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SSC MOCK TEST
(SOLUTION) 11. (C)
1. (D)
2. (B)
Similarly,
3. (C)
4. (C)
15, 49 24
2+5–1=6 5+4–3=6 Similarly, x+7–1=6 x+6=6 x=6–6 =0
14
.in
5 × (4 – 1) 7 × (8 – 1) 6 × (5 – 1) Similarly, 8 × (4 – 1)
24
20
13. (C) 14. (B)
gl
5. (B)
12. (D)
Similarly,
-c
6. (C)
sc
7. (D)
n 1
315 n 1 3 105 = n – 1 ? n = 105 + 1 = 106 Hence, 106th term is 320.
9. (A)
16. (C)
w
320 5 3
15. (C)
w
w
.s
8. (C) let the nth term of 5,8,11,14, ......... is 320. (a = 5 and d = 3) an = a + (n – 1) d 320 = 5 + (n – 1) 3
Here, (BC = AD = 10 km) Required distance = AD = 10 km 17. (B) CONCERN 18. (C) MPDRNOPRDUXRDPRDMNDRD 1 2 19. (C)
10. (A)
20. (B)
21. (A)
23. (C) Similarly,
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22. (D)
For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in And for regular updates on our website like us on facebook - www.facebook.com/cgl.ssc2014 40. (C)
24. (D) 25. (D) 41. (C)
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.in
gl
1· § 1· § 1· § 1· § = ¨1 ¸ ¨1 ¸ ¨1 ¸ ¨1 ¸ 4¹ © 4¹ © 4¹ © 4¹ © §5· = ¨ ¸ ©4¹
4
52. (D)
3. Poster 3
4. Posterity 5
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5. Posterior 4
w
1. Postage 2
36. (D) 2. Post 1
1 4
? Required value
.s
Side =
16 = 4 m Perimetre = 4 × Side =4×4 = 16 metre
will be maximum if a = b = c = d =
sc
Inhospitable Institution 3 2 34. (C) a b a b / a b a b / a b a b 35. (B) Area of the square = 16 sq. metre (Side)² = 16
14
Improvement Inadequate Incompeten t 5 1 4
-c
33. (C)
'D' is daughter of 'F' and 'F' is father of 'G'. So 'F' is father of 'D' and 'G'. Hence, 'D' is the sister of 'G'. 42. (B) 43. (D) 44. (A) 45. (D) 46. (C) 47. (A) 48. (C) 49. (C) 50. (C) MEN represents the sets of following nuumbers M – 56, 68, 75, 87, 99 E – 00, 12, 24, 31, 43 N – 57, 69, 76, 88, 95 51. (D) a + b + c + d = 1 Then value of (1 + a) (1 + b) (1 + c) (1 + d)
20
26. (C) 27. (C) 28. (B) 29. (D) (A) 95 – 82 = 13 (B) 69 – 56 = 13 (C) 55 – 42 = 13 (D) 48 – 34 = 14 30. (C) (A) 135 – 123 = 12 (B) 123 – 111 = 12 (C) 111 – 100 = 11 (D) 100 – 88 = 12 31. (D) 32. (A) Except option (A) all others are equal to 3/4.
