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QA / Exercise - 1 CEX-5301/P2S/17 / Page 1

Quantitative Aptitude – 1Algebraic Identities

Answers and Explanations

1. a x y z 3 2 5 0+ + = − − + =�

( )( )( )3 3 3x y z 3xyz 3 3 2 5 90∴ + + = = − − = .

2. a x3 – 3x2 + 3x – 1 + x3 + (x3 + 3x2 + 3x + 1) = 3x3 – 3x⇒ 3x3 + 6x = 3x3 – 3x ⇒ 9x = 0.

⇒ x = 0.

3. a Given x = 997, y = 998, z = 999� x2 + y2 – xy – yz – zx

= 12

2 2 2[2x 2y 2z 2xy 2yz 2zx]+ + − − −

= 2 2 21

[(x y) (y z) (z x) ]2

− + − + −

= 1

[1 1 4]2

+ + = 3.

4. d � a + b + c = 8⇒ (a – 4) + (b – 3) + (c – 1) = 0If x + y + z = 0, then x3 + y3 + z3 – 3xyz = 0∴ (a – 4)3 + (b – 3)3 + ((–1)3 – 3(a – 4)(a – 4)(b – 3)(c – 1) = 0.

5. a Given x = 1

aa

+ and y = 1

aa

−

∴ x + y = 2 a and x – y = 2

a

� x4 + y4 – 2x2y2 = (x2 – y2)2 = [(x + y)(x – y)]2

=

22

2 a.a

= 16.

6. b If a + b + c = 0, then a3 + b3 + c3 = 3abc

2 2 2a b cbc ca ab

+ + = 3 3 3a b c

abc+ +

= 3abcabc

= 3.

7. c a3 + b3 + c3 – 3abc = 0(a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0

21(a b c)[(a b)

2⇒ + + − 2 2(b c) (c a) ]+ − + − = 0

� a + b + c ≠ 0Then, (a – b)2 + (b – c)2 + (c – a)2 = 0⇒ a – b = b – c = c – a = 0a – b = 0 ⇒ a = bb – c = 0 ⇒ b = cc – a = 0 ⇒ c = a∴ a = b = c.

8. b � a + b + 1 = 0a3 + b3 + (1)3 = 3 × a × b × 1⇒ a3 + b3 + 1 – 3ab = 0.

9. c1 1 1 5 yz zx xy 5x y z 7 xyz 7

+ ++ + = ⇒ =

5yz zx xy 126 90

7⇒ + + = × =

( ) ( )2 2 2 2x y z x y z 2 xy yz zx+ + = + + + + +�

( ) ( )22 2 2x y z x y z 2 xy yz zx⇒ + + = + + − + +

2 2 2 2x y z 19 2 90⇒ + + = − ×

2 2 2x y z 361 180 181⇒ + + = − = .

10. a (a – b) = 4, (b – c) = –5, (c – a) = 1� a3 + b3 + c3 – 3abc= (a + b + c)[a2 + b2 + c2 – ab – bc – ca]

= 1

(a b c)2

+ + 2 2 2[2a 2b 2c 2ab 2bc 2ca]+ + − − −

= 1

(a b c)2

+ + 2 2 2[(a b) (b c) (c a) ]− + − + −

1 a 2 a 3 a 4 d 5 a 6 b 7 c 8 b 9 c 10 a11 d 12 d 13 a 14 a 15 d 16 d 17 b 18 a 19 c 20 a21 c 22 c 23 b 24 a 25 a 26 a 27 a 28 c 29 d 30 a31 b 32 d 33 b 34 b 35 a 36 b 37 c 38 b 39 c 40 d41 c 42 c 43 b 44 c 45 d 46 c 47 d 48 b 49 c 50 b

P-2 (S)

QA / Exercise - 1CEX-5301/P2S/17 / Page 2

3 3 3a b c 3abca b c

+ + −∴+ +

= 2 2 21

[(a b) (b c) (c a) ]2

− + − + −

( ) ( ) ( )2 2 214 5 1

2 = + − +

[ ]116 25 1 21.

2= + + =

11. da b c

1 a 1 b 1 c+ +

− − − = 1

⇒ a b c

1 1 11 a 1 b 1 c

+ + + + + − − − = 3 + 1 = 4

⇒ a 1 a b 1 b c 1 c

1 a 1 b 1 c+ − + − + −+ +

− − − = 4

⇒ 1 1 1

1 a 1 b 1 c+ +

− − − = 4.

12. d (a – 1)2 + (b + 2)2 + (c + 1)2 = 0⇒ a – 1 = 0 ⇒ a = 1;b + 2 = 0 ⇒ b = –2c + 1 = 0 ⇒ c = –1∴ 2a – 3b + 7c= 2 – 3 (–2) + 7 (–1)= 2 + 6 – 7 = 1.

13. a x2 + y2 + 2x + 1 = 0⇒ x2 + 2x + 1 + y2 = 0⇒ (x + 1)2 + y2 = 0⇒ x + 1 = 0 ⇒ x = –1 and y = 0∴ x31 + y35 = –1.

14. a (3a + 1)2 + (b – 1)2 + (2c – 3)2 = 0⇒ 3a +1 = 0⇒ 3a = –1b–1 = 0⇒ b = 1 2c – 3 = 0⇒ 2c = 3∴ 3a + b + 2c = –1 + 1 + 3 = 3.

15. d1a

= 1

6 5−

= 6 56 5

+−

= 6 5+

Similarly,

1b

= 1

5 2;c

+ = 2 3+

∴ 1 1 1a b c

> > ⇒ a < b < c.

16. dx 1 1a a x

= −

⇒ xa

= x aax−

⇒ x2 = x – a⇒ x – x2 = a.

17. b a2 + b2 + c2 + 3= 2a + 2b + 2c⇒ a2 – 2a + 1 + b2 – 2b + 1 + c2 – 2c + 1 = 0⇒ (a – 1)2 + (b – 1)2 + (c – 1)2 = 0⇒ a – 1 = 0 ⇒ a = 1;b – 1 = 0 ⇒ b = 1and, c – 1 = 0 ⇒ c = 1∴ a + b + c = 3.

18. a x – y = x y

7+

= xy4

= k

⇒ x – y = kx + y = 7k∴ (x + y)2 – (x – y)2 = 49k2 – k2

⇒ 4xy = 48k2

⇒ 16k = 48k2

⇒ k = 13

∴ xy = 4k = 1

43

× = 43

.

19. c a + b + c = 0⇒ a + b = – c; b + c = – a,c + a = – b

∴ a b

c+

+ b c

a+

+ c a

b+

= –1 –1 –1 = –3

ab c+ +

bc a+ +

ca b+

= –1 –1 –1 = –3∴ Expression = (–3) × (–3) = 9.

20. a1

ab

+ = 1 ⇒ ab + 1 = b

⇒ ab = b – 1 .....(i)Again,

1b

c+ = 1

1c

= 1 – b ⇒ c = 1

1 b− .....(ii)

On multiplying (f) & (ii)

abc = b 11 b

−− = –1.


21. c a2 + b2 = 5ab

⇒ 2 2a bab+

= 5

⇒ a bb a

+ = 5

On squaring both sides,

∴ 2a b

b a +

= 25

⇒ 2 2

2 2a b

b a+ + 2 = 25

⇒ 2 2

2 2a b

b a+ = 25 – 2 = 23.

22. c21

p p k4

+ +

= ( )2 22 21 1 1

p 2 p. k8 8 8

+ + − +

⇒ k2 = 21 1

k8 8

⇒ = ± .

23. bdc

= a – b

⇒ cd

= 1 a b

a b a b+=

− −

⇒ c dc d

+− =

a b a ba b a b

+ + −+ − + =

ab

(By componendo and dividendo)

⇒ 1

c d− = ab

⇒ (c – d) = ba

⇒ c2 – d2 = (c + d) (c – d) = ba

.

24. a a4 + b4 – a2b2 = 0 ... (i)We know, a6 + b6 = (a2)3 + (b2)3

= (a2 + b2) (a4 – a2b2 + b4)= (a2 + b2) × 0 = 0 [From equation (i)].

25. a1

x 3x

+ =

Cubing both sides,

331 1

x 3 xxx

+ + + = ( )33

⇒ 331

x 3 3x

+ + = 3 3

⇒ 3

31

xx

+ = 0 ⇒ x6 + 1 = 0.

Now, x18 + x12 + x6 + 1= x12 (x6 + 1) + 1 (x6 + 1)= (x12 + 1) (x6 + 1) = 0.

26. a1

x4x

+ = 32

Multiplying both sides by 2

12x

2x+ = 3

Cubing both sides,

33

1 18x 3 2x

2x8x+ + × ×

12x 27

2x × + =

⇒ 3

31

8x 3 38x

+ + × = 27

⇒ 3

31

8x8x

+ = 27 – 9 = 18.

27. a1

x y+ = 1 1x y

+ = y xxy+

⇒ (x + y)2 = xy⇒ x2 + 2xy + y2 = xy⇒ x2 + xy + y2 = 0∴ x3 – y3 = (x – y) (x2 + xy + y2)= 0.

28. c xy (x + y) = 1

⇒ 1

x yxy

+ =

Cubing both sides,

x3 + y3 + 3xy (x + y) = 3 31

x y

⇒ x3 + y3 + 3xy × 3 31 1xy x y

=

⇒ 3 31

x y – x3 – y3 = 3.

29. d4

41

xx

+ = 119

⇒ 2

221

x 2x

+ − = 119 ⇒

22

21

xx

+ = 121

⇒ 2

21

xx

+ = 11⇒ 21

x 2x

− + = 11

⇒ 21

xx

− = 9 ⇒

1x

x− = 3


Cubing both sides,

31x

x −

= 27

⇒ 3

31

x 3x

− − 1x

x −

= 27

⇒ 3

31

x 3 3x

− − × = 27

⇒ 3

31

xx

− = 27 + 9 = 36.

30. a x + y = z ⇒ x + y + (–z) = 0∴ x3 + y3 – z3 + 3xyz= x3 + y3 + (–z)3 – 3x.y (–z) = 0.

31. b1

aa

+ = 3

On cubing both sides,

331 1 1

a 3a. aa aa

+ + + = 3 3

⇒ 3

31

a 3 3a

+ + = 3 3

⇒ 3

31

aa

+ = 0 ... (i)

⇒ 6

61

a 2a

− +

= ( )223

31

a 2a

− +

= 3 33 31 1

a a 2a a

+ − + = 2.

32. d x2 + 1 = 2x (Given)

⇒ 1

xx

+ = 2 ...(i)

Expression

=

42

2

1x

xx – 3x 1

+

+ =

6

2

2

x 1

x(x 3x 1)

+

− +

= 6

2 2x 1

(x 1 3x).x

++ −

= 6

2x 1

(2x 3x)x

+− =

6

3x 1

x

+−

= 6

3x 1

x

+− =

6

3 3x 1

x x

− +

= 3

31

xx

− +

=

31 1x 3 x

x x

− + − + = – [23 – 3 × 2]= – 2.

33. b 2x

x 2x 1− + =

13

⇒ 2x 2x 1

x− +

= 3 ⇒ 1

x 2x

− + = 3

⇒ 1

xx

+ = 5

On cubing both sides

331 1

x 3 xxx

+ + + = 125

⇒ 331

xx

+ = 125 – 3 × 5 = 110.

34. b1

xx 1

++ = 1 ⇒

1(x 1)

x 1+ +

+ = 2

On squaring

22

1(x 1) 2

(x 1)+ + +

+ = 4

⇒ 2

21

(x 1)(x 1)

+ ++ = 2 ... (i)

Again, cubing 1

(x 1)(x 1)

+ ++ = 2

33

1 1(x 1) 3 (x 1)

(x 1)(x 1)

+ + + + + ++ = 8

⇒ 3

31

(x 1)(x 1)

+ ++

= 8 – 3 × 2 = 2

∴ 2

21

(x 1)(x 1)

+ + +

33

1(x 1)

(x 1)

+ + +

= 2 × 2 = 4

⇒ 5

51 1

(x 1) (x 1)(x 1) (x 1)

+ + + + ++ +

= 4

∴ 5

51

(x 1)(x 1)

+ ++

= 4 – 2 = 2.


35. a1 1a b

− = 1

a b−

⇒ b aab−

= 1

a b−⇒ (a – b) (a – b) = – ab⇒ a2 – 2ab + b2 = – ab⇒ a2 – ab + b2 = 0∴ a3 + b3 = (a + b) (a2 – ab + b) = 0.

36. b (a + b + c)2

= a2 + b2 + c2 + 2 (ab + bc + ca)⇒ 36 = 14 + 2 (ab + bc + ca)⇒ ab + bc + ca = (36 – 14) ÷ 2⇒ ab + bc + ca = 11∴ a3 + b3 + c3 – 3abc= (a + b + c)(a2 + b2 + c2 – ab – bc – ca)⇒ 36 – 3abc = 6(14 – 11) (By (i)]⇒ 36 – 3abc = 84 – 66 = 18⇒ 3abc = 36 – 18 = 18⇒ abc = 6.

