BSC Academy
1. 1 2. 1 3. 5 4. 4 6. 1 7. 4 8. 1 9. 5 Correct order-DEAFGBC 11. 4 12. 1 13. 3 14. 5 16. 1; Replace ‘in’ with ‘over’ 17. 3; Replace ‘has’ with ‘has been’ 18. 3; Replace ‘faking’ with ‘fake’ 19. 3; Replace ‘has been’ with ‘were’ 20. 2; Replace ‘had been’ with ‘has been’ 21. 1 22. 3 23. 2 24. 2 26. 1 27. 3 28. 1 29. 2 31. 4; ? (4003 × 66 – 21015) 3 = 81061
5. 3 10. 2 15. 2
41.4; Required average 5700 518 11 42. 3; Required %
=
25. 1 30. 5
ad
55 23840 = 1334 + 13112 100 34. 4; ? 8892 342 × 7 = 182 35. 1; ? (5450 50) + 645 25 + 3991 26 = 109 + 25.8 + 153.5 = 288.3 288 36. 1; I. x2 + 4x – 3x – 12 = 0 (x + 4)(x – 3) = 0 x = –4, + 3 II. y2 + 7y + 6y + 42 = 0 (y + 6)(y + 7) = 0 y = –6, –7 Hence, [x > y] 37. 3; I. 4x2 + 14x + 18x + 63 = (2x + 9)(2x + 7) = 0
210 100 4% 5700 43. 1; Highest increase is from S to T = 780 – 210 = 570 Lowest increase is from R to S = 210 – 190 = 20 570 100 2850% Required % = 20 44. 2; Required sum = 10(1 + 2 + ........+ 11) 12 = 10 × 11 × = ` 660 2 45. 1; Required % more 950 – 720 100 32% = 720 46. 1; Books sold by A = 10700 B = 9900 C = 10200 D = 10500 E = 9900 F = 8900 Hence, maximum book are sold by shopkeeper A. 47. 3; Required ratio = 9900 : 10200 = 33 : 34 48. 1; Required average 9900 8900 9900 9567 = 3 49. 2; Difference = 10500 – 9900 = 600 50. 1; Required answer
=
32. 1; ? 33489 2601 – (83)2 – 372 = 9333 – 6889 – 1369 ? = 1075 = 14446
C
Ac
33. 5; ? 23 × 58 +
em y
SBIPO-PT-A-001
–9 –7 –4.5, –3.5 2 2 II. y2 + 2y + 3y + 6 = 0 (y + 2)(y + 3) = 0 y = –2, –3 Hence, [y > x] 38. 4;
BS
x=
39.1; The series is (43)2 = 1849, (47)2 = 2209, (51)2 = 2601, (55)2 = 3025, (59)2 = 3481, (63)2 = 3969 40. 3;
SBIPO-PT-A-001
16 15800 = 2528 100 51. 4; Let the distance be ‘x’ km,
=
x x – 1 6 7.5 x = 30 km.