37. (B)
Jawaharlal Nehru Lal Bhadur Shastri 3 1
Indira Gandhi V.P. Singh 2 5
P.V. Narashimha Rao 4
53. (C)
38. (C)
39. (D)
100 – 60 ÷ 6 × 2 + 8 100 – 10 × 2 + 8 100 – 20 + 8 108 – 20 = 88
BPC = 60º (given) ? A + B + C = 180º [Angle sum property of a ' ) B + C = 180º – A (180º – MBC) + (180º – NCB) www.ssc-cgl2014.in = 180º – A
For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in And for regular updates on our website like us on facebook - www.facebook.com/cgl.ssc2014 [ B + MBC = C + NCB = 180º , linear pairs] 360º + A = MBC + NCB + 180º
90º +
1 MBC NCB 180 2
From 1st and last ratio
1 1 1 A = MBC NCB 90º 2 2 2
1 A 2
6 AN = 10 6
= PBC + PCB
36 = 3.6 cm 10 NC = AC – AN = 10 – 3.6 = 6.4 cm
AN =
[ BP & CP bisect MBC & NCB respectively] 90º +
1 A 2
= 180º – BPC
90º +
1 A 2
= 180º – 60º
1 A 2
= 120º – 90º
1 A 2
= 30º
Now,
AN 3.6 9 = = NC 6.4 16 AN : NC = 9 : 16
3
1· 1§ 1· § = ¨ x ¸ 3x . ¨ x ¸ x¹ x© x¹ ©
1
x3
56. (B)
.in
180 +
=
14
1 360ºA 2
55. (B)
20
x3
0
§
3
1· 1· § § = ¨ x ¸ 3¨ x ¸ x¹ x¹ © ©
1 Ụ
º
·2
gl
1 x ¸«¨ x ¸ 3 » = 0 ¨ x¹ » © x ¹¬«© ¼
A = 60º
w
.s
sc
-c
54. (A)
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w
[perpendicular from centre to any chord bisects the chord] 1 Similarly, CN = ND = CD = 12 cm 2
In ' AMO, OM2 = OA2 – AM2 = 132 – 52 = 169 – 25 = 144 OM = 12 cm In ' CNO, ON2 = OC2 – CN2 = 132 – 122 ON = 5 cm The distance between the two parallel ? chords = MN = MO + ON = 12 + 5 = 17 cm
2
1· § ¨x ¸ x¹ ©
– 3 =0
2
1· § ¨x ¸ x¹ ©
?
=3
1 x
x
=
1· § Now, ¨ x ¸ x¹ ©
3
4
= ( 3 )4 = 9 3
57. (C)
x
1· 1 § = 3 ¨ x ¸ = 33 x¹ x ©
x3
x3
x3
?
x3
1 x
3
1 x3 1 x3 1 x3
3.x .
1§ 1· ¨ x ¸ = 27 x© x¹
3(3) = 27
= 27 – 9 = 18 2
1· § Also, ¨ x ¸ = 32 x¹ ©
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..... (1)
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x2
?
x2
1 x
2
1 2
2.x .
1 =9 x
Now,
=7
cos T (8 tan T 5) = cos3 T u §¨ tan3 T 2 3 ·¸ cos2 T ¹ ©
..... (2)
x Multiply (1) & (2)
§ 3 1 · § 2 1 · ¨ x 3 ¸ ¨ x 2 ¸ = 18 × 7 x ¹ © x ¹ ©
x3 x
x5
x2 x3
1 x5
= 126
=
§ 1· ¨ x ¸ = 126 © x¹ x 5
1 x5 1 x5
= 61. (C)
= 126 – 3
sec 2 T(8 tan T 5) tan3 T 2 3 sec 2 T (1 tan2 T )(8 tan T 5) tan3 T 2 3(1 tan2 T) (1 22 )(8.2 5) 23 2 3(1 22 )
= 123
gl
cos T =
tan T = tan 30º
The angle of elevation of the top of the tower = 30º tan7 T . tan 2 T = 1 sin 7T.sin 2T =1 cos 7T. cos 2T
cos(7T 2T) cos(7T 2T ) =1 cos(7T 2T) cos(7T 2T)
w
w
w
cos 5 T – cos 9 T = cos9 T + cos 5 T 2 cos 9 T = 0 cos 9 T = 0
9 T = (2n + 1)
S 2
[if cos T = 0, then T =
S , where n is an integer] 2
T = (2n + 1)
S 18
Now, when n = 0 T = (2 × 0 + 1)
=
S 18
tan T = 2
S 6
62. (B)
In ' ANO, OA2 = AN2 + NO2 152 = 122 + (21 – x)2 225 – 144 = (21 – x)2 81 = (21 – x)2 21 – x = 9 x = 21 – 9 = 12 ? Again CM2 = CO2 – OM2 = 152 – 122 = 225 – 144 = 81 CM = 9 cm ? The length of the 2nd chord CD = 2CM = 2 × 9 = 18 cm 63. (B) In ' ABC, §S· ABC = 75º, ACB = ¨ ¸ ©4¹
180 = 10º [ S radius = 180º] 18
Now tan 3 T = tan 3 × 10 = tan 30º = 60. (A)
T = 30º =
.s
(2n 1)
cos T = cos 30º
-c
T = 30º
3 2
sc
59. (C)
5 u 21 5 u 21 21 = = 10 15 25 5
7sin2 T + 3cos2 T = 4 7(1 – cos2 T ) + 3cos2 T = 4 7 – 7cos2 T + 3cos2 T = 4 7 – 4 = 4cos2 T 3 = 4cos2 T
20
58. (B)
?
