = 2520 × 6600

⇒ p =2520 6600

40 252××

= 1650.

4. True, If the number 3n ends with the digit 0,then its prime factorisation contains theprime 5. But by the Fundamental Theoremof Arithmetic, there is no prime other than 3in the factorisation of 3n.

5. The required number would be the HCF of967 – 7 = 960 and 2060 – 12 = 2048.Let us find the HCF of 960 and 2048 byusing Euclid’s algorithm.Since 2048 > 960∴ 2048 = 960 × 2 + 128

960 = 128 × 7 + 64128 = 64 × 2 + 0

Since the remainder becomes zero and thedivisor at this stage is 64, the HCF of 960and 2048 is 64.Hence, the required number is 64.

6.

Clearly, 456 = 23 × 3 × 19and 360 = 23 × 32 × 5∴ HCF = 23 × 3 = 24and LCM = 23 × 32 × 5 × 19 = 6840.

7. Let us assume the contrary that 3 is arational number.

So, 3 =ab

, where a and b are coprime.

∴ 3 =2

2

ab

(Squaring both sides)

⇒ 3b2 = a2

⇒ a2 is divisible by 3⇒ a is divisible by 3 because 3 is a prime.We can write a = 3c for some integer cSubstituting a =3c in 3b2 = a2, we get

3b2 = 9c2 ⇒ b2 = 3c2

⇒ b2 is divisible by 3⇒ b is divisible by 3.Therefore, both a and b are divisible by 3.But this contradicts the fact that a and b arecoprime that is, no common factor otherthan 1.

13LAER REBMUN S

Consequently, we arrive at the result that

our assumption that 3 is rational, is wrong.

Hence, 3 is an irrational number.

8. Let a be any odd positive integer. Then, it isof the form 6p + 1, 6p + 3 or 6p + 5.Here, three cases arise.Case I. When a = 6p + 1,

∴ a2 = 36p2 + 12p + 1= 6p(6p + 2) + 1 = 6q + 1,where q = p(6p + 2).

Case II. When a = 6p + 3,∴ a2 = 36p2 + 36p + 9

= 36p2 + 36p + 6 + 3= 6(6p2 + 6p + 1) + 3= 6q + 3,

where q = 6p2 + 6p + 1.Case III. When a = 6p + 5,

∴ a2 = 36p2 + 60p + 25= 36p2 + 60p + 24 + 1= 6(6p2 + 10p + 4) + 1= 6q + 1,

where q = 6p2 + 10p + 4.Hence, a is of the form 6q + 1 or 6q + 3.

ASSESSMENT SHEET–2

1. (D) 145871250

= 11.6696.

Clearly, the decimal expansion terminatesafter four decimal places.

2. (C) LCM (p, q) = x3 y2 z3.

3. HCF × LCM = Product of the two numbers.

⇒ 9 × LCM = 306 × 657

⇒ LCM = 306 657

9×

= 22338.

4. The maximum number out of 3, 5, 15, 25,75 is 75. Therefore, the HCF of 525 and 3000is 75.

5. The denominator of 2575000

is 5000.

5000 = 5 × 103 = 5 × (2 × 5)3

= 23 × 54.

Further,2575000

= 3 3

257 257 25 10 5 10 2

×=

× × ×

=4

51410

= 0.0514.

6. Let x = 2p + 1 and y = 2q + 1∴ x2 + y2 = (2p + 1)2 + (2q + 1)2

= 4p2 + 4p + 1 + 4q2 + 4q + 1

= 4(p2 + p + q2 + q) + 2

= S + T

where S = 4(p2 + p + q2 + q) and T = 2

S is divisible by 4 and so an even integer.

T is not divisible by 4 but an even integer.

Therefore, S + T is even, as sum of any twoevens is even, and not divisible by 4.

7. Let us assume the contrary that 5 is arational number.We can take coprime a and b (say) such that

5 =ab

; b ≠ 0

⇒ b 5 = a

Square both the sides to get5b2 = a2

⇒ a2 is divisible by 5

⇒ a is divisible by 5 because if square of anumber is divisible by a prime, then thenumber is divisible by the prime.

Let us take some integer c such that

a = 5c

Square both the sides to get

a2 = 25c2

Substitute a2 = 25c2 in 5b2 = a2 to get

5b2 = 25c2

b2 = 5c2

⇒ b2 is divisible by 5⇒ b is divisible by 5Therefore, both a and b are divisible by 5.


This contradicts the fact that a and b arecoprime that is a and b have no commonfactor.∴ Our assumption is false.

So, we conclude that 5 is an irrational

number.

8. Any positive integer n can be written in theform 3q, 3q + 1 or 3q + 2.Here, three cases arise as follows:Case I. When n = 3q,

∴ n3 = (3q)3 = 27q3

∴ n3 + 1 = 27q3 + 1 = 9 × 3q3 + 1

= 9m + 1, where m = 3q3.

Case II. When n = 3q + 1,

∴ n3 = (3q + 1)3

= 27q3 + 1 + 3(3q + 1) × 3q

= 27q3 + 27q2 + 9q + 1

∴ n3 + 1 = 27q3 + 27q2 + 9q + 2

= 9 (3q3 + 3q2 + q) + 2

= 9m + 2,

where m = 3q3 + 3q2 + q

Case III. When n =3q + 2,

∴ n3 = (3q + 2)3 = 27q3 + 8 +

3 × 6q (3q + 2)

= 27q3 + 8 + 54q2 + 36q

∴ n3 + 1 = 27q3 + 54q2 + 36q + 9

= 9 (3q3 + 6q2 + 4q + 1)

= 9m, where m = 3q3 +

6q2 + 4q + 1.Hence, n3 + 1 can be expressed in the form9m, 9m + 1 or 9m + 2 for some integer m.

CHAPTER TEST

1. (D)Since 32844 = 2 × 2 × 3 × 7 × 17 × 23So, 11 is not prime factor of 32844.

2. (A)

... LCM =306 × 1314

18

= 22338.

3. (C)As, 8q is even and 6 is even, 8q + 6 is even.

4. ... 0.56125 = 56125

100000=

449800

= 449

32 25×= 5 2

4492 5×

∴ 2n × 5m = 25 × 52

n = 5, m = 2.

5. ( )22 – 9 = 2 – 2 18 9+

= 11 – 2 18

= irrational.6. Yes.

2 × 3 × 5 × 13 × 17 + 13= 13 × (2 × 3 × 5 × 17 + 1)= 13 × 511= a composite number.

7. No.Hint: Prime factors of 9n will be type of

32n, i.e., × ×3 3 ... 3 Even no.

of times.

8. 120 = 23 × 3 × 5105 = 3 × 5 × 7150 = 2 × 3 × 52

∴ HCF = 3 × 5 = 15And LCM = 23 × 3 × 52 × 7

= 8 × 3 × 25 × 7= 4200.

9. Hint:

Let 2 – 3 3 = x where x is rational.

⇒ ( )22 – 3 3 = x2

⇒ 2 + 27 – 6 6 = x2

⇒ 29 – x2 = 6 6

⇒229 –

6x

= 6 6 .

Since 6 is not a perfect square. So 6 isalways irrational.

∴ It's a contradiction.

15LAER REBMUN S

10. We know that any positive integer is of theform 3q or 3q + 1 or 3q + 2.

Case I: n = 3q⇒ n3 = (3q)3 = 9 × 3q3 = 9m⇒ n3 + 1 = 9m + 1, where m = 3q3.

Case II: n = 3q + 1⇒ n3 = (3q + 1)3

= 27q3 + 1 + 27q2 + 9q= 9q (3q2 + 3q + 1) + 1= 9m + 1

⇒ n3 + 1 = 9m + 2, wherem = q(3q2 + 3q + 1).

Case III: n = 3q + 2⇒ n3 = (3q + 2)3

= 27q3 + 8 + 54q2 + 36q

n3 + 1 = 27q3 + 54q2 + 36q + 9= 9(3q3 + 6q2 + 4q + 1)= 9m,

where m = 3q3 + 6q2 + 4q + 1.Hence, n3 + 1 can be expressed in the form9m, 9m + 1 or 9m + 2, for some integer m.

11. Length = 8.25 m = 825 cmBreadth = 6.75 m = 675 cmHeight = 4.50 m = 450 cmThe required length of the rod will be thehighest common factor of 825 cm, 675 cmand 450 cm.Now, 825 = 3 × 52 × 11

675 = 33 × 52

450 = 2 × 32 × 52

So, HCF (825, 675, 450) = 3 × 52 = 75Hence, length of the rod is 75 cm.

❑❑


8. Solving α + β = 3 and α – β = –1,we get α = 1, β = 2... Polynomial is x2 – (α + β) x + αβ⇒ p(x) = x2 – 3x + 2.

9. According to the division algorithm,p(x) = g(x) × q(x) + r(x)

⇒ x3 – 3x2 + x + 2 = g (x) × (x – 2) + (– 2x + 4)(As given in question)

⇒ g(x) =3 2 – 3 + 3 – 2

– 2x x x

xTo find g(x), we proceed as given below.

Thus, g(x) = x2 – x + 1.

10.13

− ; 32

Hint: 6x2 – 7x – 3 = 0⇒ 6x2 – 9x + 2x – 3 = 0⇒ 3x (2x – 3) + 1 (2x – 3) = 0⇒ (2x – 3) (3x + 1) = 0

... x = 32

or 13

−

... α + β = 1

3−

+ 32

= 76

= –ba

... α . β = –13

. 32

= –12

= ca

.

2Chapter

POLYNOMIALS

WORKSHEET–101. (C)

Hint: put x2 + 2x + 1 = 0 and solve for x.

2. (C) Since the given graph of y = p(x) cutsx-axis at three points, so the number ofzeroes of p(x) are 3.

3. (A)

Hint: α +β1 1+ =

α β αβ.

4. Let one zero be α, then the other one will

be 1α

.

∴ α . 1α

= –15k

⇒ k = – 15.

5. Sum of zeroes (S) =2 3

–43

+

=3 – 84 3

= 5–

4 3

Product of zeroes (P) =2 3

– ×43

=1

–2

Now, required polynomial will be

x2 – Sx + P, i.e., x2 + 54 3

x1

–2

or 4 3 x2 + 5x – 2 3 .

6. Let f (x) = 2x2 + 2ax + 5x + 10If x + a is a factor of f (x), then f (– a) = 0Therefore, 2a2 – 2a2 – 5a + 10 = 0⇒ a = 2.

7. x3 – 4x2 + x + 6Hint: If the roots are α, β and γ of a cubicalpolynomial, then the polynomial will be(x – α) (x – β) (x – γ)= (x – 3) (x – 2) (x + 1) = x3 – 4x2 + x + 6.

17YLOP MON LA SI

5. p = 2Hint: (2)3 – 3(2)2 + 3(2) – p = 0⇒ 8 – 12 + 6 – p = 0⇒ 2 – p = 0∴ p = 2.

6. Let α and β be the two zeroes off(x) = ax2 + 2x + 3a

Then, α + β = – 2a

and αβ = 3aa

= 3

According to the question,

– 2a = 3

⇒ a = – 23 .

7. Let the third zero be α, then

sum of the zeroes = – 2

3coefficient of coefficient of

xx

⇒ 2 + 3 + α = –– 61

⇒ α = 1Hence, the third zero is 1.

8. Let us divide 6x4 + 8x3 + 17x2 + 21x + 7 by3x2 + 4x + 1.

Clearly, the remainder is x + 2.Now, ax + b = x + 2Comparing like powers of x both the sides,we obtain

a = 1, b = 2.

9. We know that,Dividend = (Divisor × Quotient) + Remainder⇒ 4x3 – 8x2 + 8x + 1 = g(x) × (2x – 1) + x + 3⇒ g(x) × (2x – 1) = 4x3 – 8x2 + 7x – 2

g(x) = 3 24 – 8 7 – 2

2 – 1x x x

x+

11. Let p(x) = x4 + x3 – 34x2 – 4x + 120Given zeroes of p(x) are 2 and – 2... (x – 2) (x + 2) = x4 – 4 is a factor of p(x).We divide p(x) by x2 – 4,

x2 + – 30x

x x x3 2– 30 – 4 +120

x4 + – 34 – 4 + 120x x x3 2

x x4 2– 4–

+

–

+

–

+

– 4x x3

– 30 + 20x2 1

– 30 + 20x2 1

x2 – 4

0

... p(x) = (x2 – 4) (x2 + x – 30)

... Other zeroes of p(x) are given byx2 + x – 30 = 0

⇒ x2 + 6x – 5x – 30 = 0⇒ x(x + 6) – 5(x + 6) = 0⇒ (x – 5) (x + 6) = 0

x = 5, – 6Hence, all the zeroes are 2, – 2, 5 and – 6.

WORKSHEET– 11

1. (A)∵ p(x) = 2x2 – 2x + 1∴ Sum of zeroes = 1

Product of zeroes =12

.

2. (A)Let α = 5 and β = – 5, then the quad-ratic polynomial will be x2 – (α + β)x + αβor x2 – 25.

3. (D) Let us take option (D)p(x) = (x2 – 2) – (x2 + 3x) = – 3x – 2This is a linear polynomial.

4. For zeroes of p(x), put p(x) = 0⇒ 4x2 – 4x + 1 = 0⇒ 4x2 – 2x – 2x + 1 = 0⇒ 2x (2x – 1) – 1(2x – 1) = 0⇒ (2x – 1) (2x – 1) = 0∴ 2x – 1 = 0

∴ x = 12

,12

.


Now,

2 1x –

2 – 3 + 2x x2

4 – 8 + 7 – 2x x x3 2

4 – 2x x3 2

– 6 + 7 – 22x x

– 6 + 32x x

4 – 2

4 – 2

x

x

+–

+ –

+–

0

Hence, g(x) = 2x2 – 3x + 2.

10. 3 and 1

Hint: x2 – 3x – x + 3 = 0

⇒ x = 3 , 1

Now, sum of zeroes = 3 + 1

= – 2Coefficient of

Coefficient of xx

And product of zeroes = 3

= 2Constant term

Coefficient of x.

11. p(x) = 2 2 12x x− − ; ( )− 2 2, 3 2

Hint: For zeroes: 2 – 2 – 12x x = 0

⇒ 2 2 2 3 2 12x x x+ − − = 0

⇒ ( )– 3 2x ( )2 2x + = 0

⇒ x = – 2 2 or x = 3 2 .

WORKSHEET–12

1. (C) Sum of zeroes = – (– 5)

13

= 15

Product of zeroes =

3213

= 92

.

2. (C) Sum of zeroes = 6

⇒ 6 = –– 3

1k

∴ k =63

= 2.

3. (D) Let one zero be α, then the other one

will be 1α

.

So, α . 1α = 2

44

aa +

a2 – 4a + 4 = 0⇒ (a – 2)2 = 0⇒ a = 2.

4. (A)Let the zeroes be α, β, γ. If γ = – 1, then

αβγ = –1c

If γ = – 1, then αβ = c ...(i)Further, (– 1)3 + a (– 1)2 + b (– 1) + c = 0⇒ – 1 + a – b + c = 0⇒ c = b – a + 1 ...(ii)From equations (i) and (ii), we have

αβ = b – a + 1.5. Given polynomial is:

f (x) = x2 – px – 2p – c... α + β = pand α . β = – 2p – c... (α + 2) (β + 2) = αβ + 2 (α + β) + 4

= – 2p – c + 2p + 4= (4 – c).

6. λ = 6Hint: (α + β)2 = (α − β)2 + 4αβ.

7. x = –1 or 3; f(x) = x2 – 2x – 3Hint: x = – 1 or 3,∴ Sum of zeroes = 2Product of zeroes = – 3∴ p(x) = x2 – (α + β)x + αβ

= x2 – 2x – 3.

8. x2 – x – 474

Hint: f (x) = {x2 – (sum of roots) x + (productof roots)}

19YLOP MON LA SI

9. The number which to be subtracted is theremainder when 4x4 + 2x3 – 8x2 + 3x – 7 isdivided by 2x2 + x – 2. To find the remainder,we proceed as following.

2 2x + x –2

2 – 2x2

4 + 2 – 8 + 3 – 7x x x x4 3 2

4 + 2 – 4x x x4 3 2

– + 3 – 7x x24

– 2 4x x +24

5 11x –

––

–+ + –

+

Hence, 5x – 11 must be subtracted from4x4 + 2x3 – 8x2 + 3x – 7 so that it becomesexactly divisible by 2x2 + x – 2.

10. g(x) = x2 + 2x + 1

Hint: p(x) = g(x) × q(x) + r(x)

⇒ g(x) = ( ) – ( )

( )p x r x

q x

where, p(x) = 3x3 + x2 + 2x + 5q(x) = 3x – 5

and r(x) = 9x + 10.

11. Since x = 53

and x = – 53

are zeroes of

p(x) = 3x4 + 6x3 – 2x2 – 10x – 5, so p(x) is

divisible by 5 5 – +

3 3

x x , i.e., x2 – 53 .

Here, other two zeroes of p(x) are the othertwo zeroes of quotient 3x2 + 6x + 3Put 3x2 + 6x + 3 = 0

⇒ 3(x + 1)2 = 0⇒ x = – 1 and x = – 1

Hence, all the zeroes of p(x) are 53

,

–53

, – 1 and – 1.

WORKSHEET–131. (A)

Hint: Given polynomial can be written as:p(x) = 2x2 + 3x – 11

Sum of zeroes =–ba

Product of zeroes =ca .

2. (B) Sum of zeroes = – 99 = –ve Product of zeroes = 127 = +veIf the sum of both zeroes is negative, thenthe zeroes would be either both negativeor one negative and other one positive. Ifthe product of both the zeroes is positive,then the zeroes would be either bothpositive or both negative.Consequently, we obtain that both thezeroes are negative.

3. (D) We know that the degree of the remain-der is less than the degree of divisor.Here, degree of the divisor is 3, therefore,the possible degree of the remainder canbe any out of 0, 1 and 2.

4. Hint: Substitute x = – 2 in x2 + 2x + k = 0.5. Since α, β are the zeroes of x2 + px + q, then

α + β = –p; αβ = q

Now, + –pq

α +β1 1 = =α β αβ

And × q1 1 1 1= =α β αβ

So the polynomial having zeroes and1 1 α β

will be

p(x) = x2 – 2 1+ × = + +

px x x

q q1 1 1 1+α β α β

or p(x) = qx2 + px + 1.


6. g(x) = x2 + 2x + 7.Hint: Divide x3 + 3x – 14 by x – 2.

7. p(x) = 3x2 – 3x + 12.g(x) = x2 – x + 4

... q(x) = 3r(x) = 0.

8. 1 1, –

7 7Hint: For zeroes: 21x2 – 3 = 0

x2 = 17

... x = 17

± .

9. Since a = 2 is a zero of a3 – 3a2 – 10a + 24,therefore a3 – 3a2 – 10a + 24 is divisible bya – 2. Further the obtained quotient willprovide the other two zeroes.

Put a2 – a – 12 = 0 for other zeroes.⇒ (a – 4) (a + 3) = 0⇒ a = –3, 4Thus, the other two zeroes are – 3 and 4.

10. g(x) = x + 1.

Hint: Applying division algorithm, we get

x4 + 1 = g(x) × (x3 – x2 + x – 1) + 2

⇒ g(x) =4

3 2– 1

– + – 1x

x x x

= ( )( )( )

( )( )2

2

+1 – 1 +1

– 1 +1

x x x

x x

= x + 1.

11. Hint: +α β2 21 1

=2 2

2 2α + βα β

= ( )2

2 22α + β − αβ

α β=

2

2– 2b acc

.

OR

Let us divide x4 + 2x3 + 8x2 + 12x + 18 by x2 + 5.

Clearly, the remainder is 2x + 3.Now, px + q = 2x + 3Comparing like powers of x both the sides,we get

p = 2, q = 3.

WORKSHEET–141. (C) If a quadratic polynomial has equal

roots , then its discriminant must be zero.So, b2 – 4ac = 0⇒ b2 = 4acThis last equation holds if a and c havesame sign.

2. (D) Sum of zeroes = – 3 + 7 = 4,Product of zeroes = (– 3) × 7 = – 21A polynomial may be k(x2 – 4x – 21)where k has infinitely many real values.Hence, infinitely many number of poly-nomials can be.

3. (A) α + β = 32

, αβ = 12

... (α – β)2 = (α + β)2 – 4αβ

=94

– 2 = 14

⇒ α – β = ± 12

21YLOP MON LA SI

... α = 12

, β = 1 or α = 1, β = 12

... α + 2 =52

, β + 2 = 3 or α + 2 = 3,

β + 2 = 52

.

Hence, the required polynomial can be

x2 – 5

32

+ x +

52

× 3, i.e., x2 –112

x + 152

.

4. Let zeroes be α and β.α + β = 6, αβ = 4Using (α – β)2 = (α + β)2 – 4αβ, we get(α − β)2 = 62 – 4 × 4 = 20 ⇒ α – β, = ± 2 5Thus, the difference of zeroes is ± 2 5 .

5.5.5.5.5. Hint:α β

+β α

=α +β

αβ

2 2

= ( )2 2α + β − αβ

αβ

= 25 126− =

136

.

6. Hint: x2 – 1 = (x + 1) (x – 1)... x = – 1 or 1, both will satisfy with thegiven polynomial.... we get, p + q + r + s + t = 0 ...(i)

and p – q + r – s + t = 0 ...(ii)From (ii),

p + r + t = q + sFrom (i),

2 (q + s) = 0 ⇒ q + s = 0... p + r + t = q + s = 0.

7. No.Hint: Divide q(x) by g(x). If the remainderobtained is zero, then the g(x) is a factor ofq(x) otherwise not.

8. a = 1, b = 7Hint: Put remainder = 0 and equate coefficientof x in the remainder and constant termwith zero.

9. According to division algorithm, p(x) = g(x) × q(x) + r(x)

(i) p(x) = 6x2 + 3x + 2, g(x) = 3q(x) = 2x2 – x, r(x) = – 2

(ii) p(x) = 8x3 + 6x2 – x + 7, g(x) = 2x2 + 1q(x) = 4x + 3, r(x) = – 5x + 4

(iii) p(x) = 9x2 + 6x + 5, g(x) = 3x + 2,q(x) = 3x, r(x) = 5.

10. Given quadratic polynomial is

5 5 x2 + 30x + 8 5To find its zeroes, put

5 5 x2 + 30x + 8 5 = 0

⇒ 5 5 x2 + 20x + 10x + 8 5 = 0

⇒ 5x ( )5 4x + + 2 5 ( )5 4x + = 0

⇒ ( )+5 2 5x ( )+5 4x = 0

⇒ x = –25

or x =– 4

5

i.e., x = –2 5

5 or x = –

4 55

So, sum of zeroes =– 2 5

5 –

4 55

= – 6 55

And product of zeroes

=2 5 4 5

– × –5 5

= 85

.

Also, sum of zeroes = – 2Coefficient of

Coefficient of x

x

= – 305 5

= – 6 55

And product of zeroes = 2Constant term

Coefficient of x

= 8 55 5

= 85

.

Hence verified.OR

Hint: Let S =1 11 1

α − β −+

α + β +

P =1 11 1

α − β − α + β +


... Required polynomial p(x) = x2 – Sx + P.

11. As 32

and – 32

are the zeroes of the given

quadratic polynomial, so3

–2

x

and

32

x

+

will be the factors of that, Conse-

quently,3

–2

x

×32

x

+

, i.e.,

2 3–

2x

must be the factor of that. Let us divide

2x4 – 10x3 + 5x2 + 15x – 12 by 2 3–

2x .

2 – 10 + 8x x2

– 10 + 8 + 15 12x x x –3 2

2 – 10 + 5 + 15 – 12x x x x4 3 2

2 – 3x x4 2

3

2

–

+

–

–

+

x2 –

– 10 + 15x x3

8 12x –2

–

8 12x –2

+

0

Now, 2x4 – 10x3 + 5x2 + 15x – 12

= 2 3–

2x

(2x2 – 10x + 8)

By splitting –10x, we factorise 2x2 – 10x + 8as (x – 4) (2x – 2). So, its zeroes are givenby x = 4 and x = 1.

Therefore, the zeroes of the given poly-

nomial are 32

, – 32

, 1 and 4.

WORKSHEET–151. (B)

Hint: f (x) = x2 – px – (p + c)

(α + 1) (β + 1) = αβ + (α + β) + 1 .

2. (A)

Hint: 1 1 1 α + β + γ+ + =

αβ βγ γα αβγ .

3. (D) Let zeroes be α and β, then(α – β)2 = 144

⇒ α – β = + 12 ...(i)α + β = – p ...(ii)

αβ = 45 ...(iii)Also, we have

(α – β)2 = (α + β)2 – 4αβ144 = p2 – 180

⇒ p = ± 18.

4. Let the given linear polynomial bey = ax + b ....(i)

This passes through points (1, –1), (2, 1) and

3, 0

2

∴∴∴∴∴ – 1 = a + b ...(ii)1 = 2a + b ...(iii)

0 =32

a + b ...(iv)

Solving equations (ii) and (iii), we get a = 2,b = – 3 which satisfy to equation (iv).Consequently, using equation (i), we get

y = 2x – 3∴ Polynomial is p(x) = 2x – 3

Since p(x) = 0 if x = 32

⇒ x = 32

is zero of p(x).

5. Let us divide ax3 + bx – c by x2 + bx + c bythe long division method.

ax ab–

– + ( –abx b – ac) x c2

ax bx c3 + –

ax + abx + acx3 2

–

+

– –

++

x bx c2 + +

– –abx ab x – abc2 2

( + ) –ab b – ac x + abc c2

Put remainder = 0

23YLOP MON LA SI

⇒ (ab2 + b – ac)x + (abc – c) = 0⇒ ab2 + b – ac = 0 and abc – c = 0Consider abc – c = 0 ⇒ (ab – 1) c = 0⇒ ab = 1 or c = 0. Hence, ab = 1.

6. Hint: Let p(x) = x3 – mx2 – 2npx + np2

(x – p) is a factor of p(x)⇒ p(x) = 0 at x = p.⇒ p3 – p2m – p2n = 0⇒ p2 [(p – (m + n)] = 0⇒ p = m + n where p ≠ 0.

7. x3 – 4x2 + x + 6Hint: The required cubic polynomial is givenby (x – 3) (x – 2) (x + 1) or x3 – 4x2 + x + 6This is the required polynomial.

8. – 2, 3, 4Hint: α + β + γ = 5

αβ + βγ + αγ = –2αβγ = –24

Let αβ = 12... γ = –2... α + β = 7⇒ (α – β)2 = 1⇒ α – β = ± 1... α + β = 7 and α – β = 1⇒ α = 4

β = 3or α + β = 7 and α – β = –1⇒ α = 3

β = 4.9. f (x) would become exactly divisible by g(x)

if the remainder is subtracted from f(x).Let us divide f(x) by g(x) to get the remainder.

x 62 + + 8x

6 – 16 – 12 + 21x x x3 2

x4 + 2x x x3 2– 13 – 12 + 21

x x x4 3 2– 4 + 3– –

–

–

+

+–

–

x x2 – 4 + 3

6 – 24 + 18x x x3 2

8 – 30 + 21x x2

8 – 32 + 24x x2

2 – 3x

+

Hence, we should subtract 2x – 3 from f(x).

10. If 2 ± 3 are zeroes of p(x), then x – ( )2 3+

and x – ( )2 – 3 are factors of p(x).

Consequently ( ){ }– 2 3x + ( ){ }– 2 – 3x

i.e., (x – 2)2 – 3, i.e., x2 – 4x + 1 is factor ofp(x).Further,

x2 – 2 – 35x

– 2 – 27 + 138 – 35x x x3 2

x x x x4 3 2– 6 – 26 + 138 – 35

x x x4 3 2– 4 +– –

+

+

+

–

–

+

x x2 – 4 + 1

– 2 + 8 – 2x x x3 2

– 35 + 140 – 35x x2

– 35 + 140 – 35x x2

0

+

Clearly x2 – 2x – 35 is a factor of p(x)⇒ (x – 7)(x – 5) is a factor of p(x)⇒ x – 7 and x + 5 are factors of p(x)⇒ x – 7 = 0 and x + 5 = 0 give other zeroes

of p(x)⇒ x = 7 and x = – 5 are other zeroes of p(x).Hence, 7 and – 5 are required zeroes.

11. Hint: 2 2 4 4

2 2 2 2α β α + β + =β α α β

2 2 2 2

2 2{(α + β) − 2αβ} − 2α β=

α β .

OR

Given polynomial is:f (x) = pqx2 + (q2 – pr)x – qr

Put f (x) = 0 to find roots.pqx2 + (q2 – pr) x – qr = 0

⇒ pqx2 + q2x – prx – qr = 0⇒ qx(px + q) – r(px + q) = 0⇒ (px + q)(qx – r) = 0

⇒ x = –qp or x = r

q


Sum of roots = –qp +

rq =

2–pr qpq

= – 2

Coefficient of

Coefficient of

xx

Product of roots = – ×q rp q = –

rp = –

qrpq

= 2Constant term

Coefficient of x.


1. (C) Discriminant = 0⇒ b2 – 4ac = 0

⇒ b2 = 4ac

LHS = b2 = positive sign⇒ RHS = 4ac must be positive sign.⇒ c and a have same signs.

2. (D) Required quadratic polynomial= x2 – (sum of zeroes)x + product of zeroes

= x2 – 2 3x – 5 3 .

3. p(x)= x2 – ax – (a + 1)At x = – 1, p(x) = (–1)2 – a( –1) – (a + 1)

= 1 + a – a – 1 = 0q(x) = ax2 – x – (a + 1)

at x = – 1, q(x) = a( –1)2 – ( –1) – (a + 1)= a + 1 – a – 1 = 0

Therefore, x + 1 is the common factor ofp(x) and q(x).

4. Correct,

f(x) = x2 – p(x +1) – c = x2 – px – (c + p)∴ α + β = p; αβ = – (c + p)Now, (α + 1) (β + 1) = αβ + (α + β) + 1

= – (c + p) + p + 1= – c – p + p + 1= 1 – c.

5. Let f(x) = 6x3 + 2 x2 – 10x – 4 2

As 2 is a zero of f (x), (x – 2 ) is a factor

of f(x).

Let us divide f(x) by ( )– 2x .

6 + 7 2 + 4x x2

7 2 – 10 – 4 2x x2

6 – 10 – 4 23 2x xx + 2

6 – 6 2x x3 2

–

+

+

+

–

–

x x7 2 – 142

x4 – 4 2

x4 – 4 2

0

��

�

�

��

�

��

x – 2�

∴ f (x) = ( )– 2x ( )+ +26 7 2 4x x

= ( )– 2x ( )+ + +26 3 2 4 2 4x x x

= ( )– 2x ( ) ( )+ +3 2 4 2 1x x

Hence, ( ) ( )+ +3 2 4 2 1x x gives x = –2 2

3

or x = –12

Therefore, other two zeroes are –2 2

3 and

–2

2.

6. p(y) = y2 + 3 5

2y – 5

Here, a = 1, b = 3 5

2 , c = – 5

Discriminant

D = b2 – 4ac =2

3 52

– 4 × 1 × (– 5)

=454

+ 20 = 125

4

Now, y =

– 3 5 125– ± D 2 4

2 2 1b

a

±=

×

25YLOP MON LA SI

=±

– 3 5 5 52 2

2

=

2 522

or

– 8 522

=5

2 or – 2 5

Hence, the zeroes are 52

and – 2 5 .

7. α and β are zeroes of f (x) = x2 – x – 2

Sum of roots = α + β = – –11

= 1 ...(i)

Product of roots = αβ = –21

= – 2 ...(ii)

∴ (2α + 1) + (2β + 1) = 2(α + β) + 2

= 2(1) + 2 [Using (i)]

= 4 ...(iii)

And (2α + 1) (2β + 1) = 4αβ + 2α + 2β + 1

= 4αβ + 2 (α + β) + 1

= 4 (– 2) + 2 (1) + 1 [Using (i) and (ii)]

= – 5 ...(iv)Now, required polynomial can be given by x2 – {(2α + 1) + (2β + 1)}x + (2α + 1)(2β + 1)i.e., x2 – 4x – 5. [Using (iii) and (iv)]

8. Let us divide p(x) by 2x2 – 5.

3 + 4 + 52 xx

x x ax b8 + 10 + +3 2

6 + 8 5x a b4 – + +x x x3 2

6 – 15x x4 2

–

+

+

+

–

–

8 – 20x x3

10 + (20 + )x a x + b2

10 – 25x2

2 – 5x2

(20 + ) + 25a x + b

Here, remainder is (20 + a)x + b + 25.If the polynomial p(x) is exactly divisibleby 2x2 – 5, the remainder must be zero.∴ (20 + a)x + (b + 25) = 0Comparing the coefficients of like powersof x between both the sides, we have20 + a = 0 and 25 + b = 0⇒ a = – 20 and b = – 25.


1. (C) Sum of zeroes = – 3 2

–3

= 2

Product of zeroes = 13

.

2. (B) At x = 2, p(x) = 0, i.e., p(2) = 0∴ a (2)2 – 3 × 2 (a – 1) – 1 = 0⇒ 4a – 6a + 6 – 1 = 0

⇒ a = 52

.

3. Sum of zeroes = α + β = 5Product of zeroes = αβ = 4

Now,1α

+1β

– 2αβ =α + βαβ

– 2αβ

=54

– 2 × 4

=27

–4

.