then,
1.5x 1 6 7.5
1
BSC Academy 52. 2; Let the age of the teacher be x 714 x 34 + 1 = 22 x = 56 years
R=
r = 8% Required answer
3 33600 = ` 8400 12
= 7500 +
5 33600 = ` 14000 12
54. 5; Required probability =
52
C2 C2
20 221
22 9 (2 10 – 0.4) 0.4 = 221.76 m3 7
56. 3; P, Q and R together complete the work
=
45 30 2
45 45 15 15 30 30 2 2
work done by all in 9 days =
SP =
180 13
61. 3; Required answer
13 13 9 180 20
6 2 4 45 15
13 4 1 which is to be done by R. 20 15 12 R alone do the whole work in
C
= 1–
180 15 180 15 = 13 = = 180 days. 180 15 15 13
BS
1 180 of the work is done by R in = 15 days. 12 2 t 57.1; 241 – 6 30 24t 24 – =6 30 24t 18 = 30
t=
2
420 34 – 244 50 50 – 34 14280 – 12200 = = 130 m 16 62. 5; A + B = 14 + B + C A = C + 14 14 years 63. 5; Monthly income of Rahul 9870 100 100 = = ` 18800 75 70 64. 2; Required answer
=
work done by Q and R in 6 days =
Remaining work
6400 145 = ` 9280 100
Ac
2 15
55 x 2 – 4 = 10 60 2x 240x = 11(x2 – 4) (x – 22)(11x + 2) = 0 x = 22 km/hr 60. 2; Let CP be ` x 8320 – x = x – 4480 2x = 12800 x = 6400
ad
55. 3; Required volume =
7500 4 12 = ` 11100 100
59. 1; Let the speed of the stream in still water be x km/hr
14000 – 8400 2 100 = 66 % Required % = 8400 3 16
3050 r 4 = 4026 100
em y
53. 1; P =
58. 4; 3050 +
18 30 45 1 22 minutes 24 2 2
2 1 200 5 5 = 48 + 40 = 88 kg
= 120 ×
65. 4; Ratio of P : Q : R = 6 × 12 : 7 × 12 : 7 × 6 = 12 : 14 : 7 7 66000 = ` 14000 Share of R = 33
(66-70) 66. 4; Given: E > B A< M N >A> P Combining both, we get, E > B A > P and E > B A < N and
...(i) ...(ii)
SBIPO-PT-A-001
The part of books which is paper can never be colour. So, conclusion I follows. Again, there is no relation between book and colour. So, conclusion II doesn’t follow. 72. 1;
BS
C
Ac
ad
M >A< N We can’t compare N and B. So, conclusion I is not true. Again, we can’t compare M and N. So, conclusion II is also not true. 67. 4; Given : N=EP>Q ...(i) PS ...(ii) Combining (i) and (ii) we get, N = E P S and Q
Q=TR ...(i) QS ...(ii) Combining (i) and (ii) we get, V < P > Q = T S and SQ=TR We can’t compare T and S. So, conclusion I is not true. Again, we can’t compare S andV. So, conclusion II also is not true. 69. 5; Given : L=IJJM Conclusion I is true. Again, conclusion II is also true. 70. 4; Given : A>BC B C > D and A > B C E and D < C < L and EC
em y
BSC Academy
SBIPO-PT-A-001
From diagram II conclusion I follows. Again, conclusion II doesn’t follow because the part of book which is paper can never be colour.
73. 3;
Conclusion I doesn’t follow. Again, conclusion II doesn’t follow. [complementary pair] 74. 1;
Conclusion I is true. Again,Conclusion II does not follow.
3
BSC Academy 88. 3;
em y
75. 5;
(89-90)
ad
Conclusion I follows. Again, from diagram II conclusion II also follows. (76-80) people – cil like – mig IPL – pan Hyderabad – zil will – na/pa win – na/pa support – li players – kit match – nit 76. 4; 77. 2; 78. 2; 79. 1; 80. 4;
Days
P.M.
