MOCK TEST SBI CLERK MAINS - 04
NUMERICAL ABILITY
51. (b) profit earned by the company in the year 2006 =
1.84 + 25% of 1.84 = 2.3 lakhs
52. (a) Since the graph gives percent increase in profit, it
means the profits of companies were increasing
through the given periods. Hence both the companies
made highest profit in 2009.
53. (d) Required percentage =
15
 100 = 75
20
54. (d) Let the profit earned by company M in the year
2006 be P. Then
1.25  1.15  P = 3.63
P = 2.52 Lakhs
55. (e) Required ratio = (1.15  5 : 1.1  6) = 575 : 660 =
115 : 132
56. (c) Sum of adjacent angles of a parallelogram=180°
one angle of triangle = (2/3)×180°=120°
sum of remaining two angles = 180-120=60°
∴second largest angle = (60/12)×7=35°
57. (b)
Ajay’s score=63+30=93
Rahul’s score=93-15=78
Manish’s + Suresh’s score=63×3 - 78
=189-78=111
58. (e) Speed of the car= 588/6= 98km/hr
Speed of train = (10/7)×98=140km/hr
Distance covered by the train in 13 hours =
140×13=1820km
59. (d) Raju runs on Monday and Friday=1250×2=2500m
On Tuesday, Wednesday, Thursday and Saturday,
Raju runs= 1500×4=6000m
In one week Raju runs 6000+2500=8500m
In 3 weeks Raju runs= 3×8500=25500m=25.5km
60. (c) side of the square = 56/4=14cm
∴smallest side of the triangle = 14-8=6cm
Length of the rectangle =96/8=12cm
second largest side of the triangle = 12-4=8cm
∴largest side of the triangle =
√(6²+8²)=√(36+64)=√100=10cm
61. (c) CP= (996+894)/2= Rs 945
62. (a) (11x/100)=5236
=>x = Rs 47600(Monthly income)
total monthly amount invested by Deepti =
47600×(11+19+7)%
=47600×(37/100)=Rs 17,612
total annual amount invested by Deepti = 17612×12=
Rs 2,11,344
63. (a); Let larger number= x
x × (x − 2) = 3248
∴ x = −56, 58
i.e. larger number=58
64. (d); Let the number be 10x + y
Then, after interchanging of digit we get
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10 x + y − (10y + x) = 9
∴ x − y = 1 …………………(i)
& given x + y = 9 ………….(ii)
From (i) and (ii)
x = 8, y = 7
∴ Number= 10 × 8 + 7 = 87 or 78
65. (d); Let earning of A & B be 4x & 5x
Then,
120
100
80
5x×
100
4x×
=
6
5
i.e. we can not determined the value of x
Hence, earning’s cannot be also determined
15
3
23×23×23
66. (c); C.I.= 8000 (1 + ) − 8000 = 8000 ×
−
100
20×20×20
8000 = Rs. 4167
67. (a); Let upstream speed = 3x
Downstream speed = 5x
22.5
∴
= 22.5
3x
∴x=3
∴ downstream speed= 5 × 3 = 15
1
∴ Speed of boat in still water = (15 + 9)
2
= 12 kmph
5000×r×2
3000×4×r
68. (d);
+
= 2200
100
100
22000r = 22000 × 10
r = 10%
69. (d); x × 100 = (x − 10) × 110
100x = 110x − 1100
10x = 1100
x = 110
70. (c); For 2 year, C.I = 6000 (1 +
5
100
5 2
2
) − 6000 = 615
For 3 year, C.I. = 6000 (1 + ) − 6000 = 945.75
100
∴ Required different = 330.75 Rs.
71. (e) Both the statements together is not sufficient.
72. (d) Both statements alone is sufficient.
Let age of A is x and age of B is y.
∴ 3x = y ... (i)
From statement I:
x  10 3
 ... (ii)
y  10 7
Solving equations (i) and (ii), x = 20, y = 60
From statement II:
=
x  10 1

y  10 5
=>5x – 50 = y – 10
=>5x – y = 40 (iii)
From equations (i) and (iii), x = 20, y = 60
73. (c) Let length the train be x.
From statement I:
𝑥
Speed =
8
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From statement II:
x  50
20
Speed =
Solving this, x =
400 100
=
3 meter
12
74. (e) From I and II we find a + b = 2z
we can not find a + b from I and II together.
75. (e) Data is not sufficient.