37. c x3 + y3 + z3 – 3xyz

= 1

(x y z)2

+ +

[(x – y)2 + (y – z)2 + (z – x)2

= 1

(333 333 334)(0 1 1)2

+ + + +

= 1000.

38. b p – 2q = 4On cubing both sides,(p – 2q)3 = 64⇒ p3 – 8q3 + 3p . 4q2 – 3p2 . 2q = 64⇒ p3 – 8q3 + 12pq2 – 6p2q = 64⇒ p3 – 8q3 – 6pq (p – 2q) = 64⇒ p3 – 8q3 – 6pq × 4 = 64⇒ p3 – 8q3 – 24pq – 64 = 0.

39. c1

xx

+ = 2

⇒ x2 – 2x + 1 = 0⇒ (x – 1)2 = 0 ⇒ x = 1

∴ 17

191

xx

+ = 1 + 1 = 2.

40. d x = 3 5 2+

⇒ x – 2 = 3 5On cubing,x3 – 3x2 × 2 + 3x. (–2)2 – 23 = 5⇒ x3 – 6x2 + 12x – 8 = 5⇒ x3 – 6x2 + 12x – 13 = 0.

41. c Expression

=

2 2

3 23

(x y z)(x y 3z)

xy z

+ + + −

Putting x = 1, y = –3, z = – 1

= 3

(1 9 1)(1 3 3)

1 27 1

+ − − +× − ×

= 33− = – 1.

42. c a3 + b3 + c3 – 3abc will be minimum if a = b = 1,c = 2∴ Least value = 1 + 1 + 8 – 3 × 1 × 1 × 2= 10 – 6 = 4.

43. b x = 1

6x

+ ⇒ 1

xx

− = 6

On squaring both side,

⇒ 2

21

x 2x

+ − = 36

⇒ 2

21

xx

+ = 36 + 2 = 38

On squaring again,

441

x 2x

+ + = 1444

⇒ 4

41

xx

+ = 1444 – 2 = 1442

44. c x2 – 3x + 1 = 0⇒ x2 + 1 = 3x

⇒ 2x 1x+

= 3

⇒ 1

xx

+ = 3

∴ 6 4 2

3x x x 1

x

+ + + =

6 4 2

3 3 3 3x x x 1

x x x x+ + +

= 3

31 1

x xx x

+ + + = 3

31 1

x xxx

+ + +

= 31 1 1

x 3.x. xx x x

+ − +

1x

x + +

= 33 – 3 × 3 + 3 = 27 – 9 + 3 = 21.

45. d x2 + y2 + 1 = 2x⇒ x2 + y2 + 1 = 2x = 0⇒ x2 – 2x + 1 + y2 = 0⇒ (x – 1)2 + y2 = 0⇒ x – 1 = 0⇒ x = 1 and y = 0∴ x3 + y3 = 1 + 0 = 1.


46. c p = 99 (Given)∴ p(p2 + 3p + 3) = p3 + 3p2 + 3p= p3 + 3p2 + 3p + 1 – 1= (p + 1)3 – 1 = (99 + 1)3 – 1= (100)3 – 1 = 999999.

47. d x = 997y = 998z = 999∴ x – y = 997 – 998 = – 1y – z = 998 – 999 = – 1z – x = 999 – 997 = 2∴ x2 + y2 + z2 – xy – yz – zx

= 2 2 21

(2x 2y 2z – 2xy – 2yz – 2zx)2

+ +

= 2 2 2 2 2 21

(x y – 2xy y z – 2yz x z – 2zx)2

+ + + + +

= 2 2 21

[(x y) (y z) (z x) ]2

− + − + −

= 2 2 21

[( 1) ( 1) (2) ]2

− + − +

= 1 1

(1 1 4) 6 32 2

+ + = × = .

48. b p3 + 3p2 + 3p = 7⇒ p3 + 3p2 + 3p + 1 = 7 + 1 = 8⇒ (p + 1)3 = (2)3

⇒ p + 1 = 2 ⇒ p = 2 – 1 = 1∴ p2 + 2p = 1 + 2 × 1 = 3.

49. c t2 – 4t + 1 = 0⇒ t2 + 1 = 4t

⇒ 2t 1

t+

= 4

⇒ 1

tt

+ = 4

On cubing both sides,

31t

t +

= 43

331 1

t 3 ttt

+ + + = 64

⇒ 3

31

t 3 4t

+ + × = 64

⇒ 3

31

tt

+ = 64 – 12 = 52.

50. b (a – b)2 = a2 – 2ab + b2

x4 – 2x2 + k = (x2)2 – 2.x2.1 + k∴ k = (1)2 = 1.


Quantitative Aptitude – 2Logarithm Identities and Progressions


1. b 32 3 4log log log 64

2 3 4log (log (3log 64))=

2 3log log (3 3)= × 34 4 4( log 64 log 4 3log 4 3)= = =�

2 3log (2 log 3)= ×

2log 2 1.= =

2. c 102

10

log 729log x

log 3=

610

210

log 3log x

log 3⇒ = 10

210

6log 3log x

log 3= =

2log x 6⇒ =

6x 2 64.⇒ = =

3. c 4 3 2log log log x 0=

03 2log log x 4 1⇒ = =

12log x 3 3⇒ = =

3x 2 8.⇒ = =

4. d � alog (ab) x=

a alog a log b x⇒ + =

a1 log b x⇒ + = alog b x 1⇒ = −

b1

log ax 1

⇒ =− b

1log a 1 1

x 1⇒ + = +

−

b b1 x 1

log a log bx 1+ −⇒ + =

−

bx

log ab .x 1

⇒ =−

5. bx y 1

log (logx logy)5 2+ = +�

x y) 1log log(x.y)

5 2( +⇒ =

12x y)

log log(x.y)5

( +⇒ =

12x y

(xy)5+⇒ =

2 2x y 2xyxy

25+ +⇒ =

2 2x y 2xy25

xy+ +⇒ =

x y2 25

y x⇒ + + =

x y23.

y x⇒ + =

6. c3x 3y

log(x y) log2−+ =

3x 3yx y

2−⇒ + = ⇒ x = 5y

x5

y⇒ =

xlog log5

y⇒ =

⇒ logx – logy = log5.

7. a log2x + log4x + log1621

x4

=

2 2 21 1 21

log x log x log x2 4 4

⇒ + + =

⇒ log2x = 3 n[ logm nlogm]=�

⇒ x = 23 = 8.

1 b 2 c 3 c 4 d 5 b 6 c 7 a 8 a 9 c 10 c11 c 12 b 13 c 14 b 15 c 16 c 17 a 18 d 19 a 20 a21 b 22 b 23 c 24 a 25 a 26 a 27 d 28 a 29 b 30 c31 b 32 d 33 d 34 c 35 d 36 b 37 d 38 d 39 b 40 c41 b 42 a 43 d 44 b 45 b 46 b 47 a 48 d 49 c 50 d

P-2 (S)


8. a16 25 81

Let A 7log 5log 3log15 24 80

= + +

m

log logm lognn

= − �

⇒ A = 7(log16 – log15) + 5(log25 – log24) + 3(log81 –log80)

⇒ A = 7(log24 – log(3 × 5)) + 5(log52 – log(8 × 3) +3(log34 – log(16 × 5))

⇒ A = 7(4log2 – log3 – log5) + 5(2log5 – 3log2 – log3)+ 3(4log3 – 4log2 – log5)

⇒ A = log2.

9. c Let A = xlogy – logz.ylogz – logx.zlogx – logy

logy logz logx

logz logx logy

x y zA . .

x y z⇒ =

logy logz logz

logz logy logz

x y xA . .

x x y⇒ = logb loga( a b )=�

A 1⇒ = .

10. c Let logab = logbc = logca = k

⇒ ak = b, bk = c, ck = a

⇒ ak.bk.ck = (abc)

⇒ k = 1

∴ a = b = c.

11. cx a

1 22

log 10 log 10= −�

10 10 10log x 2log a log 100⇒ = −

2

10 10a

log x log100

⇒ =

2ax .

100⇒ =

12. b � a2 + b2 = c2 ⇒ c2 – a2 = b2 ...(i)

Let c a (c a)

1 1A

log b log b+ −= +

⇒ A = logb(c + a) + logb(c – a)

⇒ A = logb(c + a)(c – a)

⇒ A = logb(c2 – a2)

⇒ A = logbb2 (from (i))

⇒ A = 2 logbb = 2.

13. c Given series is an AP.

Sn = n

[2a (n 1)d]2

+ −

n = 15, a = 1, d = 2

n

15S [2 (15 1)2]

2⇒ = + −

= 15

[2 28]2

+ = 225.

14. b Given series is a GP.

1 1a , r

2 2= = − and Tn =

1256

−

�Tn = arn–1

1256

⇒ − = n 1

1 12 2

− × −

n 112

− ⇒ −

= –1

128

∴ n – 1 = 7

⇒ n = 8.

15. c Last term = a + (n – 1)d = 2 + (50 – 1) × 2 = 2 + 49 × 2 = 100.

16. c Let its nth term be 32.� a = 68, d = – 2∴ 32 = 68 + (n – 1) (– 2) ⇒ – 36 = – 2n + 2⇒ 2n = 38 ⇒ n = 19.

17. a 7th term ⇒ a + 6d = 84 ...(i)13th term ⇒ a + 12d = 66 ...(ii)By using equations (i) and (ii), we geta = 102, d = – 3,Then, its 30th term is= 102 + 29 × (– 3) = 102 – 87 = 15.

18. d Let the nth term be 5

.2187

� a = 15 and r = 13

, then

52187

= 15n 11

3

−

⇒ n 21

3

−

= 1

2187

n 2 71 13 3

− ⇒ =

n 2 7 n 9.∴ − = ⇒ =


19. a Let the arithmetic mean be A1, A2 and A3, then 4, A1,A2, A3, 32, are in AP

d = b an 1

−+ , here b = 32, a = 4, n = 3

∴ d = 32 43 1

−+ = 7

So, the required arithmetic means are 4 + 7 = 11, 11 +7 = 18, 18 + 7 = 25Hence, three A.M’s are 11, 18, 25.

20. a � a, b and c are in GP∴ b2 = acTaking logarithm on both sides, we getlog b2 = log (ac)

⇒ 2 log b = log a + log c ⇒ log b =log a log c

2+

log b is the arithmetic mean of log a and log c.Therefore, log a, log b and log c are in AP.

21. b x1, x2, x3 , x4, ... are in AP.x1 = 1 and x4 = a + (n – 1)d⇒ 5.5 = 1 +(4 – 1)d (Since n = 4)

⇒ d = 5.5 – 1

3=1.5.

22. b The students will follow the rule751, 701, 651, ..., 1 (an AP series)

where na a (n 1)d= + −

nHere a 1=a = 751 and d = –50∴ 1 = 751 + (n – 1) (–50)⇒ n – 1 = 15⇒ n = 16.Hence, number of rounds = 16 – 1 = 15.

23. c Let the numbers be (a – d), a, (a + d). Then,(a – d) + a + (a + d) = 21 ⇒ 3a = 21 ⇒ a = 7

(a – d) × a × (a + d) = 315 ⇒ a(a2 – d2) = 315

⇒ 7(49 – d2) = 315 ⇒ 49 – d2 = 45

⇒ d2 = 4 ⇒ d 2= ±If d = 2, then numbers are 5, 7, 9If d = – 2, then numbers are 9, 7, 5.

24. a Clearly, the numbers between 200 and 400 which aredivisible by 3 are 201, 204, 207, ... , 399.Let n be the number of terms in AP. Then,an = 399 = 201 + (n – 1) × 3 ⇒ 198 = 3n – 3

3n = 201 ⇒ n = 67

∴ Sum = ( ) ( )nn 67

a a 201 3992 2

+ = + = 20100.

25. a a6 = – 8, a9 = 64, thenar5 = – 8 ...(i)ar8 = 64 ...(ii)On dividing equation (ii) by (i), we get

8

5ar 64

8ar=

− ⇒ r3 = – 8 ⇒ r = – 2

Substituting the value of r in equation (i), we get 1

a4

=

∴ The GP is 1 1

, ,1, ...4 2

−

26. a Given sequence is 1 3 3 1

, , , , ...4 10 8 2

The reciprocal of the sequence is in AP i.e. 4,

10 8, , 2, ...

3 3

It can be written as 12 10 8 6

, , ,3 3 3 3

, ...

Here in AP, a = 4, d = 23

−

∴ 8th term, a8 = a + (n – 1)d = 4 + (8 – 1) × 23

−

= 4 – 143

= 23

−

Here 8th term of HP is 32

− .

