ssc mock test – 124 – 1 [answer with solution]sschansrajacademy.com/questions/ssc-pre_124_1...
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P-1
SSC MOCK TEST – 124 – 1 [Answer with Solution]1. (A) The young one of a baboon is called an infant while
the young one of an beaver is called a kitten.
2. (D) The series is a, ab, abc, a, ab, abc, ...
3. (D) 7 13 25 49 97 ?
+6 +12 +24 +49 +96
×2 ×2 ×2 ×2
? = 97 + 96 = 193
4. (B) The letters in the series are moving four placesback in the alphabetical series. So, the next lettersin the series will be FBX.
5. (D)
G J X U M P R O S V
+6 +6+6 +6
–6–6
6. (C) 17 150 167 1500 1667 15000
+150 +1500 +15000
16667
×10 ×10
7. (D) Gourd, Pumpkin and French Beans grow on vineswhile Jackfruit grows on trees.
8. (C) Sum of the digits of 272 is 11. While in rest of thethree option, sum of the digits of the numbers is10.
9. (C) Animals in options (A), (B) and (D) are mammalswhile hen is a bird.
10. (D)
O6 km 8 km
6 km 6 km
14 km Ravi
8 km
Rakesh
Let O be the initial position of Ravi and Rakesh.
Distance between Ravi and Rakesh
= 2 214 14 14 2 km
11. (A) The order of the height of five given persons willbe:
Anil > Karan > Mukesh and Anil > Sunil > Rakesh
Hence, Anil is the tallest among the five.
12. (B) Option (B) is the answer since the given word hasa single 'N' while TOURNAMENT uses 'N' twice.
13. (C) In the series D F J P X the number of lettersskipped between adjacent letters in consecutiveodd numbers starting from 1.
D 2 F 4 J 6 P 8 X
14. (A) Total members = 2 + 3 + 4 × 2 + 4 × 5 = 33.
15. (D) P(16 4 T(20)
L(12) 4 H(8)
A(1) 4 E(5)
N(14) 4 J(10)
E(5) 4 I(9)
T(20) 4 P(16)
Similarly,
S(19) 4 (W(23)
Q(17) 4 M(13)
U(21) 4 Y(25)
A(1) 4 W(23)
R(18) 4 V(22)
E(5) 4 A(1)
16. (A) In column 1,
52 = 25, 122 = 144 5 × 12 ÷ 2 = 30
In column 2,
72 = 49, 142 = 196 7 × 14 ÷ 2 = 49
Similarly, in column 3,
92 = 81, 162 = 256 9 × 16 ÷ 2 = 72.
17. (D) 7 15 30 59 116 229
+8 +15 +29 +57 +113
8×2–1 15×2–1 29×2–1 57×2–1
P-2
18. (B) 27 13 51 29 63 37
14÷2=7 22÷2=11 26÷2= 13
19. (A)GlovesThings
Warm
As is clear from the Venn diagram, only conclusion(III) follows.
20. (B)
21. (D) Universe contains millions of galaxies and galaxycontains millions of stars and planets.
22. (C) 1 and 5 are opposite but in option (A) they areadjacent. Similarly, 3 and 6 are opposite but inoption (B) they are adjacent. Similarly, 2 and 4are opposite but in option (D) they are adjacent.Hence, option (C) is correct.
23. (C)
24. (A)
25. (C) For matrix,
B – 45, 62
O – 25, 46
R – 22, 63
N – 33, 53
Option options,
BORN 62, 46, 22, 53.
26. (A) 30
20
8
5/2
1×5/2 ×5/2
5302
Total money spend = 5302
× 9 = 292.50
27. (C)
28. (A)
29. (A)
30. (D) Fixed price + 5 km = 350 .......... (i)
Fixed price + 20 km = 800 .......... (ii)
Eq. (i) – (ii)
15 km = 450
1 km charge = ` 30
Putting in (i)
So, fixed price = 350 – 30 × 5 = ` 200
31. (C) ATQ,45 man = 25 women9 man = 5 women
Man 5Women 9
Let, Man = 50Women = 90
Married Man = 50 × 45
100 = 22.5
Married Women = 90 × 25
100 = 22.5
Total adults who are married = 45Total adults who are not married = 140 – 45 = 95
So, % 95
140×100
= 95014
= 67.85%
32. (D) Let, B = 221, C = 1, a = 133. (A) ATQ,
CP ×37 1 51.802 100 100
CP = ` 280
34. (B)18 3
647 47
Given
Quantity Cheaked Total 47 : 3 = 50×20 ×20940 1000 gram
35. (B) After taking out 8 ltrs.Ratio will remain same to 5 : 3So, Alcohal in new mixture of 32 ltrs.