4. Using division algorithm, we haveg(x) × (x – 2) – 2x + 4 = x3 – 3x2 + x + 2

⇒ g(x) =+3 2– 3 3 – 2

– 2x x x

x

Here, at x = 2, x3 – 3x2 + 3x – 2

= 8 – 12 + 6 – 2 = 0

∴ = x3 – 3x2 + 3x – 2= (x – 2) (x2 – x + 1)

= 2( – 2)( – 1)

( – 2)x x x

x+

⇒ g(x) = x2 – x + 1.


5. Given s = 2 and p = –32

The required polynomial is given byk [x2 – sx + p]

i.e., k

2 3– 2 –

2x x , where k is any real

number.

6. Let f(x) = +24 3 5 – 2 3x x

= 24 3 8 – 3 – 2 3x x x+

= 4 ( 3 2) – 3 ( 3 2)x x x+ +

= ( 3 2)(4 – 3)x x+

To find zeroes of f (x), put f(x) = 0

3 2 0x + = or 4 – 3 0x =

– 2 – 2 3

33x = = or

34

x =

Thus, the zeroes are α = –2 3

3and β =

34

Sum of zeroes = α + β

= –2 3

3+

34

=– 5 3

12

=5 3

–4 3×

= – 5

4 3

= – 2Coefficient of Coefficient of

xx

Product of zeroes = αβ = –2 3

3. 3

4

= – 2 34 3

= 2

Constant termCoefficient of x

.

Hence verified.

7. (i) Let y = p(x)∴ y = – x2 + x + 6

The table for some values of x and theircorresponding values of y is given by

x – 2 – 1 0 1 2 3

y 0 4 6 6 4 0

Let us draw the graph of p(x) using thistable.

From the graph, it is clear that the zeroesof p(x) are – 2 and 3.

(ii) Let y = p(x)

∴ y = x3 – 4x

The table for some values of x and theircorresponding values of y is given by

x – 2 – 1 0 1 2

y 0 3 0 – 3 0

Let us draw the graph of p(x) by using thistable.

(2, 0)(–2, 0)

(–1, 3)

p x x x( ) = – 43

X� X

Y�

Y

3

2

1

–

–

–

1

2

3

–

–

–

–

–

–

3 2 1

– – – – – –

––– 1 2 3

(1, 3)–

0

From the graph, it is clear that the zeroes of p(x) are – 2, 0 and 2 .

(0,6)

(3, 0)(–2, 0)

(2, 4)(–1, 4)

(1,6)

p x x x( ) = – + + 62

X� X

Y�

Y

6

5

4

3

2

1

–

–

–

–

–

–

1

2

3

–

–

–

–

–

–

3 2 1

– – – – – –

––– 1 2 30

27YLOP MON LA SI

8. Let f (x) should be added to p(x) so that theresulting polynomial is exactly divisible byg(x). Since the degree of f (x) is less thanthat of g(x).So, f(x) may be ax + b i.e., f(x) = ax + b.Therefore, the new dividend would bex4 + 2x3 – 2x2 – 5x + 7 + ax + b,i.e., x4 + 2x3 – 2x2 + (a – 5)x + b + 7Let us divide this new dividend by g(x).

x2 + 1

x a x b2 + ( – 5) + + 7

x4 + 2x x a x b3 2– 2 + ( – 5) + + 7

x x x4 3 2+ 2 – 3– – +

+––

x x2 + 2 – 3

+ 2 – 3x x2

( – 7) + + 10a x b

Thus, the remainder obtained must be zero. ∴ (a – 7)x + b + 10 = 0⇒ (a – 7)x + (b + 10) = 0 . x + 0⇒ a – 7 = 0 and b + 10 = 0⇒ a = 7 and b = – 10Hence, f (x) = 7x – 10.

CHAPTER TEST

1. (A) ∵ α + β = – 52

, αβ = 12

∴ α + β + αβ = –2

So, option (A) is correct.

2. (B)p(x) = x2 – (α + β)x + αβ

= x2 + x – 2So, option (B) is correct.

3. (B)

Hint: p(x) = x 72

x + ∴ zeroes are given by

x72

x + = 0

⇒ x = 0 or – 72

.

So, option (B) is correct.

4. Hint: g(x) =( ) – (7 – 5 )

2p x x

x

=+ + + +3 22 4 5 7 – 7 5

2x x x x

x

= x2 + 2x + 5.

5. Hint: α + β = 24

4 5=

65

αβ = – 9 54 5

= – 94

.

6. Hint: α = – β

α + β = 0 ⇒–ba

= 0

⇒ 3 12

k + = 0 ⇒ k = –13

.

7. Answer may vary.

8. If α, β and γ are the zeroes of a cubicpolynomial f(x), then

f(x) = x3 – (α + β + γ) x2

+ (αβ + βγ + γα) x – αβγHere, α + β + γ = 4, αβ + βγ + γα = 1and αβγ = – 6∴ f(x) = x3 – 4x2 + x + 6.

9. We know thatDividend = Quotient × Divisor

+ Remainder⇒ Dividend – Remainder = Quotient

× DivisorClearly, RHS of the above result is divisibleby the divisor. Thus, if we subtract remainderfrom the dividend, then it will be exactlydivisible by the divisor.Dividing x4 + 2x3 – 13x2 – 11x + 10 byx2 – 4x + 3, we get


x2 + 6 + 8x

6 – 16 – 11 +10x x x3 2

x x x x4 3 2– 13 – 11 + 10+ 2

x x x4 3 2– 4 + 3– –

+

– –

+

––

+

6 – 24 + 18x x x3 2

8 – 29 + 10x x2

8 – 32 + 24x x2

x x +2 – 4 3

3 – 14x

Quotient = x2 – 6x + 8 andremainder = 3x – 14.Thus, if we subtract the remainder 3x – 14from x4 + 2x3 – 13x2 – 11x + 10, it will bedivisible by x2 – 4x + 3.

10. Hint: Since x = –1

3and

13

x = are zeroes.

Therefore, 1 1–

3 3x x

+

will be a

factor of p(x), i.e., 2 1–

3x is a factor of p(x).

3 – 15 + 182 xx

– 15 + 18 + 5 – 6x x x3 2

3 – 15 + 17 + 5 – 6x x x x4 3 2

3x – x4 2

–

+

–

+

–

+

– 15 + 5x x3

x18 – 62

x18 – 62

0

x2 –31

∴ Other zeroes are given by

3x2 – 15x + 18 = 0

⇒ x2 – 5x + 6 = 0

⇒ (x – 3) (x – 2) = 0

∴ x = 3, 2.

11. We have

24 3 5 – 2 3x x+ = ( )( )3 2 4 – 3x x+

So, the value of +24 3 5 – 2 3x x is zero

when,

3 2x + = 0 or 4 – 3x = 0, i.e., when x =–2

3

or x = 3

4. Therefore, the zeroes of

24 3 + 5 – 2 3x x are – 2

3 and 3

4.

Now,

sum of zeroes– 2

3+

34

=– 5

4 3

= 2

– Coefficient of Coefficient of

xx

Product of zeroes = ×

– 2 343

= – 2 34 3

= 2Constant term

Coefficient of x.

❑❑

29ENIL AUQERA T OI N S ...RIAP FO

3Chapter

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

WORKSHEET–181. (B) Since (3, a) lies on the equation

2x – 3y = 5, therefore, (3, a) must satisfythis equation.∴ 2 (3) – 3 (a) = 5

⇒ 3a = 1 ⇒ a = 13

.

2. (A) Hint: – 32

=– 9k

⇒ k = 6.

3. (C) Hint: The condition of inconsistency oftwo equations a1x + b1y = c1 and a2x + b2y = c2

is given by ≠1 1 1

2 2 2=

a b ca b c

.

4. x = 1, y = 2

Hint: Joining the given equations, we get80x + 80y = 240

or x + y = 3 ...(i)Subtracting given first equation from otherone, we get

6x – 6y = – 6or x – y = –1 ...(ii)Solving equations (i) and (ii), we obtain

x = 1, y = 2.

5. x = 3, y = 2

Hint: Let +1

x y= u,

1x y−

= v.

∴ Given equations become10u + 2v = 4 and 15u – 5v = – 2.

6. FalseLet us substitute c = 40, The givenequations become

x – 2y = 8or 5x – 10y = 40

Here,15

= –2

–10 =

840

⇒ The equations represent a pair ofcoincident lines.

⇒ The equations have infinitely manysolutions.

7. The given equations are4(2x + 3y) = 9 + 7y

and 3x + 2y = 4or 8x + 5y –9 = 0

3x + 2y – 4 = 0By cross-multiplication, we have

− 20 + 18x

= 32 + 27

y−− =

116 15−

⇒– 2x = –

– 5y = 1

1

x = – 2 and y = 5Hence, x = – 2, y = 5 is the solution of thegiven system of equations.

8. To draw a line, we need atleast twosolutions of its corresponding equations.x + 3y = 6; at x = 0, y = 2 and x = 3, y = 1.So, two solutions of x + 3y = 6 are:

x 0 3

y 2 1

2x – 3y = 12; at x = 0, y = –4 and at x = 6, y = 0So, two solutions of 2x – 3y = 12 are:

Now, we draw the graph of given systemof equations by using their correspondingsolutions given in the above tables.

x 0 6

y – 4 0


2. (A) Condition for parallel lines is

1

2

aa

= 1

2

bb

≠ 1

2

cc

⇒ 13

=– 2k

≠ – 3– 1 ⇒ k = – 6.

3. (D) As y = 2 and y = 7,both represent straight lines parallel tox-axis∴ y = 2 and y = 7 are parallel lines.Hence, the given pair of equations has nosolution.

4. The given lines to be coincident, if

12 I

k = 3

IIk

= – ( – 3)

–III

kk

Taking I and II, we havek2 = 36 ⇒ k = ± 6. ...(i)Taking II and III, we havek2 – 3k = 3k ⇒ k(k – 6) = 0⇒ k = 0 or 6 ... (ii)Using (i) and (ii), we obtain

k = 6.

5. x = 5, y = 2Hint: Adding the given equations,we get 2x + y = 12 ...(i)Subtracting the given equations,we get 3x + y = 17 ...(ii)Now, (i) – (ii) ⇒ – x = – 5

x = 5∴ from (ii ) ⇒ 3(5) + y = 17

y = 2.

6. Yes.Applying the condition

1

2

aa = 1

2

bb

= 1

2

cc

We have13

= 26

= – 3– 9

That is true.Therefore, the pair of equations is consis-tent with infinitely many solutions.

From the, graph the two lines intersect they-axis at (0, 2) and (0, – 4).

9. Let the fixed charges and change per kmbe Rs. x and Rs. y respectively.

x + 10y = 105 ...(i)x + 25y = 255 ...(ii)

Subtracting equation (i) from equation (ii),we get

15y = 150y = 10 ...(iii)

From equations (i) and (iii), we getx = 5

Now, the fare for travelling a distance of35 km

= x + 35y= 5 + 35 × 10= Rs. 355.

Fixed charge = Rs. 5Charge per km = Rs. 10Total charge for 35 km = Rs. 355.

WORKSHEET–19

1. (C) x – 5y = 5.

(2, k) lies on it.

∴ 2 – 5(k) = 5 ⇒ 5(k) = – 3

⇒ k = 3–

5.


7. x = 25552

, y =615104

Hint: The system:9x – 10y + 15 = 0 5x + 6y – 60 = 0

By cross-multiplication, we have 1

= =600 90 540 75 54 + 50

yx −− − − .

8. For equation 3x + y – 2 = 0,

For equation 2x – 3y – 5 = 0,

As the lines corresponding to the givenequations intersect each other at (1, –1),the required solution is x = 1, y = –1.

9. Let the man's starting salary and fixedincrement be x and y respectively.

According to the question,x + 4 y = 15000 ...(i)

x + 10 y = 18000 ...(ii)

Equations (i) and (ii) from the requiredpair of linear equations. On solving thesetwo equations, we will find starting salaryof x = Rs. 13000 and fixed increment ofy = Rs. 500.

WORKSHEET–20

1. (B) As the lines are intersecting each other,

3a

≠ 2–1

⇒ a ≠ –32

.

2. (D) Line x = a is parallel to y-axis and theline y = b is parallel to x-axis. These linesintersect each other at (a, b).

3. 3x – y – 5 = 0 and 6x – 2y – k = 0 have nosolution.⇒ These equations represent a piar of

parallel lines.

⇒36

= –1–2

≠ –5k

⇒ k ≠ 10.

4. No.For infinitely many solutions, the followingcondition must be satisfied.

λ2

=36

= 7

–14

But, here 36

≠ –714

as 12

≠ 1

–2

Hence, no value of ‘λ’ provides the pair ofinfinitely many solutions.

5. The given system of equations can bewritten as

ax + by – (a – b) = 0bx – ay – (a + b) = 0

By cross-multiplication,

– ( + ) – ( – ) ( )

xb a b a a b

i

= –

– ( + ) + ( – ) ( )

ya a b b a b

ii

2 21

– – ( )

a biii

⇒ Taking (i) and (iii) simultaneously,we get x = 1 and y = – 1

x 0 1

y 2 – 1

x – 2 1

y – 3 – 1


Hence x = 1, y = – 1 is the solution of thegiven system of equations.

6. x = 6, y = – 4, m = 0

Hint: Take 1x = u and 1

y= v.

7. No; (6, 0) , (4, 0)

Hint: For x + 3y = 6

For 3x + 9y = 12

Let us draw the graph of lines using thetables obtained above.

In the graph, lines are parallel. So, the pairof equations is not consistent.The lines intersect the x–axis at (4, 0) and(6,0).

8. Let the initial length be x and breadth be y.Then according to question,

xy – 9 = (x – 5) (y + 3) ...(i)and xy + 67 = (x + 3) (y + 2) ...(ii)Simplifying equations (i) and (ii), we have

3x – 5y = 6 ...(iii)2x + 3y = 61 ...(iv)

On solving (iii) and (iv),x = 17, y = 9

Hence, length of rectangle is 17 units andthat of breadth is 9 units.

WORKSHEET–21

1. (B) Hint: According to the condition ofinfinitely many solutions, we reaches at

+ 2 – 21= =

2 3 7a b a b

.

2. (C) Hint: Simplifying the given linearequations, we have7 2 8 7

– = 5, + = 15 y x y x

Now take 1 x

= u, 1 y

= v; and solve.

3. (D) Let unit's and ten's digit be x and yrespectively.

x + y = 9 ...(i)10y + x + 27 = 10x + y ...(ii)

Solving equations (i) and (ii), we have x = 6, y = 3Hence, the required number is 3 × 10 + 6,that is 36.

4. False... Equations are 5x – 5y = 3 and

10x – 10y – 3 = 0

∴ 1

2

5 110 2

aa

= = ; 1

2

– 5 1–

10 2bb

= =

1

2

31

3cc

= = ∴ 1 1 1

22 2

a b ca b c

= ≠

⇒ lines are parallel.

5. p ≠ 6

Hint: ≠3 510p

⇒ p ≠ 6

∴ p can take any value but not 6.

6. x = 4 –

5a b

a, y =

− + 45

a bb

Hint: – 3 (2 + ) + 2 ( + 2 )

xb a b b a b

= –

– 2 (2 + ) + 3 ( + 2 )y

a a b a a b

= 1

2 × 2 – 3 × 3a b a b

x 0 3

y 2 1

x 1 4

y 1 0


Take first and third terms as well as secondand third terms and solve.

7. a = 7, b = 3Hint: For infinitely many solutions,

24

=( )( )

41

aa

− −− −

= 2 15 1bb

+−

Take12

= 41

aa

−−

⇒ a = 7

and12

= 2 15 1bb

+−

⇒ b = 3.

8. Table for values of x and y as regardingequation 3x + y – 5 = 0 is

Similarly table for equation 2x – y – 5 = 0 is

Let us draw the graph of lines using thetables obtained above.

The lines intersect y-axis at (0, 5) and(0, – 5).

9. Let son's present age be x years and father'spresent age be y years.Since, the father's age is 3 years more than3 times the son's age.... y = 3x + 3⇒ 3x – y = 0 ...(i)3 years hence, father's age = (y + 3) yearsand son's age = (x + 3) years.Since, 3 years hence, the father's age willbe 10 years more than twice the son's age.... y + 3 = 2 × (x + 3) + 10⇒ 2x + 6 + 10 – y – 3 = 0⇒ 2x – y + 13 = 0 ...(ii)Subtracting equation (ii) from equation (i),we have

x = 10Substituting x = 10 in equation (i), we have

y = 3 × 10 + 3 = 33.Son's age = 10 years; father's age = 33 years.

ORLet the speed of train and bus be x km andy km respectively.According to the given conditions,

60 2404 = +

x y ...(i)

DistanceUsing Time =

Speed

And 10 100 200

4 + = +60 x y

...(ii)

Putting 1x = u and

1y = v in (i) and (ii)

and simplifying, we get15u + 60v = 1 ...(iii)24u + 48v = 1 ...(iv)

On solving equations (iii) and (iv), weobtain

u = 1

60 and v =

180

i.e., x = 60 and y = 80

Hence, the speed of train is 60 km/hr andthat of bus 80 km/hr.

x 0 1

y 5 2

x 0 1

y – 5 – 3


WORKSHEET–221. (D) The condition to be coincident for lines

ax + by + c = 0 and dx + ey + f = 0 is givenby

ad

= be

= cf

⇒ ae = bd ; bf = ce.

Note: Two lines are coincident if both theequations follow the condition of infinitelymany solutions.

2. (C) Let the required equation be ax + by + c = 0.

Then,2

a= 6

– 3 ≠

– 5c

⇒2

a=

6– 3

= k (say)

⇒ a = 2 k, b = – 3 k, c any real number

Then, 2 k – 3 k + c = 0Putting c = –1, we have

⇒ 2 kx – 3 ky – k = 0

⇒ 2 x – 3 y = 1.

3. (A) For no solutions,

12k

=3k

≠ –( – 2)

–k

k

⇒ k = ± 6

if k = 6

∴6 3 6 – 2 4 2

12 6 6 6 3= ≠ = = True

if k = – 6

– 6 – 83 412 – 6 – 6 3

= ≠ = True

∴ Required value of k, can be 6 or – 6.

4. For infinite number of solutions, we have

+2

p q =

– 3–( – 3)p q+

= – 7

–(4 )p q+

On solving+2

p q=

– 3–( – 3)p q+

and

+– 3

–( – 3)p q=

+–7

–(4 )p q,

we obtain p = – 5, q = – 1.

5. x = 1, y = 2Hint: Adding and subtracting the giventwo equations, we have

x + y = 3 ...(i)and x – y = – 1 ...(ii)Now, solve equations (i) and (ii).

6. x = a2, y = b2

Hint: Given system of linear equations maybe written as

bx + ay – ab (a + b) = 0b2x + a2y – 2 a2b2 = 0

Solve these two equations by the methodof cross-multiplication.

7. Let the two digits number be 10x + y.Since ten's digit exceeds twice the unit'sdigit by 2∴ x = 2y + 2⇒ x – 2y – 2 = 0 ...(i)Since the number obtained by inter-changing the digits, i.e., 10y + x is 5 morethan three times the sum of the digits.... 10y + x = 3 (x + y) + 5⇒ 2x – 7y + 5 = 0 ...(ii)On solving equations (i) and (ii), we obtain x = 8 and y = 3... 10x + y = 83Hence, the required two digits numberis 83.

8. Tables for equations 3x + y – 11 = 0 andx – y – 1 = 0 are respectively.

and

Let us draw the graph.

x 3 4

y 2 – 1

x 0 1

y – 1 3


From the graph, it is clear that the linesintersect each other at a point A(3, 2). Sothe solution is x = 3, y = 2.

The line 3x + y – 11 = 0 intersects they-axis at B(0, 11) and the line x – y – 1 = 0intersects the y-axis at C (0, –1). Draw theperpendicular AM from A on the y-axisintersect it at M.

Now, in ∆ABC,base BC = 11 + 1 = 12 units, heightAM = 3 units.

... ar(∆ABC) =12

× base × height

=12

× 12 × 3 = 18 sq. units

x = 3, y = 2 ; Area = 18 sq. units.

9. Speed of boat = 6 km/hr,Speed of stream = 2 km/hrHint: Let the speed of boat in still water =x km/h and the speed of stream = y km/h

12 40+ = 8

x – y x + y ...(i)

DistanceUsing Time

Speed =

16 32+ = 16

x – y x + y ...(ii)

Put x – y = u, x + y = v and solve furtherfind x and y.

OR

Let each boy receive Rs. x and the numberof boys be y. Then sum of money which isdistributed is Rs. xy.

Had there been 10 boys more, each wouldhave received a rupee less,

... (y + 10) (x – 1) = xy⇒ 10x – y = 10 ...(i)Had there been 15 boys fewer, each wouldhave received Rs 3 more,... (y – 15) (x + 3) = xy⇒ 5x – y = –15 ...(ii)Solving (i) and (ii), we getx = 5 and y = 40... xy = 200Hence, sum of money = Rs. 200And number of boys = 40.

WORKSHEET–23

1. (A) Here, = ≠– 32 9

6 – 9 – 5

... Lines are parallel.

2. (B) As the given equation are homogeneousso only solution will be x = 0 and y = 0.

3. (C)

Hint: Put 1

x = u ,

1 y

= v and solve.

4. The given equations have a unique solution

⇒ al

≠ bm

⇒ am ≠ bl.

5. The given equation can be written as6ax + 6by = 3a + 2ab ...(i)

and 6bx – 6ay = 3b – 2a ...(ii)Multiplying equation (i) by a and (ii) by band adding the results, we have

6(a2 + b2) = 3(a2 + b2)

⇒ x =12


Substituting x = 12

in equation (i), we have

62a

+ 6by = 3a + 2b

⇒ 6by = 2b ⇒ y = 13

Thus, the solution is x = 12

, y = 13

.

6. a = 5, b = 1Hint: Two linear equations a1 x + b1 y + c1 = 0and a2 x + b2 y + c2 = 0 have infinite numberof solutions if

1 1 1

2 2 2= =

a b ca b c .

7. Given system of linear equations can bewritten as

(a + 2b) x + (2a – b) y – 2 = 0

(a – 2b) x + (2a + b) y – 3 = 0By cross-multiplication,

– 3 (2 – ) + 2 (2 + ) ( )

xa b a b

i–

=– 3 ( + 2 ) + 2 ( – 2 )

( )

ya b a b

ii 1

=( + 2 ) (2 + ) – (2 – ) ( – 2 )

( ) a b a b a b a b

iii

Taking (i) and (iii), 5 – 2

=10 b a

xab

Again taking (ii) and (iii), + 1010

a by

ab=

Thus, 5 – 2

=10 b a

xab

, + 10=

10 a b

yab

is the

solution of the given system of equations.

8. Speed of rowing = 6 km/hr,Speed of current = 4 km/hr

Hint:2 × (x + y) = 20

[Time × Speed = Distance]

2 × (x – y) = 4Where, x = speed of rowing and,

y = speed of current.

ORLet fare from A to B and from A to C beRs. x and Rs. y respectively.According to the given conditions,

2x + 3y = 795 ...(i)3x + 5y = 1300 ...(ii)

Solving eqn. (i) and (ii), we obtainx = 75, y = 215

Hence, fares from A to B is Rs. 75 andfrom A to C is Rs. 215.

9. Let us make the table for the values of xand corresponding to the equation

2x + y – 8 = 0

Similarly, for the equation x – y – 1 = 0

Let us draw the graph.

x 2 4

y 4 0

x 4 3

y 3 2


From the graph, the lines intersect eachother at the point A(3,2). Therefore, thesolution is x = 3, y = 2.The lines intersect the y-axis at B(0,8) andC (0, – 1).To find the area of the shaded portion,that is, ∆ABC, draw perpendicular AMfrom A on the y-axis to intersect it in M.

Now, AM = 3 units and BC = 8 + 1 = 9 units.

... ar(∆ABC) =12

× BC × AM

=12

× 9 × 3

=272

sq. units

x = 3, y = 2; Area = 13.5 sq.units.

WORKSHEET–24

1. (C) For unique solution,

42

≠ 2p

⇒ p ≠ 4.

2. (A) In the case of no solution,

36

=–1–2

≠–5–k

⇒ k ≠ 10.

3. (D) x = 80, y = 30

Hint: x + 2y = 140, 3x + 4y = 360.

4. TrueAccording to the conditions of consistency,either

≠

2 2– 5 – 5 13 3or = =

3 3– 5 – 5 32 2

Clearly, the first condition holds. Hence,the system of equations is consistent witha unique solution.

5. For infinitely many solutions,

+3

p q=

2( – )4

p q =

–(5 – 1)–12

p

⇒ 4p + 4q = 6p – 6q and

– 12p – 12q = – 15p + 3⇒ 2p – 10q = 9and 3p – 12q = 3⇒ p = 5, q = 1.

6. x = 1, y = 1,

Hint: Take 1

3x y+ = u,

13x y−

= v

∴ given equation can be written as:

u + v =34

⇒ 4u + 4v = 3

and 12

u – 12

v = –18

⇒ 4u – 4v = – 1.

7. x =1

2−

, y =13

Hint: Put 1 x

= u and 1 y

= v.

8. Table for values of x and y correspondingto equation 4x – 5y – 20 = 0 is

Similarly for the equation 3x + 5y – 15 = 0

Let us draw the graphs for the two equations.

x 5 0

y 0 – 4

x 5 0

y 0 3


As the graphs of the two lines intersecteach other at the point A(5, 0), the requiredsolution is x = 5, y = 0.The graphs intersect the y-axis at B (0, 3)and C(0, – 4). Therefore, the coordinates ofvertices of the triangle ABC are A(5, 0),B(0, 3) and C(0, – 4).x = 5, y = 0 and (5, 0), (0, 3), (0, – 4).

9. Let speeds of two cars that start from placesA and B be x km/hr and y km/hr respec-tively.Case I: When two cars travel in same direction.Let the cars meet At C

Distance travelled by the car that startsfrom A

AC = 5 × xSimilarly distance for other car

BC = 5 × y... AC – BC = 5x – 5y⇒ 5x – 5y = 100⇒ x – y = 20 ...(i)Case II: When two cars travel in oppositedirections.Let the cars meet at D

Distance travelled by the car that startsfrom A

AD = 1 × xSimilarly distance for other car

BD = 1 × y... AD + BD = x + y

⇒ x + y = 100 ...(ii)Solving equations (i) and (ii), we get

x = 60 and y = 40Hence, speeds of two cars that start fromplace A and B are 60 km/h and 40 km/hrespectively.

WORKSHEET–25

1. (C) The given equations represent to beparallel lines if

2( – 1)3

k=

1–1

≠ –1–1

⇒ k – 1 =3

–2

⇒ k =1

–2

.

2. (D) For the point of intersection of any linewith x-axis, put y = 0... – 3x + 7 (0) = 3 ⇒ x = – 1So the required point is (– 1, 0).

3. (B) x – y = 0 ...(i)2x – y = 2 ...(ii)– + –

– x = –2 (Subtracting)

... x = 2.Further y = x = 2.

4. For inconsistency,

+ 22

k=

63

≠ 2–(3 2)

– 4k +

⇒ k + 2 = 4 and (3k + 2)2 ≠ 8

⇒ k = 2 and k ≠ 13

( )± 2 2 – 2

Hence, k = 2.

5. m ≠ 4

Hint: ≠ – 2 2 – 1m .

6. Given system of equations can be writtenas2x + 3y – 18 = 0 ...(i)x – 2y – 2 = 0 ...(ii)

Now, ≠2 31 – 2

Hence the system has unique solution.Now, by cross-multiplication on (i) and (ii),we get


– 6 – 36x

=– 4 + 18

– y=

1– 4 – 3

⇒ x = 6, y = 2Thus, the solution of given system is

x = 6, y = 2.

7. x = 5, y = –1

Hint: Take 1 1

= = + –

u, vx y x y

and solve.

8. Let Meena received x notes of Rs. 50 and ynotes of Rs. 100Since total number of notes is 25... x + y = 25 ...(i)Since the value of both types of notes isRs. 2000.

... 50x + 100y = 2000and x + 2y = 40 ...(ii)Solving equations (i) and (ii), we get

x = 10, y = 15Hence, Meena received 10 notes of Rs. 50and 15 notes of Rs. 100.

ORLet the length and breadth of rectangle bex units and y units respectively.Then area of rectangle = xy sq. unitsCase I: The length is increased and breadthis reduced by 2 units.... (x + 2) (y – 2) = xy – 28⇒ xy – 4 – 2 x + 2 y = xy – 28⇒ x – y = 12 ...(i)Case II: The length is reduced by 1 unitand breadth increased by 2 units.∴ (x – 1) (y + 2) = xy + 33⇒ xy – 2 – y + 2x = xy + 33

2x – y = 35 ...(ii)Solving equations (i) and (ii), we get

x = 23 and y = 11Hence, the length of the rectangle is 23units and the breadth is 11 units.

9. The given linear equations are4x – y – 8 = 0 ...(i)

and 2x – 3y + 6 = 0 ...(ii)

To draw the graphs of the equations (i)and (ii), we need atleast two solutions ofeach of the equations. These solutions aregiven below:

Plot the points A(0, – 8), B(2, 0), C(0, 2)and D(– 3, 0) on graph paper and join themto form the lines AB and CD as shown infigure.

The graphs of these lines intersect eachother at P(3, 4). So, unique solution:

x = 3, y = 4.

Also, the graphs meet the x-axis at D(– 3, 0)and B(2, 0).

Hence, the triangle formed by the linesis PBD with vertices P(3, 4), B(2, 0) andD(– 3, 0).

x 0 2

y = 4x – 8 – 8 0

x 0 – 3

y 2 6

3x +

2 0


WORKSHEET–26

1. (C) For coincident lines,

12

= 2k

= 714

⇒ k = 4.

2. (A) The given system of equations can bewritten asx + 2y = 140, 3x + 4y = 360Solving this system, we obtainx = 80, y = 30.

3. Adding the given equations, we have3x = 0 ⇒ x = 0

Substituting x = 0 in any of the givenequations, we get y = 0Hence, the required solution is x = 0, y = 0.

4. False

Hint: As 1

2

aa =

24

, 1

2

bb

=5

10, 1

2

cc = 6

...12

= 12

≠ 6

⇒ They are parallel.

5. a = –1, b =52

Hint: 2 – (2 + 5) 5= =

2 + 1 – 9 15a

b .

6. Put 1 x

= u and 1 y

= v in given system of

equations.u + v – 7 = 0 ...(i)

2u + 3v – 17 = 0 ...(ii)By cross-multiplication,

1= =

– 17 + 21 – 17 + 14 3 – 2u – v

⇒ u = 4, v = 3

⇒ x = 14

, y = 13

Hence, x = 14

, y = 13 is the solution of the

given system of equations.

7. x = – 2, y = 5 and m = – 1

Hint: 2x + 3y = 11 ⇒ 11 – 2=

3x

y

Substitute this value of y in 2x – 4y = – 24and solve for x.

8. The given system of linear equations is2x – y – 5 = 0 ...(i)3x + y – 5 = 0 ...(ii)

To draw the graph of equations (i) and (ii),we need atleast two solutions of each ofthe equations, which are given below:

Using these points, we are drawing thegraphs of lines as shown below:

From the graph, the lines intersect eachother at the point P(2, –1). Therefore, thesolutions is x = 2, y = –1.The lines meet the y-axis at the pointsQ(0, 5) and R(0, –5).

9. Let the fixed charge and additional chargefor each day be Rs. x and Rs. y respectively.

x 0 4

y = 2x – 5 – 5 3

x 0 – 3

y = –3 5

3x+

5 – 4


To solve the system for x and y, using themethod of cross-multiplication, we have

+– ( ) – ( – )x

q p q p p q=

+ +–

– ( ) ( – )y

p p q q p q

= 2 21

– –p q

⇒ 2 2– –x

p q=

2 2

–

– –

y

p q= 2 2

1– –p q

⇒ x = 1, y = –1.

8. The given system of equations can bewritten as

3x – 4y – 1 = 0 ...(i)

6x – 15y = 0 ...(ii)

To draw the graph, we need atleast twosolutions for each of the equations (i) and(ii), which are given below:

Let us draw the graph to use these points.

From the graph, it is clear that the lines areparallel. Hence, the given system ofequations is inconsistent.

Since saritha paid Rs. 27 for a book keptfor 7 days... x + 4y = 27 ...(i)Also, Susy paid Rs. 21 for the book keptfor 5 days... x + 2y = 21 ...(ii)Subtracting equation (ii) from (i), we get

2y = 6 ⇒ y = 3Again substituting y = 3 in equation (ii),we get

x = 15Hence, the fixed charge is Rs. 15 andadditional charge for each day is Rs 3.

ORSon's age = 10 years, father's age = 40 years.Hint: Let the present age of father and sonbe x and y years respectively.... x + 5 = 3 (y + 5)And x – 5 = 7 (y – 5).

WORKSHEET–27

1. (A) Hint: In a cyclic quadrilateral,

∠A + ∠C = 180° and ∠B + ∠D = 180°.

2. (C) Hint: Both lines are passing throughthe origin.

3. (A) For infinite number of solutions,

2p + q

=2 –

3p q

=–21–7

⇒ p + q = 6 and 2p – q = 9p = 5 , q = 1.

