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SBI PO 2015
PRELIM (TIER – I) EXAM
MODEL PAPER
SOLUTIONS
Part 1 – SOLUTIONS ENGLISH LANGUAGE
Answers (1-10)
1. (5)
2. (1)
3. (2)
4. (3)
5. (4)
6. (4) nurturing
7. (1) drive
8. (4) hurdle
9. (2) manageable
10. (4) differ from
Explanations (11 -15)
11. (2) Use ‘on’ in place of ‘by’. (lack of interest on the part
of the public)
12. (2) Use ‘booming’ in place of ‘boom’. (a booming
business fuelled)
13. (5) No error.
14. (4) Use ‘among’ in place of ‘between’. (among the
public)
15. (5) No error
Answer (16-20)
16. (4)
17. (4)
18. (5)
19. (5)
20. (1)
Answers (21-30)
21. (2)
22. (3)
23. (5)
24. (3)
25. (4)
26. (5)
27. (4)
28. (2)
29. (1)
30. (1)
Part 2 – SOLUTIONS QUANTITATIVE APTITUDE
Solutions (31-35)
31. (2)
32. (2)
33. (3)
34. (4)
35. (1)
Solutions (36-40)
36. (2)
I. 12x2 + 11x + 12 = 10x2 + 22x
2x2 - 11x + 12 = 0
2x2 - 8x – 3x + 12 = 0
2x (x – 4) – 3 (x – 4) = 0
(x – 4) (2x – 3)
x = 4, x = 1.5
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II. 13y2 – 18y + 3 = 9y2 – 10y
4y2 – 8y + 3 = 0
4y2 – 6y – 2y + 3 = 0
4y2 – 2y – 6y + 3 = 0
2y (2y – 1) 3 (2y -1) = 0
(2y – 1) (2y – 3) = 0
y = 1/2 , y = 3/2
y = 1.5, y = 3/2
So x > y
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37. (3)
I. 18/x2 + 6/x – 12/x2 = 8/x2
18 + 6x - 12/ x2 =8/x2
6 + 6x = 8
6x = 2
X = 2/6 = 0.33
II. y3 + 9.68 + 5.64 = 16.95
y3 = 16.95 – 15.32
y3 = 1.63
y = √1.63
y ~ 1.18
x<y
38. (1)
I. √1225 x + √4900 = 0
35x + 70 = 0
x = - 70/35
x=-2
II. (81)1/4 y + (343)1/3 = 0
3y + 7 = 0
y = - 7/3
y = - 233
x>y
39. (1)
I. (2)5 + (11)3 / 6 = x3
32 + 1331 / 6 = x3
x3 = 1363/6 = 227.167
II. 4y3 = - (589 ÷4) + 5y3
y3 = (589 ÷4)
y3 = 147.25
x>y
40. (4)
I. (x7/5 ÷ 9) = 169 ÷ x3/5
X10/5 = 9 x 169
X2 = 9 x 169
X = ± 3 x 13
X = ± 39
II. y1/4 X y 1/4 X 7 = 273 ÷y1/2
y = 273/7 = 39
x<y
Solutions (41-45)
41. (5) 29, 37, 21, 43, 13, 53, 5
29 – 8 = 21,
21 – 8 = 13,
13 – 8 = 5
37 +8 = 45,
45 + 8 = 53
So wrong number in the series is 43.
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42. (3) 600, 125, 30, 13, 7.2, 6.44, 6.288
6.288 x 5 – (5x5) = 6.44
6.44 x 5 – (5x5) = 7.2
7.2 x 5 – (5x5) = 11
11 x 5 – (5x5) = 30
30 x 5 – (5x5) = 125
125 x 5 – (5x5) = 600
So wrong number in the series is 13.
43. (3) 80, 42, 24, 13.5, 8.75, 6.375, 5.1875
5.1875 x 2 – (2x2) = 6.375
6.375 x 2 – (2x2) = 8.75
8.75 x 2 – (2x2) = 13.5
13.5 x 2 – (2x2) = 23
24 x 2 – (2x2) = 42
42 x2 – (2x2) = 80
So wrong number in the series is 24.
44. (2) 10, 8, 13, 15, 35, 135, 671, 4007
10 x1 – 2 = 8
8 x2 – 3 = 13
13 x3 – 4 = 35
35 x4 – 5 = 135
135 x 5 – 6 = 669
669 x 6 – 7 = 4007
So wrong number in the series is 671
45. (5)150, 290, 560, 1120, 2140, 4230, 8400
150 x2 – 10 = 290
290 x 2 – 20 = 560
560 x 2 – 30 = 1090
1090 x2 – 40 = 2140
So wrong number in the series is 1120.