27. d We have; b = a + a2 + a3 + ... ∞Clearly, R.H.S. is a geometric series with first term ‘a’and common ratio ‘a’.

ab b – ab a

1– a∴ = ⇒ =

ba

1 b⇒ =

+

Hence, the value of b

1 b+ is a.

28. a Given, r = 2, last term l = 192 and sum S = 381.Let a be the first term.

∴ lr a

Sr 1

−=− , 381 =

2 192 a2 1

× −−

⇒ a = 384 – 381 = 3.

29. b Here a = – 16, common ratio r = 12

− , then 10th term

= a10 = ar(10 – 1) = ar9

= (– 16) × 91

2 −

= (– 16) × 1

512−

=

132

.


30. c Here, 3

a4

= , r = 2

Sum of six terms, ( )66

2 13 3 189S 63 .

4 2 1 4 4

− = = × = −

31. b Let ‘a’ be the first term and ‘d’ be the common differenceof the given AP. Then,p times pth term = q times qth term.⇒ p [a + (p – 1) d] = q [a + (q – 1) d]

⇒ p [a + (p – 1) d] – q [a + (q – 1) d] = 0

⇒ a (p – q) + [p (p – 1)] – q (q – 1)] d = 0

⇒ a (p – q) + [(p2 – q2) – (p – q)] d = 0

⇒ a (p – q) + (p – q) (p + q – 1) d = 0

⇒ (p – q) [a + (p + q – 1) d] = 0∴ a + (p + q – 1) d = 0ap+q = 0

∴ The (p + q)th term of the given AP is zero.

32. d Sum = 9 ⇒ (a + ar + ar2 + ... ∞ ) = 9

⇒ a

91 r

=− … (i)

Sum of the squares = 27⇒ (a2 + a2r2 + a2r4 + .... ∞ ) = 27

2

2a

271 r

⇒ =−

… (ii)

On dividing the square of (i), by (ii), we get

( )( )22 2

2 2

9a 1 r 1 r 13 r .

27 1 r 2a1 r

− +× = ⇒ = ⇒ =−−

Putting r = 12

in (i), we get

a 99 a

1 212

= ⇒ =−

∴ The GP is 9 9 9

, , , ...2 4 8

∞ .

33. d Let ‘a’ and ‘d’ be the first term and common differenceof given AP. Then,S1 = sum of n terms

1n

S2

⇒ = [2a + (n – 1)d] … (i)

S2 = sum of 2n terms

22n

S2

⇒ = [2a + (2n – 1)d] … (ii)

and S3 = sum of 3n terms

33n

S2

⇒ = [2a + (3n – 1)d] … (iii)

Now, S2 – S1 2n2

= [2a + (2n – 1)d] – n2

[2a + (n – 1)d]

n2

= [2a + (3n – 1) d]

∴ 3(S2 – S1)3n2

= [2a + (3n – 1) d] = S3.

34. c Here, a = 72, d = 63 – 72 = –9an = 0∴ an = a + (n – 1)d⇒ 0 = 72 + (n – 1) × –9⇒ 72 = 9(n – 1) ⇒ n – 1 = 8.⇒ n = 9.

35. d Clearly, repetition takes place for each set of four terms.Hence, 507th term will be 2507, when divided by 4, gives 3 as remainder and 3rd

term is 2.

36. b Sum of n terms of a GPna(r 1)

r 1−=

− (when r > 1)

8a(3 1)6560

3 1−∴ =

−a(6561 1)

65602

−⇒ =

6560 2a a 2

6560×⇒ = ⇒ = .

37. d a, a – b, a – 2b … is an AP with first term = a andcommon difference = – bNow,t10 = a + (10 – 1) × (–b)⇒ 20 = a – 9b … (i)t20 = a + (20 – 1) (–b)⇒ 10 = a – 19b … (ii)From equation (i) – (ii),20 – 10 = a – 9b – a + 19b⇒ 10b = 10 ⇒ b = 1From equation (i),20 = a – 9b ⇒ a = 29∴ tx = 29 + (x – 1) (– 1)= 29 – x + 1 = 30 – x.

38. d Expression

= 1 1 1 1 1

.....2 6 12 20 n(n 1)

+ + + + ++

= 1 1 1 1 1

...1 2 2 3 3 4 4 5 n(n 1)

+ + + + +× × × × +

= 1 – 1 1 1 1 1 1 1 1 1

...2 2 3 3 4 4 5 n n 1

+ − + − + − + + −+

= 1 n 1 1 n

1n 1 n 1 n 1

+ −− = =+ + + .


39. b First term, a = 1n

Common difference

d = n 1 1 n 1 1

1n n n+ + −− = =

∴ nth term = a + (n – 1) d

= 1n

+ (n – 1) . 1

= 2 21 n n n n 1n n

+ − − += .

40. c Expression= 4 + 44 + 444 + .... to n terms= 4 (1 + 11 + 111 + ... to n terms)

= 49

(9 + 99 + 999 + .... to n terms)

= 49

[(10 – 1) + (100 – 1) + (1000 – 1) + ..... to n terms]

= 49

[(10 + 102 + 103 + .... to n terms) – n]

[� 1 has been added n times]

= 49

[ 10 ( 1 + 10 + 102 + .... to n terms) – n]

= ( )n10 140 4

. – n9 9 9

−

[ � 1 + 10 + 102 + .... to n terms = n10 19

−]

= ( )n40 410 1 n

81 9− − .

41. b We know that

1 + 2 + 3 + 4 + .... + n = n(n 1)

2+

∴ 1 + 2 + 3 + 4 + ..... + 1000

= 1000(1000 1)

2+

= 1000 1001

5005002× = .

42. a2 3

1 1 11 ...n

2 2 2+ + + + to n terms is a Geometric series

whose first term is 1 and the common ratio (r) is 12

.

n

na(1 r )

S1 r

−=−

=

n

nn

2 111

221.1 1

12 2

− − =−

= n n

n n 12 1 2 1

2.2 2 −

− −= .

43. d 2 + 4 + 6 + 8 + .... + 198= 2 ( 1 + 2 + 3 + .... + 99)∴ Number of terms = 99Middle term

= 99 1

50th2+ = term = 100.

Second MethodIt is an arithmetic series.a = 2, an = 198, d = commondifference = 2Number of terms = n∴ an = a + ( n – 1)d⇒ 198 = 2 + (n – 1)2⇒ (n – 1)2 = 198 – 2 = 196

⇒ n – 1 = 196

2= 98

Middle term = 99 1

2+

= 50th term

∴ a50 = 2 + (50 – 1)2= 2 + 98 = 100.

44. b � (a + b)2 – (a – b)2 = 4ab∴ (1012 + 25)2 – (1012 – 25)2

= 4 × 1012 × 25 = 1014

⇒ 1014 = 10n

⇒ n = 14.

45. b The largest number will be 6.For n = 2(n – 1).n(n + 1) = 6, for n = 3, (n – 1) (n) (n + 1) = 24 etc.

46. b Expression, 1 1 1 11 1 1 ... 1

5 6 7 100 − − − −

= 5 1 6 1 7 1 99 1 100 1

...5 6 7 99 100− − − − −

= 4 5 6 98 99

...5 6 7 99 100

× × × × ×

= 4 1

100 25= .

47. a 12 + 22 + 32 + ... + n2

= n(n 1)(2n 1)

6+ +

210 11 21 4 5 920

6 6× × × ×= + −

= 385 + 400 – 30 = 755.


48. d S = 12 – 22 + 32 – 42 + ... 102

S = (12 + 32 + 52 + 72 + 92) – (22 + 42 + 62 + 82 + 102)We know that sum of squares of first n odd naturalnumbers

2n(4n 1)3

−=

Sum of squares of first n even natural numbers

= 2

n(n 1)(2n 1)3

+ +

Hence, S = 5(4 5 5 1) 2

5(5 1)(2 5 1)3 3

× × − − × + × +

S = 5 99 2 30 11

3 3× × ×−

= 165 – 220 = – 55.

49. c 12 + 22 + 32 + 42 + ... + n2 = n(n 1)(2n 1)

6+ +

∴ 52 + 62 + ... + 102 = (12 + 22 + ... + 102)– (12 + 22 + 32 + 42)

= 10 11 21 4 5 9

6 6× × × ×−

= 385 – 30 = 355.

50. d Using formula 13 + 23 + 33 + ... + n3

= 2

n(n 1)2+

we have, 13 + 23 + 33 + ... + 103

= 2

210 11(55)

2× =

= 55 × 55 = 3025.


Quantitative Aptitude – 3Mensuration-3D (Solid Figures)


1. d Thickness of the wood = 1.5 cm =23

cm

⇒ Internal length of box3 3

482 2

= − + = 48 – 3 = 45 cm.

Internal breadth of box3 3

36 36 32 2

= − + = − = 33 cm.

Internal height of box = 30 – 3 32 2

+ = 30 – 3 = 27 cm.

∴ Volume of the box = 45 x 33 x 27 cm3.

Volume of one brick = 6 × 3 × 43

= 2

27cm3.

Let the required number of bricks be n.

∴ n = brickoneofVolumeboxtheofVolume

=

227

273345 ××

= 227

273345 ×××= 45 × 33 × 2 = 2970 bricks.

2. b � 1 hectare = 10,000 m2

Area of the ground = 1.5 × 10000 = 15000 m2

Volume of rain water collected= 15000 × 5 × 10–2 = 150 × 5 = 750 m3.

3. b Rate of flow of water 2 1000 100

60 3×= = m/min

Fall of water into the sea in one minute

1003 40

3= × × = 4000 m3 = 40,00,000 litre.

4. d The dimensions of the new room are all twice theoriginal one. Hence, the area shall be four times theoriginal one. Cost shall be four times the original costof covering the walls with paper.Hence, cost of covering the walls of new room withpaper = 4 × 480 = �1,920.

5. b Let the dimensions of the rectangular box be 5x, 6xand 11x.Total surface area = 2(lb + bh + hl)= 2(30x2 + 66x2 + 55x2)

⇒ 14798 = 302x2 2 14798

x 49302

⇒ = = ⇒ x = 7

∴ Volume of the box = 5x × 6x × 11x = 330 × x3

= 330 × 343 = 113190 m3.

6. d Volume of water to be filled in tank = 200 × 150 × 8 m3

Volume of water flow into tank per hour= 0.3 × 0.2 × 20 × 1000 m/hour∴ Time required to fill 8 m water level in tank

200 150 80.3 0.2 20 1000

× ×=× × × = 200 hours.

7. b Total area of room to be painted= 2[4 × 3 + 3 × 3 ] + 4 × 3 = 54 m2.

8. a 50 cm

20 cm

5 cm

5 cm

40 cm

10 cm

5 cm

Volume of the box = length × breadth × height= 40 × 10 × 5 = 2000 cm3.

1 d 2 b 3 b 4 d 5 b 6 d 7 b 8 a 9 d 10 a11 b 12 d 13 c 14 a 15 a 16 b 17 b 18 d 19 c 20 b21 b 22 c 23 a 24 d 25 b 26 a 27 d 28 b 29 a 30 a31 d 32 a 33 d 34 d 35 b 36 d 37 d 38 b 39 d 40 b41 d 42 a 43 c 44 c 45 b 46 c 47 b 48 a 49 d 50 a

P-2 (S)


9. d Let sides of the cuboid be l, b and h.Then, lb = 12, bh = 20, hl = 15⇒ lb × bh × hl = 12 × 20 × 15

⇒ (lbh)2 = 3600

⇒ lbh = 60 cm3.

10. a Volume of water to be filled = 50 × 44 × 7 × 10–2

Water flowing per hour out of pipe = π × (7 × 10–2)2 ×5 × 103

∴ Total required time

–2

2 2 3

50 44 7 102 hours.

227 7 10 10 5 10

7− −

× × ×= =× × × × × ×

11. b 3 cm

24 cm

18 c

m

Length of box = 24 – (2 × 3) = 18 cmWidth of box = 18 – 2 × 3 = 12 cmHeight of box = 3 cm∴ Surface area of box = 18 × 12 + 2(12 × 3 + 3 × 18)= 216 + 180 = 396 cm2.

12. d Volume of the box = (56 × 35 × 28) cm3

Volume of a soap cake = (8 × 5 × 4) cm3

∴ Number of soap cakes

56 35 28343

8 5 4

× ×= =

× ×.

13. c Let r be the radius of the hole. ∴ Thickness t + r = 10 ⇒ r = 10 – t

Internal volume ( )21V r 1000= π = ( )21000 10 tπ −

Volume of solid cylinder V = ( ) ( )210 1000π

= ( )21000 10π

( )( )

21

2

10 tV 1

V 410

− ∴ = =

⇒ 10 t 1

10 2

−= ⇒ t = 5

∴ Thickness of the shell is 5 cm.