4. False, as a + 5b = – 10.

5. False, x = 4, y = 1 does not satisfy thesecond equation.

6. No solution

Hint: 2x + 3y = 7, 6x + 9y = 11

Here, 2 3 7=

6 9 11 ≠ Parallel lines.

7. The given system of linear equations canbe written as

px + qy – (p – q) = 0qx – py – (p + q) = 0

x 3 7

y =– 3 – 1

4x

2 5

x 32

112

y = 6 158

x +3 6


9. Let the fraction be xy

On adding 1 to each of numerator and

denominator, the fraction becomes 65

... +1 6=

+1 5xy

⇒ 5x + 5 = 6y + 6⇒ 5x – 6y = 1 ...(i)Further, on subtracting 5 from each ofnumerator and denominator, the fraction

becomes 32

... – 5 3=

– 5 2xy

⇒ 2x – 10 = 3y – 15⇒ 2x – 3y = – 5 ...(ii)Solving equations (i) and (ii), we getx = 11, y = 9

Hence, the required fraction is 119

.

OR

Rs. 6000, Rs. 5250Hint: Let incomes of x and y be 8x and 7xrespectively; and expenditures of them be19y and 16y respectively.∴ 8x – 19y = 1250 ...(i)

7x – 16y = 1250 ...(ii)

WORKSHEET–28

1. (A)

Hint: – 3 3= =

12k k

k k.

2. (C) The condition that the given system ofequations has unique solution is represen-ted by

1

2

aa ≠ 1

2

bb

.

3. (B) The condition that the given system ofequations represents parallel lines is

++

2 13 1pp

=– 23

p ≠

52

⇒ 5p = – 5 ⇒ p = – 1.

4. (A) Multiplying first equation by 2 andthe other one by 3 and adding, we get

21.8x = 10.9 ⇒ x = 12

Substituting x =12

in any of the given

equations, we have y = 13

.

∴∴∴∴∴ x = 12

, y = 13

.

5. TrueThe condition for parallel lines is

26

= – 2– 6

≠ – 35

⇒⇒⇒⇒⇒13

= 13

≠ –35

The condition holds. The lines are parallel.

6. x = a2 , y = b2

Hint: Put 1x

= u and 1y = v.

7. Given system of linear equations can bewritten as:(a – b) x + (a + b) y – (a2 – 2ab – b2) = 0 (a + b) x + (a + b) y – (a2 + b2) = 0By cross -multiplication,

2 2 2 2– ( + )( + ) + ( + ) ( – 2 – ) x

a b a b a b a ab b

2 2 2 2

– =

– ( – )( + ) + ( + ) ( – 2 – ) y

a b a b a b a ab b

1=

( – ) ( + ) – ( + ) ( + )a b a b a b a b

⇒ 2 2

– 1= =

– 2 ( + )– 2 ( + ) – 4yx

b a bb a b ab

Hence, the solution of given system ofequations is

x = a + b, y = – 2 + a b

a b.


8. To draw graph of the equation, we needatleast two solutions.Two solutions of the equation4x + 3y – 24 = 0 are mentioned in thefollowing table:

Similarly, two solutions of each of theequations 2x – y – 2 = 0 and y + 4 = 0 arerespectively.

and

Using the tables obtained above, let usdraw the graph.

Observing the graph, the lines meet eachof her pairwise in A(3, 4), B(–1, – 4) andC(9, – 4).Hence, the vertices of the triangle ABCso obtained are A(3, 4), B(–1, – 4) andC(9, – 4).

area of ∆ ABC = 12

× base × height

=12

× 10 × 8 = 40 sq. unit.

9. Rs. 600, Rs. 700Hint: Let cost price of trouser be Rs. x andthat of shirt Rs. y. Then

125 110 + = 1520

100 100110 125

+ = 1535100 100

x y

x y

⇒ 25 + 22 = 3040022 + 25 = 30700

x yx y

OR

6 l of 50% and 4 l of 25%.Hint: Let x litres of 50% acid and y litres of25% acid should be mixed.

50 25 40 + = ( + )

100 100 100+ = 10

x y x y

x y

⇒ 2 = 3

+ = 10x y

x y

WORKSHEET–29

1. (C) x = 9, y = 6

Hint: x – y = 3 and 2x + 3y = 36.

2. (D)3

18p

=624

≠ 5075

⇒ 2p

= 32

≠ 23

∴ p = 3 .

3. (A) Solving 3x – 2y = 4 and 2x + y = 5, wehave x = 2, y = 1.

Now, substituting these values of x and yin y = 2x + m, we have 1 = 2 × 2 + m

∴ m = – 3.

x 0 6

y 8 0

x 0 1

y – 2 0

x 0 4

y – 4 – 4


4. x = b, y = – aHint: a2 x – b2 y = ab (a + b), ax – by = 2abSolving the equation we get x = b1, y = – a.

5. For inconsistency,

α12

=α3

≠ α

α– 3

⇒ α2 = 36 and

3α ≠ α2 – 3α⇒ α = ± 6 and α ≠ 0 or α ≠ 6 ⇒ α = –6.

6. x = 225

a, y =

−265

b

Hint: 4bx + 3ay – 2ab = 03bx + ay – 8ab = 0.

7. 3x + 2y = 800.12x + 8y = 3000; Not possibleHint: Let cost of 1 chair be Rs. x and thatof 1 table be Rs. y.∴3x + 2y = 800, 12x + 8y = 3000.

8. Let the actual prices of tea-set and lemon-set be Rs. x and Rs. y respectivelyAccording to the question,Case I: Selling price – cost price = Profit⇒ 0.95x + 1.15y – (x + y) = 7⇒ – 0.05x + 0.15y = 7 ...(i)Case II: Selling price – Cost price = Profit⇒ 1.05x + 1.10y – (x + y) = 13⇒ 0.05x + 0.10y = 13 ...(ii)Solving equations (i) and (ii), we get

x = 100 , y = 80Hence, actual prices of tea-set and lemon-set are Rs. 100 and Rs. 80 respectively.

ORThe person invested Rs. 500 at the rate of12% per year and Rs. 700 at the rate of10% per year.Hint: Let the person invested Rs. x at therate of 12% per year and Rs. y at the rateof 10 % per year

∴1012

+ 100 100

yx= 130

⇒ 6x + 5y = 6500 ...(i)

and12 10

+ 100 100

y x= 134

⇒ 5x + 6y = 6700 ...(ii)Adding and subtracting (i) and (ii), we get

x + y = 1200 ...(iii)x – y = – 200 ...(iv)

9. Two solutions of 6y = 5x + 10 are:

Two solutions of y = 5x – 15 are

Now, we draw the graph of the system onthe same coordinate areas.

(i) From the graph, we look that the twolines intersect each other at A(4, 5).

(ii) The vertices of the triangle : A (4, 5);B (–2, 0); C (3, 0).Height of ∆ABC corresponding to thebase BC,

h = 5 unitsand base, b = BC = 5 units

Now, ar(∆ABC) = 12

× b × h

= 12

× 5 × 5

= 12.5 square units.

x – 2 4

y 0 5

x 3 2

y 0 – 5


Solving equations (iii) and (v), we have

x = 32

; y = 12

.

Therefore, xy = 34

Solving equations (iv) and (vi), we have

x = 1–

4; y =

34

Therefore, xy = – 3

16.

Hence, xy = 34

or – 3

16.

6. Given equations can be written as

xa

+ yb

– (a + b) 2 = 0

2xa

+ 2

yb

– 2 = 0

Let us apply cross-multiplication methodto solve these equations.

+ +22 1

–

xa

b bb

=

2

–2 1

–

yb

a a a+ +

=

2 2

11 1

–ab ba

⇒+

2

–b xb a

= +

2––

a ya b

= 2 2

–a ba b

Taking+

2

–b xb a

= 2 2

–a ba b

and+

2––

a ya b

= 2 2

–a ba b

⇒ x = 2 2

2

( – )

( – )

a b a b

b a b and y =

2 2

2

( – )

( – )

a b a b

a a b

⇒ x = a2 and y = b2.

7. Given equations of lines are:3x + y + 4 = 0 ...(i)

and 6x – 2y + 4 = 0 ...(ii)To draw the graphs of lines (i) and (ii), weneed atleast two solutions of each equation.


1. (C) Let us check option (C).

23

x + 52

y = 23

(– 3) + 52

(– 2) = – 2 – 5 = – 7

3x – 12

y = 3 (– 3) – 12

(– 2) = – 9 + 1 = – 8.

2. (D) 4x – y = 42 ⇒ x – y = 2 ...(i) x – 2y = 8 ...(ii) – + –

y = – 6

∴ x + y = – 10

∴ from (i) ⇒ x = – 4.

3. For coincident lines

1

2

aa = 1

2

bb

= 1

2

cc

⇒ 2a b+

= 3

– 3a b+ =

74a b+

⇒ a – 5b = 0.

4. False, because the given pair of equationshas infinitely many solutions at k = 40 only.

5. Given equations are2y – x. (x + y) = 1

⇒ x + y = –1

2y x ...(i)

and (x + y)x – y = 2 ...(ii)Substituting the value of x + y fromequation (i) in equation (ii), we get

–

–1

2

x y

y x = 2

⇒ (2x – y)x – y = 21

⇒ (x – y)2 = 1⇒ x – y = ± 1⇒ x – y = 1 ...(iii)or x – y = –1 ...(iv)Substituting x – y = 1 and x – y = –1 inequation (ii), we get respectively

x + y = 2 ...(v)

and x + y = 12

...(vi)


For equation (i), two solutions are:

For equation (ii), two solutions are:

Let us draw the graphs of the lines (i)and (ii)

xa

+ yb

– 2 = 0

From the graph it is clear that the two linesintersect each other at a point, P (–1, – 1),therefore, the pair of equations consistent.The solution is x = –1, y = – 1.

8. Let the cost price of the saree and the listprice of the sweater be Rs. x and Rs. yrespectively. Now two cases arise.Case I:

Sale price of the saree = x + x × 8

100

= 108100

x

Sale price of the sweater = y – y ×10100

= 90

100y

∴ 108100

x + 90

100y = 1008

⇒ 108x + 90y = 100800 ...(i)Case II:

Sale price of the saree = x + x × 10

100

= 110100

x

Sale price of the sweater = y – y × 8

100

= 92

100y

∴ 110100

x + 92100

y = 1028

⇒ 110x + 92y = 102800 ...(ii)Adding equations (i) and (ii), we get

218x + 182y = 203600 ...(iii)Subtracting equation (i) from (ii), we get

2x + 2y = 2000or 218x + 218y = 218000 ...(iv)

(Multiplying by 109)Solving equation (iii) and (iv), we have

x = 600 and y = 400Hence, the cost price of the saree isRs. 600 and the list price of the sweater isRs. 400.


1. (B) For no solution,3

12=

72k

≠ +4 1

kk

∴3

12=

72k

⇒ k = 14.

2. (B) For infinitely many solutions,1339

= 6k

= + 4k

k

Taking1339

= 6k

⇒ k = 2.

x 0 – 3

y – 4 5

x 0 2

y 2 8


3. x = – a and y = k must satisfy both the givenequations. Let us substitute these values ofx and y in

bx – ay + 2ab = 0b (–a) – ak + 2ab = 0

⇒ –ak + ab = 0⇒ k = b.

4. Yes, because

1

2

aa

= 1

2

bb

= 1

2

cc , i.e.,

13

= 26

= – 3– 9

.

5. Given equations are:4x + 5y = 2

(5p + 2q)x + (4p – 2q)y = p + q + 3Here, a1 = 4, b1 = 5, c1 = – 2,

a2 = 5p + 2q, b2 = 4p – 2q,c2 = –p –q –3

For infinitely many solutions,

1

2

aa

= 1

2

bb

= 1

2

cc

⇒+4

5 2p q=

54 – 2p q

= –2

– – – 3p q

Taking+4

5 2p q=

+5

4 2p q

⇒ 25p + 10q = 16p – 8q⇒ 9p = – 18q⇒ p = – 2q ...(i)Also, taking

+4

5 2p q=

–2– – – 3p q

⇒ –10p – 4q = – 4p – 4q –12⇒ p = 2Substitute p = 2 in equation p = – 2q to get

q = – 1.Hence, p = 2, q = – 1.

6. Given system of equations is43x + 67y = –24 ...(i)67x + 43y = 24 ...(ii)

Adding (i) and (ii); and subtracting (i) from(ii), we get respectively.

110x + 110y = 0and 24x – 24y = 48i.e. x + y = 0 ...(iii)

and x – y = 2 ...(iv)Adding equations (iii) and (iv); andsubtracting equation (iv) from (iii), we getrespectively.

x = 1 and y = – 1.

7. One of the given equation is2x + y = 14 ...(i)

Here, at x = 0, y = 4and at x = 2, y = 0Two solutions of equation(i) are given inthe following table:

Another given equation is2x – y = 4 ...(ii)

Two solutions of equation (ii) are given bythe following table:

Let us draw the graph of the two equations(i) and (ii) by using their correspondingtables.

x 0 2

y 4 0

x 0 2

y – 4 0


From the graph, vertices of the triangleABC are A(0,4), B(0, – 4) and C(2, 0).

ar(∆ABC) = 12

× base × height

= 12

× AB × OC

= 12

× (4 + 4) × 2

= 8 square units.Thus, area of the triangle is 8 square units.

8. Let Anand had a total of x oranges; and hemakes lot A of p oranges and lot B ofremaining x – p oranges. There are twocases now.Case I:

Selling price of lot A = Rs. 23

p

Selling price of lot B = Rs. (x – p)

∴ 23

p + x – p = 400

⇒ 3x – p = 1200 ...(i)Case II:

Selling price of lot A = Rs. p

Selling price of lot B = Rs. 45

(x – p)

∴ p + 45

(x – p) = 460

4x + p = 2300 ...(ii)Add equations (i) and (ii) to get

7x = 3500⇒ x = 500Hence Anand had a total number of 500oranges.

CHAPTER TEST

1. (C) A pair of linear equations is said toconsistent, if the lines intersect each otherat a point or coincide.

2. (C) 6, 36

Hint: Let the son's age = x,father's age = y

∵ y = 6x

and y + 4 = 4(x + 4)Solve yourself.

3. (C) The lines are coincident

⇒ 36

=– 1– k

= 8

16 ⇒ k = 2.

4. Yes,For consistency,

Either 24

aa

≠ 2bb

or 24

aa

= 2bb

= ––2

ab

Here the relation24

aa

= 2bb

= ––2

aa

,

i.e.,12

= 12

= 12

holds.

⇒ The pair is consistent.

5. 21x + 47y = 11047x + 21y = 162

+ + +68x + 68y = 272

⇒ x + y = 4 ...(i)

Subtracting the given equation, we get

– 26x + 26y = – 52⇒ x – y = 2 ...(ii)Solve equation (i) and (ii) and we get

x = 3, y = 1.

6. We are given

+2xyx y =

32

...(i)

and 2 –xyx y =

–310 ...(ii)

Taking equation (i)

+2xyx y =

32

⇒ 3x + 3y = 4xy ...(iii)Now, Taking equation (ii)

2 –xyx y =

–310


⇒ – 6x + 3y = 10xy ...(iv)Multiplying equation (iii) by 2 and addingits with (iv), we get

9y = 18xy

∴ x = 12

Putting x = 12

in equation (iv), we get

⇒ – 3 + 3y = 5xy

∴ y = –32

This, x = 12

and y = –32

.

7. The given system of equations will haveinfinite number of solutions if

1–a b

= +2

a b =

+1

– 2a b

⇒ 1–a b

= +

1– 2a b

and

+2

a b=

+1

– 2a b

⇒ a + b – 2 = a – b and2a + 2b – 4 = a + b

⇒ a + b – a + b = 2 and a + b = 4⇒ b = 1 and a = 3Hence, the given system of equations willhave infinite number of solutions, if

a = 3, b = 1.

8. Let the cost price of saree be Rs. x and listprice of sweater be Rs. y.

Profit on a saree = 8%

∴ Selling price of a saree = x + 8

100x

= 108100

x

Discount on a sweater = 10%

∴ Selling price of a sweater = y – 10

100y

= 90

100y.

According to condition, we have

+90108

100 100yx

= 1008

108x + 90y = 1008006x + 5y = 5600 ...(i)

If profit on a saree is 10% and discount ona table is 8%, then total selling price isRs. 1028.

⇒ + +

810–

100 100yx

x y = 1028

⇒ +92110

100 100yx

= 1028

110x + 92y = 10280055x + 46y = 51400 ...(ii)

Multiplying eqn. (i) by 55 and eqn. (ii) by6, we get

330x + 275y = 308000 ...(iii)330x + 276y = 308400 ...(iv)

Subtract eqn. (iv) from eqn. (iii), we gety = 400

the value of y, put in eqn. (i), we havex = 600

Cost price of saree = Rs. 600List price of sweater = Rs. 400.

9. To draw the graph of a line, we arerequired atleast two solutions of itscorresponding equation.At x = 0, 5x – y = 5 gives y = –5At x = 1, 5x – y = 5 gives y = 0Thus, two solutions of 5x – y = 5 are givenin the following table:

x 0 1

y – 5 0


Similarly, we can find the solutions of eachremaining equation as given in thefollowing tables:

x 2 4

y 5 – 7

x 9 1

y – 4 0

x + 2y = 1

6x + y = 17Now, we will draw the graphs of the threelines on the same coordinate axes.

From the graph, it is clear that the linesform a triangle ABC with vertices A(1, 0),B(3, –1) and C(2, 5).

❑❑

51AIRT GN L SE

A

D F

EB C

4Chapter

TRIANGLES

WORKSHEET–331. (A) ∵ ∆ABC ~ ∆DEF

∴2

2ABDE

= 2

2BCEF

= 2

2ACDF

= ∆∆

( ABC)( DEF)

arar

Taking2

2BCEF

=∆∆

( ABC)( DEF)

arar

⇒2

2BC

(15.4)=

64121

⇒ BC = 64 × 15.4 × 15.4121

⇒ BC = 11.2 cm.

2. (D) Observing the figure, we obtain∠A = ∠R, ∠B = ∠Q, ∠C = ∠P

∴ ∆ABC ~ ∆RQP.

3. Yes.Here, 262 = 242 + 102 = 676⇒ AC2 = AB2 + BC2

∴ ∆ABC is a right triangle.

4. In ∆ABC, LM || CB

∴ =AM ALAB AC

…(i)

(Basic Proportionality Theorem)Similarly in ∆ADC,

∴ =AN ALAD AC

…(ii)

Comparing equations (i) and (ii), we have

=AM ANAB AD

. Hence proved.

5. As ∠1 = ∠2⇒ PQ = PR …(i)

QRQS

=QTPR

(Given)

⇒ QRQS

=QTPQ

[Using (i)]

and ∠1 = ∠1 (Common)∴ ∆PQS ~ ∆TQR (SAS criterion)

Hence proved.

6. 1 : 4

Hint: ar(∆DEF) = 14

ar ∆ABC

⇒∆∆

( DEF)( ABC)

arar

= 14

.

7. 13 mHint: Distance between tops = AD

∴ AD = +2 2AE DE

= 2 2(5) (12)+

= 13 m.

8. Hint: Use Pythagoras Theorem.

9. Statement: If a line is drawn parallel toone side of a triangle to intersect the othertwo sides at distinct points, the other twosides are divided in the same ratio.Proof: ABC is a given triangle in whichDE || BC. DE intersects AB and AC at Dand E respectively.

We have to prove

ADBD

= AECE

Let us draw EM ⊥ AB and DN ⊥ AC. JoinBE and CD.


2. (D) DC2 = BC2 + BD2 = BC2 +

2AB2

= BC2 + 14

(AC2 – BC2)

= 9 + 14

(25 – 9) = 9 + 4 = 13

⇒ DC = 13 cm.

3. (C)Hint: As DE || BC

∴ADDB

= AEEC

⇒−

−2 1

3x

x=

2 51

x +x −

.

4. DE || BC and DB is transversal

⇒ ∠EDA = ∠ABC(Alternate interior angles)

Similarly, ∠AED = ∠ACB

Consequently, ∆ADE ~ ∠ACB

(AA similarity)

∴2

2ADAB

=∆∆

( ADE)( ABC)

arar

⇒2

2AD

9AD=

∆( ADE)153

ar

⇒ ar(∆ADE) = 17cm2.

5. No.

Here,DPPE

= 5

10=

12

AndDQQF

= 6

18=

13

∵ DPPE

≠ DQQF

Therefore, PQ is not parallel to EF.

6. Hint: Use Basic Proportionality Theorem.

Now, ar(∆ADE) = 12

× base × height

= 12

× AD × EM ...(i)

Also, ar(∆ADE) = 12

× AE × DN ...(ii)

ar(∆BDE) = 12

× BD × EM ...(iii)

ar(∆CDE) = 12

× CE × DN ...(iv)

Dividing equation (i) by equation (iii) andequation (ii) by equation (iv), we have

∆∆

( ADE)( BDE)

arar

= ADBD

...(v)

and∆∆

( ADE)( CDE)

arar

= AECE

...(vi)

But ar(∆BDE) = ar(∆CDE) ...(vii)

(Triangles are on the same base andbetween the same parallels BC and DE)Comparing equations (v), (vi) and (vii), wehave

ADBD

= AECE

.

2nd Part:As ∠B = ∠C⇒ AB = AC⇒ AD + DB = AE + EC⇒ AD = AE (... BD = EC)

∴ADDB

= AEEC

∴ By converse of Basic ProportionalityTheorem, DE || BC.

Hence proved.

WORKSHEET– 34

1. (A) ∆BAC ~ ∆ADC

⇒BCAC

=ACDC

⇒ y2 = 16 × 4 ⇒ y = 8 cm.

53AIRT GN L SE

A

DB C

7. As AB = BC = AC

∴ AD ⊥ BC ⇒ BD = 12

BC

∴ Using Pythagoras TheoremAB2 = AD2 + BD2

⇒ AD2 = AB2 −

21BC

2

=23AB

4⇒ 4AD2 = 3AB2. Hence proved

OR

Let ABCD be a rhombusSince, diagonals of a rhombus bisect eachother at right angles,∴ AO = CO, BO = DO, ∠AOD = ∠DOC ∠COB = ∠BOA = 90°Now, in ∆AOD

AD2 = AO2 + OD2 ...(i)Similarly, DC2 = DO2 + OC2 ...(ii)

CB2 = CO2 + BO2 ...(iii)and BA2 = BO2 + AO2 ...(iv)Adding equations (i), (ii), (iii) and (iv),we haveAD2 + DC2 + CB2 + BA2

= 2(DO2 + CO2 + BO2 + AO2)

= 2

+ + +

2 2 2 2BD AC BD CA4 4 4 4

= BD2 + CA2. Hence proved

8. Hint: BD = DE = EC = 13

BC

Using Pythagoras Theorem.

9. Statement: In a triangle, if square of thelargest side is equal to the sum of thesquares of the other two sides, then theangle opposite to the largest side is a rightangle.Proof: We are given a triangle ABC with

A′C′2 =A′B′2 + B′C′2 ...(i)We have to prove that ∠B = 90°

Let us construct a ∆PQR with ∠Q = 90°such that

PQ =A′B′ and QR + B′C′ ...(ii)

In ∆PQR,PR2 = PQ2 + QR2

(Pythagoras Theorem)=A′B′2 + B′C′2 ...(iii)

[From (ii)]But A′C′2 =A′B′2 + B′C′2 ...(iv)

[From (i)]From equations (iii) and (iv), we have

PR2 = A′C′2

⇒ PR = A′C′ ...(v)Now, in ∆ABC and ∆PQR,

A′B′ = PQ [From (ii)]B′C′ = QR [From (ii)]A′C′ = PR [From (v)]

Therefore, ∆A′B′C′ ≅ ∆PQR(SSS congruence rule)

⇒ ∠B′ = ∠Q (CPCT)But ∠Q = 90°... ∠B′ = 90°. Hence proved.2nd PartIn ∆ADC, ∠D = 90°... AC2 = AD2 + DC2 = 62 + 82

= 36 + 64 = 100In ∆ABC,

AB2 + AC2 = 242 + 100 = 676and BC2 = 262 = 676Clearly, BC2 = AB2 + AC2

Hence, by converse of Pythagoras Theorem,in ∆ABC,

∠BAC = 90°⇒ ∆ABC is a right triangle.


WORKSHEET–35

1. (B)∆∆

( ADE)( ABC)

arar

= 2

2DEBC

⇒ ar(∆ADE) =

2

2

2BC

3BC

× 81 = 36 cm2.

2. (A) ∆OAB ~ ∆OCD

⇒OAOC

= OBOD

⇒ OB = 4 × 32

= 6 cm.

3. In ∆ABC, To make DE || AB, we have totake

⇒ADDC

= BEEC

⇒ ++

3 193

xx

= +3 4xx

⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12⇒ 6x = 12 ⇒ x = 2.

4. No,∵ ∆FED ~ ∆STUCorresponding sides of the similar trianglesare in equal ratio.

∴DETU

= EFST

∴ DEST

≠ EFTU

.

5. AB || PQ ⇒ APAO

=BQBO

…(i)

AC || PR ⇒ APAO

=CRCO

…(ii)

From (i) and (ii)BQBO

=CRCO

⇒ BC || QR. (By converse of BPT)

6. 1 : 2.Hint: Let AB = BC = a

... AC = 2 a

...∆∆

ABE)ACD)

ar(ar(

= 2

2ABAC

.

7. In ∆ABC and ∆AMP∠A = ∠A (Common)

∠ABC = ∠AMP = 90°

(i) ∴ ∆ABC ∼ ∆AMP, (AA criterion)

(ii) ∴ CAPA

= BCMP

.

(∴ Corresponding sides ofsimilar triangles are proportional.)

8. Hint:ar(∆ AXY) = ar(BXYC) + ar(BXYC)

⇒ 2. ar(∆AXY) = ar(BXYC) + ar(∆ AXY)= ar(∆ABC)

⇒ ∆∆

( ABC) ( AXY)

arar

= 21

As ∆ABC ~ ∆ AXY

∴

2ABAX

= ∆∆

( ABC)( AXY)

arar =

21

⇒ABAX

= 2

1⇒

BXAB

=−2 12

.

9. Hint: Prove converse of PythagorasTheorem.

WORKSHEET–361. (A)

In triangle ABC,

ADDC

= 621

= 27

BEEC

= 18 –14

14 =

27

∴ ADDC

= BEEC

⇒ DE || AB .

2. (A)∆∆( DEF)( ABC)

arar

= 2

2EFBC

⇒ ar(∆DEF) = 54 × 169

= 96 cm2.

3. (B) ∆ABD ~ ∆BCD

⇒ABBC

= BDCD

A

EBC

D

A

X

B C

Y

55AIRT GN L SE

⇒5.4BC

= 3.65.2

⇒ BC = 5.4×5.2

3.6= 7.8 cm.

4. In ∆ABC and ∆ADE,

∠BAC = ∠DAE (Common angle)∠ACB = ∠AED (Each 90°)

∴ ∆ABC ~ ∆AED (AA criterion)

AB = 2 2AC + BC = 25 + 144 = 13 cm

Now,ABAD

= BCDE

= ACAE

⇒ 133

= 12DE

= 5

AE

⇒ DE = 3613

cm and AE = 1513

cm.

5. No.Ratio of areas of two similar triangles= square of ratio of their corresponding

altitudes.

=

235

= 9

25 ≠

65

.

Hence, it is not correct to say that ratio of

areas of the triangles is 65 .

6. AE2 = AC2 + EC2 …(i)

BD2 = DC2 + BC2 …(ii)

Adding (i) and (ii), we get

AE2 + BD2 = AC2 + EC2 + DC2 + BC2

= (AC2 + BC2) + (EC2 + DC2)

AE2 + BD2 = AB2 + DE2.Hence proved.

7. In ∆AQO and ∆BPO,∠QAO = ∠PBO (Each 90°)∠AOQ = ∠BOP

(Vertical opposite angles)

So, by AA rule of similarity,∆AQO ~ ∆BPO

⇒AQBP

= AOBO

⇒AQ

9=

106

⇒ AQ = 10 × 9

6⇒ AQ = 15 cm.

ORLet the height of the tower be h metres

P

Q R40m

x

A

B C

12m

8m

∆ABC ~ ∆PQR.

⇒ ABPQ

=BCQR

⇒ 12h

= 840

⇒ h =12 × 40

8 = 60 metres.

8. Hint: As ∆AOB ~ ∆CODA B

O

D C

∆∆

( AOB)( COD)

arar

= 2

2ABCD

= ( )2

2

2 CD

CD =

41

.

9. Hint: Prove Pythagoras Theorem.For 2nd Part:∴ AB2 = AD2 + BD2 …(i)Also AC2 = AD2 + CD2 …(ii)From (i) and (ii),⇒ AB2 − AC2 = BD2 − CD2

⇒ AB2 + CD2 = AC2 + BD2.Hence proved.

WORKSHEET–371. (A) ∠M = 180° – (∠L + ∠N) (ASP)

= 180° – (50° + 60°) = 70° ∵ ∆LMN ~ ∆PQR ∴ ∠M = ∠Q ⇒ ∠Q = 70°.


2. (C) In ∆KMN, as PQ || MN,KPPM

= KQQN

⇒ KPPM

= KQ

KN – KQ

⇒ KNKQ

– 1 = PMKP

⇒ 20.4KQ

– 1 = 134

⇒ 20.4KQ

= 1 + 134

= 174

⇒ KQ = 20.4×4

17⇒ KQ = 4.8 cm.

3. (A) ∆ABC ~ ∆DEF.

4. ∵ ∆ABC ~ ∆PQR

∴ ∆∆

( PRQ)( BCA)

arar

=2

2QRBC

=

231

= 91

= 9 : 1.

5. TrueHint: Use Basic Proportionality Theorem

6. Hint:Use: ∠1 = ∠2 ∠3 = ∠4.

7. Draw EOF || AD

A D

B C

OE F

∴ OB2 = EO2 + EB2

OD2 = OF2 + DF2

∴ OB2 + OD2 = EO2 + EB2 + OF2 + DF2

= EO2 + CF2 + OF2 + AE2

[∵DF = AE, EB = CF]= (EO2 + AE2) + (CF2 + OF2)

OB2 + OD2 = OA2 + OC2.

ORJoin OA, OB and OC

In right ∆AOF,

AO2 = AF2 + OF2 ...(i)

In right ∆AOE,AO2 = AE2 + OE2 ...(ii)

From equations (i) and (ii), we have

AF2 + OF2 = AE2 + OE2 ...(iii)Similarly, we can find out that

BD2 + OD2 = BF2 + OF2 ...(iv)

and CE2 + OE2 = CD2 + OD2 ...(v)

Adding equations (iii), (iv) and (v), wearrive

AF2 + BD2 + CE2 = AE2 + CD2 + CD2 + BF2.

Hence the result.

8. ∆ABC ~ ∆PQR

⇒ABPQ

=BCQR

and ∠B = ∠Q

⇒ ABPQ

=

1BC

21

QR2

and ∠B = ∠Q

⇒ ABPQ

=BP

QM and ∠B = ∠Q

(∵ BD = DC and QM = MR)

⇒ ∆ABD ~ ∆PQM

⇒ ABPQ

= ADPM

. Hence proved.

ORHint: Show that∆ABD ~ ∆PQM

A B

D C4

3

2

1E

F

57AIRT GN L SE

9. Let the two given triangles be ABC andPQR such that ∆ABC ~ ∆PQR

∴ABPQ

= BCQR

...(i)

Let us draw perpendiculars AD and PMfrom A and P to BC and QR respectively.∴ ∠ADB = ∠AMQ = 90° ...(ii)Now, in ∆ABD and ∆PQM,

∠B = ∠Q (∵∆ABC~∆PQR)∠ADB = ∠AMQ [From (ii)]

So, by AA rule of similarity, we have∆ABD ~ ∆PQM

⇒ ABPQ

= ADPM

...(iii)

From equations (i) and (iii), we get

BCQR

= ADPM

...(iv)

Now,∆∆

( ABC)( PQR)

arar

= × ×

× ×

1BC AD

21

QR PM2

= × ×

× ×

1BC BC

21

QR QR2

[Using (iv)]

=

2BCQR ...(v)

Similarly, we can prove that

∆∆

( ABC)( PQR)

arar

=

2ABPQ

...(vi)

and( ABC)( PQR)

arar

∆∆

=

2ACPR

...(vii)

From equations (v), (vi) and (vii), we obtain

( ABC)( PQR)

arar

∆∆

=

2ABPQ

=

2BCQR

=

2ACPR

.

Hence, the theorem.Further, in the question,

( ABC)( DEF)

arar

∆∆ =

2BCEF

⇒ 64121

= 2BC

15.4 × 15.4

⇒ BC = × ×64 15.4 15.4

121

= 811

× 15.4 = 11.2 cm.

WORKSHEET–38

1. (B) Ratio of areas of two similar triangles= Ratio of squares of their

corresponding sides.= 42 : 92 = 16 : 81.

2. (A) In ∆ABC, DE || BC

⇒ABAC

=ADAE

⇒ AB = 21 × 57

= 15 cm.

3. (C) ∠M = ∠Q = 35° (Corresponding angles)

PQML

= QRMN

(Ratio of corresponding sides)

⇒ MN = 5 × 126

= 10 cm.


4 . Yes.

