SSC MOCK TEST – 118 [Answer with Solution]sschansrajacademy.com/questions/TEST PAPER 126-3 [Solution].pdf · P-2 19. (A) Since one premise is particular, the conclusion must be

P-1

1. (C) 2. (B) 3. (C) 4. (D) 5. (A)6. (B) 7. (B) 8. (A) 9. (B) 10. (D)11. (B) 12. (C) 13. (A) 14. (B) 15. (A)16. (D) 17. (A) 18. (B) 19. (A) 20. (C)21. (D) 22. (A) 23. (B) 24. (B) 25. (A)26. (B) 27. (B) 28. (C) 29. (C) 30. (C)31. (B) 32. (B) 33. (C) 34. (D) 35. (D)36. (C) 37. (C) 38. (D) 39. (C) 40. (A)41. (D) 42. (B) 43. (C) 44. (C) 45. (A)46. (A) 47. (A) 48. (C) 49. (B) 50. (B)51. (C) 52. (A) 53. (A) 54. (B) 55. (D)56. (C) 57. (B) 58. (C) 59. (C) 60. (C)61. (A) 62. (B) 63. (B) 64. (A) 65. (B)66. (A) 67. (B) 68. (B) 69. (A) 70. (A)71. (A) 72. (C) 73. (D) 74. (D) 75. (C)76. (A) 77. (C) 78. (A) 79. (D) 80. (B)81. (A) 82. (B) 83. (D) 84. (C) 85. (C)86. (A) 87. (D) 88. (A) 89. (A) 90. (D)91. (B) 92. (B) 93. (B) 94. (C) 95. (D)96. (C) 97. (D) 98. (D) 99. (B) 100.(A)

1. (C) H G F: E14 15 16 17 8 7 6 5

: : F E D: C21 22 23 24 6 5 4 3

2. (B) 583 : 488 : : 293 : 378

Here, 5 + 8 + 3 = 16

4 + 8 + 8 = 20 Difference = 4

So,

2 + 9 + 3 = 14

3 + 7 + 8 = 18

Difference = 4

3. (C) 25 × 2 = 50

So, 789 × 2 = 1578

4. (D) According to dictionary 2, 1, 4, 3, 5

5. (A) 45 – 29 = 16 63 – 44 = 19

98 – 79 = 19 87 – 68 = 19

6. (B) B E C+1

+2

Q T S+2

+1

,

G J H+1

+2

L O M+1

+2

,

7. (B) 23 90 357 1424

8. (A) Column wise –

12 + 52 = 26, 32 + 22 = 13,So,

42 + 82 = 80

9. (B) 13 17 20 22

10. (D)

Total distance = 80

11. (B)

Hence, it is clear Q is facing N.12. (C) E and G cannot occupy seats on either side of B.

13. (A)

SO,

14. (B)15. (A) By interchanging X and eq become –

16 + 4 × 2 – 21 7 = 16 + 8 – 3 = 21

16. (D)

failed in Hindi only 42 – 17 = 25%failed in English only 52 – 17 = 35%Total passed 100% – 25% – 35% = 40%

17. (A) T – R T is mother of RM × T M is brother of THence, M is maternal uncle of R.

18. (B) a d b c a a c b d / a b c d / a d b c a c b d

SSC MOCK TEST – 118 [Answer with Solution]

P-2

19. (A) Since one premise is particular, the conclusionmust be particular and should not contain middleterm. So, it follows that some heavy things aregrey in color.Hence, only I holds.

20. (C) After cutting we have length of new cube = 3unit 33 – 27 smaller cuber..when we put them in one line to form cuboidbreadth & height remain some but length become 3 × 27 = 81 cm

dimension 3cm × 3cm × 81cm21. (D)22. (A)23. (B)24. (B)25. (A) 65 – 23 + 17 – 27 + 19 – 14 = 37 i.e. number of

beads inside glass No. of beads outside glass = 65 – 37 = 28.

26. (B) Let 5 number are a, b, c, d, e. a × b × c × d × e = (HCF)4–1 ×LCM

= (4)5–1 × 27720= 256 × 27720= 7096320

27. (B)2 22 5

=

84 82 16

5 5

= 8 815 1 11

5 5

1 × 2 = 2

28. (C)

3 2 4

24

A+B8

B+C12

A+B+C6

Efficiency of A + B + C = 4 and.Efficiency of A + B = 3 so,Efficiency of C = 1Efficiency of B + C = 2 so,Efficiency of B = 1

and Efficiency of A = 2Efficiency of A + C = 3,

and they can do the same work in 243 = 8 days

29. (C) The train covers only 420 mtr. (platform's length)in (20 – 8) sec. = 12 sec.

so, speed of train 42012

mtr/ sec.