From statement I:
80
x=
×𝑧
100
From statement II:
60
y=
×𝑧×z
100
From these two equations
𝑥
𝑦
=
4
3
but cannot find x and
y separately.
76. (c) The pattern of the number series is :
50 × 1.20 = 60
60 × 1.25 = 75
75 × 1.30 = 97.5
97.5 × 1.35 = 131.625
131.625 × 1.40 = 184.275
77. (c) The pattern of the number series is :
12 × 1 + 3 × 1 = 15
15 × 2 + 3 × 2 = 36
36 × 3 + 3 × 3 = 117
117 × 4 + 3 × 4 = 480
480 × 5 + 3 × 5 = 2415
78. (b) The pattern of the number series is :
1×1+1=2
2×2+2=6
6 × 3 + 3 = 21
21 × 4 + 4 = 88
88 × 5 + 5 = 445
445 × 6 + 6 = 2676
79. (d) The pattern of the number series is :
20 + 12 = 21
21 + 22 = 25
25 + 32 = 34
34 + 42 = 50
50 + 52 = 75
80. (e) The pattern of the number series is :
600
+ 5 = 125
5
125
5
𝟑𝟎
𝟓
11
5
+ 5 = 30
+ 5 = 11
3
Y= ,
5 4
Hence, x < y
84. (c) x = 183
Y = 193
Hence, x < y
85. (c) x = 7/5, 5/3
Y = 2, 7/4
Hence, x < y
18
86. (a); In Mumbai, distance travelled by road= ×
25
3350 = 2412 km
7
In Bangalore, distance travelled by road= × 2016 =
8
1764 km
2412
So, required percentage=
× 100% = 136.73%
1764
In Chennai= 3528 ×
7
→ x=
5
9 7
49 25
81
3 7
√y = 4 , 8
9
,
49
49
→y= ,
16 64
Hence, No Relation
82. (e) x = −7√3 , 5√5
Y = 3√5, 2√5
Hence, No Relation
3 5
83. (c) x = − ,
2 7
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18
= 980 km
4
16.10×1
TA by rail in Lucknow
2
161
=
=
= 0.44 = = 2 ∶ 5
12.10×3
121×3
5
90. (b); TA earned in Delhi
11
By rail= × 432 × 12.60 = Rs. 4989.60
12
1
By road= × 432 × 15.20 = Rs. 547.20
12
TA earned in Lucknow
3
By rail= × 1024 × 12.10 = Rs. 9292.80
4
1
By road = × 1024 × 16.10 = 4121.60
4
Hence, total TA = 4989.60 + 547.20 + 9292.80 +
4121.60 = 18951.2 Rs.
33
15
1
91. (a); ? = ×
×
× 4400 = 2.3150 ≈ 2.3
16
92.
93.
94.
95.
96.
588
100
15
(a); ? = 0.5 × 0.4 × = 0.42857 ≈ 0.43
7
(b); ? = 1.264 + 6.444 + 3.222 = 10.93 ≈ 11
(b); ? = 12.25 + 18.6624 + 27.9841 = 58.8965 ≈ 59
12+10+8−1
(d); ? =
× 2.432 = 1.8125 × 2.432 = 4.408
16
2
(c); ? = 2 × (79) + 2 × (49)2 = 2 × [6241 + 2401] =
17284
2
,
25
5
In Hyderabad= 1456 × = 448 km
13
Hence, distance travelled is maximum in Chennai.
TA by road in Lucknow
89. (b); Required ratio=
+ 5 = 7.2
81. (e) √x =
3
87. (d) Distance travelled in Kolkata by rail= × 4256 =
19
672 km
So, required TA= 672 × 12.90 = Rs. 8668.8
3
88. (c); Distance covered by rail in Kolkata= 4256 × =
19
672 km
7
In Mumbai= 3350 × = 938 km
97. (b); ? =
(73 )3 ×(83 )
56
4
2
3
7
=
49×64
56
= 56
98. (d); ? ( − 1) = × 616 × 12 ×
1
3
308×21
3
1
1
?× =
?× =
3
3
16
8
− 323 − 81 =
1
16
1617
4
− 17 × 19 − 81
− 404
4
∴ ?=
4
99. (d); ?2 = 529 × 324
∴? = 23 × 18 = 414
100. (b); 33?+1 = 337.8+1.2−5
∴ ?= 9− 5 −1 = 3
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