APAQ

= 5

7.5 =

23

BPBR

= 46

= 23

Here,APAQ

= BPBR

Hence, due to the converse of BasicProportionality Theorem, AB || QR.

5. ∵ DB ⊥ BC and AC ⊥ BC

∴ DB || AC

Now, ∠DBA = ∠BAC (Alternate angles)

And, ∠DEB = ∠ACB (Each 90°)

∴ ∆BDE ~ ∆ABC (AA similarity)

BEAC

= DEBC

(Corresponding sides)

⇒ BEDE

= ACBC

. Hence proved.

6. AXAB

= 2 22

−

Hint: See worksheet-35 Sol. 8.

7. Hint: AM = 12

AB;

AL = 12

AC


8. Hint: Join AC and use Basic proportionalitytheorem.

A B

CD

E FX

9. Converse of Pythagoras Theorem: In atriangle, if square of one side is equal to

the sum of the squares of the other twosides, then the angle opposite the first sideis a right angle.

Proof: We are given a ∆ABC in which

AC2 = AB2 + BC2 ...(i)

We need to prove ∠ABC = 90°.

Let us construct a ∆PQR such that∠PQR = 90°

and PQ = AB ...(ii)

QR = BC ...(iii)

Using Pythagoras Theorem in ∆PQR, wehave

PR2 = PQ2 + QR2

∴ PR2 = AB2 + BC2 ...(iv)

[Using equations (ii) and (iii)]

From equations (i) and (iv), we have

AC = PR ...(v)

Now, in ∆ABC and ∆PQR,

AB = PQ (From (ii)]

BC = QR [From (iii)]

AC = PR [(From (v)]

So, ∆ABC ≅ ∆PQR (SSS congruence)

∴ ∠ABC = ∠PQR (CPCT)

But ∠PQR = 90° (By construction)

∴ ∠ABC = 90°. Hence proved.

2nd Part:

As AB2 = 2AC2

= AC2 + AC2

⇒ AB2 = BC2 + AC2

[... BC = AC]

⇒ ∠C = 90°.

A

B C

59AIRT GN L SE

WORKSHEET–391. (A) Let the length of shadow is x metres.

BE = 1.2 × 4 = 4.8 m

∆ABC ~ ∆DECABDE

= BCEC

⇒ 3.60.9

= 4.8 + x

x3.6x = 4.32 + 0.9x.

x =4.322.7

= 1.6 m.

2. (B) Here,(a)2 + ( )23a = a2 + 3a2

= 4a2 = (2a)2

According to the converse of PythagorasTheorem, the angle opposite to longest sideis of measure 90°.

3. (A)ADDB

= 23

⇒ AB – ADAD

= 32

⇒ ABAD

–1 = 32

⇒ ABAD

= 52

DE || BC ⇒ ∆ABC~∆ADE

∴ BCDE

= ABAD

= 52

.

4. No.In ∆PQD and ∆RPD,

∠PDQ = ∠PDR = 90°

But neither ∠PQD = ∠RPDnor ∠PQD = ∠PRDTherefore, ∆PQD is not similar to ∆RPD.

5. Hint: ∆BAC ~ ∆ADC

⇒BAAD

= ACDC

= BCAC

⇒ CA2 = BC × CD.

6.ADDB

= 54

⇒ AD

AB – AD =

54

⇒ 5AB – 5AD = 4AD ⇒ ADAB

= 59

...(i)

As DE || BC,

∆ADE ~ ∆ABC

∴DEBC

= ADAB

= 59

...(ii)

[Using (i)]∵ DE || BC and DC is a transversal

∴ ∠EDC ~ ∠BCD

(Alternate interior angles)

i.e., ∠EDF = ∠BCF ...(iii)

Similarly,

∠DEF = ∠CBF ... (iv)

From equations (iii) and (iv), we have

∆DEF ~ ∆CBF (AA similarity)

⇒∆∆

( DEF)( CFB)

arar

=

2DEBC

= 2581

.

[Using equation (ii)]

7. A

B CL

D

E M F

AB = AC; DE = DF

∴ABAC

= DEDF

= 1

⇒ABDE

= ACDF

also ∠A = ∠D

⇒ ∆ABC ∼ ∆DEF

∴∆∆

( ABC)( DEF)

arar

= 2

2ALDM

⇒ALDM

= 45

∴ Ratio of corresponding heights is 4 : 5.

A

B C

2

1

D


OR

Proof: Draw a ray DZ parallel to the rayXY.

In ∆ADZ, XY || DZ

∴ AY AX 2

= =YZ XD 3

⇒ 2YZ = 3AY ... (i)

In ∆YBC, BY || DZ

∴ YZ BD 1= =

ZC DC 1(... BD = DC)

⇒ 2YZ = 2ZC ... (ii)From (i) and (ii),

2ZC = 3AY ... (iii)Now, AC = AY + YZ + ZC

= AY + 32

AY + 32

AY = 82

AY

= 4AY

Therefore, AC : AY = 4 : 1. Hence Proved.

8. 2 5 cm

Hint: BD = 12

BC;

AE = EB = 12

AB


9. Statement : In a right triangle, the squareof the hypotenuse is equal to the sum ofthe square of the other two sides.Proof: We are given a right triangle ABCright angled at B.We need to prove that

AC2 = AB2 + BC2

Let us draw BD ⊥ AC.Now, ∆ADB ~ ∆ABC

So,ADAB

= ABAC (sides are proportional)

orAD .AC = AB2 ...(i)Also, ∆BDC ~ ∆ABC

So,CDBC

= BCAC

or CD.AC = BC2 ...(ii)Adding equations (i) and (ii) we getAD .AC + CD.AC = AB2 + BC2

AC(AD + CD) = AB2 + BC2

AC.AC = AB2 + BC2

AC2 = AB2 + BC2.Hence proved.

2nd part:

AB2 = AD2 + BD2

= AD2 + (3CD)2

= AD2 + 9CD2

= AD2 + CD2 + 8CD2

= AC2 + 8CD2

= AC2 + 8 21

BC4

1CD = BC

4 ∵

2AB2 = 2AC2 + BC2. Hence proved.

WORKSHEET– 40

1. (A)ABPQ

=ABQR

= ACPR

(∵ ∆ABC ~ ∆PQR)

⇒129

= 7x

= 10y

x = 7 × 9

12 =

214

and y = 9×10

12 =

152

.

2. (C) Required ratio = 1625

= 45

= 4 : 5.

3. 17 mHint:

N

W E

S

15 m O

P

8 m

Use Pythagoras Theorem and find OP.

61AIRT GN L SE

4. Hint:Let AB = c

AC = bBC = a

... a2 = b2 + c2

Also, ar(∆ABE) = 234

a

ar(∆BCF) = 234

c

ar(∆ACD) = 234

b .

5. See worksheet −−−−− 36 sol. 6.

6. Let ABCD be a quadrilateral of whichdiagonals intersect each other at O.It is given that

AOCO =

BODO

orAOBO

= CODO

...(i)

In ∆AOB and ∆COD,∠AOB = ∠COD

(Vertical opposite angles)

AOBO

= CODO

[From (i)]

Hence, by SAS rule of similarity, we obtain∆AOB ~ ∆COD

⇒ ∠BAO = ∠DCOi.e. ∠BAC = ∠DCAThese are alternate angles.Therefore, AB || CD and AC is transversal⇒ ABCD is a trapezium. Hence proved

ORHint:As ∠BAC = ∠EFG ; ∠ABC = ∠FEGAnd ∠ACB = ∠FGE

∴12

∠ACB = 12

∠FGE

∴ ∠ACD = ∠FGHand ∠DCB = ∠HGE∴ ∆DCA ~ ∆HGFSimilarly ∆DCB ~ ∆HGE.

7. Hint:A

B C

D

E

Prove that ∆AEB ~ ∆DEC.8. See worksheet–33, sol. 9 (Ist part).

2nd partJoin EF and join BD to intersect EF at O.

∵ AB || DC, and EF || AB,∴ AB || DC || EFIn ∆ABD, EO || AB,

DEAE

= DOBO ...(viii)

(Basic Proportionality Theorem)Similarly, in ∆BCD,

DOBO =

CFBF

...(ix)

Using equations (viii) and (ix), we obtainthe required result.

AEED

= BFFC .

WORKSHEET– 41

1. (B)DEAB

= EFBC

= DFAC

(i) (ii) (iii)

= DE + EF + DFAB + BC + CA

(iv)

⇒42

= ∆Perimeter of DEF

3 + 2 + 2.5

[Taking (ii) and (iv)]⇒ Perimeter of ∆DEF = 15 cm.


2. (A) DE || BC

⇒– 2x

x=

+ 2–1

xx

⇒ x2 – 4 = x2 – x⇒ x = 4.

3. ∆KNP ~ ∆KML

⇒ xa

=c

b c+ ∴ x =

acb c+

.

4. Hint:

A LB

D C

P

Prove that ∆ADL ~ ∆CPD.

5. Hint: 2AP = PC ⇒ AP =13

AC

Similarly, BQ = 13 BC

Use Pythagoras Theorem.

6. PQ || BC ⇒ APPB

= AQQC

= 12

∴ Also ∆APQ ~ ∆ABC

⇒ ∆∆

( ABC) ( APQ)

arar

=

2ABAP

= ( )23 = 9

= AB 3AP

∵

⇒ ∆∆

( ABC) ( APQ)

arar

− 1 = 8

⇒ ∆

( BPQC) ( APQ)

arar

=81

⇒ ∆( APQ)

( BPQC)ar

ar=

18

.

∴ Ratio of area of ∆APQ and trapeziumBPQC is 1 : 8.

ORLet the given square be ABCD.Let us draw an equilateral triangle APBand another equilateral triangle AQC onthe side AB and on the diagonal ACrespectively.

We need to prove

ar (∆APB) = 12

ar (∆AQC)

In right ∆ABC,

AC = 2 2AB + BC

= AB 2 ...(i)(∵ AB = BC)

Now, ar(∆APB) = 3

4AB2 ...(ii)

And ar(∆AQC) = 3

4 AC2

= 3

4 ( )2AB 2

= 3

2 AB2 ...(iii)

Divide equation (ii) by equation (iii), weobtain

∆∆( APB)( AQC)

arar

=

2

2

3AB

43

AB2

= 12

⇒ ar(∆APB) = 12

ar(AQC).

Hence proved.

7. Hint:

Extend AD till E such that AD = DE andsimilarly, PM = MN

63AIRT GN L SE

Prove that ∆ACE ~ ∆PRN∠∠∠∠∠1 = ∠∠∠∠∠2 ...(i)

But ∠∠∠∠∠3 = ∠∠∠∠∠5 (CPCT)∠∠∠∠∠3 = ∠∠∠∠∠4 ∠∠∠∠∠4 = ∠∠∠∠∠6

... ∠ ∠ ∠ ∠ ∠5 = ∠∠∠∠∠6 ...(ii)Adding (i) and (ii),

∠∠∠∠∠1 + ∠∠∠∠∠5 = ∠∠∠∠∠2 + ∠∠∠∠∠6⇒ ∠∠∠∠∠BAC = ∠∠∠∠∠QPR... ∆ABC ~ ∆PQR. (By SAS)

8. Let us take two similar triangles ABC andPQR such that ∆ABC ~ ∆PQR.

...ABPQ

= BCQR

= CARP

...(i)

We need to prove

∆∆

( ABC)( PQR)

arar =

2

2ABPQ =

2

2BCQR =

2

2CARP

Let us draw AM ⊥ BC and PN ⊥ QR.∵ ∆ABC and ∆PQR∴ ∠B = ∠Q ...(ii)In ∆ABM and ∆PQN,

∠B = ∠Q [From (ii)]and ∠M = ∠N (Each 90°)∴ ∆ABM ~ ∆PQN (AA criterion)

∴ABPQ =

AMPN ...(iii)

From equations (i) and (iii), we have

AMPN =

BCQR ...(iv)

Now, ar(∆ABC) = 12

× base × height

= 12

× BC × AM

And ar(∆PQR) = 12

× QR × PN

Therefore,∆∆

( ABC)( PQR)

arar

= BC × AMQR × PN

= 2

2BCQR

...(v)

[Using (iv)]From results (i) and (v), we arrive

∆∆

( ABC)( PQR)

arar

= 2

2ABPQ

= 2

2BCQR

= 2

2CARP

.

Hence the result.Further, consider the question in thefollowing figure.

∠ABO = ∠CDO and ∠BAO = ∠DCO(Alternate angles)

⇒ ∆AOB ~ ∆COD (AA rule)

⇒∆∆

( AOB)( COD)

arar =

2

2ABCD

⇒ ar(∆COD) = 84 ×

212

= ∵ CD 1

AB 2

= 21 cm2.

WORKSHEET–421. (A) ∆OBC ~ ∆ODA (AA criterion)

⇒OBOD =

OCOA =

BCDA

= 2

⇒ BC = 2DA = 2 × 4 = 8 cm.

2. (D) Let the given areas be 2x and 3x.

Required ratio = 2 : 3x x = 2 : 3 .

3. (C) x = 11 or 8

Hint: Use ODOB

= OCOA

.


4. TrueGeometrical figures which are equiangulari.e., if corresponding angles in twogeometrical figure are same, are similar.

5. In right ∆ADC,AD2 = AC2 – CD2

= (2CD)2 – CD2

[∵ AC = BC = 2 CD]= 3 CD2.

6. Hint: BMDN is a rectangle.∆∆∆∆∆BMD ~ ∆∆∆∆∆DMC

⇒ DNDM

=DMMC

⇒ DM2 = DN × MC

Also, ∆ ∆ ∆ ∆ ∆BND ~ ∆∆∆∆∆DNA.

⇒ DMDN

=DNAN

⇒ DN2 = DM × AN.

7. Let BE = 3x and EC = 4x.In ∆∆∆∆∆BCD, GE || DC∴ ∆∆∆∆∆BGE ~ ∆∆∆∆∆BDC

∴ BEBC

= GEDC

⇒ 33 + 4

xx x

= GE2AB

(∴ DC = 2AB)

⇒ GE = 67

AB ...(i)

Similarly, ∆∆∆∆∆DGF ~ ∆∆∆∆∆DBA

⇒ FGAB

= 47

⇒ FG = 47

AB ...(ii)

Adding equations (i) and (ii), we get

GE + FG = 67

AB + 47

AB

⇒ EF = 107

AB

⇒ 7 EF = 10AB. Hence proved.

8. Hint: Let AB = BC = AC = a

∴ AE ⊥ BC

⇒ BE = EC = 12

a

and BD = 13

BC = 13

a

∴ Using Pythagoras TheoremAD2 = AE2 + DE2

= AE2 + (BE − BD)2

⇒ AD2 = AE2 + BE2 + BD2 − 2.BE.BD

⇒ AD2 = AE2 + EC2 +

213

a − 2

1 12 3

a a

= AC2 + 2 2

9 3a a−

= a2 + 2 2

9 3a a− =

279a

⇒ 9AD2 = 7AB2.

9. See worksheet-39 sol. 9 (I part).

Hint: 2nd Part:

AC2 = AD2 + DC2

= AD2 + (3BD)2

= AD2 + 9BD2

= AD2 + BD2 + 8BD2

= AB2 + 8 21

BC4

⇒ 2AC2 = 2AB2 + BC2 .

WORKSHEET–43

1 . (C) 294 cm2

Hint: Prove that ∆OBP ~ ∆OAQ.

2. (A) 6 cmHint: Use AA-similarity to prove∆AOB ~ ∆COD.

3. Hint: Draw AM ⊥ BCand DN ⊥ BCAs ∆AOM ~ ∆DON

⇒∆∆

( ABC)( DBC)

arar

=

1 BC AM21 BC ON2

× ×

× ×

= =AM AODN OD

.

A

B CD E

AC

D

O

N

MB

65AIRT GN L SE

4. Hint: Use concept of similarity.

5. Draw AP ⊥ BC

A

B CD P

1 2

∴ AB2 = AP2 + BP2

= AP2 + (BD + DP)2

⇒ AB2 = AP2 + BD2 + DP2 + 2BD. DP

= AD2 + BD (BD + 2DP)

⇒ AB2 − AD2 = BD × CD. [... BP = PC]Hence proved.

6. Hint:

P ba

x

p

From figure, show x =+ab

a b.

7. ∆MDE and ∆MCB,

∠MDE = ∠MCB (Alternate angles)

MD = MC (M is mid-point of CD)

∠DME = ∠CMB(Vertically opposite angles)

∴ ∆MDE ≅ ∆MCB, (ASA criterion)⇒ DE = CB (CPCT)⇒ AE – AD = BC⇒ AE = 2BC ...(i) (∵ BC = AD)Now, in ∆LAE and ∆LCB,⇒ ∠LAE = ∠LCB (Alternate angles)

⇒ ∠ALE = ∠CLB(Vertically opposite angles)

∴ ∆LAE ~ ∆LCB (AA criterion)

⇒AEBC

= LEBL


⇒2BCBC =

ELBL

[Using equation (i)]

⇒ EL = 2BL. Hence proved.

ORHint:

As AD is median

so, AB2 + AC2 = 2(AD2 + BD2)

⇒ AB2 + AC2 = 22

2 BCAD

4 +

⇒2(AB2 + AC2) = 4AD2 + BC2 ...(i)⇒ Similarly,

2(AB2 + BC2) = 4BE2 + AC2 ...(ii)2(AC2 + BC2) = 4CF2 + AB2 ...(iii)

Add (i), (ii) and (iii),3 (AB2 + AC2 + BC2)

= 4 (AD2 + BE2 + CF2).

8. See worksheet-33 sol. 9 (Ist part).2nd Part: Draw EM || ABM is a point on CB∴ EM || AB

⇒CEAE

= CMMB ...(i)

Also DEEB

= CMMB ...(ii)

From (i) and (ii),

CEAE

= DEEB

.


WORKSHEET– 44

1. (D) BE = 3

4a ⇒ a =

43

BE

∴ AB2 + BC2 + CA2

= a2 + a2 + a2

= 3a2

=24

3 BE3

×

= 16 BE2.2. (B)

Hint: Use pythagoras theorem

3. (B)Hint: Using basic proportionality theoremwe get, x = 4.

4. Hint: In ∆ACD and ∆ABC,∠A = ∠A

∠ACD = ∠ACB = 90°⇒ ∆ACD ~ ∆ABC⇒ AC2 = AB.AD …(i)

∆BCD ~ ∆BAC⇒ BC2 = BA.BD …(ii)∴ Applying (ii) ÷ (i) gives the result.

5. Let the given parallelogram be ABCDWe need to prove that

AC2 + BD2 = AB2 + BC2 + CD2 + DA2

Let us draw perpendiculars DN on ABand CM on AB produced as shown infigure.

In ∆BMC and ∆AND,BC = AD (Opposite sides of a ||gm)

∠BMC = ∠AND (Each 90°)CM = DN (Distance between

same parallels)

∴ ∆BMC ≅ ∆AND (RHS criterion)⇒ BM = AN ...(i) (CPCT)Using result on obtuse triangle ABC

(∠ABC > 90°),AC2 = AB2 + BC2 + 2AB.BM ...(ii)

Using result on acute triangle ABD(∠BAD < 90°),

BD2 = AB2 + DA2 – 2AB.AN⇒ BD2 = CD2 + DA2 – 2AB.BM ...(iii)

[Using (i) and AB = CD]Adding equations (ii) and (iii), we have

AC2 + BD2 = AB2 + BC2 + CD2 + DA2.Hence proved.

6. Hint: AP || QB || RCUse Basic proportionality theorem.

7. See worksheet 38 sol. 9 (Ist part).8. Let us produce AD to J and PM to K so

that DJ = AD and MK = PM.Join CJ and QK.

In ∆ADB and ∆JDC,AD = JD, ∠ADB = ∠JDC, BD = CD⇒ ∆ADB ≅ ∆JDC

(SAS criterion of congruence)⇒ AB = JC ...(i) (CPCT)Similarly, we can prove that

PQ = KR ...(ii)According to the given conditions, we have

ABPQ

= ADPM

= ACPR

⇒ JCKR

=

AJ2

PK2

= ACPR

[Using (i) and (ii)]

67AIRT GN L SE

⇒JCKR

= AJPK

= ACPR

⇒ ∆AJC ~ ∆PKR (SSS criterion ofsimilarity)

⇒ ∠JAC = ∠KPR (Correspondingangles)

i.e., ∠DAC = ∠MPR ...(iii)Similarly, we can prove that

∠DAB = ∠MPQ ...(iv)Adding equations (iii) and (iv), we obtain

∠BAC = ∠QPR ...(v)Thus, in ∆ABC and ∆PQR, we have

ABPQ

= ACPR

(Given)

and ∠BAC = ∠QPR [From (v)]Therefore, ∆ABC ~ ∆PQR.

(SAS criterion of similarity)Hence proved.

ORHint: As ∆ADC is obtuse angled triangle,

AC2 = AD2 + CD2 + 2.CD.DE …(i)and ∆ADB is acute-angled triangle

So, AB2 = AD2 + BD2 − 2.DB.DE …(ii)

Add (i) and (ii), to get the result.

WORKSHEET– 45

1. (D) ∆ABC ~ ∆PQR

⇒ 20h

= 1050

⇒ h = ×50 2010 = 100 m.

2. (A) The ratio of similar triangles is equalto the ratio of squares of their correspond-ing altitudes.

∴10049

= 2

25h

⇒ h2 = ×25 49

100

⇒ h = ×25 49100

⇒ h2 =×5 710

= 3.5 cm.

3. (B) Altitude AM divides base BC in twoequal parts. That is BM = MC = 7 cm usingPythagoras Theorem in right ∆ABM,

AM = 2 225 – 7 = +(25 7) (25 – 7)

= ×32 18 = 24 cm.

4. (i) We know that diagonal of a square

= 2 × side

In square AEFG, AF = 2 AG ...(i)

In square ABCD,AC = 2 AD ...(ii)

Using equations (i) and (ii), we obtain

AFAG =

ACAD

. ...(iii)

(ii) ∠GAF = ∠DAC (Each 45°)⇒ ∠GAF – ∠GAC = ∠DAC – ∠GAC⇒ ∠CAF = ∠DAG ...(iv)From equations (iii) and (iv), we have

∆ACF ~ ∆ADG.(SAS criterion)

5 . Hint: ∵ ∠1 = ∠2∴ PQ = PR

∴ QRQS

= QTPQ

.

6. Hint: Draw AM ⊥ BC and DN ⊥ BC.7. Hint: Fig. A

C Ba

bc

p


8. Hint for 1st part: Prove Pythagoras Theorem.Hint for 2nd part: Let the given triangle isABC with

∠A = 90°Draw AD ⊥ BC

∆ABC ~ ∆DACand ∆ABC ~ ∆DBA.

9. Hint: Let the DC = AB = x

Then QC = 45

x and AP = 35

x

∆QRC ~ ∆PRA.OR

Let the given parallelogram be ABCD.We need to prove

AC2 + BD2 = AB2 + BC2 + CD2 + DA2

Let us draw perpendiculars DN on AB andCM on AB produced as shown in figure.In ∆BMC and ∆AND

BC = AD (opposite sides of aparallelogram)

∠BMC = ∠AND (Each 90°)CM = DN (Distance between

same parallels)∴ ∆BMC ≅ ∆AND (RHS criterion)⇒ BM = AN ...(i) CPCTIn obtuse triangle ABC (∠ABC > 90°),

AC2 = AB2 + BC2 + 2AB.BM...(ii)

In acute triangle ABD (∠BAD < 90°),BD2 = AB2 + DA2 – 2AB.AN

⇒ BD2 = CD2 + DA2 – 2AB.BM...(iii)

[Using (i) and AB = CD]Adding equations (ii) and (iii), we have

AC2 + BD2 = AB2 + BC2 + CD2 + DA2.Hence proved.


1. (C) In ∆ABC, PQ || BC

∴ APBP

= AQQC

∴2.4BP

= 23

⇒ BP = 3.6 cm

∴ AB = AP + BP = 2.4 + 3.6 = 6 cm.

2. (B)∆∆

( ABC)( DEF)

arar

= 2

2BCEF

⇒ 94

=

2BCEF

⇒BCEF

= 32

.

3. Let draw AM ⊥ BC and DN ⊥ BC∵ ∠AMO = ∠DNO = 90°and ∠AOM = ∠DON

∴ ∆AMO ~ ∆DNO(AA similarity)

∴AMDN =

AODO

...(i)

Now, ∆∆

( ABC)( DBC)

arar =

×

×

1BC × AM

21

BC × DN2

= AODO . [Using (i)]

4. True, because ∆BCD ~ ∆CAD⇒ CD2 = BD .AD.

5. PQ || BC and AB is transversal∴ ∠APQ = ∠ABC ...(i)

(Corresponding angles)In ∆ABC and ∆APQ,

∠BAC = ∠PAQ (Common)

69AIRT GN L SE

∠ABC = ∠APQ [From (i)]so, by AA criterion of similarity,

∆ABC ~ ∆APQ

∴∆∆

( ABC)( APQ)

arar =

2

2ABAP

Subtracting unity from both the sides, wehave.

⇒ ∆ ∆∆

( ABC) – ( APQ)( APQ)

ar arar

= 2 2

2

AB – AP

AP

⇒ ∆

(trapezium BPQC)( APQ)

arar

=

2ABAP

– 1

...(ii)It is given that

APAB

= 12

∴ PBAP

= 2

∴PBAP

+ 1 = 2 + 1

⇒PB + AP

AP= 3 ⇒

ABAP

= 3

∴

2ABAP

= 9 ...(iii)

From equations (ii) and (iii), we have

∆( APQ)(trapezium BPQC)

arar =

18

⇒ ar(∆APQ) : ar(trapezium BPQC)

= 1 : 8.6. See worksheet-36 sol. 6.7. See worksheet-33 sol. 9 (Ist part).8. We are given two triangles ABC and PQR

such that ∆ABC ~ ∆PQR.Draw perpendiculars AD and PM on BCand QR respectively.

We need to prove

∆∆

( ABC)( PQR)

arar

= 2

2ADPM

In ∆ABD and ∆PQM,∠ADB = ∠PMQ = 90°∠ABD = ∠PQM (∴∆ABC ~ ∆PQR)

∴ ∆ABD ~ ∆PQM(AA criterion of similarity)

⇒ABPQ =

ADPM

...(i)

(Corresponding sides)We know that the ratio of areas of twosimilar triangles is equal to ratio of squaresof their corresponding sides

∴ ∆∆

( ABC)( PQR)

arar

= 2

2ABPQ

...(ii)


∆∆

( ABC)( PQR)

arar

= 2

2ADPM

. Hence proved.


1. (B) 5x

= 6

6 + 2 ⇒ x =

308

, ⇒ x = 154

cm.

2. (A) ∆PQR ~ ∆CAB,

⇒ PQCA

= PRBC

= QRAB

orABQR

= BCPR

⇒ CAPQ

.

3. In ∆ABC, ∠M || AB.Using Basic Proportionality Theorem, wehave

ACAL

= BCBM

⇒ 2– 3x

x=

+2 3– 2

xx

⇒ 2x2 – 4x = 2x2 + 3x – 6x – 9⇒ – x – 9 ⇒ x = 9.

4. False, because ∆PQD and ∆PRD don’t obeyany condition of similarity.

5. Let the given right angled triangle be ABCwith ∠C = 90° such that AC = b, BC = aand AB = c.


Using Pythagoras Theorem, we have

AB2 = AC2 + BC2

⇒ c2 = a2 + b2 ...(i)Area of equilateral triangle drawn on sideBC

= 34

a2 ...(ii)

Similarly, areas of equilateral trianglesdrawn on side BC and side AB arerespectively

= 34

b2 ...(iii)

And =34

c2 ...(iv)

Sum of areas of equilateral triangles drawnon the sides BC and AC

= 34

a2 + 34

b2

[Adding (ii) and (iii)]

= 34

(a2 + b2)

= 34

c2[Using (i)]

= Area of equilateral triangle drawn on hypotenuse AB.

Hence proved.

6.

∵ ∆ABC ~ ∆PQR

∴ABPQ

= BCQR

=

1BC

21

QR2

⇒ ABPQ

= BDQM

...(i)

(a) In ∆ABD and ∆PQM,

⇒ABPQ =

BDQM [From (ii)]

∠ABD ~ ∠PQM (∵∆ABC ~ ∆PQR)So, by SAS criterion of similarity, we have

∆ABD ~ ∆PQM

⇒ ABPQ

= ADPM

(b) ∵ ∆ABD ~ ∆PQM, [From part (a)]⇒ ∠ADB = ∠PMQ⇒ 180° – ∠ADC = 180° – ∠PMR

[From figure]⇒ ∠ADC = ∠PMR. Hence proved.

7. See worksheet-38 sol. 9 (Ist part).8. BD = BE (Given)

In ∆OBD, AF ⊥ OB and BD ⊥ OB∴ AF || BD∴ ∆OAF ~ ∆OBD

⇒ OAOB

= AFBD

⇒ OAOB

= AFBE

...(i) [Using (i)]

In ∆AFC and ∆BEC,∠FAC = ∠EBC (Each 90°)∠FCA = ∠ECB

(Vertically opposite angles)So by, AA criterion of similarity,

∆AFC ~ ∆BEC

⇒ACBC

= AFBE

...(ii)

Comparing equations (i) and (ii), we have

OAOB

= ACBC

71AIRT GN L SE

⇒OAOB

= OC – OAOB – OC

⇒ OA × OB – OA × OC= OB × OC – OB × OA

⇒ (OA + OB) × OC= 2OA × OB

Dividing both sides by OA × OB × OC, weget

1OA

+ 1

OB=

2OC

. Hence proved.

CHAPTER TEST

1. (B) BC = 2 25 +12 = 13 cm

∆ABD ~ ∆CBA

⇒ABBC

= ADAC

⇒ AD = 5 × 12

13 = 60

13 cm.

2. (D) ∆∆

1

2 =

2122

PP

= 2

24050

= 1625

⇒ ∆1 : ∆2 = 16 : 25.

3. (C)ADDB

= AEEC

⇒ 1.53

= 1

EC

⇒ EC = 31.5

= 2 cm.

4. Yes.

MQ = PQ – PM= 15.2 – 5.7 = 9.5 cm

NR = PR – PN = 12.8 – 4.8 = 8 cm

Now,PMMQ

= 5.79.5

= 0.6

andPNNR

= 4.88

= 0.6

Clearly,PMMQ

= PNNR

⇒ MN || QR.

5. ∆AOB ~ ∆COD (AAA criterion ofsimilarity)

⇒AOCO

= BODO


⇒7 – 92 – 1

xx

= 9 – 8

3x

x

⇒ 21x2 – 27x = 18x2 – 16x – 9x + 8

⇒ 3x2 – 2x – 8 = 0 ⇒ (x – 2) (3x + 4)

⇒ x = 2 or x = –34

⇒ x = 2. (Negative value rejected)

6. ∴ ∆ABE ≅ ∆ACD∴ AB = AC and AE = AD (CPCT)Consider AB = AC⇒ AD + DB = AE = EC⇒ DB = EC ...(i) (∴ AE = AD)Also AD = AE ...(ii)

(Proved above)Dividing equation (ii) by equation (i), wehave

ADDB

= AEEC ...(iii)

Hence, in ∆ABC

ADDB

= AEEC

⇒ DE || BC (Converse of BasicProportionality Theorem)

⇒ ∠ADE = ∠ABC and ∠AED = ∠ACB⇒ ∆ADE ~ ∆ABC.

7. Hint:

∆PAC ~ ∆QBC ⇒ xz

= ACBC

∆RCA ~ ∆QBA ⇒ yz

= ACAB

.

8. Hint:Draw MN || AD, passing through O tointersect AB at M and DC at N.


Use Pythagoras Theorem for ∆AOM,∆BOM, ∆CON and ∆DON.

9. Hint :

∆ABC ~ ∆PQR ⇒ ABPQ

= BCQR

and ∠B = ∠Q

⇒ ABPQ

= 2BD2QM

and ∠B = ∠Q

⇒ ABPQ

= BDQM

and ∠B = ∠Q

⇒ ∆ABD ~ ∆PQM

⇒ ABPQ

= ADPM

.

73RTNI GIRTOI ODO U TC N O MN T O E T R Y

5Chapter

INTRODUCTION TO TRIGONOMETRY

WORKSHEET–50

1. (B) sin 3A = cos (A – 26°)⇒ sin 3A = sin {90°– (A – 26°)}

⇒ 3A = 90° – A + 26° ⇒ A = 116°

4⇒ A = 29°.

2. (D) 2 sin 2x

= 1 ⇒ sin 2x

=12

⇒ sin 2x

= sin 6π ⇒

2x

= 6π

⇒ x = 3π

⇒ x = 60°.

3. (B) sin θ = 2425

⇒ sin2 θ = 224

25

⇒ 1 – sin2 θ = 1 –2

22425

⇒ cos2 θ = 2

2725

⇒ cos θ = 725

Now, tan θ + sec θ = sin cos

θθ

+ 1

cos θ

=

2425725

+ 1725

= 247

+ 257

= 497

= 7.

4. tan θ = 2

sin

1 sin

θ

− θ .

5. cot 25° + tan 41°.

6. True,

LHS =cos 80sin10

°°

+ cos 59° cosec 31°

= cos(90 – 10 )

sin10° °

°+ cos 59° cosec (90° – 59°)

= sin 10°sin 10°

+ cos 59° sec 59°

= 1 + cos 59°cos 59°

= 1 + 1 = 2

Hence, the given equation is valid.