= 58
×32 = 20
Water in new mixture of = 38
×32 = 12
New quantity of water = 12 + 8 = 20So, Alcohal : Water
20 : 20= 1 : 1
P-3
36. (A) Let Distance = 120×4 = 480 km
Time taken = 12020
= 6 hrs.
for Ist side
For IInd side = 12030
= 4 hrs.
IIIrd side = 12040
= 3 hrs.
IVth side = 12060
= 2 hrs.
Average speed = total distance
total time
= 480
6 4 3 2 = 32
37. (C)
2 3
69I II
18
= To make it 34
flat = 34
× 185 = 2h 42 min
38. (A) an + bn is always divisible by a +b
if x is odd number
So,17 29 46
23 23 = 2
So, remainder = zero
39. (C) 10 k3 10242 10 10100
10 k10 30 22 10 10 2 10
1032 = 10k
k = 32
40. (D) A B 650
Relative speed (A+B) = 65010
= 65 km/h
A + B = 65 km/h .......... (i)
Now,
Distance covered by (A + B) in 8 hour
8 × 65 = 520 km
Remain distance = 650 – 520 = 130
130 km distance is covered by A in 4 hour 20 min.
speed of A = 130
4hrs20 min
= 130 313
= 30 km/hrs.
speed of B = 65 – 30 = 35 km/ hrs.
41. (A)
96
A32
C24
43
B48
2
work done by A = 4 ×3 = 12
work done by B = 2× (4+2) = 12
work done by C = 96 – (12+12) = 72
work done by
A B C
12 12 72
1 1 6 = 8
×810 ×810
4860 6480 (given)
42. (B) timep taken by pipe A to fill the tank = 48 hour.
time taken by B to likage the tank of its fulleffeciency
= 45 1002 75 = 30 hour
240
48 30
–85
A B
ATQ,
capicity = 240 × 3 × 14
= 180
Now, A + B = 5 – 8 = –3 unit/hour
Require time = 180
3 = 60 hour
P-4
Answer Key1. (A) 2. (D) 3. (D) 4. (B) 5. (D)6. (C) 7. (D) 8. (C) 9. (C) 10. (D)11. (A) 12. (B) 13. (C) 14. (A) 15. (D)16. (A) 17. (D) 18. (B) 19. (A) 20. (B)21. (D) 22. (C) 23. (C) 24. (A) 25. (C)26. (A) 27. (C) 28. (A) 29. (A) 30. (D)31. (C) 32. (D) 33. (A) 34. (B) 35. (B)36. (A) 37. (C) 38. (A) 39. (C) 40. (D)41. (A) 42. (B) 43. (D) 44. (D) 45. (B)46. (B) 47. (B) 48. (C) 49. (B) 50. (A)51. (B) 52. (C) 53. (B) 54. (D) 55. (C)56. (D) 57. (A) 58. (A) 59. (D) 60. (C)61. (B) 62. (C) 63. (A) 64. (B) 65. (A)66. (A) 67. (D) 68. (D) 69. (B) 70. (C)71. (D) 72. (C) 73. (C) 74. (A) 75. (A)76. (A) 77. (B) 78. (B) 79. (B) 80. (C)81. (C) 82. (C) 83. (C) 84. (B) 85. (C)86. (A) 87. (A) 88. (C) 89. (B) 90. (B)91. (B) 92. (A) 93. (A) 94. (C) 95. (C)96. (D) 97. (B) 98. (A) 99. (D) 100.(C)
43. (D)
A
B
C
P T70° 45°
65°
70°
PAC + BAC + BAT = 180°
PAC = 180° – 65° – 45° = 70°
PAC = ABC = 70° (alternate angles)
44. (D) EF || DC (given)
EGF ~ CGD ( by A–A)
EG EFGC CD
5 EF
10 18
EF = 9cm.
45. (B) A
CB D
P R Q
APR ~ ABD ( by A–A)
AP ARAB AD
3 4.55 AD AD = 7.5 cm
46. (B) cos2 + cos2 = 2
= = 0º
tan7 + sin8 = 0
47. (B) cot 90º = 0
hence 0
48. (C) Let, a = s , b = s, c = 0
Now, 2 2 2 2
2 2
s s s s s 0 ss s 0
= 1
49. (B) put n = 1
sum = 1–12
= 12
So, option (B)
50. (A) As rate is same-annualy, so
Time will be doubled
Amount Principle
1331 1000
Taking square root
11 10
So 1
10 × 100 = 10% half yearly
or
20% per annualy
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