Mock Test 75 With Answer Key

==================================================================================

1

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

MEERUT VARANASI ROHTAK PANIPAT SONEPAT BAHADURGARH PATNA AGRA

1. (A) and

Similarly,

2. (*) For complete question see Hindi.

3. (B)

4. (B)

5. (B) 3 : 27 : : 4 :

3 3 4 4

6. (C)

7. (C)

Similarly,

8. (C)

9. (D)

10. (C)

11. (B) Author composes Novel. Similarly,Fashion Designer designs Dress.

12. (C) Light travels in the form of Rays.Similarly, Voice travels in the form ofWaves.

13. (B) Cloth is made of Thread. Similarly, Grillis made of Wire.

14. (A) Except Tomato, all others growunderground.

15. (D) All others refer to differentiation, butoption (D) refers to grouping.

16. (A) After rearranging the letters of the words,

we get the name of flowers, except option(A).

(A) ORANGE (B) JASMINE

(C) ROSE (D) LOTUS

17. (B) (A) (B)

(C) (D)

18. (B) (A) (B)

(C) (D)

19. (B) (A) (B)

(C) (D)

20. (B) Number 27 is a complete cube.

21. (D) All others are synonyms.

22. (C) All others are synonyms.

23. (B) Situation of right angle is 2 times in an

hour, but in 12 hours, the situation of

right angle is 22 times.

24. (A) Let total number of students = x

Then, number of students between

Nishtha and Deeksha = 4

x

4

(7 14)x

x - + =

4

21x

x - =

214

xx =-

321

4

x=

x = 28

25. (D) As per question, order in terms of height

is as follows:

Roshan > Rita > Ujjawal > Ravi

26. (A) In the given series, the order asked is as

follows:

Even digit 9 Even digit

896 5359683 496 5269737 294 13794173

498 45397615319574 296 85329574 894 51

SSC MOCK TEST -75 (SOLUTION)

==================================================================================

2



27. (D)

28. (B)

29. (C)

30. (B) In the given series, each term is obtained

by the sum of previous two terms. Hence,

next term = 8 + 13 = 21

31. (D)

32. (C) First letter of the first term and second

letter of the second term are in

alphabetical order. Similarly, first letter

of the third term and second letter of the

fourth term are in order.

33. (A)

34. (A) K L M N / K L M N / K L M N / K L M N

35. (A) r t u s / r t u s / r t u s / r t u s

36. (B) n s t n s t n s t n s t n s t n s

37. (B) (25 + 20) 3 = 15

(20 + 7) 3 = 9

(7 + 29) 3 = 12

Similarly,

(29 + 25) 3 = 18

38. (C) 2 + 4 + = 4 + 16 = 20 and

5 + 2 + = 25 + 4 = 29

Similarly,

7 + (?) = 65

(?) = 65 - 49 = 16

? = 16 = 4

39. (D) (4 + 9 + 6) 17 = 19 17 = 2

(5 + 8 + 9) 20 = 22 20 = 2

(7 + 9 + 9) 23 = 25 23 = 2

Similarly, (9 + 7 + ?) 19 = (13+?) 19 = 2

? = 19 + 2 13 = 8

40. (D) If third day of the month is Monday then

7th day will be Friday and 21st day will also

be Friday.

5th day from 21st = 21st + 4 = 25th day of

the month. Clearly, 25th day will be Tuesday.

41. (C) From the figures obtained after tossing thedice, as per rule-1 we know that side with1 opposite to 6, side with 3 opposite to 4and side with 5 opposite to 2 would beinscribed.Only figure (C) of the given figures in theoptions confirms the same rule.

42. (A)43. (D) In this statements the middle term is not

present. Hence no valid conclusion can bedrawn from these.

44. (B)

Since, minute hand points towards north-east direction.

\ hour hand will point towards eastdirection.

45. (A) My sisters mother = My mother My mothers husband = My father My fathers sister = My bua My buas son = My cousin (from fathers

side)= Cousin (Paternal brother)

46. (B)47. (B)48. (D)49. (D)50. (B)51.(B) After division,

2

5= 0.40,

3

7= 0.43,

2

9= 0.22 ,

11

7= 1.57,

13

9= 1.44

By arranging in descending order.