7. 24

cot 30° + 2

1sin 60°

– cos2 45°

=( )2

4

3+ 2

1

32

–2

12

=43

+ 43

– 12

=8 8 3

6+ −

= 136

.

8. sin (x + y) = 1 and cos (x – y) = 3

2⇒ sin (x + y) = sin 90° and cos (x – y) = cos 30°⇒ x + y = 90° and x – y = 30°

Adding and subtracting, we get respectively

2x = 120° and 2y = 60°

i.e., x = 60° and y = 30°.

9. cosec A = 10

sin A =1

cosec A =

110

cos A = 21 – sin A = 1

1 –10

= 310

tan A =sin Acos A

= 13

cot A =1

tan A = 3


sec A =1

cos A =

103

.

10. Hint: RHS =2 2 sin A 1 – cos A

=1 – cos A 1 – cos A

= (1 – cos A) (1 + cos A)

1 – cos A

OR

Hint: LHS =2

2

sin (1 – 2 sin )(2cos – 1 )

θ θθ

= sin × cos 2 cos × cos 2

θ θθ θ

=sincos

θθ

.

WORKSHEET–511. (A) sin (θ + 36°) = cos θ

⇒ sin (θ + 36°) = sin (90°– θ)⇒ θ + 36° = 90° – θ ⇒ 2θ = 54°⇒ θ = 27°.

2. (C)Hint: Divide numerator and denominatorby cos θ.

3. (D) sec θ = 54

⇒ sec2 θ = 2516

⇒ sec2 θ – 1 = 2516

– 1

⇒ tan2 θ = 9

16 ⇒ tan θ =

34

.

4. (A)Hint: ∠A = 30º, ∠B = 90º, ∠C = 60º.

5. sec θ + tan θ = (1 sin )1 sin

cos cos cos + θθ+ =

θ θ θ

=2 2 2 2

2

11 sin

1 sin1

ab ab

a b ab

++ θ += =

− θ −−

= b a b a

b ab+ a b a+ +=

−−.

6. sin A = 725

, cos A = 2425

,

sin C = 2425

and cos C = 725

.

7. cos 60° + sin 30° cot 30°tan 60° + sec 45° cosec 45°

−−

=

1 1 + – 3

2 23 2 – 2+

= 1 – 3 3

3 3×

= 3 – 33

.

8. Given expression

=cot tan (90° ) sec (90° ) cosec

sin cos (90° ) + cos sin (90° )θ − θ − − θ θ

θ − θ θ − θ

= cot cot cosec cosec

sin sin + cos cos θ θ − θ θ

θ θ θ θ

= 2 2

2 2

cot cosec

sin + cos

θ − θθ θ

= cosec2 θ – 1 – cosec2 θ[∵ sin2 θ + cos2 θ = 1]

= – 1.

9. 12

Hint: Draw ∆ ABC with AB = BC = AC = a (say)Draw AD ⊥ BC

∴ ∠BAD = ∠DAC = θ = 30°and BD = DC = a/2

∴ sin θ = BD /2AB

aa

= =12

⇒ sin 30° = 12

.

10. LHS =

1tan tan

1 1 tan1tan

θ θ+− θ−

θ

=( )

2tan 1tan 1 tan 1 tan

θ+

θ − θ − θ

=( )

21 1tan

tan 1 tan θ − θ − θ


=31 tan 1

tan 1 tan

θ − θ − θ

= tan θ + cot θ + 1

=2tan 1

1tan

θ ++

θ =

2sec1

tanθ

+θ

= 1 + sec θ cosec θ = RHS

OR

LHS =cot A – cos Acot A + cos A

=

cos Acos Asin A

cos Acos A

sin A

−

+

=

1cos A – 1

sin A1

cos A 1sin A

+

=

11sin A

1 + 1sin A

− = cosec A – 1

cosec A + 1 = RHS

WORKSHEET–52

1. (A)Hint: tan 5º = cot 85º; tan 25º = cot 65º.

2. (C) 8 tanx = 15

⇒ tan2 x =22564

⇒ sec2x – 1 =22564

sec2 x =28964

⇒ sec x =178

⇒ cos x = 817

Now, sin x – cos x = 21 – cos x – cos x

=64 8

1 – –289 17

=15 – 8

17

=7

17.

3. (A)Hint: sec θ = cosec 60°

⇒ cos θ = sin 60° ⇒ θ = 30°

... 2cos2 30°– 1 = 32× 1

4− =

31

2− =

12

4. (D)Hint: (1 + sin θ) (1 – sin θ) = cos2 θ

= 2 1

sec θ.

5. (A)Hint: sec 4A = cosec (90º – 4A).

6. Hint: cos (90º – θ) = sin θ, sin (90º – θ) = cos θ.

7.(1 sin )(1 – sin )(1 cos )(1 – cos )

+ θ θ+ θ θ

= 2

21 – sin1 – cos

θθ

= 2

2cossin

θθ

= cot2 θ = 27

8

= 4964

.

OR2 2

2 2cosec + cot cosec – sec

θ θθ θ

= ( )2 2

2 2

1 cot cot

1 cot 1 tan

+ θ + θ+ θ − + θ

=2

2 21 2cot

cot – tan+ θ

θ θ

=1 2 3

13 –

3

+ ×=

783

=218

.

8. sin2 30° + sin2 45° + sin2 60° + sin2 90°

= 21

2

+2

12

+2

32

+ ( )21

= 1 1 3

+ +4 2 4

+ 1

= 1+ 2 + 3 + 4 10

=4 4

= 52

.

9. Hint:

LHS = 1 sin 1 sin ×

1 + sin 1 sin − θ − θ

θ − θ

= 2(sec tan )θ − θ


ORHint:

LHS =2cos A sin A

1 tan A sin A cos A+

− −

=2 2cos A sin A

–cos A sin A cos A – sin A−

=2 2cos A sin A

cos A sin A−−

= cos A + sin A.

10. Hint: 1 – 2sin2 θ = 2cos2 θ – 1 = cos2 θ – sin2 θ .

WORKSHEET–53

1. (A) tan θ = 34

= Perpendicular

Base

BC = 2 23 4+

= 25 = 5

cos θ = BaseHypotenuse

= 45

∴

41 1 cos 15= =

41 + cos 91 + 5

−− θθ

.

2. (C)Hint:1 + tan θ + sec θ

1 + cot + cosec 1= 1 + + sec =

cot cotθ θ

θθ θ

3. (B)Hint: A + B = 90º; A – B = 30º.

4. (A) tan 2θ = cot (θ + 9°)⇒ tan 2θ = tan [90° – (θ + 9°)]⇒ 2θ = 90° – θ – 9° ⇒ 3θ = 81°⇒ θ = 27°.

5.2

coscot

1 cos

θθ =

− θ.

6. True.Hint: A6 + B6

= (A2 + B2) [(A2 + B2)2 – 3A2 B2].

7. Hint:

LHS =1 2sin cos + 1 2sin cos

(sin + cos )(sin cos )− θ θ θ θ

θ θ θ − θ+

.

8. LHS = cosA sin A 1cosA sin A 1

− ++ −

Dividing Numerator and Denominator bysin A, we get

= cotA 1 cosecAcotA 1 cosecA

− ++ −

= ( ) ( )2 2cotA + cosecA cosec A – cot A

cotA 1 cosec A

−

+ −

= (cosecA + cot A) [1 cosecA + cot A]

cot A cosecA + 1−

−= cosec A + cot A = RHS.

9. Given expression

=2sin 68cos22

°°

– 2cot155 tan75

°°

–3 tan 45° tan 20° tan 40° tan50° tan70

5°

=2 sin (90° 22°)

cos 22−°

– 2 cot (90° 75°)

5 tan 75°−

–

( ) ( )3 1 tan 90 – 70 tan 90 – 50

tan 50 tan705

× × ° ° ° °° °

=2cos 22cos 22

°°

–2 tan 755 tan 75

°°

–3cot 70 cot 50 tan 50 tan70

5° ° ° °

= 2 – 25

–

1 13 tan 50 tan70

tan 70 tan 505

× × ° °° °

= 2 – 25

– 35

=10 2 3

5− −

= 55

= 1.

A 4

3

C

5

B

�


10. Given expression

= 8 3 cosec2 30°. sin 60°. cos 60°. cos2 45°.sin 45°. tan 30°. cosec3 45°.

= 8 3 × 21

sin 30°. sin (90° – 30°).

cos (90° – 30°) cos2 (90° – 45°). sin 45°.

sin 30cos 30

°° 3

1sin 45°

= 8 3 × 21

sin 30°× cos 30°. sin 30°. sin2 45°.

sin 45°. 3

sin 30 1cos 30 sin 45

° ×° °

= 8 3 × 2sin 30°. sin 30°

sin 30°

× cos 30cos 30

°°

×2

3sin 45° sin 45°

sin 45°

= 8 3 × 1 × 1 × 1 = 8 3 .

OR

Hint: sec2 θ = 22

1 1216

xx

+ +

... sec2 θ – 1 = 22

1 1216

xx

+ −

... tan2 θ = 21

4x

x −

⇒ tan θ = ±1

4x

x −

.

WORKSHEET–54

1. (B) As sin A = 34

,

let BC = 3x and CA = 4x

∴ AB = ( )2 24 – (3 )x x = 7 x

Now, tan A = BCAB

= 37xx

= 37

.

2. (C) 22 tan 30

1 tan 30°

+ ° = 2

12

3

11

3

×

+

=

23

43

= 3

2.

3. (D)

Hint: tan x = 158

⇒ sin x = 1517

, cos x = 817

∴ sin2 x – cos2 x = 225289

– 64289

= 161289

.

4. (A) sin 30 tan 45 – cosec 60sec 30°+ cos 60° + cot 45°

° + ° °

=

1 21–

2 32 1

123

+

+ + =

3 2 3 – 44 3 2 3

++ +

=3 3 – 43 3 4+

× 3 3 – 43 3 – 4

= 27 16 – 24 327 – 16

+ = 43 – 24 311

.

5. ∵ A + B + C = 180º

∴ LHS = C + A 180º – B cot = cot

2 2

= cot (90º – B2

) = tan B2

= RHS.6. Yes.

Hint: Both sides =725

.

7. LHS = (cosec A – sin A) (sec A – cos A)

=1 1

– sin A – cos Asin A cos A

=21– sin A

sinA ×

21 – cos Acos A

=2 2cos A sin A

sin A cos A×

= sin A cos A .... (i)


RHS =1

tan A cot A+ =

1sin A cos Acos A sin A

+

= 2 2sin Acos A

sin A cos A+= sin A cos A ....(ii)

(∵ sin2 A + cos2 A = 1)From equations (i) and (ii) we obtainLHS = RHS.

8. 7 sin2 θ + 3(1 – sin2 θ) = 4Let sin θ = x∴ 7x2 + 3 – 3x2 = 4

⇒ 4x2 = 1 ⇒ x2 = 14

⇒ x = ± 12

∴ sin θ = 12

or sin θ = 1

2−

sin θ = –12

is not possible as θ is acute.

⇒ cosec θ = 2 ∴ cos θ = 3

2

∴ sec θ + cosec θ = 2

23

+ . Hence proved

9. 1Hint: cos (40º + θ) = sin {90° – (40° + θ)}

= sin (50° – θ)and cos 40° = sin 50°.

10. LHS = m2 – n2 = (tan θ + sin θ)2

– (tan θ – sin θ)2 = 4sin θ tan θ ...(i)RHS.

= 42

22

sin= 4 – 1 = 4 sin sec –1

cosmn

θ θ θθ

= 4 sin θ tan θ .... (ii)From (i) and (ii), LHS = RHS.

WORKSHEET–55

1. (A) Required value = 25 64 36 8

2 –100 100 6

+ ×

= 25 × 1

300 (192 + 216 – 400)

=1

12 × 8 =

23

.

2. (C) sin θ = 21 – cos θ = 1 0.36− = 0.8

And tan θ = sincos

θθ

= 0.80.6

= 43

Now, 5 sin θ – 3 tan θ = 5 × 0.8 – 3 ×43

= 0

3. (D)Hint: Divide numerator and denominatorby sin A.

31 +1 + cot A 4=

31 – cot A 1 –4

.

4. (C) sec A =23

⇒ sec A = sec 30° ⇒ A = 30°

⇒ A + B = 90° ⇒ B = 90°– 30° = 60°

Now, cosec B = cosec 60° = 23

5. (A) Given expression

= ( )

( )

222

2

1 23 + 4 3 5 0

2 3

2 2 – 3

+ + ×

+

= 3 2 4

4 – 3+ +

= 9.

6. False.Hint: ∠A = 30º, ∠B = 60º.

7.sec – cosec sec + cosec

θ θθ θ

=

1 1 –

cos sin 1 1

+ cos sin

θ θ

θ θ

= sin cos sin + cos

θ − θθ θ

=( )

( )

1sin cos

sin1

sin + cos sin

θ − θ ×θ

θ θ ×θ

=1 – cot 1 + cot

θθ

=

31 –

43

1 + 4

= 17

. Hence proved.


8. cos sin cos + sin

θ − θθ θ

= cos sin cos + sin

θ − θθ θ

= cot – 1cot + 1

θθ

.

(Dividing numerator and denominator bysin θ)

=– 1

1

pqpq

+ =

–p qp q+

.

9. Given expression

= 2 2sin 35° cos 55°

cos 55° sin 35° +

– 2cos 60°

= ( )2sin 35°

cos 90° 35° −

+ ( )2cos 55°

sin 90° 55° −

– 2cos 60°

= 2sin 35°

sin 35°

+ 2cos 55°

cos 55°

– 2cos 60°

= 1 + 1 – 2 cos 60°

= 1

2 22

− × = 2 – 1 = 1

ORGiven expression

= cos 58°sin 32°

+ sin 22°cos 68°

– cos 38°cosec 52°

tan 18° tan 35° tan 60° tan 72° tan 55°

= ( )cos 58°

sin 90° – 58° +

( )sin 22°

cos 90° – 22°

– ( )

( )( )

cos 38°cosec 90° – 38°tan 18° tan 35° tan 60° tan 90° – 18°

tan 90° – 35°

= cos 58°cos 58°

+ sin 22°sin 22°

– cos 38° sec 36°

tan 18° tan 35° tan 60° cot 18° cot 35°

= 2 –

1cos 38° ×

cos 38°tan 18° tan 35° tan 60° ×

1 1×

tan 18° tan 35°

= 2 – 1

tan 60°

= 2 – 13

= 2 3 – 1

3 ×

33

= 6 – 3

3.

10. tan A = n tan B

⇒ cot B = tanA

n and sin A = m sin B

⇒ sin B = 1sin A

m ⇒ cosec B =

sin Am

... cosec2 B – cot2 B = 1

⇒2 2

2 2sin A tan Am n− = 1

⇒2 2 2

2cos A

sin Am n− = 1

⇒ m2 – 1 = (n2 – 1) cos2 A

⇒2

21

– 1mn

−= cos2 A.

Hence proved

OR

Consider an equilateral triangle PQR inwhich PS ⊥ QR. Since PS ⊥QR so PS bisects ∠P as wellas base QR.We observe that ∆PQS is aright triangle, right-angled atS with ∠QPS = 30° and ∠PQS = 60°.For finding the trigonometric ratios, we needto know the length of the sides of thetriangle. So, let us suppose PQ = x

Then , QS = 12

QR = 2x

and (PS)2 = (PQ)2 – (QS)2 = x2 – 2

4x

= 23

4x

∴ PS = 34

x


(i) cos 60° =QSPQ

= 2x

x =

12

(ii) sin 60° =PSPQ

=

32

x

x =

32

(iii) tan 30° =QSPS

= 232

x

x =

13

.

WORKSHEET–56

1. (B) b2x2 + a2y2 = b2a2 cos2 θ + a2b2 sin2 θ= a2b2.

2. (A) A = 90° – 60° = 30°∴ cosec A = cosec 30° = 2.

3. (C) tan θ = 125

⇒ 1 + tan2 θ = 1 + 2

2125

⇒ sec θ = 135

Now, 1 + sin 1 – sin

θθ

=

1 + sin cos

1 sin cos

θθ

− θθ

= sec tansec – tan

θ + θθ θ

=

13 125 5

13 12–

5 5

+ =

25515

= 25

tan θ =125

sin θ =2

1 1=

cosec 1 cosθ + θ

=2

1

51

12 +

= 1213

.

4. (A) sin 29°cos 61°

+ 2cos 27°

sin 63°

– 4 cos2 45°

= ( )sin 29°

cos 90° 29°− + ( )

2cos 27°

sin 90° 29°

−

– 4

× 2

12

= 1 + 12 – 42

= 0.

5. (B) Given expression

= 44 21 1

2 2

+ –3 ( )

221

12

− –

23

2

= 4 1 1

16 4 +

– 3 1

– 12

– 34

= 14

+ 1 – 32

+ 3 – 34

= 174

– 94

= 84

= 2.

6.1

tan A +

sin A1 + cos A

= cos Asin A

+ sin A

1 + cos A

= ( )2 2cos A + cos A + sin A

sin A 1 + cos A

=( )

1 + cos Asin A 1 + cos A

= cosec A

= 2.

7. ∵ sin θ = 34

∴ cosec θ = 43

∵ cos θ = 21 – sin θ = 9

1–16

= 74

∴ sec θ = 47

and cot θ = 73

Now, LHS

= 2 2

2cosec – cot

sec – 1θ θ

θ=

16 7–

9 916

– 17

=

9997

= 79

= 73

= RHS. Hence Proved.


8. Hint: LHS =1

1 cos A–

sin A sin A

– 1

sin A

=sin A

1 – cos A –

1sin A

=( )

2sin A – 1 + cos Asin A 1 – cos A

=( )

21 – cos A – 1 + cos Asin A 1 – cos A

=( )( )

cos A 1 – cos Asin A 1 – cos A

= cot A.

OR

Hint: Using a3 + b3 = (a2 + b2 – ab) (a + b)we get

3 3sin cos.sin cos

sin cosθ+ θ

+ θ θθ+ θ

=2 2 .(sin cos )(sin cos sin cos )

sin cosθ + θ θ + θ − θ θ

θ + θ+ sin θ ⋅ cos θ

= . .1 sin cos sin cos 1− θ θ + θ θ = .

9. ( )( )( )( )

2 + 2 sin 1 – sin 1 + cos 2 – 2 cos

θ θθ θ

= ( )( )( )( )

2 1 + sin 1 – sin 2 1 + cos 1 – cos

θ θθ θ

= 2

21 – sin1 – cos

θθ

= 2

2cossin

θθ

= cot2 θ

= 215

8

= 22564

.

10. Hint: p2 – 1 = sec2 θ + tan2 θ + 2 sec θ tan θ – 1= 1 + tan2 θ + tan2 θ + 2 sec θ tan θ – 1= 2 tan θ (tan θ + sec θ)Similarly p2 + 1 = 2 sec θ (tan θ + sec θ).

WORKSHEET–57

1. (B)Hint: x + y = 2 cot A

x – y = 2 cos A

∴2

x yx y

− +

= sin2 A

and2

2x y−

= cos2 A

∴ sin2 A + cos2 A = 1.

2. (A)

Hint: (x + 1)2 = x2 + 52

3. (A) tan A = 13

= tan 30°

⇒ ∴ ∠A = 30°

∴ ∠C = 180° – A – B = 180° – 120°

= 60°Now, sin A cos C + cos A sin C= sin 30° cos 60° + cos 30° sin 60°

= 12

× 12

+ 3

2 ×

32

= 1.

4. (C) cos α = 12

⇒ cos α = cos 60°

⇒ α = 60°

tan β =13

⇒ tan β = tan 30°

⇒ β = 30°.

Now, sin (α + β) = sin (60° + 30°) = sin 90° = 1.

5. (B) tan 1° tan 2°.... tan 43° tan 44° tan 45°tan 46° tan 47°..... tan 88° tan 89°

= (tan 1° tan 89°)(tan 2° tan 88°)....(tan 43° tan 47°)(tan 44° tan 46°) tan 45°

= (tan 1° cot 1°)(tan 2° cot 2°)....(tan 43°

cot 43°)(tan 44° cot 44°) tan 45°

= (1) × (1) × .... × (1) × (1) × tan 45°

= (1 × 1 ×....× 1 × 1) × tan 45°

= 1 × 1 = 1.

6. Given expression

= tan 50° + sec 50°

cot 40° + cosec 40° + cos 40° cosec 50°


= tan 50° + sec 50°

cot (90° 50°) + cosec (90° 50°)− − +

cos 40°cosec (90° – 40°)

= tan 50° + sec 50°tan 50° + sec 50°

+ cos 40°. 1

cos 40°

= 1 + 1= 2.

7. LHS= tan (A – B) = tan (60° – 30°) = tan 30°

= 13

.

RHS

= tan A tan B

1 + tan A tan B−

⋅ =

tan 60° tan 30°1 tan 60° tan 30°

−+ ⋅

=

13 –

31

1 + 3 . 3

=

3 – 13

1 1+ =

23

2

= 13

= LHS. Hence verified.

8. RHS = 6 2

6 2 2 sin 3 sin 1.+ + 1cos cos cos

θ θθ θ

= 6 6 2 2

6

sin + cos + 3sin coscos

θ θ θ θθ

= ( )32 2

6

sin cos

cos

θ + θθ

6 2 2= sec = LHS. [ sin +cos = 1]θ θ θ∵

ORHint: Numerator ofLHS = tan θ + sec θ – (sec2 θ – tan2 θ)= (tan θ + sec θ) – (tan θ + sec θ) (sec θ – tan θ)= (tan θ + sec θ) (1 – sec θ + tan θ).

9. cos θ + sin θ = 2 cos θ

Squaring both sides, we getcos2 θ + 2 cos θ sin θ + sin2 θ = 2 cos2 θ⇒ 2 cos2 θ – cos2 θ – 2 cos θ sin θ = sin2 θ⇒ cos2 θ – 2 cos θ sin θ = sin2 θAdding sin2 θ to both sides , we have

⇒ sin2 θ + cos2 θ – 2 cos θ sin θ = sin2 θ + sin2 θ⇒ (cos θ – sin θ)2 = 2 sin2 θ

⇒ cos θ – sin θ = 2 sin θ Hence proved.

10. Hint: l tan θ + m sec θ = n ...(i) × l′l′ tan θ – m′ sec θ = n′ ...(ii) × l

⇒ ll′ tan θ + ml′ sec θ = nl′l′ l tan θ – m′ l sec θ = n′ l

– + – (m′ l + ml′) sec θ = nl′ – n′ l

⇒ sec θ = nl n lm l ml

' '' '

−+

Similarly, tan θ = nm mnlm ml

' '' '

++

.

WORKSHEET–58

1. (D) Given expression

= 2 2

2 2

cos (90 70 ) cos 70

sec (90 40 ) cot 40

− ° + °° − ° − °

+ 2 {cosec2 58° – cot 58° tan (90° – 58°)}

= 2 2

2 2sin 70° + cos 70°

cosec 40° – cot 40°+ 2 (cosec2 58° – cot2 58°)

= 11

+ 2(1) = 1 + 2 = 3.

2. (A) sec 5A = cosec (A – 36°)⇒ sec 5A = sec {90° – (A – 36°)}

5A = – A + 126° ⇒ A = 21°.

3. (D) Given expression

= sin2 5° + sin2 10° ... + sin2 40° + sin2 45°+ sin2 50° + ... + sin2 80° + sin2 85° + sin2 90°

= cos2 85° + cos2 80° + .... + cos2 50° +21

2

+ sin2 50° + .... + sin2 80° + sin2 85° + (1)2

= (cos2 85° + sin2 85°) + (cos2 80° + sin2 80°)

+ .... + (cos2 50° + sin2 50°) + 12

+ 1

= (1 + 1 + .... 8 terms) + 12

+ 1


= 8 +12

+ 1 = 19

2.

4. (A) tan 3x = 12

.12

+12

= 12

+12

= 1

⇒ tan 3x = tan 45° ⇒ x = 453

° = 15°.

5. (B) cosec A = 2 ⇒ sin A = 12

cos A = 21 – sin A = 1

1–2

= 12

tan A = 1, cot A = 1

Now, ( )

2 2

2 2

2 sin A + 3 cot A

4 tan A – cos A =

12 × + 3 × 12

14 1

2 −

= 42

= 2.

6. TrueHint:

a cos θ + b sin θ = 4 ...(i) × sin θa sin θ – b cos θ = 3 ...(ii) × cos θ

⇒ a cos θ sin θ + b sin2 θ = 4 sin θa sin θ cos θ – b cos2 θ = 3 cos θ

– + –

b = 4 sin θ – 3 cos θSimilarly, a = 4 cos θ + 3 sin θ∴ a2 + b2 = 16 sin2 θ + 9 cos2 θ – 12 sin θcos θ + 16 cos2 θ + 9 sin2 θ + 12 sin θ cos θ

= 16 + 9 = 25.

7. (a2 − b2) sin θ + 2ab . cos θ = a2 + b2

Divide by cos θ

(a2 − b2) tan θ + 2ab = 2 2

cosa b+

θ⇒ (a2 − b2) tan θ + 2ab = (a2 + b2) . sec θ

= (a2 + b2 . 21 tan+ θSquaring both sides:(a2 − b2)2 tan2 θ + 4 a2b2 + 4 ab (a2 − b2) tan θ

= (a2 + b2)2 (1 + tan2 θ)

= (a2 + b2)2 + (a2 + b2)2 tan2 θ

[(a2 − b2)2 − (a2 + b2)2] tan2 θ + 4 a2b2 + 4 ab(a2 − b2) tan θ − (a2 + b2)2 = 0

⇒ – 4a2b2 tan2 θ + 4ab (a2 – b2) tan θ − a4 − b4

+ 2a2b2 = 0− 4a2b2 tan2 θ + 4ab (a2 − b2) tan θ

− (a2 − b2)2 = 04a2b2 tan2 θ − 4ab (a2 − b2) tan θ + (a2 − b2)2 = 0⇒ [2ab tan θ − (a2 − b2)]2 = 0⇒ 2ab tan θ = a2 − b2

⇒ tan θ = 2 2

2a b

ab−

.

8. Hint: Use (a2 + b2)3 = a6 + b6 + 3a2b2 (a2 + b2).

9. LHS

=( )( )

3 31+ cot A + tan A sin A – cos A

sec A – cosec A

=( )

3 3

cos A sin A1 + + sin A – cos A

sin A cos A1 1

–cos A sin A

=

2 2

3 3

1(sin A cos A + cos A + sin A)

sin A cos A(sin A cos A)

1(sin A cos A)

sin A cos A

−

−

= (sin2 A + cos2 A + sin A cos A)

= sin2 A . cos2 A = RHS. Hence proved.

10. m = cosec θ – sin θ = 1sin θ

– sin θ

=21– sin

sinθ

θ =

2cossin

θθ

n = sec θ – cos θ = 1cosθ

– cos θ

=21 – cos

cosθ

θ = 2sin

cosθθ

Now, LHS = ( ) ( )2 2

2 23 3m n mn+

=

24 2 3

2cos sin

cossin

θ θ× θθ +

22 4 3

2cos sinsin cos

θ θ× θ θ


= ( )2

3 3cos θ + ( )2

3 3sin θ

= cos2 θ + sin2 θ = 1 = RHS.OR

LHS= (1 + cot A – cosec A)(1 + tan A + sec A)

= cos A 1

1sin A sin A

+ − sin A 1

1cos A cos A

+ +

= sin A + cos A –1sin A

× cos A + sin A + 1cos A

= ( )2 2sin A + cos A – 1

sin A cos A

= 2 2sin A + 2 sin A cos A + cos A – 1

sin A cos A

= 2 sin A cos Asin A cos A

= 2

= RHS. Hence proved.


1. (B) cos 30° = BCAC

⇒ 32

= 10x

⇒ x = 20

3 ⇒ x =

20 33

cm

sin 30° =ABAC

⇒ 12

=3

20 3y ×

⇒ y = 10 3

3cm.

2. (D) tan 2θ = cot (θ + 15°) ⇒ tan 2θ = tan [90°– (θ + 15°)].

3. We know: sin θ = cos (90° – θ)so, given expression

= { }2cos(90 35 )cos 55

° − °°

+ 2cos 55

cos(90 35 )°

° − ° – 2 cos 60°

= 2cos 55

cos 55°

° +

2cos 55cos 55

° °

– 2 × 12

= 1 + 1 – 1 = 1.

4. True, because LHS = tan 60° = 3 and

RHS = 22 tan 30

1 – tan 30°

° =

231

13

+ = 3 .

5. 2 (sin6 A + cos6 A) – 3 (cos4 A + sin4 A) + 1= 2{(sin2 A + cos2 A)3 – 3sin2 A cos2A

(sin2 A + cos2 A)}– 3{(sin2 A + cos2 A)2 – 2sin2 A cos2 A} + 1 [∵ (a + b)3 = a3 + b3 + 3ab(a + b) and (a + b)2 = a2 + b2 + 2ab]

= 2(1 – 3sin2 A cos2 A) – 3(1 – 2sin2 Acos2 A) + 1

[∵ sin2 A + cos2 A = 1]= 2 – 6 sin2 A cos2 A – 3 + 6 sin2 A cos2 A

+ 1 = 0

6. 2sin x + cos y = 1 (Given)⇒ 2sin x + cos y = 20

⇒ sin x + cos y = 0 ... (i)2 2sin cos16 x y+ = 4 (Given)

⇒ ( )2 2sin cos24

x y+ = 4

⇒ ( )2 22 sin cos4

x y+= 41

⇒ 2(sin2 x + cos2 y) = 1 ... (ii)Substituting cos y = – sin x from (i) in (ii),

we get 2(sin2 x + sin2 x) = 1 ⇒ sin2 x = 14

⇒ sin x = ± 12

When sin x = – 12

, cos y = 12

When sin x = 12

, cos y = – 12

Hence, sin x = – 12

, cos y = 12

or sin x = 12

,

cos y = – 12

.

7. We know thatsin (90°– θ) = cos θ, tan (90°– θ) = cot θ, sec(90°– θ) = cosec θ


Now, 2 2

2 2sec (90 – ) – cot2(sin 25 sin 65 )

° θ θ° + °

+ 2 2 2

2 22cos 60 tan 28 tan 62

3(sec 43 – cot 47 )° ° °

° °

= ( ){ }

2 2

2 2

sec (90 ) cot

2 sin 25 cos 90 25

° − θ − θ° + ° − °

+ ( ){ }

2 2 2

2 2

2cos 60 tan 28 tan (90 28 )

3 sec 43 cot 90 43

° + ° ° − °

° − ° − °

= ( )2 2

2 2

cosec cot

2 sin 25 cos 25

θ − θ

° + °

+ 2 2 2

2 2

.2cos 60° + tan 28 cot 28

3(sec 43 tan 43 )

° °° − °

=

12 11 4

2 1 3 1

× ×+

× ×

= 1 12 6

+ = 23

.

8. Given equation are: sin θ + cos θ = p ... (i)sec θ + cosec θ = q ... (ii)Squaring both the sides of equation (i),we getsin2 θ + cos2 θ + 2sin θ cos θ = p2

Subtract unity from both the sides to getp2 – 1 = 2sin θ cos θ ... (iii)Equation (ii) can be written as

q =1

cos θ +

1sin θ

⇒ q =sin cossin cos

θ + θθ θ

... (iv)

From equations (iii) and (iv), we get

q (p2 – 1) =sin cossin cos

θ + θθ θ

× 2sin θ cos θ

⇒ q (p2 – 1) = 2(sin θ + cos θ)⇒ q (p2 – 1) = 2p. Hence proved.


1. (A) Given expression= sin 25° cos (90° – 25°) + cos 25°

sin (90°– 25°)= sin2 25° + cos2 25° = 1.

2. (B) 2sin – cos2sin cos

θ θθ + θ

=

2sin cos–cos cos

cos2sincos cos

θ θθ θ

θθ +θ θ

= 2 tan – 12 tan 1

θθ +

=

42× – 1

34

2× 13

+ = 5

11.

3.cos 45

sec 30 cosec 30°°

° + =

12

22

3+

=( )

12

2 1 3

3

+ =

12

× ( )

3

2 1 3+

=( )

( )( )3 3 – 1

2 2 3 1 3 – 1

×

+ =

3 – 34 2

×22

= 3 2 – 6

8.

4. False, because cos2 23° – sin2 67° = 0, 0 isnot a positive value.

5. LHS =cos A

1+ sin A +

1 sin Acos A+

=( )

( )22cos A + 1 + sinA

1 + sin A cosA

=( )

2 2cos A + 1 + sin A + 2sin A1 + sin A cos A

= ( )2 2 sin A

1 sin A cos A+

+ =

( )( )

2 1 sin A

1 sin A cos A

++

=2

cos A = 2 sec A



6. Let us construct a triangle ABC in whichAB = BC = AC = a (say). Draw AD ⊥ BC.AD bisects BC

⇒ BD = DC = 2a

AD bisects ∠BAC⇒ θ = 30°In right angled ∆ABD.

AD2 = AB2 – BD2 = a2 – 2

2a

= a2 – 2

4a =

234a

⇒ AD = 3

2 a

Now, in ∆ ABD,

tan θ = BDAD

⇒ tan 30° = 2

32

a

a

⇒ tan 30° = 13

.

