rahul k. gupta ssc mock test - 58_2

Centres at:MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER JAIPUR GURGAON

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1

1. (A) Genuine and Authentic are synonymous.

Similarly, Mirage and Illusion are

synonymous.

2. (B)

3. (A)

Similarly,

4. (C) 5 (5 1) = 21

Similarly,

7 (7 1) = 43

5. (B) Sorrow is antonym of Joy. Similarly

Pleasure is antonym of Pain.

6. (C) Surgeon uses forceps. Similarly,

Blacksmith uses hammer.

7. (D)

8. (B) 4 4 4 = 12

Similarly,

5 5 5 = 20

9. (A) Except option (A), the product of remaining

options are same.

10. (C)

11. (C)

12. (D)

13. (D) Except Kidney, all others are external

organs.

14. (D) Animals are different from Jungle, forest

and Woods.

15. (C) (A) (B)

(C) (D)

16. (A) (A)

(B)

(C)

(D)

17. (B)

SSC MOCK TEST 58 (SOLUTION)

18. (C) There are two alternating series. The firstseries consists of squares of consecutivenumbers while the second seriesconsists of consecutive numbers.

(1) = 1, (2) = (4), (3) = 9, (4) = 16, (5) = 252 3 4 5 6

19. (A)

20. (A)

21. (B) In each subsequent term the last letterbecomes the first letter.

22. (D)23. (A)24. (B)25. (A)26. (B)27. (D)

28 - 29:

28. (D) Ramu is facing west.29. (A) Here, DE = BE then

AE = BE - AB= 4 - 2 = 2

So, Ramu is 2 km away from the starting point.

30. (A)

31. (C) and

Therefore,

32. (C) The required number 9 is common to thecircle, rectangle but outside the square.

33. (B) LOTUS


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2

34. (A) and

Therefore

35. (D)

36. (A)

37. (B) 200/220/200/220/200

38. (D) 836 + 112 = 948 948 3 = 316

Similarly,

213 + 420 = 633 633 3 = 211

39. (A) The age of Ram = 8 years. Geeta = 6 years

The age of Kamal = 6 5 = 30 years.

40. (A) C is father of A and B.D is aunt of A and B.E is grandmother of A and B.Therefore, B is either grandson orgranddaughter of E. Considering the givenalternatives we may select option (A) asthe answer.

41. (C) and

42. (B) (16 4) 6 2 + 8 = 30

(16 4) 6 2 + 8 = 30

4 6 2 + 8 = 30

24 2 + 8 = 30

43. (B) 8 2 = 16 and 8 4 = 329 2 = 18 and 9 4 = 36

So, 10 2 = 20 and 10 4 = 4044. (C) 18 6 3 = 324 and

15 5 4 = 30011 8 x = 528

x = 6811

528

45. (A) Child Illness Hospital Doctor Medicine 1 3 5 2 4

46. (C) 825643 < 834562 < 842563 < 852463 3 4 2 5< 86425

1

47. (D)

Similarly

48. (C) 179, 246, 358

49. (C)

There are 28 triangles are in the givenfigure- 1,2,3,4,5,6,7,8,9,10,11,12,(1,12),(10,11), (11,8), (8,9), (9,10), (6,7), (3,4), (2,3),(2,5), (4,5), (1,2,5), (9, 10,12), (5,4,6), (9,8,7),(2,5,9,10) and (4,5,9,8)

50. (A)

Suresh's rank fromthe begning39 (17 + 7 1)

= 39 (24 1)= 39 23= 16th

51.(C) The required greatest number= H.C.F. of (2406 6) & (1814 4)= H.C.F. of 2400 & 1810= 10

52. (A) Given that,

M.P. of the article = 280& Discount allowed by the shopkeeper = 6%

S.P. of the article = (100 -6)% of 280

= 94% of 280

= 94

100280 = 263.20

also, given that net profit = 5% S.P. = (100 + 5)% of C.P.

263.2 = 105% of C.P.