14. a Number of turns = Length of cylinder 42

= = 140 Diameter of wire 0.3

Length of wire in each turn22 49

= 2 = 154 cm 7 2

× ×

Length of total wire = 154 × 140 cm = 215.6 m .

15. a Area of curved surface = 2πrh

= 2 × π × 21 × 90 227

π = � = 11880 cm2.

16. b Volume of each marble 34

r3

= π3

4 1.43 2

= π

Volume of water that rises in the cylinder = πR2h

27

. (5.6)2

= π ×

Total number of marbles

7 75.6

2 24 1.4 1.4 1.43 2 2 2

π × × ×=

× π × × ×

= 150.

17. b Volume of water through circular pipe = 21 1r hπ

Volume of water through cistern = 22 2r hπ

Time = volume of cistern

volume of water/sec through pipe

22 221 1

r h

r h

π=

π

5 5 210 10 5

3100 100 18

× ×=× × ×

5 18 12 1003

× × ×= = 6000 sec.

60003600

= = 1 hour 40 minutes.

18. d Volume of water column = n × volume of a sphericalmarble (where, n is the total number of sphericalmarbles submerged).

32 4 2.1

(9) 4.2 n3 2

⇒ π × × = × × π ×

9 9 4.2 3 9 9 2n 231.

2.1 2.1 0.742 2

× × × × π × ×⇒ = = ≈π × × ×

19. c Volume of earth dug out = 2 3(14) 210 mπ × ×

Volume of earth dug outRaise in field level

Area of rec tangular field=�

22 14 14 210 22 14410 cm.

7 140 225 75× × × ×= = ≈× ×


20. b 2 r 22π =

222 r 22

7⇒ × × =

7r

2∴ =

Volume of cylinder = 2r hπ

322 7 712 462 cm

7 2 2= × × × =

21. b If the height of increased water level be h cm, then

2 34r h R

3π = π

412 12 h 6 6 6

3⇒ × × = × × ×

4 2 6 6h 2 cm

12 12

× × ×⇒ = =

×

22. c Initial area of the cylinder = 2r hπ

Volume of the new cylinder = 2(1.1r) 1.1hπ ×

= 1.331 2r hπ

∴ Incresae in area = (1.331 – 1) 2r hπ

= 0.331 2r hπ

∴ Percentage increase = 2

2

0.331 r h100 33.1%

r h

π× =

π

23. a Area of the curved surface = 1

4623

× = 154 sq. cm.

22 rh 2 r 462∴ π + π =

2154 2 r 462⇒ + π =

22 r 462 154 308⇒ π = − =

2 308 308 7r 49

2 2 22

×⇒ = = =

π ×

r 49 7 cm⇒ = =

24. d Volume of water drawn from cylinder

2 22 35 35r h h

7 2 2= π = × × ×

322 35 35h 11000 cm

7 2 2∴ × × × =

11000 7 2 2h

22 35 35

× × ×∴ =

× ×

80 311 cm.

7 7= =

25. b Circumference of the base of the cone = 2π r = 44 m⇒ r = 7 m, height = 9 m∴ Volume of air contained in the conical tent

= 31

πr²h = 722

31 × × 7 × 7 × 9 = 22 × 21 = 462 m3.

26. a R = h = 3r

21V R h

3= π

21V (3r) (3r)

3⇒ = π

2 31(9r 3r) 9 r .

3= × π × = π

27. d Let the radius of their base be ‘r’ and the height be h.

Slant height of the cone 2 2l h r= +Then, 2πrh = πrl

⇒ 2 22 rh r h r π = π +

⇒ 2 22h h r= +

⇒ 4h2 = h2 + r2

r3 i.e. 3 :1.

h⇒ =

28. b We have r R

= h H

or R 3

= r 2

, V1 = πr²h,

V2 = π R²H

22

21

V R HV r h

∴ = = 827

23

23

23 =××

32

27V 800 2700 cm

8⇒ = × =

Additional water = 2700 – 800 = 1900 cm3.

29. a Volume of cone 21r h

3= π

16 6 24

3= × π × × ×

Volume of sphere 34r

3= π

Hence, 34

r3

π 16 6 24

3= × π × × ×

⇒ r = 6 cm.


30. a Volume of cone

21 1 22r h 1 7

3 3 7= π = × × ×

22

3= cu.cm.

Volume of cubical block= 10 x 5 x 2 cm3. = 100 cm3.Wastage of wood

322100 cm

3 = −

3300 22 278 cm

3 3

−= =

∴ % Wastage =

278278 23 100 92 %

100 3 3× = =

31. d Let height and radius both of a cylinder change byx%, then volume changes by

2 3

2

3x x3x %

100 100

+ +

3 20 20 20 20 203 20 %

100 10000

× × × × = × + + = (60 + 12 + 0.8)% = 72.8%

32. a Volume of the cone 21r h

3= π , new height = h + 100%h

∴ Percentage increase in volume = 100%

33. d Radius of larger sphere = R units

∴ Its volume = 34R

3π cu. units

Volume of smaller cone

31R

3= π cubic units

Volume of smaller sphere

31R

3= π

3 34 1r R

3 3∴ π = π

33

3

R Rr r

4 4⇒ = ⇒ =

∴ Surface area of smaller sphere∴ Surface area of larger sphere

2 24 r : 4 R= π π 2 2r : R=

( )2

22 33

R: R 1: 4

4

= =

( )2 41

2 331: 2 1: 2 = =

34. d Area of the curved surface = rlπwhere,

2 2 2 2l r h (32) (60) 4624 68 cm= + = + = =

Area of the curved surface

22rl 32 60

7= π = × ×

∴ Total cost of painting

22 135 32 60

7 10000= × × × × = �23.94 approx.

35. b Volume of the copper sphere = Volume of the wireLet ‘l’ be the length of the wire. Then,

34

× π × 6 × 6 × 6 = π × (0.4)2 × l

⇒ l = 16.0

21634

×π

×π = 1,800 cm = 18 m.

36. d Surface area of sphere = 4πr2

Volume of sphere = 34

r3

π

2 344 r r

3∴ π = π ⇒ r = 3 unit.

34Volume (3) 36 unit

3∴ = π = π

Now, let the radii of the 4 spheres be x, 2x, 2x and 4x.

∴ Total volume = 3 3 3 34

x (2x) (2x) (4x)3

π + + +

34r (1 8 8 64)

3= π + + + =

34(81x )

3π

3436 81 x

3∴ π = π × × 3 27 1

x81 3

⇒ = =

⇒

131

x3

=

Therefore, radius of largest sphere = 3

4unit.

3


37. d Volume of hemisphere 32r

3= π

So,

31

32

2r 6.43

2 21.6r3

π=

π

⇒3

1

2

r

r

364 4

216 6 = =

⇒ 1

2

r

r4 26 3

= =

i.e. 2 : 3.

38. bcurved surface area of hemisphere

Ratiocurved surface area of cone

=

2

2 2

2 r

r r h

π=π +

[ r h]=�

2 :1.=

39. d Volume of bigger sphere 34(8)

3= π

Let n be the number of smaller spheres.

Then, 3 34 4

n (4) (8)3 3

× π = π

n 8.⇒ =

40. b Volume of sphere = Volume of wire

⇒ 2 2(r) l (0.1) 3600π × = π × ×

Let radius of sphere be r.

Then, 34

r 36 r 3cm.3

π = π × ⇒ =

41. d Volume of the sphere = 34r

3π

or 3 34 88r (14)

3 21π = ×

or 3 34 22 4 22

r (14)3 7 3 7

× × = × ×

or r = 14The curved surface of the sphere = 4πr2

2224 14 14 2464 cm .

7= × × × =

42. a Let the volume be 8x3 and 27x3

⇒ Their radii are 2x and 3x∴ The ratio of their surface area= 4x2 : 9x2 = 4 : 9

43. c Original surface area of sphere = 24 rπ⇒ Surface area of sphere = 4π(2r)2

= 16πr2 = 4 × 4πr2 = 4 (original surface area)

44. c Surface area of sphere = 24 rπ

2224 r 346.5

7∴ × × =

24 22 r 346.5 7⇒ × × = ×

2 346.5 7r 27.5625

4 22

×⇒ = =

×

r 27.5625 5.25 cm∴ = =

45. b Lateral surface area of prism = Perimeter of base ×height= 4 × 8 × 6 = 192 cm2.

46. c Volume of a pyramid 1

base area height3

= × ×

136 3 basearea 9

3⇒ = × ×

⇒ Base area 212 3 cm=

�Base area 23a

4= (where a is one side of the

base of the pyramid)

23a 12 3

4∴ =

2a 48⇒ =

a 4 3 cm⇒ =

∴ Perimeter of the base of the pyramid

3 4 3 12 3 cm.= × =


47. b

A B

CD

E

F

Height of the triangle = 2 215 8+

225 64 289= + == 17 cm∴ Area of the lateral surface of pyramid= 4 × Area of triangle

= 4 × 1

2 × base × height

= 4 × 1

2 × 16 × 17 = 544 cm2.

48. a Area of the base of prism = 3

4 × 6 × 6 = 9 3 cm2.

∴ Volume = Area of base x height

81 3 9 3 height⇒ = ×

81 3height 9 cm.

9 3⇒ = =

49. d Total surface area= Perimeter of base × height + 2 × area of base

= 36 × 10 + 2 × 3

4× 12 × 12

= 360 + 72 3

= 72 (5 + 3 ) cm2.

50. a Area of the tetrahedron

= 1

3 × area of base × height

Area of the base = 3

4 × (side)2

3 9 33 3

4 4= × × = cm2

Now, length of the perpendicular In the equilateraltriangle

22 3 9 3 3

3 9 cm2 4 2

= − = − =

Height =

2 23 3 3

2 2

−

27 36 cm

4 4= − =

∴ Required area = 1 9 3

63 4

× × 39 2 cm .

4=


Quantitative Aptitude – 4Geometry - Quadrilaterals, Circles & Coordinate Geometry


1. c

A

E

C

D 40º

B

80º 60º

60º

40º40º

40º

40º

In ∆BEC, BEC∠ = 80º and EBC∠ = 60º.

ECB 40º∴ ∠ = , ADB ACB∠ = ∠ = 40º (Angles in

the same segment). Also, BDC∠ = 40º (given).

∴ BD bisects ADC∠ . AB and BC subtend 40º at

the circumference.∴ AB = BC.

DA subtends 40º at C (... BEC∠ = 80º, EDC∠ = 40º)

DC subtends 60º at A (... DAC DBC 60°∠ = ∠ = ).

∴ DA ≠ DC. ACB∠ = 40º and DCA∠ = 40º.

∴ AC bisects ∠ BCD.

2. d A D

B CE F

5 cm

9 cm

Area of trapezium ( )1AD BC DF

2= + ×

⇒ 35 ( )15 9 DF

2= + ×

⇒ 35 = 7 × DF⇒ DF = 5 cm

Now, BC EF

FC2−= and EF = AD = 5 cm

⇒9 5

FC 2cm2−= =

2 2CD DF FC∴ = +

2 25 2= + 25 4 29 cm.= + =

3. b

P R

S

Q

O

90 °

120 °

6 cm

PQR 120PQO = 60

2 2∠ °∠ = = °

In ∆POQ,∠OPQ = 180º – 90º – 60º = 30º

sin ∠OPQOQPQ

=

⇒ OQ = PQ sin 30º 1

6 32

= =

∴ QS = 2(OQ) = 2 × 3 = 6 cm. {�OS = OQ}

4. b Sum of the angles of a regular polygon of n sides= (2n – 4) × 90º∴ (2n – 4) × 90º = 1080º⇒ 2n – 4 = 1080 ÷ 90 = 12∴ 2n = 12 + 4 = 16⇒ n = 8.

1 c 2 d 3 b 4 b 5 d 6 c 7 b 8 b 9 d 10 a11 c 12 a 13 b 14 c 15 a 16 a 17 a 18 a 19 c 20 a21 a 22 d 23 c 24 c 25 b 26 a 27 b 28 c 29 b 30 c31 a 32 d 33 b 34 b 35 b 36 d 37 c 38 c 39 d 40 d41 c 42 b 43 d 44 d 45 b 46 c 47 b 48 a 49 c 50 c

P-2 (S)


5. d If the number of sides of regular polygon be n, then− × ° = °(2n 4) 90

150n

⇒ 3(2n – 4) = 5n⇒ 6n – 12 = 5n ⇒ n = 12.

6. c Sum of interior angles of regular polygon = (2n – 4)× 90ºWhere n = number of sides∴ (2n – 4) × 90º = 1440º

⇒ 2n – 4 = 1440

90 = 16

⇒ 2n – 4 = 16 ⇒ 2n = 16 + 4 = 20

⇒ n = 202

= 10.