7. ( a2 – b2) sin θ + 2ab cos θ = a2 + b2 (Given)Divide both sides by cosθ to get( a2 – b2) tan θ + 2ab = (a2 + b2) sec θSquaring both sides, we get(a2 – b2)2 tan2 θ + 4a2 b2 + 4ab(a2 – b2) tan θ

= (a2 + b2)2 sec2 θ

⇒ (a2 – b2)2 tan2 θ – (a2 + b2)2 tan2 θ + 4ab

(a2 – b2) tan θ – (a2 + b2)2 + 4a2b2 = 0

(∵ ∵ ∵ ∵ ∵ sec2 θ = 1 + tan2 θ)⇒ – 4a2b2 tan2 θ + 4ab (a2 – b2) tan θ

– (a2 – b2)2 = 0⇒ – 4a2b2 x2 + 4ab (a2 – b2) x – (a2 – b2)2 = 0

where x = tan θThis is a quadratic equation in x.Here, discriminant,

D = ( ) ( )2 22 2 2 2 2 2 2 216 4 4a b a b a b a b− − × −

= 0

∴ x = 2 2 2 2

2 2

4 ( ) 022 ( 4 )

ab a b a baba b

− − − −=

× −

⇒ tan θ = 2 2

2a b

ab− . Hence proved.

8. Since ABC is a acute angled triangleso, ∠A < 90°, ∠B < 90° and ∠C < 90°.Also ∠A + ∠B + ∠C = 180° ....(i )

sin (A + B – C) = 12

°(Given)

⇒ sin (A + B – C) = sin 30°⇒ ∠A + ∠B – ∠C = 30° .... (ii )Similarly, ∠B + ∠C – ∠A = 45° .... (iii )Add equations (ii ) and (iii ) to get

2∠B = 75° ⇒ ∠B = 137

2°

Subtract equation (ii) from equation (i) to get2∠C = 150° ⇒ ∠C = 75°

Subtract equation (iii) from equation (i) to get

2∠A = 135° ⇒ ∠A = 1

672

°

Thus, ∠A = 1

672

°, ∠B =

137

2°

and ∠C = 75°.

CHAPTER TEST

1. (A) x = sec

2θ

and 1x

= tan

2θ

∴ 22

12 –x

x

= 2 2 2sec tan

–4 4

θ θ

= 2 2sec – tan

24

θ θ

= 12

.

2. (C) Hint. ( )

2 2

2 2

cos 20° + cos 70

2 sin 59 sin 31

°

+ ° =

2k

( )2 2

2 2

sin 70 cos 70

2 sin 59 cos 59

° + °° + °

= 2k

1 22 k

= ⇒ k = 4.

3. (C) sin4 θ + cos4 θ = 1 + 4k sin2 θ cos2 θ⇒ (sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ

= 1 + 4k sin2 θ cos2 θ⇒ 2 sin2 θ cos2 θ (–1 – 2k) = 0

⇒ – 1 – 2k = 0 ⇒ k = – 12

.


4. tan θ = 4⇒ tan2 θ + 1 = 42 + 1

⇒ sec2 θ = 17

∴ 110

(tan2 θ + 2 sec2 θ) =1

10(16 + 2 × 17)

= 5.5. False.

Suppose A = 30° and B = 60°Then, LHS = tan (A + B) = tan (30° + 60°)

= tan 90°⇒ LHS = undefined .... (i)and RHS = tan A + tan B = tan 30°

+ tan 60°

=13

+ 3 = 1 3

3+

= 43

⇒ RHS = a real number .... (ii)From results (i) and (ii), it is clear that thegiven identity is false.

6. –17

Hint: cos 55° = cos (90° – 35°) = sin 35°cos 70° = sin 20°

and tan 5° = cot 85°.

7. 134

.

Hint: sin 30° = 12

= cos 60°, sin 60° = 3

2,

cos 45° = 12

= sin 45°, sin 90° = 1.

8. sin θ + cos θ = a

Square both sides.

sin2 θ + cos2 θ + 2 sin θ cos θ = a2

⇒ 2 sin θ cos θ = a2 – 1

⇒ sin θ cos θ = 2 – 12

a... (i)

Now, sin6 θ + cos6 θ = ( )32 2sin cosθ + θ

– 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)

= 13 – 322 – 1

2a

(1)

[Using equation (i)]

= 1 – ( )223– 1

4a =

2 24 – 3 ( – 1)4a

.

Hence proved.

9. LHS =( )( )

2

2sec tan – 1

sec tan 1

θ + θθ + θ +

=2 2

2 2sec + tan + 2 sec tan – 1sec + tan + 2 sec tan + 1

θ θ θ θθ θ θ θ

=( )

( )2 2

2 2

sec 1 + tan + 2 sec tan

sec + tan + 1 2 sec tan

θ − θ θ θ

θ θ + θ θ

=2 2

2 2

tan + tan + 2 sec tan

sec + sec + 2 sec tan

θ θ θ θθ θ θ θ

= 2 tan (tan + sec )2 sec (sec + tan )

θ θ θθ θ θ

=tan sec

θθ

= tan θ cos θ

=sin cos

θθ

. cos θ = sin θ = RHS.

Hence proved.OR

sin A – sin B cos A – cos Bcos A + cos B sin A + sin B

+

=

( )( )( )( )

( )( )

sin A – sin B sin A + sin Bcos A – cos B cos A cos B

cos A + cos B sin A + sin B+ +

= ( )( )2 2 2 2sin A – sin B cos A – cos B

cos A + cos B sin A + sin B+

=( )

( )( )

2 2 2 2sin A cos A) (sin B cos B

cos A + cos B sin A + sin B

+ − +

= ( )( )1 – 1

cos A + cos B sin A + sin B

= 0 which is an integer.

❑❑


6Chapter

STATISTICS

WORKSHEET–621. (B)

Hint: 3 Median = Mode + 2 Mean.2. (C) Since the mode is 7

∴ 2k – 1 = 7 ⇒ k = 4.3. In such case, mean will increase by 3.

∴ New mean = 18 + 3 = 21.

4. Hint: Mean = i i

i

f xf

ΣΣ

.

5. Class Frequency ( f ) Cumulativeinterval Frequency (cf )

0-8 8 88-16 10 18

16-24 16 3424-32 24 5832-40 15 7340-48 7 80

N = 80

For median class, N 80

=2 2

= 40

In the cumulative frequency column, 58 isjust greater than 40.

So, 24-32 is the median class.

Here, l = 24, cf = 34, f = 24, N2

= 40, h = 8

Using formula:

Median = l +

N–

2 ×cf

hf

= 24 +40 – 34

× 8 = 2624

Hence, median of the given distribution is 26.6. 36.25

Hint: Here maximum class frequency is 32.So, the modal class is 30-40.

Now, l = 30, f1 = 32, f0 = 12, f2 = 20, h = 10Use the formula:

Mode = l + 1 0

1 0 2

–2 – –

f ff f f

× h.

7. Hint:Production No. of Production c.f.

yield farms yield

50-55 2 more than 50 10055-60 8 more than 55 9860-65 12 more than 60 9065-70 24 more than 65 7870-75 38 more than 70 5475-80 16 more than 75 16Total 100 more than 80 0

∴ For more than ogive plot following points.(50, 100), (55, 98), (60, 90), (65, 78), (70, 54),(75, 16), (80, 0).

WORKSHEET–63

1. (B) 2. (B) median.3. The given distribution can be represented as:

Marks obtained No. of students

0-10 510-20 320-30 430-40 340-50 6

More than 50 42

Clearly, the frequency of the class 30-40 is 3.

4. Let us rewrite the given table with cumu-lative frequencies.

Class interval f cf

0-5 10 105-10 15 25

10-15 12 3715-20 20 5720-25 9 66

N = 66

89TATS LTSI SCI

∵ N = 66

∴ N2

= 33

∴ Median class = 10-15Mode class = 15-20

Required sum = 10 + 15 = 25.5. In the given distribution, maximum class

frequency is 20, so the modal class is 40-50.Here, lower limit of modal class: l = 40Frequency of the modal class: f1 = 20Frequency of the class preceding the modalclass: f0 = 12Frequency of the class succeeding the modalclass: f2 = 11Size of class: h = 10Using the formula:

Mode = l + 1 0

1 0 2

–2 – –

f ff f f

× h

= 40 + 20 – 12

× 102 × 20 – 12 – 11

= 40 + 4.70 = 44.70.

Hence, mode of the given data is 45 cars.

6. Let us rewritten the table with class intervals.

Class interval f cf

36-38 0 038-40 3 340-42 2 542-44 4 944-46 5 1446-48 14 2848-50 4 3250-52 3 35

N = 35

We mark the upper class limits on x-axisand with a suitable scale cumulativefrequencies on y-axis with a suitable scale.We plot the points (38, 0); (40, 3); (42, 5);(44, 9); (46, 14); (48, 28); (50, 32) and (52, 35).These points are joined by a free handsmooth curve to obtain a less than type ogiveas shown in the figure.

To obtain median from the graph:

We first locate the point corresponding to

N2

= 352

= 17.5 students on the y-axis. From

this point, draw a line parallel to the x-axisto cut the curve at P. From the point P,draw a perpendicular PQ on the x-axis tomeet it at Q. The x-coordinate of Q is 46.5.Hence, the median is 46.5 kg.

Let us verify this median using the formula.

Median = l +

N2

cf

f

−

× h

= 46 + 17.5 14

14−

× 2

= 46 + 714

= 46 + 0.5

= 46.5 kg.

Thus, the median is the same in bothmethods.

7. In the given data, the class intervals areformed by exclusive method. But we can


convert the series into inclusive form as:

Let the assumed mean be a = 57 and h = 3

Now, using the formula:

Mean = a + i i

i

f uf

ΣΣ

× h

= 57 + 25400

× 3

= 57.19.

WORKSHEET–64

1. (A)

Hint: 5 + 8+ 3+ 2N 9= =

2 2.

2. (D) We haveMode = 3 Median – 2 Mean

⇒ 45 = 3 Median – 2 × 27⇒ Median = 33.

3. (C) mid-points of the classes.

4. Required number of athletes= 2 + 4 + 5 + 71 = 82.

5.

Let us assumed mean be a = 52 and

h = 20

Using the formula:

Mean = a + i i

i

f uh

fΣ

×Σ

⇒ 50 = 52 + 1 2– 2.8 – 1.1 + 0.920

120f f

×

⇒ 1.1 f1 – 0.9 f2 = 9.2 ...(i)

But

68 + f1 + f2 = 120

⇒ f1 + f2 = 52 ...(ii)

Solving (i) and (ii), we obtain

f1 = 28 and

f2 = 24.

6. Let us convert the given data into less thantype distribution.

Class f Lifetimes cfinterval (in hrs.)

0-20 10 less than 20 10

20-40 35 less than 40 45

40-60 52 less than 60 97

60-80 61 less than 80 158

80-100 38 less than 100 196

100-120 29 less than 120 225

We mark the upper class, limits along thex-axis with a suitable scale and the cumu-lative frequencies along the y-axis with asuitable scale. For this, we plot the pointsA(20, 10), B(40, 45), C(60, 97), D(80, 158),E(100, 196) and F(120, 225) on a graph paper.These points are joined by a free handsmooth curve to obtain a less than type ogiveas shown in the following figure.

91TATS LTSI SCI

7. The given distribution can be againrepresented with the cumulative frequenciesas given below:

Class fi xi cf fi xiinterval

100-120 12 110 12 1320

120-140 14 130 26 1820

140-160 8 150 34 1200

160-180 6 170 40 1020

180-200 10 190 50 1900

50 7260

Mean:

Mean = i i

i

f xf

ΣΣ

∵ Σ fi = 50 and Σfixi = 7260

∴ Mean =7260

50= 145.20.

Hence, the mean is Rs. 145.20

Median:

Median = l + N2

cf

f

−

× h

∵ N = 50, 2N

= 25, f = 14, cf = 12,

l = 120 and h = 20

∴ Median = 120 + 25 12

14−

× 20

= 120 + 18.57 = 138.57

Hence, the median is Rs. 138.57.

Mode:

Mode = l + 1 0

1 0 22f f

f f f−

− − × h

∵ l = 120, f1 = 14, f0 = 12

f2 = 8 and h = 20

∴ Mode = 120 + 14 12

2 14 12 8−

× − −× 20

= 120 + 408

= 125

Hence, the mode is Rs. 125.

8. Given that median is 28.5. It lies in the classinterval 20-30, so 20-30 is the median class.

Further N = 60

⇒ x + y + 45 = 60

⇒ x + y = 15 .... (i)

Median = l +

N–

2cf

hf

×


⇒ 28.5 = ( )30 – 5 +20 + ×10

20 x

⇒ x = 8 ...(ii)


x = 8 and y = 7.

WORKSHEET–65

1. (B)Hint:

Class interval Frequency Cumulative(C.I.) ( f ) Frequency

0-10 4 410-20 4 820-30 8 1630-40 10 2640-50 12 3850-60 8 4660-70 4 50

2. (A)Hint:Draw a line parallel to the x-axis at the point

y =402

= 20. This line cuts the curve at a

point. From this point, draw a perpendi-cular to the x-axis. The point of intersectionof this perpendicular with the x-axis deter-mines the median of the data.

3. The given distribution can also be representedas follows:

Class interval Frequency

0-10 3

10-20 9

20-30 15

30-40 30

40-50 18

50-60 5

As the maximum frequency is 30, the modalclass is 30-40.

4.

C.I. fi xi fixi

1-3 9 2 183-5 22 4 885-7 27 6 1627-10 17 8.5 144.5

Σfi = 75 Σfixi = 412.5

Mean = 412.5

5.575

i i

i

f xf

Σ= =

Σ.

5. In the given distribution, the classes are inthe inclusive form. Let us convert it into

exclusive form by subtracting 163 162,

2− i.e.,

0.5 from lower limit and adding the same toupper limit of each class.

Class interval f

159.5-162.5 15

162.5-165.5 118

165.5-168.5 142

168.5-171.5 127

171.5-174.5 18

Here, the maximum frequency is 142.∴ l = 165.5, fl = 142, f0 = 118, f2 = 127, h = 3Now,

mode = l + 1 0

1 0 22f f

f f f−

− − × h

= 165.5 + 142 118

284 118 127−

− − × 3

= 165.5 + 1.85 = 167.35

Hence, the modal height of the students is167.35 cm.

6. The given data may be re-tabulated by thefollowing manner with correspondingcumulative frequencies.

93TATS LTSI SCI

Heights (cm.) No. of Girls CumulativeC.I. ( f ) frequency

(cf )

Below 140 4 4140-145 7 11145-150 18 29150-155 11 40155-160 6 46160-165 5 51

N = 51

Now, N = 51. So, N2

= 25.5.

This observation lies in the class 145-150.Then l = 145, cf = 11, f = 18, h = 5

Now, median = l +

N –

2 ×

cfh

f

= 145 + 25.5 – 11

× 518

= 149.03.Hence, the median height of the girls is149.03 cm.

7. Class Mid- Fre- ui =fiuiinterval values quency

(xi) (fi)

10-30 20 540

220

− = − – 10

30-50 40 820

120

− = − – 8

50-70 60 = A 120

020

= 0

70-90 80 2020

120

= 20

90-110 100 340

220

= 6

110-130 120 260

320

= 6

Σfi = Σfiui = 50 14

Let assumed mean be A = 60h = 20

Mean = A + h × i i

i

f uf

Σ Σ

= 60 + 20 × 1450

= 60 + 5.6 = 65.6Hence, the required arithmetic mean is 65.6.

8. 69.5.Hint: Change the given distribution intoless than type and more than typedistributions. For drawing the ‘less thantype’ ogive, take upper class limits andcorresponding cumulative frequencies; anddrawing the ‘more than type’ ogive takelower class limits and correspondingcumulative frequencies.

ASSESSMENT SHEET– 11

1. (A) Here, a = 25, h = 10.

∴ x = a + h i i

i

f uf

Σ Σ

= 25 + 10 20

100 = 27.

2. (B) Sum of 11 numbers = 11 × 35 = 385

Sum of first 6 numbers = 6 × 32 = 192

Sum of last 6 numbers = 6 × 37 = 222

∴ 6th number = 192 + 222 – 385 = 29.

3. The modal class is 30-40.

h = 10, f1 = 32, f0 = 12, f2 = 20, l = 30.

Mode = l + 1 0

1 0 22f f

f f f−

− − × h

= 30 + 32 12

64 12 20−

− − × 10

= 30 + 6.25

= 36.25.

Aixh−


4. False, the reason is given below:N = 5 + 15 + 30 + 8 + 2 = 60

∴N2

= 30

Since 30 lies in the interval 30-60, so themedian will lie in 30-60.

5. xi fi fixi

3 5 156 2 127 3 214 2 8

p + 3 4 4p + 128 6 48

Σfi = 22 Σfixi = 4p + 116

Mean = i i

i

f xf

ΣΣ

⇒ 6 =4 116

22p +

⇒ 132 = 4p + 116

⇒ 4p = 16 ⇒ p = 4.

6. Class interval Frequency( f )

0-20 420-40 640-60 1860-80 880-100 14

The class corresponding to the maximumfrequency is 40-60. So, 40-60 is the modalclass.

Mode = l + 0

1 0 22if f

f f f−

− − × h

Here, l = 40, f1 = 18, f0 = 6, f2 = 8 and h = 20

∴ Mode = 40 + 18 6

2 18 6 8−

× − −× 20

= 40 + 12 20

22×

= 50.91.

7. We notice classes are continuous. We formcumulative frequency table by less thanmethod.

Marks Number Marks cf Point(C.I.) of stu- less

dents than

0-10 5 10 5 (10, 5)10-20 8 20 13 (20, 13)20-30 10 30 23 (30, 23)30-40 9 40 32 (40, 32)40-50 6 50 38 (50, 38)50-60 7 60 45 (60, 45)

On plotting these points on a graph paperand joining them by a free hand smoothcurve, we get a curve called less than ogive.

8. f1 + f2 = 25 ... (i)

as median is 32 which lies in 30-40

So median class is 30-40.

∴ l = 30; h = 10; f = 30; N = 100

c.f. = 10 + f1 + 25 = f1 + 35.

∴ Median = l +(N/2 – )cf

f× h

⇒ f1 = 9 ∴ from (i) ⇒ f2 = 16.


1. (D) x1 + x2 + ................ + xn = n × x

⇒ 1 2x xk k

+ + .......... + nx nx

k k=

(Dividing through by k)

95TATS LTSI SCI

⇒ 1 2 .........

x x xnk k k

n

+ + +=

xk

(Dividing throught by n)

⇒ Required mean =xk

.

2. (B) The first ten prime numbers are:

2, 3, 4, 7, 11, 13, 17, 19, 23, 29.

Median =11 13

2+

= 242

= 12.

3. Mean = i i

i

f xf

ΣΣ

⇒ 15 =

5 6 10 15 6 2010 25 5

6 6 10 5

k

k

× + × + × +× + ×

+ + + +

⇒445 10

27k

k++

= 15 ⇒ k = 8.

4. False, because the values of these threemeasures depends upon the type of data, soit can be the same.

5. Let us use the assumed mean method tofind the mean of the given data.

Marks No. of Class di = fidi(C.I.) students mark xi – 35( fi ) (xi)

0-10 4 5 –30 –12010-20 6 15 –20 –12020-30 8 25 –10 –8030-40 10 35 0 040-50 12 45 10 12050-60 30 55 20 600

Σfi = 70 Σfidi = 400

Here, assumed mean, a = 35

Now, required mean = a + i i

i

f df

ΣΣ

= 35 + 40070

= 35 + 5.71 = 40.71.

6. Since mode = 36, which lies in the classinterval 30-40, so the modal class is 30-40.∴ f1 = 16, f0 = f, f2 = 12, l = 30 and h = 10.

Now, mode = l + 1 0

1 0 22f f

f f f−

− − × h

⇒ 36 = 30 + 16

32 12f

f−

− − × 10

⇒ 610

= 1620

ff

−−

⇒ 120 – 6f = 160 – 10 f⇒ 4f = 40 ⇒ f = 10.

7. The cumulative frequency table for the givendata is given below:

Marks No. of Cumulative(C.I.) students frequency

( f ) (cf )

0-10 10 1010-20 f1 10 + f120-30 25 35 + f130-40 30 65 + f140-50 f2 65 + f1 + f250-60 10 72 + f1 + f2

N = 75 + f1 + f2

Clearly, N = 75 + f1 + f2But N = 100∴ f1 + f2 = 25 .... (i)

∴N2

= 50.

The median is 32 which lies in the class30-40.So, l = 30, f = 30, cf = 35 + f1, h = 10.Using the formula:

Median = l +

N2

cf

f

−

× h

⇒ 32 = 30 + 150 3530

f− −

× 10

⇒2

10 = 115

30f− ⇒ 75 – 5f1 = 30

⇒ f1 =75 30

5−

⇒ f1 = 9

Substitute f1 = 9 in equation (i) we get9 + f2 = 25 ⇒ f2 = 16

Hence, f1 = 9 and f2 = 16.


8. We prepare the cumulative frequency tableby less than method as given below:

Scores Fre- Score Cumu- Pointquency less lative

than fre-quency

( f ) ( f )

200-250 30 250 30 (250, 30)250-300 15 300 45 (300, 45)300-350 45 350 90 (350, 90)350-400 20 400 110 (400, 110)400-450 25 450 135 (450, 135)450-500 40 500 175 (500, 175)500-550 10 550 185 (550, 185)550-600 15 600 200 (600, 200)

We plot the points given in above table ona graph paper and then joint them by freehand smooth curve to draw the cumulativefrequency curve by less than method.Similarly can be drawn the cumulativefrequency curve by more than method. Weprepare the corresponding frequency table.

Scores Fre- Score Cumula- Pointquency more tive

(f) than fre-quency

(cf)

200-250 30 200 200 (200, 200)250-300 15 250 170 (250, 170)300-350 45 300 155 (300, 155)350-400 20 350 110 (350, 110)400-450 25 400 90 (400, 90)450-500 40 450 65 (450, 65)500-550 10 500 25 (500, 25)550-600 15 550 15 (550, 15)

We plot the points given in this last table onthe same graph and join them by free handsmooth curve to draw the cumulativefrequency curve by more than method.Median: The two curves intersect each othera point. From this point, we draw aperpendicular on the x-axis. The foot of thisperpendicular is P(375, 0). The abscissa ofthe point P, i.e., 375 is the required median.Hence, the median is 375.

CHAPTER TEST1. (C) Let us rewrite the given distribution in

the other manner.

Marks No. of students

0-10 310-20 920-30 1530-40 3040-50 1850-60 5

Clearly, the modal class is 30-40.

2. (A) Let Σfi = N

Σ(fixi – x ) = Σfixi – N x

= N Ni if x

xΣ −

= N ( x – x ) = 0.

97TATS LTSI SCI

3. Hint: First, find the cumulative frequencytable and N = 13 + 10 + 15 + 8 + 11 = 57

∴ N2

= 28.5.

4. Monthly income No. of(in Rs.) families

10000-13000 1513000-16000 1616000-19000 1919000-22000 1722000-25000 18

Hence, required number of families is 19.

5. No, because an ogive is a graphical repre-sentation of a cumulative frequency distri-bution.

6. Yes; as we knowmode = 3 median – 2 mean⇒ 3 median = mode + 2 mean

⇒ Median =13

mode + 23

mean

= mode –23

mode +23

mean

= mode +23

(Mean – Mode).

7. C.I. xi fi ui = fiui

Aixh−

800-820 810 740

220

− = − – 14

820-840 830 1420

120

− = − – 14

840-860 850 190

020

= 0

860-880 870 1520

120

= 15

880-900 890 940

220

= 18

Σfi = 64 Σfiui = 5

Let assumed mean beA = 850h = 20

Mean = A + i iu ffi

Σ Σ

× h

= 850 + 564

× 5

= 850 + 0.391

= 850.391.Hence, the required mean is 850.391.

8. Mode = l + 1 0

1 0 22f f

f f f−

− − × h

Here, l = 30, f1 = 45,

f0 = 30, f2 = 12, h = 10

∴ Mode = 30 + 45 30

90 30 12−

− − × 10

= 30 + 3.125

= 33.125 marks.9. 31.5 marks.

Hint:

Classes No. of Cumulativestudents frequency

0-10 5 510-20 8 1320-30 6 1930-40 10 2940-50 6 3550-60 6 41

Draw the ogive by plotting the points:

(10, 15), (20, 13), (30, 19), (40, 29), (50, 35)

and (60, 41). Here N2

= 20.5. Locate the

point on the ogive whose ordinate is 20.5.The x-coordinate of this point will be themedian.

❑❑

Solutions toSolutions toSolutions toSolutions toSolutions toPPPPPRACTICE PAPERSRACTICE PAPERSRACTICE PAPERSRACTICE PAPERSRACTICE PAPERS

(SUMMATIVE ASSESSMENTS)[FIRST TERM]

99C EP R C TA I P A E RP S

Practice Paper-1

SECTION-A

1. (D) ... 145871250

= ×14587 81250 8

= 11669610000

= 11.6696.

2. (C) ... LCM (x, y) = HCF( , )

x×yx y

= =1800150

12.

3. (B) Let the zeroes of ax3 + bx2 + cx + d beα, β and γ. We are given γ = 0.

∴ αβ + βγ + γα = ca

⇒ αβ + 0 + 0 = ca

αβ = ca .

4. (D) 1

2

aa = 6

2 = 3, 1

2

bb

= –3–1

= 3, 1

2

cc =

109

⇒ i.e., 1

2

ab

= 1

2

bb

≠ 1

2

cc

⇒ The given lines are parallel.

5. (B) ∠D = ∠Q and ∠E = ∠R

⇒ ∆DEF ~ ∆QRP (AA rule of similarity)

⇒ DEPQ

≠ EFRP

.

6. (A) ( ) ( )+ °3 1 3 – cot 30

= ( ) ( )+3 1 3 – 3

= ( )( )+3 3 1 3 – 1 = 2 3 .

7. (C) 9α < 90° ⇒ α < 10°⇒ α is an acute angle.

cos 9α = sin α

⇒ cos 9α = cos –2π α

⇒ 9α = –2π

α

⇒ α = π20

∴ tan 5α = tan π4

= 1.

8. (B) Median.

9. (B) Consider,

3 sin θ = cos θ

⇒θθ

sincos

= 13

⇒ tan θ = 13

⇒ θ = 30°

∴ cos θ = cos 30° = 3

2.


= 2 2

2 2

sin 35°+sin (90° – 35°)cos 35°+cos (90° – 35°)

+ sin2 63°

+ cos 63°sin (90° – 63°)

= 2 2

2 2

(sin 35°+ cos 35°)(cos 35°+ sin 35°)

+ (sin2 63° + cos2 63°)

= 11

+ 1 (∵ sin2 θ + cos2 θ = 1)

= 2.

SECTION-B

11. True, because out of any two consecutivepositive integers, one is even and theother one is odd; and the product of aneven and an odd is even.

12. No, if two zeroes are α and β of polynomialx2 + kx + k, then

α + β = – k and α . β = k⇒ 2α = – k and α2 = k (when α = β)

⇒ α = –2k

and α2 = k.

PRACTICE PAPERS


⇒2

4k

= k (Comparing both)

⇒ k2 = 4k ⇒ k2 – 4k = 0k(– k – 4) = 0 ∴ k = 4, 0.

13. For infinitely many solutions,

1339

= 6k

= +4k

k

⇒1339

= 6k

and

1339

= + 4k

k⇒ k = 2 and 3k = k + 4⇒ k = 2 and k = 2, i.e., k = 2.

14. Yes, because converse of Basic propor-tionality theorem is applicable here as

PAAQ

= PBBR

= 23

.

15. Let the given pole be AB and CD withtheir feet B and D respectively (see figure).

PD = AD = 6 m∴ CP = 11 m – 6 m = 5 mAnd AP = BD = 12 mNow, in right-angled triangle ACP,

AC2 = CP2 + AP2

⇒ AC = 2 25 + 12 = 169 = 13

Hence, distance between tops of the polesis 13 metres.

16. False, because the range of sin θ is [–1, 1]

but a + 1a ≥ 2.

ORSee worksheet-50, sol. 6.

17. xi fi fixi

3 6 185 8 407 15 1059 p 9p

11 8 8813 4 52

Σfi = p + 41 Σfixi = 9p + 303

Mean = ∑∑

i i

i i

f xf x

⇒ 7.5 = ++

9 30341

pp

⇒ 7.5p + 307.5 = 9p + 303⇒ 1.5p = 4.5∴ p = 3.

18. The maximum class frequency is 20.⇒ The modal class is 15-20

Class f cf

0-5 10 105-10 15 25

10-15 12 3715-20 20 5720-25 9 66

N = 66

N2

= 662

= 33

cf just more than 33 is 37.

⇒ Median class is 10-15Now, required sum = 15 + 10 = 25.

SECTION-C

19. On the contrary let us assume that 2 3 –

3 2 is a rational number. Then, we cantake coprime a and b such that


ab

= 2 3 – 3 2

⇒2

2ab

= 12 + 18 – 12 6

(Squaring both the sides)

⇒ 12 6 = 30 – 2

2ab

⇒ 6 = 2 2

230 –

12b a

b.

Since, a and b are integers, therefore, RHSof this last equation is rational and so LHSmust be rational.

But this contradicts the fact that 6 isirrational.This contradiction has arisen due to

incorrect assumption that 2 3 – 3 2 isa rational number.

So, we conclude that 2 3 – 3 2 is anirrational number.

20. Let a be any odd positive integer. Then itis of the form 6m + 1, 6m + 3 or 6m + 5,where m is an integer.Here, 3 cases arise.Case I: When a = 6m + 1,a2 = (6m + 1)2 = 36m2 + 12m + 1

= 12m (3m + 1) + 1= 6q + 1, where q = 2m (3m + 1).

Case II: When a = 6m + 3,a2 = (6m + 3)2 = 36m2 + 36m + 9

= 36m2 + 36m + 6 + 3= 6(6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m + 1.

Case III: When a = 6m + 5,a2 = (6m + 5)2 = 36m2 + 60m + 25

= 36m2 + 60m + 24 + 1= 12(3m2 + 5m + 2) + 1 = 6q + 1, where q = 2(3m2 + 5m + 2).

Hence, a2 is of the form 6q + 1 or 6q + 3.OR

See worksheet-3, sol. 9.

21. Since – 5x is a factor of

f (x) = x3 – 3 5 x2 + 13x – 3 5 ,

so as f (x) may be rewritten.

f(x) = x3 – 3 5 x2 + 13x – 3 5= x3 – 5 x2 – 2 5 x2 + 10x

+ 3x – 3 5

= x2 ( )– 5x – ( )2 5 – 5x x

+ ( )3 5x –

= ( )– 5x ( )2 – 2 5 + 3x x

To find zeroes of f(x), put f(x) = 0.

( )– 5x ( )+2 – 2 5 3x x = 0

⇒ – 5x = 0 or x2 – 2 5 + 3 = 0

⇒ x = 5 or

x = ± × ×2 5 20 – 4 1 3

2

⇒ x = 5 or x = ±2 5 2 22

⇒ x = 5 or

x = 5 + 2 or 5 – 2

Hence all the zeroes of f (x) are 5 ,

5 + 2 and 5 – 2 .

22. The given pair of equations may be re-written as

+2x y

xy=

23

; 2 –x y

xy = 10

–3

i.e,1x

+ 1y

= 43

; – 1x

+ 2y

= 10

–3

Adding this last pair, we get3y

= – 2

⇒ y = 3

–2


Substituting y = 32

– in the first equation

of last pair, we get

1x +

1– 3

2

= 43

⇒ 1x –

23

= 43

⇒1x = 2 ∴ x =

12

Hence, x = 12

, y = 3

–2

is the required

solution.

23. Let the length of each side of the givenequilateral triangle be a, then

AB = BC = CA = a ...(i)

∴ BD = 3a ...(ii)

Draw AP⊥BC to meet BC at P. P will bethe mid-point of BC, that is

BP = 2a

...(iii)

∴ DP = BP – BD =2a

–3a

=6a

...(iv)

[Using (ii) and (iii)]

Now, in right-angled triangle APB,

AP2 = AB2 – BP2

⇒ AP2 = a2 – 2

4a

⇒ AP2 = 23

4a ...(v)

Also, in right-angled triangle APD,

AD2 = AP2 + DP2

⇒ AD2 = 23

4a

+ 2

36a =

2 22736

a a+

[From (iv) and (v)]

⇒ 36AD2 = 28a2 ⇒ 9AD2 = 7a2

⇒ 9AD2 = 7AB2. Hence proved.

24. ∠BAD = 90° – ∠CAD(∵ ∠BAC = 90°)

= 90° – (90° – ∠ACD)(∵ ∠ADC = 90°)

⇒ ∠BAD = ∠ACD ...(i)∠BDA = ∠ADC = 90° ...(ii)

Using equations (i) and (ii) in ∆ABD and∆CAD, we have

∆ABD ~ ∆CAD(AA rule of similarity)

⇒ BDAD

= ADCD

(Corresponding parts)⇒ BD . CD = AD2.


25.θ θθ + θ

cos – sincos sin

= +

1 – 3

1 3

⇒

cos sin–

cos cos 1 – 3sincos 1 3

cos cos

θ θ θ θ

=θθ + +

θ θ

⇒ 1 – tan1 tan

θ+ θ

= +

1 – 3

1 3

⇒ tan θ = 3 ∴ θ = 60°.