263.2 = 105

100 C.P.

C.P. = 263.2

100105

= 250.66

53.(C) 98035.0 + 0.1233

= 3598 35

9900

+1233 12

9900

= 3563 1221

9900

= 0.48.32


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3

54. (D) 0.20 A , 0.25B, 0.30C

20 A , 25B, 30C

4A, 5B, 6C = K (say)

So,

A : B : C

K

4 :

K

5 :

K

6

= K

604

K60

5

K60

6

= 15 : 12 : 10

55. (C) As per rule

(an + bn) is divisible by (a + b) when n is odd

(1635 + 3035) is divisible by (16 + 30 ) i.e. 46

(1635 + 3035) will also be divisible by each

and every factor of 46

(i.e. by 1, 2 and 23 also)

(1635 + 3035) is also divisible by 23

remainder = 0

56. (B) Let 1st part = x

So, 2nd part = 90 x

A.T.Q., 1

5 of x :

1

6 of (90 x) = 2 : 3

3

2

6

)90(5 x

x

6

5 90

x

x = 2

3

18x = 10 (90 -x)

18x = 900 - 10x

28x = 900

x = 32.14

57. (B) Average age of all the boys = Sum of the

ages of all the boys Total number of boys

15 10 5 12

15 5

= 150 60

15 5

= 210

20

= 1

102

yrs.

58. (C) Let C.P. of the article = x.

A.T.Q,

(S.P. at 30% profit ) (S.P. at 25% profit) = 20

130

100

x

125

100

x= 20

5

100

x= 20

5x = 2000

x = 400

59. (B) Average speed = Total distance Travelled

Total time taken

Total Distance = 5 + 5 + 5 + 5

= 20 km.

Total Time =5

30+

5

60+

5

90 +

5

150

= 900

305hrs.

Average speed = 20

305/900= 59.02 km/hr.

60.(B) Let the population 2 yrs. ago be x

Then,

59400 = x120

100

80

100

x = 59400 100

12 8

x = 61875

61.(C) Let the original radius = r

and the original height = h

So, the original volume = 2r h

Now, the new radius = 130

r100

= 13

r10

and the new height = 130

h100

= 13

h10

So, the new volume =

213

r10

13

h10

=

2

10

13

r

13h

10

= 2197

10002r h

So, change in volume

New volume original volume

= 2197

10002r h 2r h

= 2197 1000

1000

2r h

= 1197

10002r h

So,% increase in volume =

=1197

1000

2

2

r h 100

r h

= 119.7%


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4

62. (C)1

yx

=1

yx

11

1 = x (1 y)

x = 1

1 y

1z

y=1

z = 1

1y

y

yz

1

1

z=

1

y

y

x +1

z =

1

1 y + 1

y

y

= 1

1 y +( )

1

y

y

= (1 )

1

y

y =1

63. (A) Each angle of a regular polygon = 1

5 6 90

(2 4) 90 n

n= 108

(Where 'n' is number of sides)

180 n 360 = 108 n

72 n = 360

n = 360

72= 5

64. (C) R = S.I. 100

P T

Time = 4 yrs. 3 month

416 100

514000

12

4 yrs. 3

12yrs. =

51

12yrs.

= 5140

12416

= 2.45%

65. (C) Let B invests x

then, A invests 3x

Let B invests for y yrs.

then A invests for 4y yrs.

Ratio of their investment

3x 4y : x y = 12xy : xy = 12 : 1

Also, as we know that

Ratio of investment = Ratio of profit

If the profit got by B = 8000

Then, Profit got by A = 12 8000

= 96,000

Total profit = 8000 + 96,000

= 1,04,000

66. (B) 212121 xzzyyx

1 2 2 1 2 1. . . x x y y z z

2 1 2 1 2 1 x y z

xyz

67. (A) Let the volume of cylinder = 2r h

New volume =

2

2

r 2h =

2

2hr

New volume

previous volume=

2

2

/2

r h

r h=

1

2 = 1 : 2

68. (C)

Distance between A & B at 10 AM= 340 51= 289 km

Relative speed of A & B = (51 + 68)km/hr. = 119 km/hr.

Time = Dis tance

Speed =

289

119= 2hr. 26 min.

Time they meet = 10 + 2hrs. 26min= 12.26 pm

69. (B)

A centroid divides each median in the ratioof 2 : 1

AG : GD = 2 : 1

AG

AD=

2

3

AG

21=

2

3

AG = 2

3 21 = 14cm


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5

70. (A)

Orthocentre is the point B

AC2 = AB2 + BC2

AC2 = 122 +52

AC2 = 169

AC = 13

circumcentre is the mid point of AC.

Distance between orthocentre and cirumcentre

= BD = AD = 1

2 AC

=1

2 13 = 6.5 cm

71. (B) (a, b) lies on the line x = y 3

a = b 3 (i)Again (a + 2, b + k) lies on the line x = y 3

a + 2 = b + k 3 (ii)From (i) and (ii)

b 3 + 2 = b + k 3

b b 3 + 3 + 2 = k

k = 2

72. (C) L.C.M. of 3

1hr.,

2

3hr.,

11

3hr. and 1hr. 40min.