11

7,

13

9,3

7,2

5,2

9

52.(B) 17 14 13 21 6 3

= 17 14 13 18

= 17 14 5

= 253. (D) Let the number be 4x and 7x

Then, 4 8

7 4

x

x

=

1

416x 32 = 7x + 4 9x = 36

x = 4Sum of number = 16 + 28

= 44

==================================================================================

3



54. (B) Let each instalment be ` x.

8 1

100

xx

+

8 2

100

xx

+ x = 6384

2

25

xx

+4

25

xx

+ x = 6384

27

25

x+

29

25

x+ x = 6384

81

25

x= 6384

x = 6384 25

81

x = ` 1970. 37

55. (C) Let the sum be ` x and rate of interest per

annum be r

4514 = x

3

1100

r

(i)

5002 = x

6

1100

r

(ii)

divide equation (ii) by equation (i)

5002

4514=

3

1100

r

41

37=

3

1100

r

Substitute the value in eq (i)

4514 = x 41

37

4514 37

41

= x

x = ` 4073. 609 = 4073. 60

56. (C) Let the original number of students inthree classes be 3x, 5x and 7x .

3 10

5 10

x

x

=

5

7

21x + 70 = 25x + 50 20 = 4x 5 = x

Total number of students originally

3x + 5x + 7x = 15x15 5 = 75 students

57. (D) Percentage loss = 20 20

100

= 4%

58. (B) Each friend invested = ` x

Amount invested by three firends = 3x

If two more friends invest ` x each then

total amount = 3x + 2x = 5x

Amount saved by 3 friends = Amountinvested by 2 friends (2x)

3 14400 = 2x

x = 14400 3

2

x = 21600

Total investment of 3 friends= 3 21600

= ` 6480059. (B) Let the number be a, b, c

3

a b c = 35

a + b + c = 105a = 3b , b = 5c , c = b/5

3b + b + 5

b= 105

15b + 5b + b = 105 5 21b = 105 5

b = 25

a = 3b = 3 25a = 75

60. (B) y = 125

100136 = 170

x = 65% of y = 65

170100

= 221

2= 110.50

61. (C) Let the original number of workers = x x 96 = (x 16) 108 96x = 108x 16 108 16 108 = 12x 16 9 = x 144 = xNo. of workers = 144

62. (A)

So, Original mixture = (4 + 7) 8

2 + 8

= 44 + 8 = 52 litresInitial quantity of water in mixture

= 52 7

4 7

= 364

11 litres.

63. (B) Time taken to cross platform= (12 9) sec= 3 sec

Length of platform = 95 m

Speed of train = 95

3m/s

= 95 18

3 5

km/hr

= 114 km/hr.

==================================================================================

4



64. (D)

Time taken to fill the tank = 36

5 hours

After leakage time taken to fill the tank

= 36 4

5 5 = 8 hours

Time taken to empty the tank

=72

1 = 72 hours.

65. (C) Let speed upstream be x km/ hr and thespeed downstream be y km/hr.

24

x+

36

y = 14 (i)

25 45

x y = 15 (ii)

Multiplying eq. (ii) by 4 and (i) by 5.

Subtract (i) from (ii)

20

x= 10

x = 2 km/hr

putting x = 2, in eq. (i)

24

2+

36

y = 14

36

y= 2

y = 18 km/hr.

Speed of the current = 1

2(18 2)

= 8 km/hr

Speed of man in still water = 1

2(18 + 2)

= 10 km/hr.

Alternate Method :

24 () + 36 () = 14 hour 2 12 18 2

2 12.5 18 2.5

25 () + 45 () = 15 hourx + y = 18

x y = 2

Speed of man in still water = 18 2

2

=10 km/hr

66. (C) Present age of father = x

Present age of son = (38 x ) years

ATQ,

(x 4) (38 x 4) = 4(x 4)

34 x = 4

x = 30 years

Present age of father = 30 years

Present age of son = (38 30) years

= 8 years

67. (B) Let the principal be ` x and rate of C.I. be

r % per annum.