420 1812 5

126 km/hr

30. (C) The average weight of two students who left the

class is = 45 + 48 1502 100

= 48.5 kg weight loss of remaining

15048 student is 48 × kg1000

31. (B) Using angle bisector theoremIn triangle DAC

DA AEDC EC

.....(i)

In triangle BAC

AB AEBC EC

.....(ii)

from (i) and (ii)

AB ADBC CD

32. (B) 12 m = 18 w = 14 days.12 m = 18 w2 m = 3 w ..... (i)question is 8 m + 16 w = ? daysFrom eq (i)

2 m = 3w 8 m = 12 w

so, 8 m + 16 w = ? days12 w + 16 w = ? days

28 w = ? days 18 w = 14 days

1 w — = 14 × 18 days

28 w — = 14 18

28

days

= 9 days33. (C)34. (D)

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35. (D) Area of square = 6 × 6 = 36 cm2

Area of trangle = 36 cm2

so,12

× AB × h = 36

h

9 cm BA

C

12

× h = 36

h = 8 cm36. (C) The curved surface of cylindrical pillar = 264 m2

2rh = 264 rh = 42Volume of cylindrical pillar

r2h = 924 cm3

227 × 42 × r = 924

r = 7so, h = 6 ( hr = 42)so, 24 : h = 2× 7 : 6

= 14 : 6= 7 : 3

37. (C) A : BAmount 3500 : 9000

Time 12 7 (join after 5 min)

Profit Ratio 42000 : 63000

2R : 3R

[If 2R = 42000 then, 3R = 63000]

38. (D)

4×– 20%

C.P, = Rs. 100 108 = S.P, +8%

640×160(C.P , Rs. 80 Rs. 112 S.P 2 2+40%

Profit

S.P1 = 108 × 160 = Rs. 1728039. (C) a cosθ – b sin θ = C

squaring both sidesa2 cos2 θ + b2 sin2 θ – 2ab sin θ cosθ = C2

zab cosθ sin θ = – c2 + a2 cos2 θ + b2 sin2 θNow (a sin θ + b cosθ )2 = a2 sin2 θ + b2 cos2 θ +2ab sin θ cosθ= a2 sin2 θ + b2 cos2 θ + – a2 cos2 θ + b2 sin2 θ= a2 (sin2 θ + cos2 θ ) + b2 (cos2 θ + sin2 θ ) – C2

(a sin θ + b cosθ )2 = a2 + b2 – c2

a sin θ + b cosθ = ± 2 2 2a + b – c

40. (A) A

B60º

C

h

30ºx 5M

tan 30º = h

x+5

13

= h

x+5 — (i)

tan 60º = hx

3 x = h — (ii)By (i) of (ii) equation

x + 53

= 3 x

x + 5 = 3x

x = 52

M

Then h = 5 3

241. (D) sin2 θ + cos2 θ + sec2 θ + cosec2 θ + sec2 θ +

cosec2 θ – 21 + sec2 – 1 + cosec2 θ – 1 + sec2 θ + cosec2 θ1 + tan2 θ + cot2 θ + 1 + tan2 θ + 1 + cot2 θ3 + 2 tan2 θ + 2 cot2 θ

3 + ( 2 tan θ )2 + 2

2tanθ

– 2 × 2 2 + 4

7 + 2

22 tan θtanθ

Then Minimum value = 742. (B) As AoB is equilateral triangle, and it's and side is

equal to side of square.

60º

60º60º

D C

BAAs, AB = CD (sides of square)

AB = OB (side of equilateral triangle)

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so, CD = OB — (ii)so, frum above

result, triangle BOC is isosc triangle.BCO = 75º, COB = 75ºSimilarly angle DOA = 75ºso, DOC = 360 – (75 + 75 + 60) = 150º

43. (C) R.I. = 10% per annual.Consider P. amount (P.A.) = Rs. 1000

1000

100 10010010 3rd year interest101 6

1072.60

110 1

10

110

I II III

10 = 121

×

So, P.A. = 1000 × 1

10 = Rs. 600

44. (C)

AP T75º

BC

BAT = 75ºBAC = 45ºUsing alternate segments anglesBAT = ACBso, ABC = 180 – (75 + 45º) = 60ºsum of all the angles in a triangle = 180º

45. (A) Let a = 12

, b = 12

, C = 1, d = 1

thana2 + b2 = 1

12

+ 12

= 1

and b2 + c2 + d2 = 5/2

12

+ 1 + 1 = 52

So (ac – bd)2 + (ad + bc)2 =2 21 1 1 11– 1 1 1

2 2 2 2

= 0 + 42

= 246. (A) Using option (A)

We can easily figure out the answer

A

B C

47. (A)

57123

57 E

A

D

B CHere AED and ABC are similar

AED = DBC = 57ºGiven AE = 10 cm

AD = 12 cmAB = 20 cm

So,AE ADAB AC

10 1220 AE EC

12 110 EC 2

= 24 = 10 + EC = EC = 14 cm48. (C) a2 + b2 + 2b + 4a + 5 = 0

a2 + 4a + (2)2 + b2 + 1 + 2b = 0(a + 2)2 + (b + 1)2 = 0Let (a + 2)2 = 0

a = –2and (b + 1)2 = 0

b = – 1

then a + ba + c =

– 2 + 1– 2 – 1

= 13

49. (B)Expenditure on Re nt 100Expenditureon fuel

= 14% 1009%

= 156 (approx)

50. (B) 100% income = Rs. 250001% income = Rs. 250amount spent on rent and food

= [14% + 45%] × 250= 59% × 250= Rs. 14750