Solutions (46 – 55)
46. (1) Let the amount be Rs. x
According to the question:
(3/9) x – (2/14) x = 40
x (1/3 – 1/7) = 40
x = (21 X 40)/4
x = 210 Rs.
47. (3) According to the question, Rachita buys highest
number of pastries than cookies and then ice – cream
She has to buy 9 units of each.
But she has to buy total 32 units and each item of one
is must.
Pastries > Cookies > Ice – cream
13 or 12 10 or 11 9
Hence, she can buy either 10 or 11 units of cookies.
48. (3) Bus fare of first 5 km is = Rs. x
So bus fare of 1 km is = Rs. x/5
According to the question:
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5 X (x/5) + 182 X 13 = 2402
x + 2366 = 2402
x = 36
x = Rs. 36
49. (1) Let three even numbers are x , (x + 2), (x + 4)
According to the question:
x (x + 2) (x + 4) = 4032
x (x + 4) = 252 Eq (I)
(x + 2) 252 = 4032 Eq (II)
(x + 2) = (4032/252) = 16
x = 16 -2 = 14
x = 14
So numbers are 14, 16 and 18.
Now five times of the second number will be = 16 x 5
= 80
50. (1) Sum of ages of 4 members of a family 5 yr ago =
94 yr
Their present age with daughter = 94 + 4 x 5 = 114 yr
Now, their present age with daughter – in – law = 92
yr
Difference between daughter and daughter – in – law
= 114 – 92 = 22 yr
51. (4) Total ball in the bag = 13 + 7 = 20
n(S) = 20C2 = (20!/2! 8!) = (20 x 19 x 18!)/ 2 x 18! =
190
and for same colour of ball
n(E1) = 13C2 = (13!/2! 11!) = (13 x 12 x 11!)/ 2 x 11! =
(13 x 12)/2 = 78
and n(E2) = 7C2 = (7!/2! 5!) = (7x 6 x 5!)/ 2 x 5! =
(7x6)/ 2 = 21
Required probability for same colour ball = P(E) =
[n(E1) + n(E2) ]/ n(S) = (78 + 21)/190
= 99/190
52. (2) Akash scored in subject A = 73 marks
Akash scored in subject B = 56% of 150 = 84
Akash scored in subject C = x marks
Maximum marks in all the three subject is = 150
Total marks = 150 x 3 = 450
Now according to the question:
Marks obtained in subject (A + B + C) = 54% of total
marks
73 + 84 + x = 54% of 450
x = 86, hence Akash scored 86 marks in subject C.
53. (4) Area of Square = 1444m2
Side of the square = √1444 = 38m
Now, according to the question,
Breadth of the rectangle = 38 x 1/4
Length of the rectangle = 38 x 3/4
Area of rectangle = Length x Breadth
= 38 x 1/4 X 38 x 3/4 = 1444 x (3/16) m2
Now, difference between the Area of Square and
rectangle
= 1444 - 1444 x (3/16)
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= 1444 (1 – 3/16) = 1173.25 m2
54. (5) Share of A = 4/11 x 73689 = Rs. 26796
Share of B = 7/11 x 73689 = Rs. 46893
Required difference = 2 x share of B – 3 x share of A
= 2 x 46893 – 3 x 26796
= Rs. 13398
55. (4) In one day, work done by:
A + B = 1/20 (i)
B + C = 1/30 (ii)
A + C = 1/40 (iii)
Adding (i) and (ii) above, work done by
A + 2B + C = 1/20 + 1/30
2B = 1/20 + 1/30 – 1/40
B = 7/240
putting (B = 7/240) in (i) A does 5/240 work in one
day, similarly putting the same value in (ii) C does
1/240 work in one day.
Required ratio = (1/240) : (5/240) = 1:5
Solutions (56- 60)
56. (3) Number of mobiles phones sold in the month of
July = 45000 x 17/100 = 7650
Mobile phones sold by company A = (8/15) x 7650 =
4080
Mobile phones sold by company B = 7650 – 4080 =
3570
Now number of mobiles phones sold in the month of
December = 45000 x (16/100) = 7200
Mobile phones sold by Company A = (7/16) x 7200 =
3150
Mobile phone sold by company B = 7200 – 3150 =
4050
So required ratio = 3570 : 4050 = 357 : 405 = 119 :
135
57. (3) Number of mobile phones sold in the month of
November = 45000 x (12/100) = 5400
Mobile phone sold by company A = (7/15) x 5400 =
2520
Mobile phones sold by company A at a discount =
35% of 2520 = 882
Mobile phones sold by company A without a discount
= 2520 – 882 = 1638
58. (4) Number of mobile phones sold in the month of
October = 45000 x (8/100) = 3600
Mobile phone sold by company B = (5/12) x 3600 =
1500
Total profit earned by company B = 1500 x 433 = Rs.