=L.C.M. of 20 min 40min. 80 min. and 100 min.

= 400 min. = 400

60hr. = 6

2

3hrs.

So, required time = 12 :00 Noon + 62

3hrs.

= 6 : 40 PM.

73. (C)

PR : RQ = 1 : 3

and PR = 3cm

RQ = 9 cm

PQ = PR + RQ

= 3 + 9 = 12 cm

PQ = 1

2 BC (By the mid point theorem)

BC = 2PQ

= 2 12 = 24 cm

74. (C) 250 , 340, 430, 520,= (25)10 , (34)10, (43)10, (52)10

= (32)10 , (81)10, (64)10, (25)10

Greatest number = (81)10 = 340

75. (B) PQ + QR = 4 + 6 = 10cmPQ + QR = PR

P, Q, R are collinearNo circle is passingthrough P, Q and R.

76. (A) circumference of the wheel = 2 r

= 2 22

7 1.4

Distance = 22km = 22000 m

Number of rotations = Dis tance

circumference

= 22000 7

2 22 1.4

= 2500

77. (D) 2cos2 + 3sin2 =2cos2 + 3(1 cos2 )

= 2cos2 + 3 3cos2= 3 cos2

Maximum value of cos =1

Minimum value of = (2cos2 + 3sin2 )

or (3 cos2 ) = 3 (1)2

= 3 1 = 278. (C) Let the distance between two places be 'k' km

speed downstream = (x + y) km/ hr.speed upstream = (y x) km/hr.Then,

A.T.Q., k k

zx y y x

k

22 xy

yxxy= z

k = y

xyz

2

)( 22

79. (A) cot tan

cot tan

=2 (cot =

1

tan)

1tan

tan1

tantan

= 2

1 + tan2 = 2 2 tan2 3 tan2 = 1

tan2 = 1

3

tan = 1

3 = tan1

1

3

= 30

sin = sin30 = 1

2


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6

80. (C)

In ABM

tan 60 = AB

BM

AB = x 3 m

InCDM

tan 30 = CD

MD

CD = (120 -x) 1

3As, AB = CD

x 3 = 1

3 (120 -x)

3x = 120 - x

4x = 120

x = 30

Height of each pole = 30 3 m

81. (B)120sin135cos

120sin135cos

=)3090sin()4590cos(

)3090sin()4590cos(

= 30cos45sin

30cos45sin

1 3

22

1 3

22

=

2 3

2

2 3

2

=

2 3

2 3

= 32

32

82. (D) Let the 3rd number be 'x'

1st number = 65% of x = 65

100

x

2nd number = 60% of x = 60

100

x

Difference = 5

100

x

2nd number is smaller than the first number by

5

100

65

100

x

x100% = 7.7%

83. (D) Let the speed of the 2nd train = x km/hr.

Distance = speed time

Then, (140 + 110) = (x + 24) 5

18 20

250 = (x + 24) 5

910

5 = (x + 24 ) 1

9

(x + 24) = 45 x = 45 24 = 21 km/hr.

84. (B) Tank filled by both the taps in 1 hrs.

= 1 1

9 12

4 3

36

7

36 part

In 1hr., the part of the water tank filled 7

36 part

In 1

42

hrs., the part of the water tank filled

= 7

36

9

2=

7

8part

85. (C) Surface area of cube = 6 (side)2

= 6 (30)2

= 6 900= 5400 sq. feet

So, required cost to paint the outer surface

of the cube = 5400 1

30 50

= 9,000

86. (B) Let the two consecutive even numbers bex and (x + 2).Now,(x + 2)2 x2 = 92x

2 + 4x + 4 x2 = 92

4x = 92 4

4x = 88

x = 22

The sum of the two consecutive even numbers

= 22 + 24 = 46

87. (C) sin2 (30 + A) sin2 (30 A)

= sin (30 + A + 30 A). sin (30 + A 30 + A)

[ sin2 A sin2 B = sin (A + B) sin (A B)]

= sin60 . sin2A

= 3

2sin2 A

88. (B) Let the two number be 4x and 3x

L.C.M. = 12x

12x =180

x = 180

12= 15

Smaller number = 3x = 3 15 = 45


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7

89. (C) Weight of the student who left the class

= 60 50 59 49.8 = 61.8 kg

90. (B) Since 2, 3, 7, 11 are prime number and

the given expression is 210 310 721

1129

So, the number of prime factors in the given

expression is- (10 + 10 + 21 + 29) = 70

91. (C) side of one square = 52

4= 13 cm

[ perimeter = 4 side]

side of other square = 48

4 = 12 cm

A.T.Q.