7. b AD||BC ⇒ AD||BQ

E

Q

BA

D C

Point B is the mid-point of AE.∴ Q is the mid-point of DE,In ∆s DQC and BQE,∠DQC = ∠BQE∠DCQ = ∠QBE∠CDQ = ∠QEB∴ Both triangles ∆ DQC and ∆ BQE are similar.

∴ DQQE

= CQBQ

= 1 : 1.

8. b D

A

C

BO

∠ACB = 90º (Angle of semi-circle)∠CAB = 30º∴ ∠CBA = 180º – 90º – 30º = 60ºAgain, ∠ADC + ∠ABC = 180º∴ ∠ADC = 180º – 60º = 120º

9. dD C

BA

The sum of opposite angles of a concylic quadrialteral= 180º∴ ∠A + ∠C = ∠B + ∠D = 180º

10. a

CD

A B

P

Q

AB = BC = CD = DA{ABCD is a rhombus}

DP = = =1 1 1AB BC CD

2 2 2=

1DA

2In ∆s APQ and BCQ,

= = ∠ =P QCB; A QBC : Q Q

∴ ∆ APQ and ∆ BCQ are similar.

∴ +AB BQ

BQ =

+AD DPBC

⇒ +AB1

BQ = =

3BC 32

BC 2

⇒ ABBQ

= − =3 11

2 2

⇒ BQAB

= 21

⇒ 2 : 1.

11. cD G C

F

BA

H

E

O a

The length of the diagonal of ABCD is 2a.By midpoint theorem, the length of each side of square

EFGH is 1

2a a.2

× =

12. a Let O be the centre of the circles. If AB touches theinner circle at D, then OD is perpendicular to AB and Dis the midpoint of AB.∴BD2 = OB2 – OD2 = 132 – 52 = 144 ⇒ BD = 12 cm.∴ AB = 24 cm.

A B

13 cm

5 cm

O

D


13. b We have, ∆TQR similar to ∆PTR,(∠R is common ∠QTR = ∠TPQ, alternate segment)

TQ PT PT × QR = TR =

QR TR TQ⇒ ⇒ = cm5.4

463 =×

.

14. c

O

A

C

D

B

M x6 cm

8 cm

2 cm

r

L

r

Let r be the radius of the circle. Then,

In AMO,∆ 82 + x2 = r2 ... (i)

In CLO,∆ 62 + (2 + x)2 = r2 ... (ii)

Solving (i) and (ii), we get x = 6 cm and r = 10 cm∴ Diameter = 2 × 10 = 20 cm.

Alternate:

2 2 2 2r 6 r 8 2 r 10 cm.− − − = ⇒ =

15. a ABC is a tangent to circle at B. BD is a chord throughB. Hence, ∠DBC = ∠BED = 36º (Angle between thechord and tangent is equal to the angle in the alternatesegment.)∠DBE + ∠BED + ∠BDE = 180º (Sum of the angles ofa triangle)

BDE 180 {36 62 } 180 98 82° ° ° ° ° °∴∠ = − + = − = .

16. a

A

P

BN

∠APB = 90º ; BN = xAP = 8 cm ; BP = 6 cmAB2= AP2 + BP2

= 64 + 36 = 100⇒ AB = 10 cm

Now, PNA PNB 90∠ = ∠ = ° {� N is the foot of theperpendicular drawn from the point P on AB}⇒ BP2 = BN2 + NP2

and AP2 = AN2 + NP2

∴ AP2 – BP2 = AN2 – BN2

⇒ 64 – 36 = (10 – x)2 – x2

⇒ 28 = 100 + x2 – 20x – x2

⇒ 72x 3.6 cm.

20= =

17. a A

B CD

EAD is the bisector of ∠BACAB BD

.AC DC

=

18. a C

P

D

A

5 cm2 cm

B

∆BDP ~ ∆ACPAC AP 5

.BD BP 2

= =

19. c A

O

P B∠PAO = ∠OBP = 90ºand ∠AOB = 120º⇒ ∠APB = 60º {�Sum of all angles of a quadrilateralis 360º}Also, line PO bisects ∠APB equally in two parts.

APO 30∴∠ = °APB 60

2 :1.APO 30

∠∴ = =∠

20. a C

TS

A B

R

OQ3 2°

P

∠OQA = ∠OPA = 90º∠QOP + ∠QAP = 180º⇒ ∠QOP = 180º – 32º = 148º∠QOP = ∠SOR = 2∠RTS {�Angle made by the arcon the circumcircle is half the angle made at the cen-tre}

∴ ∠RTS = 148

74 .2

= °


21. a MBA and MDC are secants to the circle from the samepoint M.∴ MB × MA = MD × MC⇒ 4 × (MB + BA) = 5 × MC⇒ 4 × (6 + 4) = 5 × MC⇒ 4 × 10 = 5 × MC⇒ MC = 8 cm∴ CD = MC – MD = 8 – 5 = 3 cm.

22. d

P Q T R S

X

l

Join XT perpendicular to line � .QT = RT ...(i)PT = ST ...(ii)(Perpendicular from the centre of a circle to a chordbisects the chord.)Subtracting (i) from (ii), we getPT – QT = ST – RT⇒ PQ = SRHence, SR = 3 cm.

23. c Let each side of the triangle be a cm.

A

CDB

E

GR

r

Then, 2 2AD AB BD= −

2 aa

4

2 = −

3acm

2=

3 2AG a

2 3∴ = × =

�

�

cm

Let radius of circumcircle be R cm and radius of incirclebe r cm.Then, R = AG

aR cm.

3∴ =

and 2 2r AG AE= −2 2a a

3 4= −

acm

2 3=

∴ πR2 – πr2 = 44

2 222 a a44

7 3 12

⇒ − =

2a4

⇒7

4422

= ×

2 7a 44 4 56

22⇒ = × × =

∴Required area of the triangle 23a

4=

356

4= × 214 3 cm .=

24. cP

R

A

B

O

QS

OP = OR = 5 cm, PQ = 6 cm and RS = 8 cm.

Since OA PQ⊥ and OB RS⊥ , then

PA = QA = 3 cm and RB = SB = 4 cm.In ∆OAP, right-angled at A,OP2 = OA2 + PA2

⇒ 52 = OA2 + 32 ⇒ OA2 = 25 – 9 = 16

⇒ OA = 4 cm

Similarly, in OBR,∆ OR2 = OB2 + RB2

⇒ 52 = OB2 + 42 ⇒ OB2 = 25 – 16 = 9

⇒ OB = 3 cmNow AB = OA – OB = 4 – 3 = 1 cm.

25. b OQ = OR = OP (Radii of the same circle)∴ ∠ORQ = ∠OQR = 40ºSimilarly, ∠OPQ = ∠OQP = 30º∠PQR = ∠OQP + ∠OQR = 40º + 30º = 70ºSince ∠POR = 2∠PQR = 2 × 70º = 140º (Angle at thecentre is twice the angle at circumference).

26. a

A B

C

O

DLet A and B be the centres of the two circles.

CD AB.⊥

5AO OB and CO OD

2∴ = = =

Since AC = 5 cm


2 2OC AC OA= −

22 5

52

= −

2525

4= − 75 5 3

cm4 2

= =

∴ Common chord CD = 2(CO) 5 3 cm.=

27. b

A B

C

D

O

3cm 3cm

45°

Let AB be the chord of length 6 cm.∠ABC = 45º

⇒ ∠ABO = 45º { OBC 90 }∠ = °�

BD = 3cm

cosBD

OBDOB

∠ =

⇒ cos 45º = 3

OB ⇒

1 3OD2

=

⇒ OB = 3 2 cm.

28. c

A BD

O

4 cm5 cm

∆ADO is a right-angled triangle, right-angled at D.So, AD2 + OD2 = OA2

AO = 5 cm, OD = 4 cmHence, AD = 3 cm.AB = 2(AD) = 6 cm.

29. b Let ABCD be a square and AB = 2 cmAP = PB = BQ = QC = CR = RD= DS = SA = 1 cm

A D

CB

S

Q

P R

1 cm

1 cm

1 cm

1 cm

1 cm1 cm

1 cm 1 cm

Area of the portion bounded by the 4 arcs= Area of sq. ABCD – 4 × (Area of quadrant)

( )212 2 – 4 1

4 = × × × π ×

= (4 – π) sq. cm.

30. c Transverse common tangent

P S

R Q

= − +2 21 2a (r r )

= − +2 2(24) (5 3)

= −576 64 = 512

= 16 2 cm

31. aA

C

B

DLength of the direct common tangent

= − −2 21 2(dis tance between two centres) (r r )

= − −2 2(13) (7 2)

= − = =169 25 144 12 cm.

32. d D A E

B C

DE | | BC∴ ∠DAB = ∠ABCAlso ∠ACB = ∠DAB[angle between the chord and tangent is equal tothe angle made by the chord in the alternatesegment]In ∆ABC∠ABC = ∠ACB⇒ AB = AC = 17 cm.


33. b

B

A

O P

∠AOB = 110º∠OAP = 90º∠OBP = 90º∴ ∠APB + ∠AOB + ∠OAP + ∠OBP = 360º⇒ ∠APB + 110º + 90º + 90º = 360º⇒ ∠APB = 180º – 110º = 70º

34. b A

B C

32 + 42 = 52 (Pythogorean Triplet]∆ ABC is a right angled triangle.∠ B = 90º = angle at the circumferenceSince, Diameter of circle = 5 cm∴ Circum radius = 2.5 cm.

35. b Simply, check through options & find PythagoreanTriplet

A

B CO

182 + 242 = 302

∴ Sides = 18, 24, 30

36. d P

QR

2 + r

r

3 – r

r

r

r

r

3 – r

2 + r

PR2 = PQ2 + PR2 = 32 + 42 = 25

∴ PR = 25 = 5 cm

r = Area of triangle

Semiperimeter of triangle

=

× ×= =+ +

13 4 62 1cm

3 4 5 62

[Or, in the above figure, QR = r + 2 + r = 4 ⇒ r = 1]

37. cA

BC

D

P

O

Join CB.∠AOC + ∠BOD= 2(∠ABC + ∠BCD)In ∆BCP∠PBC + ∠BCP = ∠BPDexterior angle = sum of interior opposite angles

⇒ ∠BPD = (50 40)

2+

= 45.

38. c

O

A

B

C

D

P

∠ACP = 1

AOD 10º2

∠ =

∠CAB = 1

COB 15º2

∠ =

In ∆APC∠BPC = ∠ACP + ∠CAP = 25º(Exterior angle)

39. d P

Q R

O

∠QOR = 110º∠OPR = 25º∴ ∠QPR = 110º ÷ 2 = 55º


OR = OP∴ ∠OPR = ∠PRO = 25º

∴ ∠OQR = ∠ORQ = = °7035

2∴ ∠PRQ = 25º + 35º = 60º

40. d A

CB

D

GE

H

FAE = AHBE = BFGC = FCGD = HD⇒ AE + BE + GC + GD= AH + BF + FC + HD⇒ AB + CD = AD + BC

⇒ 6 + 3 = AD + 7.5

⇒ AD = 9 – 7.5 = 1.5 cm

41. c

(0 , 3 )y = 3

(4 , 0 )

C

A B X

x = 43x + 4y = 12

Y

Area of triangle1

(AB)(AC)2

=1

(4)(3)2

= = 6 sq. units.

42. b

(3 , 0 )

(0 , 4 )

A

BC O

Y

X

∠AOB = 90º∴ AB = diameter {if angle made by chord on the circleis 90º, then the chord is the diameter of the circle}

2 2AB 3 4 5⇒ = + =∴ radius = 2.5.

43. d

(0 , 4 )

x – y + 1 = 0

(2 , 0 )

y

A

C x

2x + y = 4

(x , y )1 1

(0 , 1 )

D

(–1, 0)

B

Required altitude = ADSince point A is the intersection of linesx – y + 1 = 0 and 2x + y = 4∴ Solving the two equations, we getx1 = 1 and y1 = 2∴ AD = 2 units.

44. d Y

X

Q (5 , 4 )

P(2 , 0 )

2 2PQ (5 2) (4 0)= − + −

⇒PQ 9 16 5= + =Area of circle with radius PQ = π(PQ)2

= π(5)2 = 25π sq. units.

45. b Equation of line is x = 3y – 7.Points (a, b) and (a + 3, b + k) are lying on the line.Then,a = 3b – 7 ⇒ a – 3b = –7and a + 3 = 3(b + k) – 7⇒a + 3 – 3b – 3k + 7 = 0⇒a – 3b + 7 + 3 – 3k = 0⇒3 – 3k = 0 {� a – 3b + 7 = 0}⇒ k = 1.