9x = x

2

1100

r

(3)2 =

2

1100

r

3 = 1100

r

100

r = 2

r = 200%

68. (D) Since, Raman takes thrice times of Ajay

So, Ratio of work done by Raman and Ajay = 1 : 3

Now, one day's work =1

12 (Raman + Ajay)

Work done by Raman in 1 day = 1

4

1

12=

1

48

Total required days = 48 days

69. (C)Formula;

First number Second number = H.C.F. L.C.M.

Second number = 72

12244

Second number = 68

70. (C) Let the total number of students = 100

Number of students who failed in Maths

or English or both = 48 + 37 19 = 66

Number of students who passed in both

subjects = 100 66 = 34

Required percentage =100

34 100 = 34%

==================================================================================

5



71. (A) Let the cost price be = ` 100

Marked price = ` 136

Selling price = 75 136

100

= ` 102

Profit = ` (102 100) = ` 2

Profit percentage = 2%

72. (B) Let the cost price = ` 100

Selling price = 100 + 100

5 100

= ` 105

Marked Price = 100 75

100 = ` 140

If he gain ` 5 than Marked Price = ` 140

If he gain ` 15 than Marked Price

= ` 140 3 = ` 420

Marked price = ` 420

73. (C) a : b : c

2 : 5 : 5

4 : 4 : 3

8 : 20 : 15

c

a=

15

8

2

2

a

c=

64

225

Using compendendo and dividendo rule

22

22

ca

ca

=

22564

22564

=

161

289

74. (A) Let the parallel sides be 4x and 7x metres

We know that = 1

2 (distance between parallel

sides sum of parallel sides)

Area of trapezium = 1

2(4x + 7x) 28

1386 = 11x 14

1386

154= x

x = 9

The smaller parallel side

= 4x = 4 9

= 36 m

75. (B) Net change = (x + y +100

xy)%

+ve shows increase and ve sings show

decrease

Net change = 25 + 20 25 20

100

= 5 5

= 10%

= 10% decrease

76.(C) Let the inner radius be r meters outer

radius R = (r + 9) m

2 17

2 14

R

r

17

14

R

r

9 17

14

r

r

3r = 14 9

3r = 126 m

r = 42 m

Outer radius (R) = 42 + 9 = 51m

diameter of the outer circle = 2 51 = 102 m

77.(B)If height and radius both of a cone change

be x %, the volume changes by

=

2 2

2

33 %

100 100

x xx

= 3 25 25 25 25 25

3 25 %100 100 100

=75 25

75 %4 16

=1200 300 25

%16

=1525

%16

= 95.3125%

78. (B)Surface area of a small cube

= 6 (edge)2

= 6 4 cm2

Surface area of the large cube

= 6 82

= 6 64 cm2

Requried ratio = 6 4

6 64

= 1 : 16

==================================================================================

6



79.(B)

AD = DB = 2

x

In ABC

AB2 + BC2 = AC2

x2 + BC2 = y2

BC2 = y2 x2

In CDB

CD2 = DB2 + BC2

CD2 = 2

4

x+ y2 x2

CD2 = 2 24 3

4

y x

sin = DB

CD= 2 2

2

4 3

4

x

y x

sin = 2 22

4 3

x

y x

80.(C)

sin A sin B + cos A.cos B

= sin 30 sin 45 + cos 30 cos 45

= 1 1 3 1

2 22 2

=22

13

81.(A)The coordinates of centroid are

x = 1 2 3

3

x x x , y =

1 2 3

3

y y y

1 = 32 4

3

x , 1 =

33 2

3

y

x3 = 3 , y

3 = 2

co-ordinates of third point are (-3, 2)

82.(B) Required area

= Area of four walls + area of the base

= 2 2.45 (7+ 5)+ 75

= 4.90 (12)+ 35

= 93.80 m2

83.(B)

AB = height of building = 125 m

CD = height of tower = h m

In ABC

tan 60 = AB

AC

AC = 125

3m

In ABC

tan 30= h

AC

1

3=

h

125

3

h = 125

3m

h = 41.66 m

84. (D)