=
.in
?
1
5 x
5 x
2
=
14
x5
8sinT 5cos T sin T 2cos3 T 3cos T 3
º
º
§ 180 · ¸ = 45º = ¨ © 4 ¹
1 3
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ABC + ACB + BAC = 180º [Angle sum property of a ' ]
For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in And for regular updates on our website like us on facebook - www.facebook.com/cgl.ssc2014 75º + 45º + BAC = 180º BAC = 180º – 120º = 60º
% Discount =
S = radius 3
(1101)2
= 1212201
Now, 121.2201 =
=
71. (C) 1212201 10000 1212201
10000
1101 = 11.01 100 p, q, r are in GP. q r = q p
? 72. (C)
GP is always same)
16 x 79 x = 7x 43 x (79 + x ) (7 + x ) = (16 + x ) (43 + x ) 553 + 86x + x2 = 688 + 59x + x2 86x – 59x = 688 – 553 27x = 135 x =5 73. (C) Total expenditure during the year = 2200 × 3 + 2550 × 4 + 3120 × 5 = Rs. 32400 Total saving = Rs. 1260 Total income = Rs. 32400 + 1260 = 33660
20
pr
HCF a & b = 12
a and b must be multiplies of 12 (e.g.
gl
66. (D)
=
=
Then,
q2 = pr q
5 18
l 10 = b 8 l:b =5:4 Let x be added to each of the number 7, 16, 43, 79
( The common ratio in a
=
14
180 × 100 = 10% 1800
5 18 10l + 10b = 18l 10b = 8l
=
65. (C)
l 2(l b )
=
.in
64. (B)
1800 1620 × 100 1800
1 1 2 1 1 = = work 6 12 12 4
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1 work is completed by A & B in 1 day 4
1 work is completed by A & B in
1 1 4
= 4 days 68. (B) = = 69. (B)
100 × 10 – 100 + 2000 ÷ 100 100(10 – 1) + 20 100 × 9 + 20 = 920 Total property given by the person
70. (D)
1 1 1 5 10 4 19 = = 4 2 5 20 20 Marked price = 120% of 1500
=
120 × 1500 = Rs. 1800 100 SP = 108% of 1500
=
Average monthly income
33660 12 = Rs. 2805
=
74. (A)
w
=
w
.s
sc
-c
12, 24, 36, 48, ...) Also, a > b > 12 So, the smallest possible values of a & b will be as follows :b = 2 × 12 = 24 a = 3 × 12 = 36 So, Answer is 36, 24. 67. (D) Work done by A & B in 1 day
Suppose 'm' stand for man and 'b' stand for boy ATQ, 12(3m + 4b) = 10(4m + 3b) 36m + 48b = 40m + 30b 18b = 4m Now, 36m + 48b 18 b + 48b 4 = 162b + 48b = 210b 210 boys complete the work in 1 day ? 1 boy complete the work in 210 days
= 36 ×
18 § · ? 2m + 3b = ¨ 2 u 4 b 3b ¸ © ¹
12 boys complete the work = =
108 = × 1500 = Rs. 1620 100
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210 days 12
70 35 = days 4 2
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% Discount = =
80. (C)
10 18 × 3 5 = 12 km/hr Let P = Rs. 100 CI for Ist year = 5% of 100 = Rs. 5 CI for the 2nd year = 5 + 5% of 5 = Rs. 5.25 When CI is Rs. 5.25, P = Rs. 100
6000 5500 × 100 6000
=
500 × 100 6000
25 1 %= 8 % 3 3 76. (D) CP of 10 cycles = 10 × 500 = Rs. 5000 Repairing charge = Rs. 2000 Net CP = Rs. 7000 SP = Rs. 750 × 5 + 550 × 5 = 3750 + 2750 = 6500 SP < CP
=
81. (C)
7000 6500 × 100 7000
gl -c
72 × 100 3600 = 2% Let the total number of students be x No. of students passed in Eng.