46. c y

x ’

y ’

x

C

B D O

A

(1 ,0) (3 ,0)

(0 ,1)(0 ,2)

x = 0 is the equation of y-axis.y = 0 is the equation of x-axis.Putting x = 0 in x + y = 1, y = 1Putting y = 0 in x + y = 1, x = 1Putting x = 0 in 2x + 3y = 63y = 6 ⇒ y = 2


Putting y = 0 in 2x + 3y = 62x = 6 ⇒ x = 3OB = 1; OA = 1OD = 3; OC = 2Required area = ∆OCD – ∆OAB

= 1 1

3 2 1 12 2

× × − × ×

= 1

32

− = 1

22

sq. units.

47. b 2x + 1 = 0 ⇒ x = 12

−

and 3y – 9 = 0 ⇒ y = 3

∴ 1,3

2−

48. a 3x + 2y = 18 ... (i)3y – 2x = 1 ... (ii)Equation (i) × 2 + (ii) × 3 gives,

6x 4y 36

6x 9y 3

13y 39

+ =− + =

=

⇒ y = 3Putting y = 3 in (ii)3y – 2x = 1 ⇒ x = 4∴ (p, q) = (4, 3)and hence, p + q = 7

49. c Y

Y ’

X ’ X

P

O

Q

Equation of a straight line parallel to x-axis ; y = aHere, a = – 3∴ Equation is : y = – 3

50. c 3x + 4y = 10 ... (i)– x + 2y = 0⇒ x = 2y∴ From equation (i),3 × 2y + 4y = 10 ⇒ 10y = 10

⇒ y = 1010

= 1

∴ x = 2∴ (a, b) = (2, 1)∴ a + b = 2 + 1 = 3


Quantitative Aptitude – 5Trigonometric Ratios and Identities


1. b3sin62 sec 42cos28 cosec48

° °−° °

=( ) ( )3cos 90 62 cosec 90 42

cos28 cosec48

° − ° ° − °−

° °

= 3cos28 cosec 48cos28 cosec 48

° °−° ° = 3(1) – 1 = 2.

[� sin (90º – θ ) = cos θcosec (90º – θ ) = sec θ ]

2. c tan5 tan25 tan30 tan45 tan65 tan85° ° ° ° ° °

( )( )tan5 tan85 tan25 tan65 tan30= ° ° ° ° ° [� tan 45 = 1]

( ) ( )tan5 tan 90 5 tan25 tan 90 25 tan30 = ° ° − ° ° ° − ° °

( )( )tan5 cot5 tan25 cot25 tan30= ° ° ° ° °

1 11 1 .

3 3= × × =

3. a tan1 tan2 tan3 ...tan89° ° ° °

( ) ( ) ( )tan 90 89 tan 90 88 tan 90 87

...tan87 tan88 tan89

= ° − ° ° − ° ° − °

° ° °

cot89 cot88 cot87 ...tan87 tan88 tan89= ° ° ° ° ° °

( )( )( )cot89 tan89 cot88 tan88 cot87 tan87

...(cot 44 tan44 )tan45

= ° ° ° ° ° °

° ° °= 1 × 1 × 1 ... × 1 × 1 = 1.

4. c ( ) ( )4 4 2 2

2

2 cos 60 sin 30 tan 60 cot 45

3sec 30

° + ° − ° + °

+ °

( ) ( )24 4 2 21 1 2

2 3 1 32 2 3

= + − + +

( )1 1 42 3 1 3

16 16 3 = + − + + ×

1 12 4 4 .

8 4= × − + =

5. c2

1 sin 1 sin 1 sin

1 sin 1 sin 1 sin

+ θ + θ + θ× =− θ + θ − θ

2

1 sin 1 sin 1 sincos cos coscos

+ θ + θ θ= = = +θ θ θθ

sec tan .= θ + θ

6. c

A

12

√3

sinAcot A

1 cosA+

+

13 3 121 13 2 3

12

= + = +++

2 3 3 1

2.2 3

+ += =+

Alternate method:

cos A sinA(1 cos A)sinA (1 cos A)(1 cos A)

−++ −

2

cosA sinA(1 cosA)sinA 1 cos A

−= +−

cosA 1 cosAsinA+ −= = cosec A = 2.

1 b 2 c 3 a 4 c 5 c 6 c 7 c 8 d 9 d 10 a11 b 12 c 13 a 14 c 15 a 16 a 17 a 18 b 19 b 20 c21 c 22 b 23 a 24 c 25 b 26 b 27 a 28 c 29 c 30 b31 a 32 d 33 a 34 d 35 a 36 b 37 c 38 c 39 d 40 d41 a 42 b 43 a 44 b 45 a 46 d 47 d 48 b 49 c 50 d

P-2 (S)


7. c 3tan

4θ =

22 3

sec 1 tan 14

⇒ θ = + θ = +

9 51

16 4= + =

1 4cos

sec 5⇒ θ = =

θ

4 111 cos 15 5 .

4 91 cos 915 5

−− θ∴ = = =+ θ +

8. d2 2 25sin 30 cos 45 4 tan 602sin30 cos60 tan45

° + ° + °° ° + °

( )22 21 1

5 4 32 2

1 12 1

2 2

+ + =× × +

5 112

4 21

12

+ +=

+

5 2 484

1 22

+ +

= +55

.6

=

9. d3

A2ππ < < exists in III quadrant.

Hence, sin A is negative and tan A is positive.

2 1 3sinA 1 cos A sinA 1

4 4= ± − ⇒ = − − = −

3 2sin A , cosecA

2 3= − = −

andsinA 3 2

tanA 3cos A 2 1

= = − × − =

2 2 44 tan A 3cosec A 4 3 3 8.

3∴ − = × − × =

10. a sin(A – B) = 12

⇒ sin(A – B) = sin30º⇒ A – B = 30º ... (i)

and cos(A + B) = 12

⇒ cos(A + B) = cos60º⇒ A + B = 60º ...(ii)From (i) and (ii), we getA = 45º and B = 15º.

11. b cos(40º + x) = sin30º⇒ cos(40º + x) = sin(90º – 60º)⇒ cos(40º + x) = cos60º ⇒ 40º + x = 60º⇒ x = 20º.

12. c ( )24 2 2 2cot A cot A cot A cot A+ = +

2 2 2[cosec A 1] (cosec A 1)= − + −

( )2 21 cot A cosec A+ =�

4 2 2cosec A 2cosec A 1 cosec A 1= − + + −

4 2cosec A cosec A.= −

13. a Using 2sin sin cos( ) cos( )α β = α − β − α + β ,

we get 2sin 3A sin A= cos(3A – A) – cos(3A + A)= cos2A – cos4A.

14. c Using C D C DsinC sinD 2sin cos

2 2− + − =

andC D C D

cosC cosD 2sin sin2 2+ − − = −

We get,

A B A B2sin cos

sinA sinB 2 2A B A Bcos A cosB 2sin sin

2 2

− + − =

+ −− −

A Bcot .

2+ = −

15. a( )

( )22 2 2

2 2 42

2sin cos(sin2 ) 4sin cos

(1 cos2 ) 4cos2cos

θ θθ θ θ= =+ θ θθ

22

2sin

tan .cos

θ= = θθ

16. asin (cos 1) sin (cos 1)sin (cos 1) sin (cos 1)

θ − θ − θ − θ −×θ + θ − θ − θ −

2 2

2 2

sin cos 2cos 1 2sin (cos 1)

sin cos 2cos 1

θ + θ − θ + − θ θ −=θ − θ + θ −

2

2 2cos 2sin 2sin cos

2cos 2cos

− θ + θ − θ=− θ + θ

2(1 cos ) 2sin (1 cos )2cos (1 cos )

− θ + θ − θ=θ − θ

2(1 sin )2cos

+ θ=θ

1 sin.

cos+ θ=

θ


17. a Given sin2 α = cos3 α

cot6 α – cot2 α 6

26

coscot

sin

α= − αα

42

6

sincot

sin

α= − αα

22

1cot

sin= − α

α= cosec2 α – cot2 α = 1.

18. b (1 + tanθ + sec θ)(1 + cot θ – cosec θ)

sin 1 cos 11 1

cos cos sin sinθ θ = + + + − θ θ θ θ

cos sin 1 sin cos 1cos sin

θ + θ + θ + θ − = θ θ

2(sin cos ) 1sin cosθ + θ −=

θ θ

1 2sin cos 12.

sin cos+ θ θ −= =

θ θ

19. bA A

sin A 2sin cos2 2

=

2 A1 cos A 2cos

2+ =

2

A A2sin cossinA A2 2 tan .

A1 cos A 22cos2

∴ = =+

20. c We have,

2

2 2

2 2

cos1

cot 1 sincot 1 cosec

θ −θ − θ=θ + θ

2 2 22 2

2

cos sin sincos sin

1sin

θ − θ θ= × = θ − θθ

cos2 .= θ ( )2 2cos2A cos A sin A= −�

21. c We have, sin(60 ) sin(60 ) 2cos60 sin° + θ − ° − θ = ° θ

[ 2cos A.sinB sin(A B) sin(A B)]= + − −�

12 sin sin .

2= × × θ = θ

22. btanA tanB

tan(A B)1 tanA tanB

++ =−

We have, 7tan A

8= and

1tanB

15=

7 1105 88 15tan(A B)

7 1 120 718 15

+ +∴ + = =−− ×

113tan(A B) 1 tan

113 4π⇒ + = = = tan 1

4π =

�

A B .4π∴ + =

23. a 63 14’ 51”

'51

63 1460

= ° +

'17

63 1420

= ° +

'280 17

6320

+ = °

'297

6320

= °

29763

20 60

° = + × 99

63400

° = +

25299400

° =

c c25299 281

.400 180 8000

π π = × =

24. c un = cosnα + sinnα∴ u6 = cos6α + sin6α(cos2α)3 + (sin2α)3 = (cos2α + sin2α)3 – 3cos2αsin2α(sin2α + cos2α)[� a3 + b3 = (a + b)3 – 3ab(a + b)]= 1 – 3 cos2α sin2αu4 = cos4α + sin4α.(cos2α)2 + (sin2α)2

= (cos2α + sin2α)2 – 2cos2α . sin2α= 1 – 2cos2αsin2α∴ 2u6 – 3u4 + 1= 2(1 – 3 sin2α cos2α) – 3(1 – 2sin2α cos2α) + 1= 2 – 3 + 1 = 0.

25. b (1 + sec 20º + cot 70º)(1 – cosec 20º + tan 70º)= (1 + sec 20º + tan 20º)(1 – cosec 20º + cot 20º)[� tan (90º – θ) = cot θ; cot(90º – θ) = tan θ ]

1 sin201

cos20 cos20° = + + ° °

1 cos201–

sin20 sin20° + ° °

1 cos20 sin20cos20

+ ° + °=°

sin20 1 cos20sin20° − + °×

°

( )2cos20 sin20 1

sin20 cos20

° + ° −=

° °

2 2cos 20 sin 20 2sin20 cos20 1sin20 cos20

° + ° + ° ° −=° °

= 2. [� sin2θ + cos2θ = 1]


26. b 2 sin α + 15 cos2 α = 7⇒ 2 sin α + 15(1 – sin2 α) = 7⇒ 2 sin α + 15 – 15 sin2 α = 7⇒ 15 sin2 α – 2 sin α – 8 = 0⇒ 15 sin2 α – 12 sin α + 10 sin α – 8 = 0⇒ 3 sin α (5 sin α – 4) + 2(5 sin α – 4) = 0⇒ (3 sin α + 2)(5 sin α – 4) = 0

⇒ sin α = 45

{because sin α ≠ 23

− }

cos α = 16 3

125 5

− =

∴cot α cossin

α=α

335 .

4 45

= =

27. a sin 2θ = 12

= sin 30º

⇒ 2θ = 30º ⇒ θ = 15º∴ cos(75º – θ) = cos(75º – 15º)

= cos 60º = 1

.2

28. c P sinθ = 32

P cosθ 12

=

On squaring and adding,

P2 sin2θ + P2 cos2θ 3 14 4

= +

⇒ P2 (sin2θ + cos2θ) = 1⇒ P2 = 1⇒ P = 1.

29. c A and B are complementary angles.So, A + B = 90º ⇒ A = 90º – Bsin A = sin (90º – B) = cos Bcos A = cos (90º – B) = sin Btan A = tan (90º – B) = cot Bcot A = cot (90º – B) = tan BGiven,sin A cos B + cos A sin B – tan A tan B + sec2 A – cot2 B= sin2 A + cos2 A – cot B tan B + sec2 A – tan2 A= 1 – 1 + 1 = 1.

30. b sec θ + tan θ = 3

1 sin3

cos cosθ⇒ + =

θ θ

⇒ 1 + sin θ = 3 cosθ

⇒ 1 + sin2 θ + 2sinθ = 3cos2 θ

⇒ 1 + sin2 θ + 2sinθ = 3 – 3sin2 θ⇒ 4sin2 θ + 2sin θ – 2 = 0

⇒ 2sin2 θ + sin θ – 1 = 0

1sin

2θ = or sin θ = –1.