222 2

1 1 3 2 1

2 163 2 3

x

1 3 4

3 2 4 3

x =

1

16

4 6 9 16

13

x =

1

16

4 13

12

x =

1

16

16x + 52 = 3

16x = 49

x = 16

49

==================================================================================

7



85. (D)

PQRS is an Rhombus. Diagonals of an

rhombus bisects each other at 90

In POQ

cos 60 = PQ

PO

1

2=

PO

30

PO = 15 cm (i)

So, PR = 2 OP = 30 cm

sin 60 = PQ

OQ

2

3=

30

OQ

OQ = 15 3 cm

So, QS = 2 OQ = 30 3 cm

86. (B)PQRS is a cyclic quadrilateral and sum of

opposite angles = 180

65 + B = 180

B = 115

In POS

PO = OS (radius of the circle)

OSP = 65

OPS + PSO +POS = 180

POS = 180 (65 + 65)

= 180 130

= 50

50 + A = 180

A = 130

(A, B) = (130, 115)

87. (D) Let the number of sides be n,

n (n 1)n

2

= 20

2n n 2n

2

= 20

n2 3n 40 = 0

n2 8n + 5n 40 = 0

n(n 8) + 5(n 8) = 0

n = 8, n -5

number of sides = 8

88. (C) Since these 3 points are collinear.

Area = 0

0 = 1

2[x

1 (y

2-y

3) + x

2(y

3-y

1) +x

3(y

1- y

2)]

=1

2[2 (8 2) + k(2 3) + 7(3 8)]=0

=1

2[12 + k (-1) 35 ] = 0

23 k = 0

k = 23

89. (A) x + 3 + 1

5x= 0

add 2 on both sides,

(x + 5) + 1

5x= 2

Formula-

3

3

3

1 1 13x x x

x xx

(x + 5)3 + 31

5x= (2)3 3(2)

= 2

90. (C)

3

2

1

12 1

xx

x x =

2

2

2

1

12 1

x xx

x x

x + 1

x= 6 x2 + 1 = 6x

=

2

2

1

6 12

x xx

x x

=

2

2

1

6

x

x =

6

34

= 3

17

==================================================================================

8



91.(C) x = 10 1 10 1

10 1 10 1

x =

2

10 1

10 1

=

10 1

3

x2 x 1 =

2

10 1

9

10 1

3

-1

=10 1 2 10

9

10 1

3

-1

=11 2 10 3 10 3 9

9

= 10 1

9

= 9

110

92.(B)(xn + an) is divisible by (x + a) for odd value of n

15316

17=

153(17 1)

17

Last term using binomial theorem

=17

)1()17( 1530

= 1

17

= 16

Remainder = 16

93.(A)6 4a

a+

6 4b

b+

6 4c

c= 0

6 4 6 4 6 40

a b c

a a b b c c

6 4

a+ 6

4

b+ 6

4

c= 0

18 = 4 1 1 1

a b c

9

2=

1 1 1

a b c

94.(C) = 36 = 36

180

radians

=5

radians

We know,

r = S

Q(S = distance)

r = 65

5

r = 65 5 7

22

r = 103.40 metres

95.(C) Volume of right prism = Area of base h

base is an equilateral triangle

Area = 3

4(6)2 = 9 3 cm2

Volume = 9 3 14 = 126 3 cm3

Lateral surface of prism

= (perimeter of base) ht

perimeter = 18

L.S.A = 18 14= 252cm2

Volume = 126 3 cm3

L.S.A. = 252cm2

96.(B) Slum population of A in 1991

= 38

87.6100

lakh

= 33.288 lakh = 33 lakh

97. (D) Difference

= 18 % of 28.2 lakh 16% of 22.6 lakh

= 18

28.2100

16

22.6100

lakh

= 5.076 lakh 3.616 lakh

= 1.46 lakh

98.(C) Total slum populations

= 28% of 64.8 lakh + 22% of 48.3 lakh

= 28

64.8100

+ 22

48.3100

= 18.144 lakh + 10.626 lakh

= 28.77 lakh

99.(D) Highest slum population is 38%.