Required % =
w
.s
78. (C)
82. (C)
1 % 7
sc
77. (D)
90x 100
n(M) =
85x 100
P=
100 × 420 5.25 = Rs. 8000
22 × 11 × 11 7 = 760.57 cm2 5 years ago sum of ages of P & Q = 30 yrs Total age of P, Q & R at present = 20 × 3 = 60 yrs. Age of R at present = 60 – (30 + 10) = 60 – 40 = 20 yrs. Age of R 10 years later = 20 + 10 = 30 yrs. Let the CP of the goods = Rs. x SP of the goods = 132% of x
83. (A)
=2×
Now,
88% of MP MP
n(E M) = 150
n(E M) = n(E) + n(M) – m(E M) x
90x 85x = + – 150 100 100
150
=
150
=
75x x x
100 5.25
r = 11 cm
w
n(E) =
When CI is Rs. 420
90x 100
w
= 90% of x =
P = Rs.
2 CSA of hemisphere = 2Sr
500 = × 100 7000
= 7
When CI is Re. 1
20
% loss
1 10 = m/s 3 3
Speed
=
?
= 3
79. (D)
.in
75. (C)
1 days 2
14
= 17
=
132x 100
=
132x 100
=
132x 100 × 100 88
=
3x 2
3x x 2 Required % = × 100 x
90x 85x – x 100
x 2 = x × 100
175x 100x 100 = 150 × 100
=
150 u 100 75 = 200
=
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1 × 100 = 50% 2
For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in And for regular updates on our website like us on facebook - www.facebook.com/cgl.ssc2014 TSA of the rectangular parallelopiped = 2(lb + bh + lh) = 2[12 × 6 + 6 × 6 + 12 × 6] = 2[72 + 36 + 72] = 360 cm2 88. (B) The expression = x4 – 2x2 + k is a perfect square. If its discriminant = 0 b2 – 4ac = 0 (–2)2 – 4 × 1 × k = 0 4 – 4k = 0 ? k =1
84. (C)
=
a2
=
3 a 2
.in
89. (C)
= 6 3
a = 12 cm Perimeter of ' ABC = 3 × 12 = 36 cm Area of verandah = Area of the hall including verandah – Area of the hall only = (25 + 3.5 + 3.5) × (15 + 3.5 + 3.5) – 25 × 15 = 32 × 22 – 25 × 15 = 704 – 375 = 329 m2 Cost @ of Rs.27.50 per m2 = 329 × 27.50 = Rs. 9047.50 Let 'r' be the base radius & 'h' is the height of the cone.
14
3 a 2
a2 4
Co-ordinates of points A = (0, 1) B = (1, 0) C = (3, 0) D = (0, 2) Area of quad. ABCD = ar( ' ABC) + ar( ' ACD)
[ Area of a ' =
.s
86. (A)
sc
-c
gl
85. (C)
AB2 BM2
20
Now,
=
then, l =
=
w
h2 r 2 TSA of cone = S rl + S r2
w w
= S r[ h2 r2 + r]
=
= S r [ r2 r2 + r] [ h = r]
= 2
TSA of the hemisphere = 3Sr2
87. (B)
Sr2( 2 1) 3Sr2
=
2 1 : 3
Length of the parallelopiped 'l ' b h
3 5 = sq. unit 2 2
1 sq. unit 2 Shortcut method:Required area = ar( ' OCD) – ar( ' OBA)
= S r2 ( 2 1 )
=
l l [0 – 1 + 3] + [0 + 3 + 0] 2 2
=1+
= S r ( 2r r )
2 1 3
l [0(0 – 0) + 1(0 – 1) + 3(1 – 0)] + 2 l [0(0 – 2) + 3(2 – 1) + 0(1 – 0)] 2
= S r. h2 r2 + S r2
Required ratio =
l [x (y – y ) + x2(y3 – y1) + x3(y1 – y2)] 2 1 2 3
=
l l ×2×3– ×1×1 2 2
6 1 5 1 = = 2 sq. unit 2 2 2 2 Each exterior angle of a regular polygon
=
= 6 + 6 = 12 cm = 6 cm = 6 cm
90. (D)
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=
360º x
For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in And for regular updates on our website like us on facebook - www.facebook.com/cgl.ssc2014 360º x x =5 Sum of all interior angles of the polygon of 5 sides = (2 × 5 – 4) × 90 = 6 × 90 = 540º radius of cylinder r1 o r2 o radius of cone h1 o height of cylinder h2 o height of cone
% decrease in production in 1999-2000