31. a 2 – cos2 θ = 3 sin θ cos θDividing by cos2 θ,2sec2 θ – 1 = 3tan θ⇒ 2(1 + tan2 θ) – 1 = 3tan θ⇒ 2tan2 θ – 3tan θ + 1 = 0

⇒ tan θ = 12

or tan θ = 1.

But tan 1θ ≠ { sin cos }θ ≠ θ�

1tan .

2∴ θ =

32. d sin θ + cos θ = 2 cos(90 )° − θ = 2 sinθ

Dividing both sides by sin ,θ

1 c ot 2+ θ =

cot 2 1.⇒ θ = −

33. a cos A + cos2 A = 1cos A = 1 – cos2 A = sin2ASquaring both sides,cos2A = sin4A∴ sin2A + sin4A = cosA + cos2A = 1.

34. d A.M. ≥ G.M.

2 22 24sec 9cos

4sec 9cos2

α + α⇒ ≥ α × α

2 24sec 9cos 2 6sec cos⇒ α + α ≥ × α α

2 2Least value of 4sec 9cos 12.∴ α + α =

35. a2 2sin A 1 cos A sin A cos A

1 11 cos A sinA 1 cos A 1 sin A

+− + − + −+ − +

2 2 2

2

1 cosA sin A 1 cos A sin A1 cosA sinA(1 cosA)

1 sinA cos A1 sinA

+ − − −= ++ −

+ −++

cosA(1 cosA) sinA(1 sinA)0

1 cosA 1 sinA+ += + +

+ += cosA + sinA.

36. b 1 – 2 sin2 θ + sin4 θ= (1 – sin2 θ)2 = (cos2 θ)2 = cos4 θ


37. c sin θ = 0.7

∴ cos θ = − θ = −2 21 sin 1 (0.7) = 0.51

38. c x.sin 60º tan 30º= sec 60º cot 45º

⇒ x × × = ×3 12 1

2 3

⇒ x = 2 × 2 = 4

39. d + θ + − θ1 11 sin 1 sin

2 2

= ( )+ ° + − °11 sin60 1 sin60

2

= + + −

1 3 31 1

2 2 2

= ( )+ + −12 3 2 3

2 2

= ( )1 14 2 3 4 2 3

2 2 2× + + −

= ( ) ( ) + + −

2 213 1 3 1

4

= ( )+ + −13 1 3 1

4

= = = °2 3 3cos30

4 2

= θ

cos2

40. d tan2 θ = 1 –e2

∴ sec θ + tan θ cosec θ= sec θ + tan2 θ tan θ cosec θ

= sec θ + tan2 θ sin 1cos sin

θθ θ

= sec θ + tan2 θ= sec θ (1 + tan2 θ)

= ( )1

2 221 tan (1 tan )+ θ + θ

= ( ) ( )3 3

2 22 21 tan 1 1 e+ θ + − = ( )−3

2 22 e

41. a x sin 60º tan 30º – tan2 45º= cosec 60º cot 30º – sec2 45º

⇒ 3 1

x. 12 3

−

= ( )× −22

3 23

⇒ − = − =x1 2 2 0

2

⇒ x

1 x 22

= ⇒ = .

42. b Expression

= ° + ° − °

° + °

2 2 2

2 2cos 60 4sec 30 tan 45

sin 30 cos 30

= + −

221 24 1

2 3

[� sin2 θ + cos2 θ = 1]

= + −1 161

4 3

= + − =3 64 12 55

12 12.

43. a 3 sin θ + 5 cos θ = 5 ...(i)Let 5 sin θ – 3 cos θ = x ...(ii)On squaring and adding both equations(3 sin θ + 5 cos θ)2 + (5 sin θ – 3 cos θ)2 = 52 + x2

⇒ 9 sin2 θ + 25 cos2 θ + 30 sin θ cos θ + 25 sin2 θ + 9cos2 θ – 30 sin θ cos θ = 25 + x2

⇒ 9 sin2 θ + 9 cos2 θ + 25 cos2 θ + 25 sin2 θ = 25 + x2

⇒ 9 (sin2 θ + cos2 θ) + 25 (cos2 θ + sin2 θ) = 25 + x2

⇒ 9 + 25 = 25 + x2

⇒ x2 = 9 ⇒ x = ± 3.

44. b tan x = sin 45º cos 45º + sin 30º

= 1 1 1 1 1

12 2 22 2

+ = + =

∴ tan x = tan 45º ⇒ x = 45º

45. a sin 3A = cos (A – 26º)⇒ cos (90º – 3A) = cos (A – 26º)⇒ 90º – 3A = A – 26º⇒ 90º + 26º = 3A + A⇒ 4A = 116º

⇒ A = 116

294

° = °


46. d Expression

= + θ − θ+− θ + θ

1 sin 1 sin1 sin 1 sin

= + θ + θ − θ − θ+− θ + θ + θ − θ

(1 sin )(1 sin ) (1 sin )(1 sin )(1 sin )(1 sin ) (1 sin )(1 sin )

= + θ − θ+− θ − θ

2 2

2 2(1 sin ) (1 sin )

1 sin 1 sin

= + θ − θ+

θ θ

2 2

2 2(1 sin ) (1 sin )

cos cos

= + θ − θ+

θ θ1 sin 1 sin

cos cos

= + θ + − θ =

θ θ1 sin 1 sin 2

cos cos = 2 sec θ.

47. d tan2 A + cot2 A – sec2 A, cosec2 A= tan2 A + cot2 A – (1 + tan2 A) (1 + cot2 A)= tan2 A + cot2 A – (1 + tan2 A + cot2 A + cot2 tan2 A= tan2 A + cot2 A – 1 – tan2 A – cot2 A – cot2 A tan2 A= – 1 –1 = – 2

[tan A cot A = 1]

48. b 4 cos2 θ – 4 cos θ + 1 = 0⇒ (2 cos θ – 1)2 = 0⇒ 2 cos θ – 1 = 0⇒ 2 cos θ = 1

⇒ cos θ = = °1cos 60

2

⇒ θ = 60º∴ tan (θ – 15º) = tan (60º – 15º) = tan 45º = 1

49. c 3 tan θ = 3 sin θ

⇒ θ = θθ

sin3 3sin

cos

⇒ = θ3 3cos

⇒ θ = =3 1cos

3 3

∴ sin θ = − θ21 cos

= − =1 21

3 3

∴ sin2 θ – cos2 θ = −

2 22 13 3

= − =2 1 13 3 3

50. d A

B C

60°

∠B = 90º∠A = 60º∠C = 180º – 90º – 60º = 30º

cos C = BCCA

⇒ cos 30º = BCCA

⇒ 3

2 =

BCCA

= 3 : 2 .


Quantitative Aptitude – 6Trigonometric Identities (Cont.), Heights and Distances


1. b Maximum value of a sin θ + b cos θ = +2 2a b

∴ Maximum value of 2 sin θ + 3 cos θ = 2 22 3 13+ = .

2. b tan θ = 34

∴ cot θ = 43

� cosec2 θ – cot2 θ = 1

⇒ cosec2 θ = + θ21 cot

= + = + =

24 16 251 1

3 9 9=

53

.

3. c cosecθ – cotθ = 72

...(i)

cosec2θ – cot2θ = 1⇒ (cosecθ + cotθ) (cosecθ – cotθ) = 1⇒ cosecθ + cotθ

= 1 2

cosec cot 7=

θ − θ ...(ii)

On adding both equations.

2cosecθ = +7 22 7

= + =49 4 53

14 14

⇒ cosecθ = 5328

4. c 5 tan θ = 4 ⇒ tanθ = 45

∴ θ − θθ + θ

5sin 3cos5sin 2cos

Dividing numerator and denominator by cos θ.

=

sin 3cos5

cos cossin 2cos

5cos cos

θ θ−θ θθ θ+θ θ

=

45 35 tan 3 5

45 tan 2 5 25

× −θ − =θ + × +

= − =+

4 3 14 2 6

5. b 2 cosec2 23º × cot2 67º – sin2 23º – sin2 67º – cot2

67º= 2.cosec2 (90º – 67º) × cot2 67º – sin2 23º – sin2 90º– 23º) – cot2 67º= 2 sec2 67º × cot2 67º – sin2 23º – cos2 23º – cot2

67º= 2 cosec2 67º – 1 – cot2 67º= cosec2 67º = sec2 23º

2 2

sec(90 ) cosec ;

sin(90 ) cos &

cosec (67 ) 1 cot 67º

° − θ = θ

° − θ = θ ° − =

�

6. d Expression

= cosec2 18º – °2

1

cot 72

= cosec2 18º – tan2 72º[� tan θ. cot θ = 1]= cosec2 18º – tan2 (90º – 18º)= cosec2 18º – cot2 18º = 1

[� tan (90º – θ = cotθ; cosec θ – cot2θ = 1]

7. c 4 tan2θ + 9 cot2θ= (2 tanθ – 3 cotθ)2 + 2 × 3 × 2∴ Minimum value = 12[� (2 tanθ – 3 cotθ)2 ≥ 0].

8. b sin 7x = cos 11x= sin (90º – 11x)⇒ 7x = 90º – 11x⇒ 18x = 90º⇒ x = 5º∴ tan 9x + cot 9x= tan 45º + cot 45º= 1 + 1 = 2

1 b 2 b 3 c 4 c 5 b 6 d 7 c 8 b 9 d 10 b11 b 12 a 13 b 14 b 15 d 16 c 17 a 18 b 19 c 20 b21 c 22 a 23 b 24 a 25 b 26 d 27 a 28 c 29 b 30 b31 a 32 b 33 d 34 b 35 b 36 d 37 d 38 b 39 a 40 c41 d 42 c 43 b 44 b 45 b 46 b 47 a 48 a 49 a 50 c

P-2 (S)


9. d cos 1º, cos 2º, cos3º, ..... cos 90º ..... cos 100º= 0 [cos90º = 0]

10. b secx = cosecy⇒ cosx = siny

⇒ sin π − =

x siny2

⇒ y = π − x2

⇒ x + y = π2

∴ sin (x + y) = sin π = 12

11. b cos2θ + cos4θ = 1⇒ cos4θ = 1 – cos2θ = sin2θ⇒ tan2θ = cos2θ∴ tan2θ + cos4θ = cos2θ + cos4θ = 1

12. a sec2θ + tan2θ = 7⇒ 1 + tan2θ + tan2θ = 7⇒ 2 tan2θ = 7 – 1 = 6

⇒ tan2θ = 3 ⇒ tanθ = 3 ⇒ θ = 60º.

13. b sinθ + cosecθ = 2

⇒ sinθ + =θ

12

sin⇒ sin2θ – 2 sinθ + 1 = 0⇒ (sinθ – 1)2 = 0⇒ sinθ = 1 ⇒ cosecθ = 1∴ sin100θ + cosec100θ= 1 + 1 = 2.

14. b cos2α + cos2β = 2⇒ 1 – sin2α + 1 – sin2β = 2⇒ sin2α + sin2β = 0⇒ sin2α = 0 & sin2β = 0⇒ α = β = 0∴ tan3α + sin5β = 0.

15. d cosθ + secθ = 2

⇒ cosθ + =θ

12

cos⇒ cos2θ – 2 cosθ + 1 = 0⇒ (cosθ – 1)2 = 0⇒ cosθ = 1⇒ secθ = 1∴ cos6θ + sec6θ = 1 + 1 = 2

16. c tan 7θ tan 2θ = 1

⇒ tan 7θ = = θθ

1cot 2

tan2⇒ tan 7θ = tan (90º – 2θ)⇒ 7θ = 90º – 2θ⇒ 9θ = 90º ⇒ θ = 10º

∴ tan 3θ = tan 30º = 1

3

17. aθ − θ =θ + θ

2sin cos1

cos sinDividing numerator and denominator by sin θ.