Area of third square= (13)2 (12)2

= 169 144

= 25 sq cm.

side of third square = 25 = 5 cm

Its perimeter = 4 5 = 20 cm92. (C) Radha's total percentage expenditure

= (30 + 15 + 25 + 10)% = 80%

Percentage savings = 100 80 = 20%

Now, 20% of her monthly salary = 2500

Her monthly salary = 2500

20100

= 12,500

93. (C) Total distance covered = 50 15 = 750 km

The required time to cover the same distance

at 25 km/hr. = 750

25= 30 hrs.

94. (D)

Let the height of the hill be 'h' meter

InCDB

tan 30 = CD

DB

DB = CD

tan30

DB = CD

1/ 3 = 3 CD

DB = 3 100

= 100 3 m

Again in ADB

tan 45 = AB

BD

AB = 100 3 m 1

AB = 100 3 m

h = 100 3 m

95. (C) Required quantity

= 2

of 1983

3

of 917

= 132 39

= 93

96. (C) Increase in expenditure = (280 180)

= 100 thousand

So% increase = 100

180100 =

500

9= 55.5%

97. (C) The required% = 1110

10360100 = 10.71%

98. (B) The required% = 1580

10010360

= 15.25%

99. (A) It is clear from the table that the required

item is food.

100. (A)


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8

SSC MOCK TEST 58 (ANSWER KEY)

1. A

2. B

3. A

4. C

5. B

6. C

7. D

8. B

9. A

10. C

11. C

12. D

13. D

14. D

15. C

16. A

17. B

18. C

19. A

20. A

21. B

22. D

23. A

24. B

25. A

26. B

27. D

28. D

29. A

30. A

31. C

32. C

33 B

34. A

35. D

36. A

37. B

38. D

39. A

40. A

41. C

42. B

43. B

44. C

45. A

46. C

47. D

48. C

49. C

50. A

51. C

52. A

53. C

54. D

55. C

56. B

57. B

58. C

59. B

60. B

61. C

62. C

63. A

64. C

65. C

66. B

67. A

68. C

69. B

70. A

71. B

72. C

73. C

74. C

75. B

76. A

77. D

78. C

79. A

80. C

81. B

82. D

83. D

84. B

85. C

86. B

87. C

88. B

89. C

90. B

91. C

92. C

93. C

94. D

95. C

96. C

97. C

98. B

99. A

100. A

101. C

102. C

103. C

104. D

105. C

106. B

107. B

108. B

109. D

110. B

111. D

112. D

113. D

114. B

115. B

116. D

117. A

118. A

119. A

120. D

121. C

122. B

123. C

124. B

125. D

126. D

127. B

128. B

129. C

130. B

131. B

132. A

133. C

134. C

135. D

136. C

137. B

138. A

139. A

140. A

141. C

142. C

143. A

144. A

145. D

146. A

147. C

148. B

149. C

150. D

Note:- If your opinion differs regarding any answer, please message the mock

test and question number to 8860330003

124; Mohammad Bin Tuglaq

151 (C); Ago in place of 'before'

152 (B); Change 'will'

153 (B); Change 'have' into 'has'

154 (C); Remove 'into'

155 (A); Remove 'very'

151. C

152. B

153. B

154. C

155. B

156. C

157. A

158. C

159. B

160. A

161. B

162. C

163. C

164. D

165. A

166. C

167. A

168. D

169. C

170. B

171. B

172. C

173. A

174. C

175. D

176. C

177. B

178. C

179. D

180. A

181. A

182. C

183. D

184. A

185. B

186. D

187. D

188. B

189. D

190. B

191. C

192. D

193. D

194. C

195. A

196. D

197. D

198. D

199. C

200. C

Correction of Mock test-57

1. (C)

40. (*) Birth Death Childhood Infancy 1 4 3

Adolescence Adulthood Old age Death 5 6 7 2

102. (C) 104. (*)

112. (C) 126. (D)

128. (B) 131. (B)

136. (B) 146. (A)

139. (C) 154. (B)

Correction of Mock test-56

169 (C) 180. (C)

rahul k. gupta ssc mock test - 58_2

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