649500
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59. (5) From sol. No. 56 –
Number of mobiles phones sold by company A in the
month of July = 4080
Number of mobiles phones sold by company A in the
month of December = 3150
: Required % = (4080/3150) x 100 = 129.53% ~ =
130%
60. (1) Number of mobiles phones sold in the month of
August = 45000 x (22/100) = 9900
Number of mobiles phones sold by company B in
August = (5/9) x 9900 = 5500
Total number of mobiles phones sold in the month of
September = 45000 x (25/100) = 11250
Number of mobiles phones sold by company B in
September = (2/5) x 11250 = 4500
So total number of phones sold by company B in
August and September = 5500 + 4500 = 1000
Solutions (61-65)
61. (1) Required average earning = Rs (3.34 + 5.83 +
1.69)3 = Rs. 3.62 lakhs
62. (2) Required respective ratio = 2.79 : 9.45 = 31 : 105
63. (5) Required increase percentage = (9.45 –
8.42/8.42) x 100 = 12.23 ~ 12%
64. (4) D’s earning increased consistently from the year
2005 to the year 2010
65. (3) Required percentage = (1.44 + 7.84)/5.53 x 100 =
167.81 ~ 168%
Part 3 – SOLUTIONS REASONING
Solutions (66-69)
After carefully study the given input and various steps of
arrangement, we may conclude that in each step numbers
are arranged in descending order from left to right while
words are arranged in opposite English alphabetical
order from right to left and the same process is continued
till the rearrangement ends.
Input: 84 why shit 14 32 not best ink feet 51 27 vain
68 92
Step I: 14 84 why shit 32 not ink feet 51 27 vain 68 92
best
Step II: 27 14 84 why shit 32 not ink 51 vain 68 92 best
feet
Step III: 32 27 14 84 why shit not 51 vain 68 92 best feet
ink
Step IV: 51 32 27 14 84 why shit vain 68 92 best feet ink
not
Step V: 68 51 32 27 14 84 why vain 92 best feet ink not
shit
Step VI: 84 68 51 32 27 14 vain 92 best feet ink not shit
why
Step VII: 92 84 68 51 32 27 14 best feet ink not shit why
vain
66. (5) There is no such step
67. (4) Step V: 68 51 32 27 14 84 why vain 92 best feet
ink not shit
Hence ‘best’ would be at 5th position from right in
Step V.
68. (2) Step VII (last step) 92 84 68 51 32 27 14 best
feet ink not shit why vain
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Hence three element (best, 14, 27) are between ‘feet’
and ‘32’.
69. (3) Step IV: 51 32 27 14 84 why shit vain 68 92 best
feet ink not
Hence ‘why’ is sixth from the left in Step IV>
Solutions (70 – 75)
70. (2)
71. (3)
72. (4)
73. (1)
74. (5)
75. (2)
Solutions (76-78)
76. (5): From statement I,
Deep is sister of Shilpa who has two siblings.
From statement II,
Kaushal is the only son of Deepa’s parents. So by
analyzing both the statement we can conclude Shilpa
has one brother.
77. (3): From statement I,
Hence D is immediate right of B.
From Statement II,
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Hence D is immediate right of B.
78. (3): From statement I,
85. (1) Only I follow
Hence Sudha is facing South direction.
From Statement II,
Hence Sudha is facing South direction
Answers: (86-90)
86. (4)
87. (4)
88. (4)
89. (2)
90. (5)
Solutions (91 – 95)
Solutions (79 – 80)
79. (4): C
80. (1) ABE
Solutions (81-85)
81. (2) Only II & III follow
82. (5) None of these
83. (4) Only I & III follow
91. (4)
92. (5)
93. (1)
94. (4)
95. (3)
96. (5) Either + or ÷
97. (4) Only three pairs in the word ‘AC’, ‘AE’, ‘CE’–
SEARCHES
Solutions (98 – 100)
Codes are as follows:
he – pa
is – ro
there – la
the – ji
at – ma
station – fu
98. (2) pa
99. (3) fu ji zo
100.(2) ta ki ro
84. (2) Only III follows
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waiting – ta
train – zo
this – bi/vi
a – vi/bi
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