− θ =θ +

2 cot1

cot 1

⇒ 2 – cot θ = cot θ + 1 ⇒ 2 cot θ = 1 ⇒ cot θ = 12

18. b sin θ + cosec θ = 2

⇒ sin θ + =θ

12

sin⇒ sin2 θ – 2 sin θ + 1 = 0⇒ (sin θ – 1)2 = 0⇒ sin θ = 1∴ cosec θ = 1∴ sin9 θ + cosec9 θ = 1 + 1 = 2

19. cθ θ= =sin cos 1

x y k⇒ x = k sin θ; y = k cos θ∴ x2 + y2

= k2 (sin2 θ + cos2 θ) = k2

⇒ k = +2 2x y∴ sin θ – cos θ

= −− =x y x y

k k k=

−

+2 2

x y

x y

20. b Expression = −θ − θ θ1 1

cosec cot sin

= θ − θ − θθ − θ

2 2cosec cotcosec

cosec cot

= cosec θ + cot θ – cosec θ = cot θ

[cosec2 θ – cot2 θ = 1 & θ1

sin = cosec θ]

Method-2:

1 11 cos sin

sin sin

−θ θ−θ θ

= 2sin 1 sin 1 cos

1 cos sin sin (1 cos )θ θ − + θ− =

− θ θ θ − θ

= − θ − + θ

θ − θ

21 cos 1 cossin (1 cos )

= cos ( cos 1) cos

cot .sin(1 cos ) sin

θ − θ + θ= = θ− θ θ

21. c tan θ – cot θ = 0⇒ tan θ = cot θ = tan (90º – θ)⇒ θ = 90º – θ ⇒ 2θ = 90º ⇒ θ = 45º∴ sin θ + cos θ

= sin 45º + cos 45º = + = =1 1 22

2 2 2


22. a tan θ + cot θ = 2

⇒ tan θ + =θ

12

tan

⇒ θ + =θ

2tan 12

tan

⇒ tan2 θ + 1 = 2 tan θ⇒ tan2 θ – 2 tan θ + 1 = 0⇒ (tan θ – 1)2 = 0⇒ tan θ – 1 = 0⇒ tan θ = 1 = tan 45ºθ = 45º.

23. b A

B C

=AB 2BC 1

⇒ AB = 2k, BC = k

∴ AC = + =2 2 2(2k) k 5k = 5k

∴ sin A + cot C = +BC BCAC AB

= +k k2k5k

= ++ =1 1 2 5

25 2 5

24. a Expression

= ° °+ − °° °

2sin43 cos198cos 60

cos47 sin71

= ° ° + − × ° − ° ° −

2sin43 cos19 18

cos(90 43 ) sin(90 19) 2

= ° °+ − ×° °

sin43 cos19 18

sin43 cos19 4

[sin (90º – θ) = cos θ; cos (90º – θ) = sin θ]= 1 + 1 – 2 = 0

25. b Expression= cot 9º cot 27º cot 63º cot 81º= cot 9º cot 27º cot (90º – 27º) cot (90º – 9º)= cot 9º cot 27º tan 27º tan 9º

[tan (90º – θ) = cot θ; cot (90º – θ) = tan θ]= cot 9º tan 9º cot 27º tan 27º

[tan (90º – θ) = 1 (tan θ, cot θ = 1]

26. d A

B C

Rope

Po le

8 m

30°

In ABC,∆

ABsin30

AC° =

1 82 AC

⇒ = ⇒ AC = 16 m.

27. a A

BO30 ° 60 °

C x

h

200 mLet h be the height of tower AB.In ∆ABC,

h htan60 3

x x° = ⇒ = ...(i)

In ∆ABO,

h 1 htan30

200 x 200 x3° = ⇒ =

+ + ...(ii)

From (i) and (ii), we get

h 100 3= m.

28. c A

B C D

360 m

30 °45 °

45 ° 30 °

Let AC be light horse and B and D be two points on theopposite banks of a riverIn ∆ABC,

360tan 45

BC° = 360

BCtan 45

⇒ =°

= 360 m

Similarly, in ∆ACD,360

tan 30CD

° =

360 360CD 360 3

1tan 303

⇒ = = =°

CD 360 3 m=

∴ Width of the river = BC + CD

= 360 + 360 3 = ( )360 3 1+

= 360 × 2.732 = 983.52 m.


29. b45°

60°

45°A

B D

E

C

60°

x

7 m 7 m

Let AB be building and CD be tower.In ∆AEC,

xtan45

AE° =

x1 AE x BD

AE⇒ = ⇒ = =

In ∆CDB,CD

tan60BD

° =

x 73 x 9.56 m

x+⇒ = ⇒ =

∴ Height of the tower = x + 7 = 9.56 + 7 = 16.56 m.

30. b A

BC

D90° – α α

9 ft

16 ft

h

Let h be the height of the tower.Then,

AB htan

BD 16α = =

⇒ tan (90º – α)ABBC

=

⇒ cot α = h9

tan α cot α h h

16 9= ×

⇒2h

1144

=

⇒ h = 12 feet.

31. a A

B C

DE

90 m h

F

30°

60°

x

Let AB be cliff and CD be tower of height h.In ∆ABC,

tan 60º AB 90

3BC x

= ⇒ =

90x 30 3

3⇒ = = m.

In ∆ADE,

tan 30º AE 1 90 hED 3 30 3

−= = =

⇒ 90 – h = 30⇒ h = 90 – 30 = 60 m.

32. b

90°– θ θP

x20 m

10 3 m

Q

R

S

Let RS be building.

In RSQ,∆

tan θ 10 3x 20

=+

...(i)

In RSP,∆

tan (90º – θ) = 10 3

x

⇒ cot θ = 10 3

x...(ii)

From (i) and (ii),

10 3x 20+

x

10 3=

⇒ x2 + 20x – 300 = 0⇒ x = 10 or x = –30 (x can not be negative)∴ x = 10 m.


33. d A

B C60°

h

36 mLet the height of the wall (AC) be h m.

htan 60

36∴ ° =

⇒ h = 36 × tan 60º

⇒ h 36 3=

⇒ h = 36 × 1.732

⇒ h = 62. 353 ≈ 62.35 m.

34. b A

B

C

D

E30°

60°

56m

Bu i

ldin

g

Po le

In ABD,∆

ABtan 60

BD° =

AB 56BD

tan 60 3⇒ = =

°56 56

3 m.33

= =

35. b

B C D45° 60°

A

30 m

h

Let the height of the tower be AD m.

In ABD,

ADtan 45

30 CD

∆

° =+

⇒ AD = 30 + CD ...(i)

In ACD,∆

ADtan 60

CD° =

AD ADCD

tan 60 3⇒ = =

° ...(ii)

From (i) and (ii), we getAD = CD + 30

ADAD 30

3⇒ = +

⇒ ( )AD 3 1 30 3− =

30 3AD

3 1⇒ =

−

( )AD 15 3 3⇒ = + = 15 × 4.732 = 70.98 m.

36. d B

A6 0 °θ

xD C

2 x

h

Let AB be the flag staff and let θ be the angle betweenthe sun rays and the ground at the time of longershadow.

In ∆ABC,

tan 60º = hx

h

3 h 3xx

⇒ = ⇒ =

In ∆ABD,

htan

3xθ =

3xtan

3x⇒ θ =

1tan 30 .

3⇒ θ = ⇒ θ = °

37. d x Bird

30°60 °

S B NBoy

Bird

50 3 m

Let distance covered in 2 min be x.

x 50 3= cot30º + 50 3 cot 60º = 200 m

∴ Speed of bird 200

m/min2

=

200 606km/hr.

1000 2= × =


38. b A

B C30 m30 °

Let AC be the broken part of the tree BAC.

tan 30ºAB ABBC 30

= =

1 AB303

⇒ =

⇒ AB = 30

3 10 3=

AB 1sin30

AC 2= ° =

AC 10 3 2 20 3⇒ = × =∴ Total height = AB + AC

10 3 20 3 30 3 m.= + =

39. a

B C

D E30°

30°

60°

60° 60 – h

hh

d

A

Let the height of tower be ‘h’ m and distance betweenthe building and the tower be ‘d’ m.

In ABC,∆

60tan60

d° =

603

d⇒ =

60d

3⇒ =

In ADE,∆

60 h 60 htan30 3

d 60− −° = = ×

1 60 h3

603

−⇒ = ×

6060 h

3⇒ − =

h 40 m⇒ =∴ Height of tower = 40 m.

40. c A

O4 5 °dQ

P1 0 m

h

3 0 °

Let the height of tower be ‘h’ m and distance of foot oftower from point P be ‘d’ m.In APO,∆

htan45

d° = h d⇒ =

In AQO,∆h h

tan3010 d 10 h

° = =+ +

1 h10 h3

⇒ =+ 10 h h 3⇒ + =

10 3 1h

3 1 3 1

+⇒ = ×− +

( )10 3 1

3 1

+=

−( )5 3 1= +

∴ Height of tower ( )5 3 1 m.= +

41. d

C

60 °D

A

B

30 °

3125

mx

A and C ⇒ position of planesBC = 3125 mLet AC = x metreIn ∆ABD,

tan 60º = ABBD

⇒ 3 = +3125 x

BD⇒ BD =

+3125 x

3In ∆ BCD,

tan 30º = BCBD

⇒ 1

3 = +

31253125 x

3

⇒ 3(3125) = 3125 + x ⇒ 9375 = 3125 + x⇒ x = 9375 – 3125 ∴ x = 6250 metre.


42. c A

B D30°45°

C 60 m

AB = Tower = h metre∠ADB = 30º∠ACB = 45ºCD = 60 metreBC = x metreFrom ∆ABC,

tan 45º = ABBC

⇒ 1 = hx

⇒ h = x

From ∆ ABD,

tan 30º = ABBD

⇒1

3 = +

hx 60

⇒1

3 = +

hh 60

⇒ 3h = h + 60 ⇒ −3h h = 60

⇒ ( )−h 3 1 = 60

⇒ h = −60

3 1=

( )( )( )

+

− +

60 3 1

3 1 3 1 = ( )+30 3 1 metre

43. b

30°

A

B

C

DAB = Post = 15 feetThe post breaks at point C.BC = x feet⇒ AC = CD = (15 – x) feet∠CDB = 30ºFrom ∆ BCD,

sin 30º = BCCD

⇒ 12

= −x

15 x⇒ 2x = 15 – x ⇒ 3x = 15 ⇒ x = 5 feet.

44. b A

B C

x

AB = Tower = x unitsBC = Shadow = 3 x units

tan (∠ACB) = ABBC

= =x 1

3x 3 = tan 30º

∴ ∠ACB = 30º

45. b64

= h

50 (Assuming ‘h’ be the height of flag pole)

⇒ h = ×50 64

= 75 feet

46. b Q

CBA

P

30°60°

1500 3 m

P & Q are the positions of the plane∠PAB = 60º; ∠QAB = 30º

PB = 1500 3 metre.

In ∆ABP,

tan 60º = BPAB

⇒ 3 = 1500 3

AB⇒ AB = 1500 metre

In ∆ACQ,

tan 30º = CQAC

⇒ AC = 1500 × 3 = 4500 metre

PQ = BC = AC – AB = 4500 – 1500 = 3000 metre⇒ 3000 m travelled in 15 sec.

∴ Speed of plane = 300015

= 200 m/sec.

47. a

Q

A

BP

90 °– θθ

h

AB = Tower = h unitsLet, ∠AQB = θ ∴ ∠APB = 90º – θPB = a; BQ = bFrom ∆ AQB,

tanθ = ABBQ


⇒ tan θ = hb

... (i)

From ∆ APB

tan(90º – θ) = h

PB

⇒ cot θ = ha

... (ii)

By multiplying (i) and (ii)

tanθ. cotθ = ×h hb a

⇒ h2 = ab ⇒ h = ab

48. a A

B CD60°

60°

30°

30°

400 m

BC = River = 400 metreAD = Height of plane = h metreBD = x metre (let)∴ CD = (400 – x) metreFrom ∆ABD,

tan 60º = ADBD

⇒ 3 = hx

⇒ h = 3x metre

⇒ x = h

3metre .... (i)

From ∆ACD,

tan 30º = ADCD

⇒ 1

3 = −

h400 x

⇒ 3h = 400 – x

⇒ 3h = − 1400

3[From equation (i)]

⇒ + h3h

3 = 400 ⇒

+3h h

3 = 400

⇒ 4h = 400 3 ⇒ h = 400 3

4= 100 3 metre = 100 × 1.732 = 173.2 metre

49. a CD = h metre, AB = 2h metre

90° – θ θ

C

B DO

A

2hh

x m etre

OB = OD = x2

metre

From ∆OCD,

tan θ = hx2

= 2hx

... (i)

From ∆OAB,

tan (90º – θ) = ABBO

⇒ cotθ = 2hx2

= 4hx

... (ii)

Multiplying both equations,

tan θ. cot θ = ×2h 4hx x

⇒ x2 = 8h2

[� tanθ cotθ = 1]

⇒ h2 = 2x

8 ⇒ h =

x

2 2 metre

50. c

C

60°

D

A

B45°

5000

m

∠ACB = 60º∠DCB = 45ºAB = 5000 metreAD = x metre∴ From ∆ABC,

tan 60º = ABBC

⇒ BC = 5000

3metre

From ∆DBC,

tan 45º = DBBC

⇒ DB = BC = 5000

3∴ AD = AB – BD

= − 50005000

3= −

1

5000 13

m.

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Algebraic Identities Answers and Explanationss3.amazonaws.com/media.careerlauncher.com/acads_download...QA / Exercise - 1 CEX-5301/P2S/17 / Page 1 Quantitative Aptitude – 1 Algebraic