26. In ∆ABC,

tan A =1

2 2=

BCAB

∴ AC = 2 2(1) (2 2)+

= 1 8+ = 3

sin A = 13

, cos A = 2 23

,

sin C = 2 23

, cos C = 13

Now, sin A. cos C + cos A. sin C

= 13

×13

+ 2 23

× 2 23

=19

+89

= 1.


27. We will use the step-deviation method.

Let a = 45. Here h = 10

Mean = a + i i

i

f uh

f ∑

× ∑

= 45 + 2985

× 10

= 45 + 3.41

= 48.41.

28. The given data is

C.I. 0-5 5-10 10-15 15-20 20-25 25-30 30-35

f 10 15 30 80 40 20 5

From the table, maximum occuringfrequency is 80. So, modal class is 15-20.

∵ Mode = l + 1 0

1 0 22f f

f f f − − −

× h

Here, l = 15, f1 = 80, f0 = 30, f2 = 40, h = 5

∴ Mode = 15 + − − −

80 30160 30 40

× 5

= 15 + 25090

= 17.78

Hence, modal size is 17.78 hectares.

OR


SECTION-D

29. Let f (x) = 3 2 x2 + 13x + 6 2

= 3 2 x2 + 9x + 4x + 6 2(Spliting middle term)

= 3x( 2 x + 3) + 2 2 ( 2 x + 3)

= (3x + 2 2 ) ( 2 x + 3)To find the zeroes of f (x), we have

3x + 2 2 = 0 or 2 x + 3 = 0

⇒ x = – 2 2

3 or

– 32

∴ Zeroes of the given polynomial are

– 2 23

and – 3

2.

Now, sum of zeroes =– 2 2

3+

– 32

=– 133 2

= – 2

Coefficient ofCoefficient of

xx

Product of zeroes =– 2 2

3 ×

– 32

=6 23 2

= 2

Constant termCoefficient of x

.

Hence proved.

30. Let speed of the train be x km/hr and thatof the bus be y km/hr.

DistanceSpeed

= Time

Case I: According to question, we get

⇒60x

+ 300 – 60

y= 4

⇒60x

+ 240

y= 4

⇒ 15x

+ 60y

= 1 ⇒ 1x

+ 4y

= 1

15...(i)

Marks(C.I.)

No. ofstudents

( fi)

Class-mark

xi

di = xi – 45

ui=

10id fiui

0-10 5 5 – 40 – 4 – 2010-20 4 15 – 30 – 3 – 1220-30 8 25 – 20 – 2 – 1630-40 12 35 – 10 – 1 – 1240-50 16 45 0 0 050-60 15 55 10 1 1560-70 10 65 20 2 2070-80 8 75 30 3 2480-90 5 85 40 4 20

90-100 2 95 50 5 10

Σfi = 85 Σfiui = 29


Case II: According to the given conditions,we get

300 –100100x y

+

= 4 +

1060

⇒100

x +200y =

256

⇒4x +

8y =

16

Dividing by 4, we get

1x +

2y =

124

... (ii)

Use equation (i) – equation (ii),

4y –

2y =

115

–124

2y =

8 – 5120

= 3

120

∴ y = 120×2

3= 80 km/hr

Put y = 80 in equation (i), we get1x +

480

= 115

⇒ 1x =

115

–120

= 4 – 360

= 160

∴ x = 60 km/hr.Hence, speed of the train = 60 km/hrand speed of the bus = 80 km/hr.


31. We are given a triangle ABC in which aline PQ parallel to BC is drawn to intersectthe sides AB and AC at P and Q respectively.We need to prove

APPB

= AQQC

Join PC and QB. Draw QM ⊥ AB andPN ⊥ AC.

Now,area of a triangle

= 12

× base × height

∴ ar(∆APQ) = 12

× AP × MQ ...(i)

Also ar(∆APQ) = 12

× AQ × NP ...(ii)

Comparing equations (i) and (ii), we get12

× AP × MQ = 12

× AQ × NP

⇒ APAQ

= NPMQ

...(iii)

Further, ar(∆BPQ) = 12

× PB × MQ ...(iv)

And ar(∆CQP) = 12

× QC × NP ...(v)

But triangles BPQ and CQP are on thesame base PQ and between the sameparallels PQ and BC, so their areas mustbe equal.

i.e., ar(∆BPQ) = ar(∆CQP)

⇒12

× PB × MQ = 12

× QC × NP

[Using equations (iv) and (v)]

⇒ PBQC

= NPMQ

...(vi)

From, equations (iii) and (vi), we get

APAQ

= PBQC

⇒ APPB

= AQQC

. Hence proved.

32. sin θ + cos θ = 3 (Given)

⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 3(Squaring both the sides)

⇒ 1 + 2 sin θ cos θ = 3⇒ sin θ cos θ = 1 ...(i)

Now, tan θ + cot θ =sin coscos sin

θ θ+

θ θ

= 2 2sin cos

sin cosθ + θ

.θ θ=

11

[Using (i)]

= 1. Hence proved.


33. Let us take LHS of the given identity.2 2

2 2 2tan cosec

tan –1 sec – cosecθ θ

+θ θ θ

=

2

2

2

2

sincos

sin– 1

cos

θθ

θθ

+ 2

2 2

1sin

1 1–

cos sin

θ

θ θ

=

2

2 2

2 2 2 2

2 2 2

sin 1cos sin

sin – cos sin – coscos sin .cos

θθ θ

+θ θ θ θ

θ θ θ

= 2 2

2 2 2 2

sin cos

sin – cos sin – cos

θ θ+

θ θ θ θ

= 2 2

2 2

sin cos

sin – cos

θ + θθ θ

= 2 21

sin – cosθ θ [... sin2 θ + cos2 θ = 1]

= RHS. Hence proved.OR

See worksheet-55, sol. 10.34. Let us prepare the cumulative frequency

table by more than method as given below:

We plot the points mentioning in the tablesuch that lower class limits are on thex-axis and the cumulative frequencies areon the y-axis. By joining these points byfree hand smooth curve, We obtain morethan type ogive as shown in the adjoininggraph.

To obtain median from graph:Draw a line parallel to x-axis passing

through y = 502

= 25. This line meets the

ogive at (68.2, 25).∴ Median = 68.2.

Practice Paper–2

SECTION-A

1. (C) ... 3 3 3

343

2 × 5 ×7=

3 3

343

2 × 5 × 343

= 1

1000= 0.001.

2. (B) ... p(x) = 3x2 – (2x + 1) x + 3= x2 – x + 3

∴ (α + β) – αβ = ––11

–31

= 1 – 3 = – 2.

3. (A) As( ABC)( DEF)∆∆

arar

= 2

2

BC

EF(Result on areas of similar triangles)

⇒54

( DEF)∆ar=

2

2

3

4 ⇒ ar(∆DEF) = 96 cm2.

Productionyield

(in kg/ha)fi

Productionyield

(in kg/ha)more thanor equal to

50-55 2 50 50 (50, 50)55-60 6 55 48 (55, 48)60-65 8 60 42 (60, 42)65-70 14 65 34 (65, 34)70-75 15 70 20 (70, 20)75-80 5 75 5 (75, 5)

Total 50

Pointcf

D

E F

A

B C


4. (C) As cosec2 θ – cot2 θ = 1⇒ 16 – 3k2 = 1

⇒ k2 = 153

⇒ k = 5 .

5. (C)o o

o

cos (90 – 70 )

sin 70+

2 coscos

θθ

=2k

⇒o

o

sin 70sin 70

+ 2 =2k

⇒ 3 =2k

⇒ k = 6.

6. (D) ∵ tan 45° = 1

∴ 2

2

1 – tan 45°

1 + tan 45°=

1 – 11 + 1

= 0

= tan 0°.

7. (B) Let us prepare the cumulative fre-quency table from the given data.

Class Frequency Cumulativeinterval frequency

10-15 4 415-20 7 1120-25 20 3125-30 8 3930-35 1 40

N = 40

The cumulative frequency just greater

thanN2

= 20 is 31 and its corresponding

class is 20-25. Hence 20-25 is the medianclass.

8. (D) 0.

9. (D) The condition for the line parallel is:

∵ 1

2

aa

= 1

2

bb

= ≠ 1

2

cc

⇒ 32

= 25c ≠ 2

1

⇒ c = 154

.

10. (B) As sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – sin {90° – (45° – θ)

[∵ sin (90° – α) = cos α]

= sin (45° + θ) – sin (45°+ θ) = 0.

SECTION-B

11. False, because 3 or 2 3 is an irrationaland sum or difference of a rationalnumber and an irrational number is anirrational number.

12. No, since the discriminant is zero for

k = ±12

.

13. The condition for infinitely many solutions

is 1

2

aa = 1

2

bb

= 1

2

cc

.

⇒ 35

= – ( + 1)1 – 2

aa

=2 – 1

3b

b

⇒ 35

=– ( + 1)1 – 2

aa

and 35

=2 – 1

3b

b

⇒ 3 – 6a = – 5a – 5 and 9b = 10b – 5

⇒ a = 8 and b = 5.

14. Yes, because ∆PBCand ∆PDE aresimilar by SAS rule.

asBPDP

=PCPE

=12

and

∠BPC = ∠DPE.

15. In the given ∆ABC, DE || ABSo by the Thales Theorem,

⇒ CDAD

= CEBE

⇒ + 33 +19xx

=3 + 4

xx

⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12⇒ 6x = 12⇒ x = 2.


16. We have the identity:sin2 θ + cos2 θ = 1

⇒2

2

a

b+ cos2 θ = 1 (∵ sin θ = a

b)

⇒ cos θ = 2

21 –

a

b=

2 2–b ab

.

17. The empirical relationship among the threemeasures of central tendency is:Mode = 3 Median – 2 Mean

= 3 × 55 – 2 × 50= 165 – 100 = 65.

ORSee Assessment sheet-11, sol. 4.

18. Let us convert the given less than typedistribution to normal distribution.


0-10 310-20 12 – 3 = 920-30 27 – 12 =1530-40 57 – 27 =3040-50 75 – 57 =1850-60 80 – 75 = 5

From the table, the maximum frequencyis 30 and its corresponding class is 30-40.Therefore, the modal class is 30-40.

SECTION-C

19. To prove that the pair of numbers(847, 2160) is coprime by using Euclid’salgorithm, we have to prove that thehighest common factor of the pair is 1.Since 2160 > 847∴ 2160 = 847 × 2 + 466Since the remainder 466 ≠ 0.∴ 847 = 466 × 1 + 381Since the new remainder 381 ≠ 0.∴ 466 = 381 × 1 + 85Since the new remainder 85 ≠ 0.∴ 381 = 85 × 4 + 41Since the new remainder 41 ≠ 0.∴ 85 = 41 × 2 + 3Since the new remainder 3 ≠ 0.∴ 41 = 3 × 13 + 2Since the new remainder 2 ≠ 0.

∴ 3 = 2 × 1 + 1Since the new remainder 1 ≠ 0.∴ 2 = 1 × 2 + 0Since, the remainder has now become zero,the divisor at this stage is 1, the HCF of847 and 2160 is 1.


20. We have time =distance

speedTime taken by Abhay to cover onecomplete round

=36012

= 30 hours

Time taken by Ravi to cover one completeround

= 36015

= 24 hours

Abhay and Ravi reach the starting pointrespectively after 30 hours and 24 hours,and their respective multiples. Therefore,they will meet again at the starting pointafter the time given by least commonmultiple of 30 hours and 24 hours Let usdetermine the LCM of 30 hours and 24hours.

30 = 2 × 3 × 5,24 = 23 × 3

⇒ LCM = 23 × 3 × 5 = 120Hence, the required time is 120 hours.

21. Let α and β be the zeroes of 6x2 + x + k.∴ α2 + β2 =(α + β)2 – 2αβ

= 21

– – 26 6

k

= 1

–36 3

k

But it is given that α2 + β2 = 2536

∴ 2536

= 136

–3k ⇒

3k

= – 2436

⇒ k = – 2.OR



22. To solve a system of equations graphicallywe need atleast two solutions of eachequation.Two solutions of the equation 2x – y = 2are given in the following table:

x 0 3

y – 2 4

Two solutions of the equation 4x – y = 8are given in the following table:

x 2 1

y 0 – 4

Let us draw the graph of the two givenequations.

From the graph, it is clear that the twolines intersect each other at the point (3, 4).Hence, the solution is x = 3, y = 4.

23. In right-angled triangle PQS,PS2 = PQ2 + QS2

⇒ QS2 = PS2 – PQ2

⇒2QR

4= PS2 – PQ2

⇒ QR2 = 4PS2 – 4PQ2 ... (i)In right-angled triangle PQR,

PR2 = PQ2 + QR2

= PQ2 + 4PS2 – 4PQ2 [Using (i)]⇒ PR2 = 4PS2 – 3PQ2. Hence proved.

24. Draw RM ⊥ PQ.In ∆PRS, ∠PSR > 90°

⇒ PR2 = PS2 + RS2 + 2PS.SM ...(i)[Using result on obtuse-angled triangle]

In ∆QRS, ∠QSR < 90°

⇒ QR2 = RS2 + SQ2 – 2SQ.SM ...(ii)

[Using result on acute-angled triangle]

Add equations (i) and (ii) to get

PR2 + QR2 = 4PS2 – 2SM(PS – SQ)

= 42PQ

2

(∵ PS = SQ)

⇒ PR2 + QR2 = PQ2. Hence proved.


25. sin θ = 34

... (i) Given

∵ sin2 θ + cos2 θ = 1∴ cos2 θ = 1 – sin2 θ

⇒ cos2 θ = 1 –23

4

[Using (i)]

⇒ cos2 θ = 716

⇒ sec2 θ = 167

... (ii)

Let us take LHS of the given equation.

LHS =2 2

2

cosec – cotsec – 1

θ θθ

= 2

1

sec – 1θ

[... cosec2 θ – cot2 θ = 1]

=1

16– 1

7

=79

= 73

[Using (ii)]



26. We have sin 60° = 32

, cos 60° = 12

,

sec 30° =2

3, cosec 30° = 2

∴ 2 2

2 2

3 + sin 60° + cosec 30°

5 + cos 60° + sec 30°

=

( )2

2

22

33 + + 22

1 25 + +2 3

=

33 + + 4

41 4

5 + +4 3

Multiply Num. and Deno. by 12,

=36 + 9 + 4860 + 3 + 16

=9379

.

27. Let us prepare the cumulative frequencytable from the given data.

C.I. fi xi di=xi–A fi di

0-10 7 5 – 20 – 14010-20 10 15 – 10 – 10020-30 15 25 0 030-40 8 35 10 8040-50 10 45 20 200

Σfi = 50 Σfidi= 40

Mean = A +∑∑

i i

i

f df

= 25 + 4050

= 25 + 0.8 = 25.8.

28. Let us convert the more than typedistribution to the normal distribution.


0-20 320-40 740-60 2060-80 1580-100 5

We observe from the table that the value20 is the maximum frequency. So, themodal class is 40-60.

Now, mode = l + 1 0

0 2×

2 i

f – fh

f f f – –

Here,l = 40, fi = 20, f0 = 7, f2 = 15, h = 20

∴ Mode = 40 +20 – 7

× 2040 – 7 – 15

= 40 + 26018

= 40 + 14.44

= 54.44 marks.

SECTION-D

29. Let f (x) = x4 + x3 – 9x2 – 3x + 18

It is given that – 3 and 3 are two ofzeroes of f (x).

⇒ x + 3 and x – 3 are the factors of f(x)

⇒ (x + 3 ) (x – 3 ) is a factor of f(x)

⇒ x2 – 3 is a factor of f (x)

� �x2 – 3 x x x4 2+ – 9 – 3 + 18x3 x2 + – 6xx x4 2– 3

– +

– 6 – 3 + 18x x2x3

– 3xx3

– +

– 6 + 18x2

+ –– 6 + 18x2

0

∴ p(x) = (x2 – 3) (x2 + x – 6)To find other zeroes of p(x)Let x2 + x – 6 = 0⇒ (x+ 3) (x – 2) = 0⇒ x = – 3 and 2

So all zeroes are ± 3 , 2, – 3.Hence, the zeroes of f(x) are ± 3 , 2, – 3.

30. Let the speed of rowing in still waterand the speed of the current be u km/hrand v km/hr respectively.The speed of rowing in downstream

= (u + v) km/hr.The speed of rowing upstream

= (u – v) km/hr.


Using the formula:

DistanceSpeed

= Time

According to first condition of the question,

18+u v

+12–u v

= 3 ... (i)

According to second condition of thequestion,

36+u v

+40–u v

= 8 ... (ii)

Let us put,

1+u v

= x and 1–u v

= y such that

equations (i) and (ii) reduce to

18x + 12y = 3 ... (iii)And 36x + 40y =8 ... (iv)Equations (iii) and (iv) form a pair oflinear equations.Multiply equation (iii) by 2 and subtractthe result from equation (iv) to get

16y = 2 ⇒ y = 18

Substitute y = 18

in equation (iv) to get

36x = 8 – 5 ⇒ x = 1

12

∵ x = 1+u v

∴ u + v = 12

∵ y = 1–u v

∴ u – v = 8

This last system gives u = 10 and v = 2Hence, the speed of the rowing in stillwater = 10 km/hr and the speed of thecurrent = 2 km/hr.

31. We are given two triangles ABC and PQRsuch that ∆ABC ~ ∆PQR.

∴ ABPQ

=BCQR

=CARP

...(i)

We have to prove

( )( )

ABC

PQR

∆∆

ar

ar=

2

2

AB

PQ=

2

2

BC

QR=

2

2

CA

RP

Draw AM ⊥ BC and PN ⊥ QRIn ∆ABM and ∆PQN,∠B = ∠Q (∵ ∆ABC ~ ∆PQR)∠M = ∠N (Each 90°)So, ∆ABM ~ ∆PQN

(AA similarity criterion)

∴ ABPQ

=AMPN

⇒ BCQR

=AMPN

...(ii) [Using (i)]

Since area of a

triangle =1

× base × height2

∴( )( )

ABC

PQR

∆∆

ar

ar=

1×BC× AM

21

× QR × PN2

=BCQR

×AMPN

Using equation (ii), we get

( )( )

ABC

PQR

∆∆

ar

ar =

2

2

BC

QR... (iii)

From equations (i) and (iii), we obtain

( )( )

ABC

PQR

∆∆

ar

ar=

2

2

ABPQ

=2

2

BCQR

=2

2

CA

RP.

Hence proved.

32. Consider, left hand side of the given equation,

tan cot+

1 – cot 1 – tanθ θ

θ θ=

1tan tan

+1 1 – tan1 –

tan

θ θθ

θ

=( )

2tan 1+

tan – 1 tan 1 – tanθ

θ θ θ

=( )

3tan – 1tan tan – 1θ θ


=( ) ( )

( )

2tan – 1 tan + 1 + tan

tan tan – 1

θ θ θ

θ θ

[∵ a3 – b3 = (a – b) (a2 + b2 + ab]

= 1

tan + +1tan

θθ

= sin cos

+ +1cos sin

θ θθ θ

= 2 2sin + cos

+ 1sin cos

θ θθ θ

= sec θ. cosec θ + 1

= RHS. Hence proved.OR

See Assessment sheet-9, sol. 8.

33. m = cosec θ – sin θ (Given)

⇒ m =1

sin θ– sin θ =

21 – sinsin

θθ

⇒ m =2cos

sinθ

θ... (i)

(∵ 1 – sin2 θ = cos2 θ)

⇒ m2 =4

2

cos

sin

θθ

... (ii) (Squaring)

Further, n = sec θ – cos θ (Given)

⇒ n =1

– coscos

θθ

⇒ n =21 – cos

cosθ

θ

⇒ n =2sin

cosθθ

... (iii)

⇒ n2 =4

2

sin

cos

θθ

... (iv)

Multiplying equations (ii) and (iii), weget

m2n = 4 2

2

cos sin×

cossinθ θ

θθ= cos3 θ

∴ (m2n)2/3 = cos2 θ ... (v)Multiplying equations (i) and (iv), we get

n2m =4

2

sin

cos

θθ

×2cos

sinθ

θ = sin3 θ

∴ (n2m)2/3 = sin2 θ ... (vi)

Adding equations (v) and (vi), we get(m2n)2/3 + (n2m)2/3 = cos2 θ + sin2 θi.e., (m2n)2/3 + (n2m)2/3 = 1.

Hence proved.

34. Let us prepare the cumulative frequencydistribution from the given distribution.

Class Frequency Cumulativeinterval frequency

(C.I.) ( f ) (c f )

0-10 5 510-20 x 5 + x20-30 20 25 + x30-40 14 39 + x40-50 y 39 + x + y50-60 8 47 + x + y

N = 47 + x + y

Since 27 is the median, therefore, 20-30is the median class.∴ l = 20, cf = 5 + x, f = 20, h = 10Given: N = 68Using the formula:

Median = l +

N–

2 ×cf

hf

⇒ 27 = 20 +

68– 5 –

2 ×1020

x

⇒ 27 = 20 + 29 –

2x

⇒ 29 – x = (27 – 20) × 2 ⇒ x = 15

From the table,N = 47 + x + y But N = 68 (Given)∴ x + y + 47 = 68⇒ 15 + y + 47 = 68 (Substituting x = 15)⇒ y = 6Thus, x = 15, y = 6.



Practice Paper–3

SECTION-A

1. (D) There are infinitely many realnumbers of both types rational and

irrational between 3 and 5 .2. (D) Decimal representation of an irrational

number is always non-terminating, non-repeating.

3. (B) Product of zeroes = 6a

= 4

⇒ a = 64

= 32

.

4. (B) For infinite number of solutions:3k

=5

10, i.e., k = 6.

5. (A)∆ABC ~ ∆DEF

∠B = ∠E =180° – (40° + 65°) = 75°.

6. (C) AC2 = BC2 – AB2

⇒ AC2 = 2 – 1 = 1AC = 1

cosec C = BCAB

=2

1= 2 .

7. (C)sin

1 cosθ

+ θ = sin 1 – cos

×1 + cos 1 – cos

θ θθ θ

= ( )

2

sin 1 – cos

sin

θ θθ

=1 – cos

sinθ

θ.

8. (D) Using empirical formula,2 Mean = 3 Median – Mode

⇒ 2 Mean = 3 × 500 – 400

⇒ Mean =1100

2 = 550.

9. (A) A + B + C = 180°(Angle sum property of a triangle)

Now, tan B C

2+

=

180 – Atan

2°

=A

tan 90 –2

° =

Acot

2.


= cos (40° + θ) – cos {90°– (50°– θ)}

+ 2 2

2 2

cos 40 cos (90 – 40 )

sin 40 sin (90 – 40 )

° + ° °° + ° °

= cos (40°+ θ) – cos (40°+θ)

+ 2 2

2 2

cos 40 sin 40

sin 40 cos 40

° + °° + °

= 0 + 11

= 1.

SECTION-B

11. 8n can be rewritten as 23n. Clearly, theprime factor of 8n is only 2. To end withthe digit 0, one of the prime factors of 8n

must be 5. Hence, 8n cannot end with thedigit zero for any n∈N.

12. True, because we find the remainder zerowhen 3x4 + 5x3 – 7x2 + 2x + 2 is dividedby x2 + 3x + 1.

13. Infinite number of solutions because thesystem obeys the following condition:

1

2

aa

= 1

2

bb

= 1

2

cc

, i.e., 13

=– 3– 9

=– 3– 9

.

14. Yes, because converse of Basic Propor-tionality Theorem holds as

PMMQ

=PNNR

=32

.

15. In ∆AOD and ∆COB,

AOOC

=DOOB

=12

and∠AOD = ∠COB⇒ ∆AOD ~ ∆COB

∴ADBC

= 12

⇒ 4

BC=

12

⇒ BC = 8 cm.


16. No.

For θ = 30°, tan θ =13

and cot θ = 3

tan2 θ + cot2 θ =13

+ 3 =103

≠ 2.

17. Since the maximum frequency is 8, somodal class is 4-8.

∴ l = 4, f1 = 8, f0 = 4, f2 = 5, h = 4

Now, mode = l + 1 0

1 0 2

–×

2 – –

f fh

f f f

= 4 + 8 – 4

× 416 – 4 – 5

= 4 + 2.29 = 6.29.


18. Let us prepare cumulative frequency table:

C.I. f cf

60-70 2 270-80 5 780-90 12 1990-100 31 50

100-110 39 89110-120 10 99120-130 4 103

N = 103

N2

=1032

= 51.5

Cumulative frequency just greater than51.5 is 89. So, median class is 100-110.

SECTION-B

19. Let us assume on the contrary that 2 is

a rational number. Then 2 can be written

as 2 = ab

, where a and b are coprime

and b ≠ 0.

2 = 2

2ab

(Squaring)

⇒ a2 = 2b2 ... (i)

⇒ a2 is divisible by 2 ...(ii)⇒ a is divisible by 2 ...(iii)[If a prime (here 2) divides d2, then thesame prime divides d, where d is a positiveinteger.]⇒ a = 2c⇒ a2 = 4c2 ... (iv)

From (i) and (iv), we get4c2 = 2b2 ⇒ b2 = 2c2

⇒ b2 is divisible by 2⇒ b is divisible by 2 .... (v)

From results (ii) and (v), we have a and bboth are divisible by 2.But this contradict the fact that a and b arecoprime. This contradiction has arisenbecause of our incorrect assumption that

2 is a rational number. Thus, we conclude

that 2 is an irrational number.


20. Let a be a positive integer. Then it is ofthe form 3q, 3q + 1 or 3q + 2.Now, three cases arise.

Case I: a = 3q

∴ a3 = (3q)3 = 27q3 = 9 (3q3) = 9m,

where m = 3q3.

Case II: a = 3q + 1

∴ a3 = (3q + 1)3

= 27q3 + 1 + 27q2 + 9q

= 9 (3q3 + 3q2 + q) + 1

= 9m + 1,

where m = 3q3 + 3q2 + q.

Case III: a = 3q + 2

∴ a3 = (3q + 2)3

= 27q3 + 8 + 54q2 + 36q

= 9 (3q3 + 6q2 + 4) + 8= 9m + 8,

where m = 3q3 + 6q2 + 4.Hence, a3 is of the form 9m, 9m + 1 or9m + 8.


21. p(t) = t2 – 15To obtain zeroes of p(t), put p(t) = 0i.e., t2 – 15 = 0

t2 – ( )215 = 0

⇒ ( ) ( )15 – 15+t t = 0

⇒ t = – 15 , 15

So, zeroes of p(t) are – 15 and 15

Sum of zeroes = – 15 + 15 = 0

=– 01

= 2

– Coefficient of

Coefficient of

t

tProduct of zeroes

= – 15 15× = – 15 = –151

=Constant termCoefficient of t

.

Hence verified.22. The given system of equations can be re-

written as 4x + 3y – 48 = 0

40x – 6y – 192 = 0Applying the method of cross multipli-cation to solve the system.

⇒– 576 – 288

x =

–– 768 + 1920

y=

1– 24 – 120

⇒– 864

x =

– 1152y

= 1

– 144

⇒ x = 864144

and y = 1152144

⇒ x = 6 and y = 8.

23. We are given a square of side length a.Then length of its diagonal will be a 2 .We know that area of an equilateral

triangle of side length x is 234

x .

∴ Area of the equilateral triangle describedof the side of the square.

Aside =23

4a ... (i) ⇒ a2 =

43

Aside

And area of the equilateral triangledescribed on the diagonal of the square

Adiagonal =3

4(a 2 )2 = 23

2a

⇒ Adiagonal =3

2×

side

4A

3[Using (i)]

⇒ Aside = diagonal1

× A2

. Hence proved.

24. In the figure drawn,AB || DC and ∆AED~ ∆BEC.∆ADC and ∆BDCboth are on the same base DC and liebetween same parallels AB and DC.So,

ar(∆ADC) = ar(∆BDC)⇒ ar(∆AED) + ar(∆DEC)

= ar(∆BEC) + ar(∆DEC)⇒ ar(∆AED) = ar(∆BEC) (i)Now,

2

2( AED) (AD)

=( BEC) (BC)

arar

∆∆

(∵ ∆AED ~ ∆BEC)

⇒ 1 = ( )( )

2

2

AD

BC[From (i)]

⇒ AD = BC. Hence proved.

25. Given expression

=4 41 1

22 2

+ – ( ) ( ){ }2 23 1+ + 3

223

= 2 1 1

16 16 +

– ( )3 1+ + 3 × 43

= 14

– 4 + 4 = 14

.

26. Consider left hand side of the givenequation.

LHS = (cosec A – sin A) (sec A – cos A)

=1

– sin Asin A

1– cos A

cos A

=2 21 – sin A 1 – cos A

.sin A cos A


=2 2cos A sin A

.sin A cos A

= sin A cos A

Also, taking right hand side,

RHS =1

tan A cot A+ =1

sin A cosAcosA sin A

+

= 2 2

sin A cos A

sin A cos A+= sin A cos A

Hence, LHS = RHS.

ORSee worksheet-51, solution-9.

27. Let us use step-deviation method toobtain the mean.

C.I. fi xi di= xi–a ui=id

hdiui

0-20 17 10 – 40 – 2 – 3420-40 p 30 – 20 – 1 – p40-60 32 50 0 0 060-80 24 70 20 1 2480-100 19 90 40 2 38

Σfi = 92+p Σfiui=28 – p

Here a = 50, h = 20

Mean = a + × ∑

∑ i i

i

f uh

f

⇒ 50 = 50 + 28 –92 +

pp

⇒ 28 – p = 0 ∴ p = 28.

28. We prepare cumulative frequency tablefrom the given data.

C.I. Frequency Cumulative( f ) Frequency

(cf )

0-8 8 88-16 10 18

16-24 16 3424-32 24 5832-40 15 7340-48 7 80

N = 80

Here, N = 80 ∴ N2

= 40

Cumulative frequency just more than 40is 58. So 24-32 is the median class.∴ l = 24, cf = 34, f = 24, h = 8

∴ Median = l +

N–

2 ×

cfh

f

= 24 +40 – 34

24

× 8 = 24 + 4824

= 26.

ORSee Assessment sheet-12, sol. 6.

SECTION-D

29. Let f (x) = 3x4 + 6x3 – 2x2 – 10x – 5

x = 53

and x = – 53

are the zeroes of f (x)

⇒5

–3

x and 53

+

x are factors of f (x)

⇒5

–3

x53

+

x = x2 –

53

is a factor of f (x)

Let us divide f (x) by x2 –53

To obtain other two zeroes,put the quotient = 0

i.e., 3x2 + 6x + 3 = 0


⇒ 3 (x2 + 2x + 1) = 0⇒ 3 (x + 1)2 = 0⇒ x = – 1 or – 1

Hence, all the zeroes of f(x) are ± 53

,

– 1, –1.

30. Let the original fraction be xy .

On adding 1 to both the numerator and

the denominator of xy , it becomes 4

5

i.e., + 1+ 1

xy

=45

, i.e., 5x + 5 = 4y + 4

i.e., 5x – 4y = –1 ...(i)

On subtracting 5 from both the numerator

and the denominator of xy , it becomes

12

i.e., – 5– 5

xy

=12

, i.e., 2x – 10 = y – 5

i.e., 8x – 4y = 20 ...(ii)

Subtracting equation (i) from equation(ii), we get

3x = 21 ⇒ x = 7Substituting x = 7 in equation (i), we get

5 × 7 – 4y = –1 ⇒ y = 9

Hence, the required fraction is79

.

ORSee worksheet-27, sol. 9 (OR).

31. Pythagoras Theorem:In a right triangle, the square of thehypotenuse is equal to the sum of thesquares of the other two sides.Proof: We are given, a ∆ABC in which∠A = 90°We need to proveBC2 = AB2 + AC2.Draw AD ⊥ BC(see figure)In ∆ABC and ∆DBA,

∠ABC = ∠DBA (Common)∠BAC = ∠BDA (Each 90°)

So, ∆ABC ~ ∆DBA(A A criterion of similarity)

Therefore,ABBD

=BCAB

or AB2 = BD . BC ...(i)

Similarly, ∆ABC ~ ∆DAC

Therefore, ACDC

=BCAC

AB2 = DC . BC ...(ii)

Adding equations (i) and (ii), we get

BD . BC + DC . BC = AB2 + AC2

or (BD + DC) . BC = AB2 + AC2

or BC . BC = AB2 + AC2

(∵ BC = BD + DC)

or BC2 = AB2 + AC2.

Hence proved.OR


32. We havex3 = sec A – cos A

=1

– cos Acos A

= 21 – cos A

cos A

∴ x =1

2 3sin Acos A

Similarly, y3 = cosec A – sin A

=21 – sin A1

– sin Asin A sin A

=

y =

12 3cos A

sin A

LHS = x2 y2 (x2 + y2)

= x4y2 + x2y4

=

4 21 12 23 3sin A cos A

cos A sin A


or x + 1

4x – x +

14x

= 2x or 1

2x.

Hence proved.

34. We prepare the cumulative frequencytable by less than method as given below:

Marks Frequ- Marks cf Pointency less

than

0-10 4 10 04 (10, 4)

10-20 10 20 14 (20, 14)

20-30 16 30 30 (30, 30)

30-40 22 40 52 (40, 52)

40-50 20 50 72 (50, 72)

50-60 18 60 90 (60, 90)

60-70 8 70 98 (70, 98)

70-80 2 80 100 (80, 100)

We take marks on the x-axis andcumulative frequency on the y-axis andthen plot the points mentioned in thetable. On joining these points by freehand smooth curve, we get less thanogive.