It is present in A.

100.(C)Slum population of city B

= 34 % of 78.4 lakh

=34

100 78.4 lakh

= 26.65 lakh

= 26.5 lakh

==================================================================================

9



SSC MOCK TEST 75 (ANSWER KEY)

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

test and question number to 8860330003

1. A

2. *

3. B

4. B

5. B

6. C

7. C

8. C

9. D

10. C

11. B

12. C

13. B

14. A

15. D

16. A

17. B

18. B

19. B

20. B

21. D

22. C

23. B

24. A

25. D

26. A

27. D

28. B

29. C

30. B

31. D

32. C

33. A

34. A

35. A

36. B

37. B

38. C

39. D

40. D

41. C

42. A

43. D

44. B

45. A

46. B

47. B

48. D

49. D

50. B

51. B

52. B

53. D

54. B

55. C

56. C

57. D

58. B

59. B

60. B

61. C

62. A

63. B

64. D

65. C

66. C

67. B

68. D

69. C

70. C

71. A

72. B

73. C

74. A

75. B

76. C

77. B

78. B

79. B

80. C

81. A

82. B

83. B

84. D

85. D

86. B

87. D

88. C

89. A

90. C

91. C

92. B

93. A

94. C

95. C

96. B

97. D

98. C

99. D

100. C

101. B

102. C

103. C

104. C

105. C

106. A

107. D

108. D

109. D

110. B

111. C

112. D

113. C

114. A

115. C

116. D

117. B

118. B

119. A

120. B

121. C

122. C

123. C

124. B

125. A

126. A

127. A

128. B

129. B

130. C

131. A

132. C

133. B

134. D

135. A

136. C

137. D

138. C

139. B

140. A

141. B

142. D

143. B

144. C

145. C

146. D

147. D

148. A

149. D

150. D

151. A

152. B

153. *

154. D

155. A

156. C

157. B

158. A

159. C

160. B

161. D

162. D

163. *

164. C

165. A

166. D

167. D

168. C

169. B

170. C

171. D

172. D

173. B

174. A

175. D

176. D

177. C

178. B

179. B

180. D

181. C

182. D

183. C

184. D

185. C

186. D

187. A

188. B

189. C

190. A

191. C

192. C

193. B

194. D

195. C

196. A

197. D

198. A

199. D

200. A

151. (A) The sentence is in the passive voice.

Hence add been after had.

152. (B) Been should be used after Never be-

cause the main verb is missing [Had +

Never + V3] Past Perfect tense requires

V3 after Had

153. (*)

154. (D) No error.

155. (A) Use of in place of from.

Correction

163. (A) Add appear after advertisement.

Corrections Mock Test - 74

20. (A)

45.(D) There are 10 triangles in

the given digram.

ABG, AGE, BGC,

ABE, ABC, AFE,

CDE, BCE, CEG,

ACE

==================================================================================

10



MEANINGS IN ALPHABATICAL ORDER

Word Meaning in English Meaning in Hindi

Abysmal Extremely bad

Apparel dress

Apprehend to arrest/to notice and understand

Aversion dislike

Boom to make a loud sound

Complement that completes

Comprehend to understand

Dingy dirty and dark

Disband to break up an organization

Disperse to move in different direction

Eclectic taken from different sources

Embroil to involve in conflict

Flounder to move in an awkward way

Hieroglyph a written character that looks

like a picture

Humorous causing laughter

Hypnotic tending to cause sleep or

holding attention

Impede to hinder

Impel to cause strong desire to do something

Innovation a new idea

Invigilate to watch

litterateur a literary person or a professional writer

Maxim a saying

Pace speed

Vigour energy

mock test 75.pdfPicture.pdfPicture 001.pdfPicture 002.pdfPicture 003.pdfPicture 004.pdfPicture 005.pdfPicture 006.pdfPicture 007.pdfPicture 008.pdfPicture 009.pdfPicture 010.pdfPicture 011.pdfPicture 012.pdfPicture 013.pdf

Unlock-SSC%2075.pdf

Documents

Mock Test 75 With Answer Key