72 =
r1 : r 2 =
3 : 2 & h 1 : h2 =
3
95. (D)
2: 3
Sr22h 2
97. (A)
14
w
93. (A) % increase in production of wheat from
w
1000 500 × 100 = 100% 500 % decrease in production from 1996-97
1997–1998 =
94. (C)
=
600 500 × 100 600
=
100 50 × 100 = % 600 3
99. (C)
=
10 × 360 = 36º 100 100. (B) Let the total amount spent on buying clothes & house maintenance = Rs. x Then, (12.5 – 10)% of x = 1500
=
1500 u 100 2.5 = Rs. 60000 Amount spent for House maintenance = 10% of 60000 = Rs. 6000
2 % 3 % decrease in production in 1995-96
= 16
=
1 % of 50,000 2
25 × 50000 200 = Rs. 6250 Amount spent for study of children & food together. = 55% of 35000 = 0.55 × 35000 = Rs. 19250 Angle of pie chart representing the expenditure on entertainment
20
gl
The production of wheat reached maximum in 1999.
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92. (C)
3
.s
? Required Ratio = 3 3 : 2
2
sc
2u 2
u
98. (A)
-c
3u 3 u 3
2
30 u 2 12 = 25 5 Reqd. Ratio = 12 : 5. ? Amount spent on buyinig clothes
= 12
§ 3 ·2 2 3 = ¨ ¨ 2¸ ¸ u 3 © ¹
3 3
30 1 12 2
=
§ r1 · § h1 · = 3¨¨ ¸¸ u ¨¨ ¸¸ © r2 ¹ © h2 ¹
=
=
96. (C)
2
=
200 1 × 100 = 13 % 1500 3 Total production from 1995 to 1998 = 1200 + 600 + 500 + 1000 = 3300 quintals
=
Sr12h1
= 1
1500 1300 × 100 1500
.in
91. (B)
=
1200 600 × 100 = 50% 1200
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x
=
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Remove 'will'
177.(C);
Change 'is' into 'was'
178.(B);
Change 'to pay' into 'paying'
179.(B);
Change 'hardly' into 'hard'
180.(A);
Add 'an' before 'umbrella'
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176.(B);
126. C 127. C 128. B 129. B 130. D 131. A 132. A 133. D 134. C 135. C 136. C 137. A 138. C 139. C 140. C 141. B 142. A 143. A 144. C 145. A 146. D 147. C 148. B 149. B 150. A
151. C 152. A 153. C 154. D 155. B 156. B 157. C 158. C 159. C 160. B 161. C 162. B 163. C 164. B 165. B 166. D 167. C 168. B 169. C 170. A 171. D 172. D 173. B 174. C 175. A
.in
101. A 102. A 103. D 104. C 105. D 106. C 107. C 108. A 109. B 110. B 111. A 112. A 113. A 114. C 115. A 116. C 117. C 118. D 119. C 120. D 121. A 122. A 123. A 124. D 125. D
20
76. D 77. D 78. C 79. D 80. C 81. C 82. C 83. A 84. C 85. C 86. A 87. B 88. B 89. C 90. D 91. B 92. C 93. A 94. C 95. D 96. C 97. A 98. A 99. C 100. B
gl
D D C A B B C B C A C B B B C D D B B D C C C A C
-c
51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75.
sc
C C B D C D A C C B D B C D C C B D A D C A C C C
.s
26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50.
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D B C C B C D C A A C D C B C C B C C B A D C D D
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1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.
(ANSWER KEY)
14
SSC MOCK TEST
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176. B 177. C 178. B 179. B 180. A 181. B 182. C 183. C 184. C 185. D 186. B 187. C 188. B 189. B 190. A 191. B 192. A 193. A 194. C 195. A 196. D 197. A 198. B 199. D 200. B