Further, we prepare the cumulativefrequency table by more than method asgiven below:

Marks Frequ- Marks cf Pointency more

thanor

equalto

0-10 4 0 100 (0, 100)

10-20 10 10 96 (10, 96)

20-30 16 20 86 (20, 86)

30-40 22 30 70 (30, 70)

40-50 20 40 48 (40, 48)

50-60 18 50 28 (50, 28)

60-70 8 60 10 (60, 10)70-80 2 70 2 (70, 2)

+

2 41 12 23 3sin A cos A

cos A sin A

=

4 22 23 3sin A cos A

cos A sin A

+

2 42 23 3sin A cos A

cos A sin A

=

18 4 3

4 2

sin A cos A

cos A sin A

×

+

14 8 3

2 4

sin A cos Acos A sin A

×

= ( ) ( )1 1

6 63 3sin A cos A+

= sin2A + cos2 A = 1 = RHS.Hence proved.

33. sec θ = x +1

4x... (i) (Given)

or sec2 θ =21

4 +

xx

(Squaring)

or 1 + tan2 θ = x2 + 21 1

216+

x

or tan2 θ = x2 + 21 1

–216x

or tan2 θ =21

–4

xx

or tan θ = ±1

–4

xx

... (ii)

Add equations (i) and (ii) to get

sec θ + tan θ = x + 1

4x + x –

14x


We will plot the points mentioned in thistable on the same graph. On joining thesepoints by free hand smooth curve, weget more than ogive.

105

100

95

90

85

80

75

70

65

60

55

50

45

40

35

30

25

20

15

10

5

010 20 30 40 50 60 70 80 90 100

(0, 100)

(10, 96)

(20, 86)

(30, 70)

(40, 48)

(50, 28)

(60, 10)

(70, 2)

more than ogive

less than ogive

(80, 100)(70, 98)

(60, 90)

(50, 72)

(40, 52)

(30, 30)

(20, 14)

(10, 4)(39, 0)

Y

X

Median: The abscissa of the point ofintersection of both the ogivesdetermines the median of the givendistribution. To find such abscissa, wedraw a perpendicular from the pointof intersection of both the ogives tothe x-axis, which meet the axis at (39, 0).Hence the required median is 39marks.

Practice Paper–4

SECTION-A

1. (D)2 32 3

−+

= 2 3 2 32 3 2 3

− −×+ −

=2 3 2 6

2 3+ −

−

= 2 6 5−

As 2 6 5− is irrational, it has non-termi-nating, non-repeating decimal form.

2. (D) As p2 and p3 are common factors.

∴ HCF (x, y) is 2 3n lp p .

3. (B) p(x) = (x – 1)2 + 2x + 1 = x2 + 2

Sum of zeroes = 01

= 0;

product of zeroes = 21

= 2.

4. (C) 2 + k = 1 and p + 3 = 2, i.e., k = – 1 andp = – 1. So, p + k = – 2.

5. (C) ADBD

= AECE

= 13

⇒ DE || BC

⇒ ∆ADE ∼ ∆ABC

∴ DE ADBC AB

=

⇒7.5x

= 3.514

⇒ x = 158

cm.


6. (D) If x = 30°,3 cos 30° – 4 cos3 30°

= 3 ×3

2 – 4 ×

3 38

= 0.

7. (A) The value of cosec θ is minimum at anangle where sin θ is maximum as sin θ ismaximum if θ = 90°.∴ sin 90° = 1So minimum value of cosec θ = 1.

8. (B) 52 is the median as the abscissa of thepoint of intersection of the two ogivesdetermines the median of the data.

9. (A) We know that sin (90° – θ) = cos θ,cos (90° – θ) = sin θ, tan (90° – θ) = cot θ∴ Given expression

=sin cos tansin cos tan

θ θ θθ θ θ

+cotcot

θθ

= 1 + 1 = 2.

10. (B) tan 2θ = cot (θ + 12°)⇒ tan 2θ = tan {90° – (θ + 12°)}

[∵ tan (90° – α) = cot α]⇒ tan 2θ = tan (78° – θ)⇒ 2θ = 78° – θ⇒ θ = 26°.

SECTION-B

11. The required highest number will be theHCF of 120, 224 and 256.120 = 23 × 3 × 5; 224 = 25 × 7; 256 = 28

Therefore, HCF = 23 = 8.12. Let the required polynomial be f(x).

Then f (x) = k[x2 – (sum of zeroes) x +product of zeroes]

= k[x2 – (–5 + 2)x + (–5)(2)]= k(x2 + 3x – 10),

k being a real number.f(x) is not a unique polynomial as k isany real number.

13. Condition for infinite number of solutions:

1

2

aa = 1

2

bb

= 1

2

cc

i.e.,2 3

4k + = 2 1 4( 1)

=3 3

k k+ − ,

i.e., k =52

.

14. Yes.

In ∆ ABC,

ADDB

=AE 1EC 3

=

So, by converse of

Thales Theorem, DE || BC.

15. No.In ∆AOB and ∆DOC,∠AOB = ∠DOC(Vertically opposite angles)

AO OBDO OC

≠

as 5 63 10

≠ , i.e., 5 33 5

≠

Therefore, ∆AOB is not similar to ∆DOC.

16. Yes.Let us take left hand side of the givenequation.

LHS =2

2

1 + sin

cos

θθ

= 2

1cos θ

+2

2

sin

cos

θθ

= sec2 θ + tan2 θ= (1 + tan θ)2 + tan2 θ = 1 + 2 tan2 θ= RHS.


17. Let us convert the given more than typedistribution to the normal distribution.


0-20 320-40 740-60 2060-80 15

80-100 5


The maximum valuable frequency is 20.So, modal class is 40-60.

∴ l = 40, f1 = 20, f0 = 7, f2 = 15, h = 20

Now, mode = l + 1 0

1 0 22f f

f f f−

− − × h

= 40 + 20 740 7 15

− − −

× 20.

= 40 + 14.44

= 54.44 marks.

18. As N = 50

∴ N2

= 25

∴ Draw a line parallel to x-axis from y = 25which meet to ogive at points P(40, 25).

∴ Abscissa of P is 40.

⇒ Median = 40.

SECTION-C

19. Let us assume to the contrary that 3 4 5−is rational. Then we can take integers aand b ≠ 0 such that

ab

= 3 – 4 5 ,

i.e., 4 5 = 3 – ab

i.e., 5 =3

4b a

b−

RHS of this last equation is rational as aand b are integers, but LHS of it isirrational. This is an incorrect statementdue to our wrong assumption that 3 4 5−is rational.

So, we conclude that 3 4 5− is

irrational.

20. Any positive integer is either of the form3q, 3q + 1 or 3q + 2.

There are 3 cases now:

Case I: When n = 3q, n + 1 = 3q + 1 andn + 2 = 3q + 2.

Here, only n is divisible by 3.

Case II: When n = 3q + 1, n + 1 = 3q + 2and n + 2 = 3q + 3 = 3(q + 1)

Here, only n + 2 is divisible by 3.

Case III: When n = 3q + 2,

n + 1 = 3q + 3 = 3(q + 1)

and n + 2 = 3q + 5 = 3(q + 1) + 2Hence only n + 1 is divisible by 3.


21 By the division algorithm,Dividend = Divisor × Quotient +

Remainder∴ 6x3 + 8x2 – 3x + 8 = g(x) × (3x + 4) +

6x + 20

⇒ g(x) =3 26 8 9 12

3 4x x x

x+ − −

+

=22 (3 4) 3(3 4)

3 4x x x

x+ − +

+

⇒ g(x) =2(3 4)(2 3)

3 4x x

x+ −

+⇒ g(x) = 2x2 – 3.

22. To draw a line, we need atleast twosolutions of its corresponding equation.

x 0 2 x 1 0

y – 4 0 y 0 – 1

Two solutions of Two solutions of

2x – y = 4 x – y = 1

From the graph, the two lines intersecteach other at the point A(3, 2).

∴ x = 3 and y = 2

Shaded region is ∆ABC.

∴ Height of ∆ABC = 3 unitsAnd its base = BC = 3 units.


∴ ar(∆ABC) =12

× 3 × 3

=92

square units.


23. Let us draw MN parallel to AB, whichpasses through P. So, AM = BN andDM = CN.

From right-angled triangles APM, BPN,CPN, DPM; we have respectivelyPA2 = PM2 + AM2 .... (i)PB2 = PN2 + BN2 .... (ii)PC2 = PN2 + CN2 .... (iii)PD2 = PM2 + DM2 .... (iv)From equations (i) and (ii),PA2 – PB2 = PM2 – PN2 .... (v)From equations (iii) and (iv),PC2 – PD2 = PN2 – PM2 .... (vi)Add equations (v) and (vi) to getPA2 + PC2 = PB2 + PD2.

Hence proved.

24. Let median AD passesthrough the point O on PQ.To prove: PO = QOProof: In ∆APO and ∆ABD,

∠PAO = ∠BAD (Common angle)

APO = ABDand AOD = ADB

∠ ∠ ∠ ∠

Correspondingangles

∴ ∆APO ~ ∠ABD (By AAA similarity)

⇒POBD

= AOAD

...(i)

Similarly, In ∆AQO and ∆ACD,

⇒QOCD

= AOAD

...(ii)


POBD

= QOCD

⇒ PO = QO (... BD = CD)Hence, median AD also bisects PQ.

Proved.OR


25. cos sin 1 3cos sin 1 3

θ − θ −=θ + θ +

Using componendo and dividendo, we get

cos sin cos sincos sin cos sin

θ − θ + θ + θθ − θ − θ − θ

= 1 3 1 31 3 1 3

− + +− − −

⇒2 cos2 sin

θ− θ

= 22 3−

⇒ cot θ = 13

⇒ cot θ = cot 60°

⇒ θ = 60°.

26. LHS = 2 4 2 42 1 2 1

cos cos sin sin− − +

θ θ θ θ

= 2 sec2 θ – sec4 θ – 2 cosec2 θ+ cosec4 θ


= 2(1 + tan2 θ) – (1 + tan2 θ)2

– 2(1 + cot2 θ) + (1 + cot2 θ)2

= (1 + tan2 θ) × (2 – 1 – tan2 θ)

– (1 + cot2 θ) × (2 – 1 – cot2 θ)

= 1 – tan4 θ – (1 – cot4 θ)

= cot4 θ – tan4 θ = RHS.Hence proved.

27. We convert the given data of less thantype to the normal distribution.

Marks fi xi di = fi dixi – 25

0-10 5 5 – 20 – 100

10-20 11 15 – 10 – 11020-30 19 25 0 030-40 30 35 10 30040-50 15 45 20 300

Σfi = 80 Σfidi = 390

Let us use the assumed mean method withassumed mean a = 25.

Now, mean = a + i i

i

f df

ΣΣ

= 25 + 39080

= 25 + 4.875 = 29.88 marks.

28. We prepare the cumulative frequency tablefor the given data.

Lifetimes Frequency Cumulative(in hrs.) (f) frequency

(cf )

1500-2000 24 24

2000-2500 86 110

2500-3000 90 200

3000-3500 115 315

3500-4000 95 410

4000-4500 72 482

4500-5000 18 500

N = 500

Here, h = 500

∵ N = 500, ∴ N2

= 250.

So, cf = 200, f = 115, l = 3000.

Median = l +

N2

cf

f

−

× h

= 3000 +250 200

115−

× 500

= 3000 + 217.39

= 3217.39 hours.

SECTION-D

29. Let f (x) = 6x4 + 8x3 – 5x2 + ax + bf (x) is divisible by 2x2 – 5.

⇒ We obtain the remainder as zero whenf (x) is divided by 2x2 – 5.Now we divide f (x) by 2x2 – 5.

2

2 4 3 2

4 2

_____________________________

3 2

3

_________________________________2

2

_______________________________

__________

3 4 5

2 5 6 8 56 15

8 10

8 20

10 ( 20)

10 25

( 20) 25

x x

x x x x ax bx x

x x ax b

x x

x a x b

x

a x b

+ +− + − + +

−− +

+ + +−

− +

+ + +−

− +

+ + +___________

As remainder = 0,

(a + 20)x + b + 25 = 0

i.e., (a + 20)x + (b + 25) = 0x + 0

i.e., a + 20 = 0 and b + 25 = 0

i.e., a = – 20 and b = – 25.

30. Let the present ages of father and his sonare x years and y years respectively.According to the given conditions:

x + y = 65


After 5 years,the father’s age = (x + 5) yearsAfter 5 years,

the son’s age = (y + 5) yearsTherefore,x + 5 = 2 (y + 5)i.e., x – 2y = 5Thus, the required pair of linear equations is

x + y = 65 .... (i)x – 2y = 5 .... (ii)

Subtracting equation (ii) from equation (i),we get

3y = 60⇒ y = 20Substituting y = 20 in equation (i), we get

x + 20 = 65⇒ x = 45Thus, present age of father = 45 years andpresent age of his son = 20 years.

31. We are given a ∆ABC in which a linePQ || BC intersects the sides AB at P andAC at Q.

We need to prove AP AQPB QC

= .

Draw QM ⊥ AB and PN ⊥ AC.

Join PC and BQ.

Proof: Area of a triangle

= 12

× Base × Height

∴ ar(∆APQ) = 12

× AP × QM

= 12

× AQ × PN

⇒ AP × QM = AQ × PN

⇒ QMPN

= AQAP

... (i)

Since ∆BPQ and ∆CQP are on the samebase PQ and between the same parallelsPQ and BC, therefore, their areas shouldbe equal.i.e., ar(∆BPQ) = ar(∆CQP)

⇒ 12

× PB × QM = 12

× QC × PN

⇒ QMPN

= QCPB

... (ii)

Form equations (i) and (ii), it is clear that

AQAP

= QCPB

,

i.e.,APPB

= AQQC

.

Hence proved.

32.cot tan

+1 – tan 1 – cot

x xx x

= 1 + sec x . cosec x

LHS

=cot tan

+1 – tan 1 – cot

x xx x

=

cos sinsin cos

+sin cos

1 – 1 –cos sin

x xx x

x xx x

=cos cos

×sin cos – sin

x xx x x

+ sin sin

×cos sin – cos

x xx x x

=2cos

sin (cos – sin )x

x x x+

2sincos (sin – cos )

xx x x

=3 3cos – sin

.sin cos (cos – sin )x x

x x x x

=2 2.(cos – sin )(cos cos sin sin )

.sin cos (cos – sin )x x x x x x

x x x x+ +

=.1 cos sin

.sin cosx x

x x+

= 1.sin cosx x

+ 1

= 1 + sec x . cosec x = RHS.Hence proved.


33. It is given thatsin θ + cos θ = a

Squaring both the sides, we getsin2 θ + cos2 θ + 2 sin θ cos θ = a2

But sin2 θ + cos2 θ = 1 .... (i)∴ 1 + 2 sin θ cos θ = a2

⇒ 2 sin θ cos θ = a2 – 1

⇒ sin θ cos θ =2 12

a −.... (ii)

Cubing both the sides of equation (i), wegetsin6 θ + cos6 θ

+ 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) = 1⇒ sin6 θ + cos6 θ + 3 (sin θ cos θ)2 = 1

[Using (i)]

⇒ sin6 θ + cos6 θ + 3 22 1

2a −

= 1

⇒ sin6 θ + cos6 θ = 1 – 3 × 2 2( 1)

4a −

⇒ sin6 θ + cos6 θ = 2 24 3( 1)

4a− −

.

Hence proved.OR

Considera sin3 α + b cos3 α = sin α . cos α

⇒ a sin α . sin2 α + b cos α . cos2 α= sin α . cos α

⇒ b cos α . sin2 α + b cos α . cos2 α= sin α . cos α

(... a sin α = b cos α)⇒ b sin2 α + b cos2 α = sin α⇒ b (sin2 α + cos2 α) = sin α∴ b = sin α ...(i)Again, a sin α = b cos α⇒ a . b = b cos α [From (i)]∴ a = cos α ...(ii)Now, squaring and adding equations (i)and (ii), we get

b2 + a2 = sin2 α + cos2 α∴ a2 + b2 = 1. Hence proved.

34. We prepare a table for less than series withcorresponding cumulative frequencies andpoints.

Marks Fre- Marks Cumu- Pointquency less lative

than frequency

0-10 5 10 5 (10, 5)

10-20 8 20 13 (20, 13)

20-30 6 30 19 (30, 19)

30-40 10 40 29 (40, 29)

40-50 6 50 35 (50, 35)

50-60 5 60 40 (60, 40)

We take upper limits on the x-axis andcumulative frequencies on the y-axis. Thenwe plot the points on the graph paper. Byjoining these points by free hand smoothcurve, we obtain less than ogive as shownin the above figure.

OR



Practice Paper–5

SECTION-A

1. (D) 2 and 5.

2. (B) Let us assume that x + y is rationalnumber and let x + y = z; when z isrational.

⇒ x + y + 2 xy = z2

⇒ 2 xy = z2 – x – y

xy = 2 – –

2z x y

which given a contradiction as LHS isirrational but RHS is rational.

⇒ x + y is an irrational number.

3. (C) Coincident lines is given by

1

2

aa

= 1

2

bb

= 1

2

cc

Here, a1 = p, b1 = q, c1 = – 4a2 = 4, b2 = 3, c2 = – 5

Now,4p

= 3q

= – 4– 5

⇒4p

= 3q

= 0.8

⇒ p = 3.2, q = 2.4

Therefore, 3p + q = 12.

4. (A)318p

= 624

≠ 5075

⇒ p = 6 183 24

⇒ p = 3 .

5. (B) Consider 3A = 90°

∴ A = 30°

So, sin 30° – cos 2 × 30°

= sin 30° – cos 60°

= 12

–12

= 0.

6. (B)DE EF FDAB BC CA

+ ++ +

=EFBC

⇒ Perimeter of ∆DEF = 42

× (3 + 2 + 2.5)

⇒ Perimeter of ∆DEF = 15 cm.

7. (C) cos A + cos2 A = 1⇒ cos A = 1 – cos2 A⇒ cos A = sin2 A⇒ cos2 A = sin4 A (Squaring)⇒ 1 – sin2 A = sin4 A

sin4 A + sin2 A = 1.

8. (C) Relation among mean, median andmode is given by the following impiricalformula:Mode = 3 Median – 2 Mean.

9. (C) tan θ = sincos

θθ

= pq

.

(... sin θ = p, sec θ = 1q

or cos θ = q)

10. (B) In right triangle ADC,

AC = 2 23 4+ = 5

sin A = CDAC

= 45

.

cos C = CDAC

= 45

Now,

sin A cos C = 4 4×

5 5=

1625

.

SECTION-B

11. True, because product of an evennumber and an odd number is an evennumber.


12. Let p (x) = ax2 + bx + cSo, y = ax2 + bx + c should be satisfied by(–1, 0), (0, –3) and (4, 0)Therefore, 0 = a – b + c; – 3 = c;

0 = 16a + 4b + c

⇒ c = – 3, a = 34

, b = –94

.

Hence, p(x) =34

x2 –94

x – 3

⇒ p(x) = 34

(x2 – 3x – 4)

This is the required expression.

13. For infinite number of solutions,

1

2

aa

= 1

2

bb

= 1

2

cc

i.e.,22a

= 5+a b

=189

,

i.e., a = 2 and a + b = 10i.e., a = 2, b = 8.

14. In ∆ADP, and ∆EBQ,AD = EB (Given)

∠ADP = ∠EBQ, (Corresponding anglesas BQ || DP and AB is transversal)

∠DAP = ∠BEQ (Corresponding angles)Therefore, ∆ADP ≅ ∆EBQ (ASA criterion)∴ DP = BQ.

Also, DP || BQ as DP || BC.So, DPQB is a parallelogram.Therefore, PQ || DB. Hence,PQ || AB.

15.ABDF

= BCEF

= CADE

= 12

⇒ ∆ABC ~ ∆DFE⇒ ∠B = ∠FBut ∠B = 60°, so, ∠F = 60°.

16. True,

sin B = ACAB

sin Q = PRPQ

sin B = sin Q

⇒ ACAB

= PRPQ

⇒ AC ABPR PQ

=

⇒2 2

2 2AC ABPR PQ

= = 2 2

2 2

AB – AC

PQ – PR

⇒2 2 2

2 2 2AC AB BC

= =PR PQ QR

⇒ AC AB BC= =

PR PQ QR

∴ ∆ACB ~ ∆PRQ

⇒ ∠B = ∠Q. OR

p2 – q2 = (a cot θ + b cosec θ)2

– (b cot θ + a cosec θ)2

= a2 cot2 θ + 2ab cot θ cosec θ + b2 cosec2 θ– b2 cot2 θ – 2ab cot θ cosec θ – a2 cosec2 θ

= – a2 (cosec2 θ – cot2 θ)

+ b2 (cosec2 θ – cot2 θ)

= – a2 (1) + b2 (1)= b2 – a2.


17. xi 15 17 19 20 + p 23

fi 2 3 4 5p 6 Σfi=15+5p

fixi 30 51 76 100p+5p2 138 Σfixi = 295+100p+5p2

Mean = ∑∑

i i

i

f xf

⇒ 20 = 2295 100 5

15 5+ +

+p pp

⇒ 300 + 100p = 295 + 100p + 5p2

⇒ 5p2 = 5⇒ p = ± 1.

But frequency cannot take negative value.So, p ≠ –1.Hence, p = 1.

18. Table for cumulative frequency:

Class FrequencyCumulativefrequency

0-10 5 510-30 15 2030-60 30 5060-80 8 5880-90 2 60

N = 60

Modal class is 30-60.

Class mark = 30 60

2+

= 45

Since N2

= 30 so, median class is 30-60.

⇒ Class mark = 45Required sum = 45 + 45 = 90.

SECTION-C

19. Let a be any positive integer. We knowthat any positive integer is either of theform 2q or 2q + 1 for some integers q.∴ a = 2q or 2q + 1Case I: When a = 2q, a + 1 = 2q + 1.

∴ a (a + 1) = 2q × 2q + 12q (2q +1) = 2r, where r = q (2q + 1)

So, a (a + 1) is divisible by 2.Case II: When a = 2q + 1,

a + 1 = 2q + 2 = 2 (q + 1)∴ a (a + 1) = 2 (2q + 1) (q + 1)

where r = (2q + 1) (q + 1)So, a (a + 1) is divisible by 2.Hence, multiplication of any two conse-cutive positive integers is divisible by 2.

20. Ram, Ravi and Nitin will meet next afterthe time given by the LCM of 5 days, 24days and 9 days.

Now, we find out the LCM of 5, 24 and 95 = 5; 24 = 23 × 3; 9 = 3 × 3∴ LCM = 23 × 3 × 3 × 5 = 360

They met last on Sunday. So, it will beSunday after 7n days, where n is a naturalnumber.

So, it will be Sunday after 357 days.Therefore, it will be Wednesday after 360days. Hence, they will meet on nextWednesday.

OR

See worksheet-2, Sol. 9.

21. Let zeroes are α, β, γLet αβ = 8 ... (i)Also we know α + β + γ = 9 ... (ii)

αβ + αγ + βγ = 26αβγ = 24

∴ 8 (γ) = 24⇒ γ = 3(ii) ⇒ α + β = 6 ... (iii)

(i) ⇒ β = 8α

∴ Use if in (iii) we get

α + 8α = 6

⇒ α2 – 6α + 8 = 0(α – 4) (α – 2) = 0

⇒ α = 4 ⇒ α = 2.


If α = 4 ⇒ β = 2 and if α = 2⇒ β = 4

∴ Zeroes are 2, 4 and 3.

22. Put 1+x y

= u and 1–x y

= v in the given

system of equation. we get

10u + 2v = 415u – 5v = – 2

i.e., 5u + v = 2 ... (i)15u – 5v = – 2 ... (ii)

Multiply equation (i) by 5 and add theresult to (ii)

40u = 8 or u = 15

Substitute u = 15

in equation (i)

v = 1

u = 15

and v = 1 give x + y = 5

and x – y = 1

On solving, we get

x = 3 and y = 2

Hence, x = 3, y = 2 is the required solution.

23. In ∆ABC, DE || BC∵∆ABC ~ ∆ADE

∴( )( )

ABCADE

∆∆

arar

=2

2ABAD

...(i)

Again, DE || BC

D E

A

B C

∵ ar (∆ADE) = ar ( BCED)∴ ar (∆ABC) = 2 ar (∆ADE)

⇒ ( )( )

ABCADE

∆∆

arar

= 2 ... (ii)

From equations (i) and (ii), we get

ABAD

= 2

1

Let AB = 2 x and AD = x,

then from the figure,

BD = 2 x – x = ( )2 – 1 x

Now,BDAB

= ( )2 – 1

2

x

x

= 2 – 1

2×

22

= 2 – 2

2.

24. In right-angled ∆ABC,AB2 + BC2 = AC2 ...(i)

(By Pythagoras Theorem)In ∆ABN,

AN2 = AB2 + BN2

= AB2 + 2BC

2

(... N is the mid-point of BC)

= AB2 +14

BC2

A

M

B N C

⇒ 4 AN2 = 4AB2 + BC2 ...(ii)

Similarly, in ∆CBM,

4 CM2 = AB2 + 4 BC2 ...(iii)

Adding equations (ii) and (iii), we get

4AN2 + 4 CM2

= (4AB2 + AB2) + (BC2 + 4BC2)

= 5 AB2 + 5BC2

⇒ 4(AN2 + CM2)

= 5 (AB2 + BC2)

= 5 AC2. [From (i)]Hence proved.


25. LHS =tan sec –1tan – sec 1

θ + θθ θ +

= ( )2 2tan sec – sec – tan

tan – sec 1

θ + θ θ θ

θ θ +

= ( ) ( ) ( )tan sec – sec tan sec – tan

tan – sec 1

θ + θ θ + θ θ θθ θ +

= ( ) ( )tan sec 1 – sec tan

tan – sec 1

θ + θ θ + θθ θ +

= tan θ + sec θ = sincos

θθ

+ 1

cos θ

=1 sin

cos+ θ

θ


26. cos 582

sin 32 ° °

– 3cos 18 cosec 52

tan 15 tan 60 tan 75 ° ° ° ° °

=( )cos 90 – 32

2sin 32

° ° °

– 3( )

( )cos 90 – 52 cosec 52

tan 90 – 75 3 tan 75

° ° °

° ° × × °

= 2sin 32sin 32

° °

– 2

1sin 52 ×

sin 521

cot 75 × 3 ×cot 75

° ° ° °

= 2 × 1 –1

3 ×3

= 2 – 1 = 1.

OR

LHS = cos 2θ = cos (2 × 30°) = cos 60° = 12

RHS = 2

21 – tan1 tan

θ+ θ

= 2

21– tan 301 tan 30

°+ °

=

2

2

11 –

3

11

3

+

=

11 –

31

13

+

=

2343

= 24

= 12

Thus, LHS = RHS. Hence proved.

27. First, we prepare the cumulative frequencytable as given below:

Class Frequency Cumulativeinterval ( f ) frequency (cf )

85-100 11 11100-115 9 20115-130 8 28130-145 5 33

N = 33

∵ N = 33 ∴ N2

= 16.5

Cumulative frequency just greater than16.5 is 20. So, median class is 100-115.

∴ cf = 11, f = 9, l = 100, h = 15

Now, median = l +

N–

2cf

hf

×

= 100 +16.5 – 11

159

× = 100 + 9.17

= 109.17

Hence, the median speed is 109.17 km/hr.

28. Since, the maximum frequency is 41, sothe modal class is 10000-15000.∴ l = 10000, f1= 41, f0 = 26, f2 = 16, h = 5000

Now, mode = l + 1 0

1 0 2

–2 – –

×

f fh

f f f

= 10000 +41 – 26

500082 – 26 – 16

×

= 10000 +75000

40

= 10000 + 1875 = 11875.Thus, the monthly modal income isRs. 11875.


ORLet a = 50; h = 20

C.I. fi xi ui =–ix ah

fiui

0-20 4 10 – 2 – 820-40 10 30 – 1 – 1040-60 28 50 0 060-80 36 70 1 36

80-100 50 90 2 100

128 Σfiui=118

x = a + h i i

i

f uf

∑ ∑

= 50 + 20 ×118128

= 50 + 29516

= 18.4 ≅ 68.4.

SECTION-D

29. Given that p (x) = a (x2 + 1) – x (a2 + 1)i.e., p (x) = ax2 – (a2 + 1)x + aTo find zeroes of p (x), put p (x) = 0.i.e., ax2 + a – a2x – x = 0i.e., (ax2 – a2x) – (x – a) = 0i.e., ax (x – a) – 1 (x – a) = 0i.e., (x –a) (ax –1) = 0

i.e., x = a, 1a

Thus, a and 1a

are the zeroes of p (x).

Sum of zeroes = 1a a+ =

2 1aa+

= –( )2– 1a

a+

= 2

Coefficient of–

Coefficient of

x

x

Product of zeroes = a ×1a = a

a

= 2Constant term

Coefficient of x.

Hence proved.OR

See worksheet-12, Sol. 10.

30. Let the speeds of the cars be x km/hrand y km/hr.

We know that

Time × Speed = Distance

When the cars run in the same direction.

80 kmA B

x km/hr y km/hr

Difference of the distances covered bythe two cars = 80 km

8x – 8y = 80

i.e., x – y = 10 ... (i)

When the cars run in the oppositedirections:

80 kmA B

x km/hr y km/hr

1 × x + 1 × y= 80 or x + y = 80 ... (ii)To solve equations (i) and (ii), addingthem and subtracting them respectively,we get 2x = 90 and 2y = 70i.e., x = 45 and y = 35Hence, the speed of the cars are 45 km/hr and 35 km/hr.

31. See worksheet-34, Sol. 9 (1st part).OR

See Assessment sheet-7, Sol. 8.

32. We haveq sin θ = p and p cos θ = q

⇒ sin θ = pq

and cos θ = qp

(i)6

pq

+6

qp

= sin6 θ + cos6 θ

= (sin2 θ)3 + (cos2 θ)3

= (sin2 θ + cos2 θ)3 – 3 sin2 θ . cos2 θ(sin2 θ + cos2 θ)

[... a3 + b3 = (a + b)3 – 3ab (a + b)]

= (1)3 – 3 .2

2

pq

.2

2

q

p. 1

= 1 – 3 = – 2. Proved.

(ii)6

6

pq

+ 6

6

qp

= – 2


⇒12 12

6 6

p q

p q

+= – 2 ⇒ p12 + q12 = – 2p6 q6

⇒ p12 + q12 + 2p6 q6 = 0

⇒ (p6 + q6)2 = 0 ⇒ p6 + q6 = 0. Proved.

33. We have to prove

sec – 1sec 1

θθ +

+sec 1sec – 1

θ +θ

= 2 cosec θ

LHS =

sec – 1 sec – 1sec 1 sec – 1

θ θ×

θ + θ+

sec 1 sec 1sec – 1 sec 1

θ + θ +×

θ θ +

= ( )2

2

sec – 1

sec – 1

θθ

+ ( )2

2

sec 1

sec – 1

θ +θ

= ( )2

2

sec – 1

tan

θθ

+ ( )2

2

sec 1

tan

θ +θ

= sec – 1 sec 1

tan tanθ θ +

+θ θ

=sec – 1 sec 1

tanθ + θ +

θ

= 2 sectan

θθ

=

2cossincos

θθθ

= 2

sin θ

= 2 cosec θ = RHS. Hence proved.

34. To draw the less than type and morethan type ogives, we prepare thecumulative frequency table by less thanand more than methods as given below:Less than type cumulative frequencytable:

Marks No. of Marks c f pointstudents less

than

0-10 7 10 7 (10, 7)10-20 10 20 17 (20, 17)20-30 23 30 40 (30, 40)30-40 51 40 91 (40, 91)40-50 6 50 97 (50, 97)50-60 3 60 100 (60, 100)

More than type cumulative frequencytable:

Marks No. of Marks c f pointstudents more

thanor

equalto

0-10 7 0 100 (0, 100)

10-20 10 10 97 (10, 97)

20-30 23 20 83 (20, 83)

30-40 51 30 60 (30, 60)

40-50 6 40 9 (40, 9)

50-60 3 50 3 (50, 3)

100

95

90

85

80

75

70

65

60

55

50

45

40

35

30

25

20

15

10

5

010 20 30 40 50

(0, 100)

(10, 97)

(20, 83)

(40, 9)

(50, 97)

(40, 91)

(30, 40)

(20, 17)

(10, 7)

Y

X60

(60, 100)

(50, 3)

(30, 60)

(32, 0)

More than ogive

Less than ogive


We plot the points as given in both ofthe tables on a graph, taking marks onthe x-axis and the cumulative frequencieson the y-axis. On joining these points byfree hand smooth curve, we obtain theless than and more than type ogives asshown in the figure.Median: The abscissa of the point ofintersection of the two ogives determinesthe median of the given data. To obtain

the coordinates of this point ofintersection, we draw a perpendicularfrom this point on the x-axis. The abscissaof the foot of this perpendicular is therequired median. Here the coordinatesof the foot of the perpendicular are (32,0),where 32 is the approximate value.Hence, the required median is nearly 32marks.

data

New Stamp

maths me n mine sol

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