Country
Mon
A
Pakistan
Tue
B
Australia
Wed
C
Sri Lanka
Thu
D
Nepal
Fri
E
India
Sat
F
England
Sun
G
Bangladesh
Ac
(81-85)
89. 5;90. 1; total distance = 25 + 40 + 60 + 90 = 215 m (91-95)
(86-88) 86. 2;
83. 1;
84. 1;
BS
87. 4;
82. 2;
C
81. 3;
4
85. 2;
91. 5; 92. 5; (96-100)
93. 3;
94. 1;
95. 4;
96. 1;
98. 2;
99. 2;
100. 3;
97. 2;
SBIPO-PT-A-001
SBIPO-PT-002
1. 4; Replace ‘policy’ with ‘policies’
37. 3; I. (6x – 1) (3x – 1) = 0 x = 1/6, 1/3 II. (8y – 3) (6y – 5) = 0
2. 4; Replace ‘fleet’ with ‘fleeting’ 3. 5; No error
3 5 , 8 6 x
y=
4. 3; Add ‘the’ before ‘very’ 5. 3; Replace ‘its’ with ‘their’ 6. 1
7. 5
8. 2
9. 4
10. 3
11. 1
12. 1
13. 5
14. 2
15. 4
16. 1
17. 5
18. 4
19. 1
20. 4
38. 5; I. (x + 4) (x – 9) = 0 x = –4, 9 II. (y – 8) (y – 9) = 0 y = 8, 9 No relationship between ‘x’ and ‘y’
(21–25): E A C B D 21. 1
22. 5
23. 2
24. 3
25. 3
26. 4
27. 1
28. 4
29. 2
30. 5
39. 4: I. x2 = 49/169 x =
31. 1; 121 × 8 – 268 = ? × 9 + 241 y=
?=
32. 4; ? =
700 – 241 = 51 9
81 4981 – 49 26
33. 1; ? = (4389 ÷ 627) ×
7 13
9–2 7 = 13 13
xy 40. 3; I. (x – 8) (x – 13) = 0 x = 8, 13 II. (y – 14) (y – 15) = 0 y = 14, 15 x
= 160
4 + 161 = 165 7
50 680 100 × 41. 5; ? 480 – 100 560
34. 1; The series is : 15 × 4 – 1 = 59 69 59 × 4 – 1 = 235
=
235 × 4 – 1 = 939
480 – 340 56
× 10 = 25
42. 1; 1000 ÷ 25 + 275 = 315 43. 2; ? 99 × 10 + 12 = 1002 44. 4; Let the number be 10x + y.
939 × 4 – 1 = 3755 35. 4; The series is : 22 + 13 = 23
x 4 From A, x + y = 10 and y = 1
23 + 23 = 31 3
31 + 3 = 58 58 + 43 = 122 121
5y 10 y 2, x 8
36. 4; The series is :
x From B, xy = 16 and y = 4
120 × 2 – 20 = 220 220 × 2 – 40 = 400 380
y=2; x= 8
400 × 2 – 60 = 740
Either from (A) or (B) ; the number is 82.
740 × 2 – 80 = 1400 1
45. 1; P + Q + R = 4800
5
C3
1 From A, P = (Q + R) 3P = Q + R 3
51.2; Required probability =
4P = 4800 P = ` 1200
52.4; Required probability = 1 –
15
C3 9
1 From B, Q = (P + R) 2Q = P + R 2
4
53. 1; Required probability =
Thus, we can’t find the share of P from statement B.
C3
15
C3
10 6 2 15 14 13 91 1–
C3 5 C3 15
C3
12 53 65 65 4 10 35 13
46. 2; Let the total number of votes polled be x. From statement B,
=
82% of x – (100 – 82) % of x = 1488
=
2 65
54. 3; Let the average expenditure be ` x
or, (82% – 18%) of x = 1488 or,
14 35 13
13 60 (x 130) =x 14
64 x = 1488 100
780 + 130 = 14x – x x = ` 70 100 or, x = 1488 × = 2325 64
Total expenditure = 13 × 60 + 70 + 130 = ` 980 55. 1; Let the money added be ` x
47. 4; Rahul’s present age = 76 – 8 = 68 year Wasim’s present age = 68 – 16 = 52 year
24800 8 4 (24800 x) 10 4 – = 5184 100 100
7 × 52 = 28 year 13 Manav’s father present age = 28 + 27 = 55 year 48. 1; Let the length of the train be l m and bridge be y m. Manav’s present age =
24800 × 40 + 40 x – 24800 × 32 = 518400 40 x + 24800 × 8 = 518400 40 x + 198400 = 518400 x = ` 8000
l y = 30 l + y = 750 m .......... (i) 5 90 18
56. 2; Let the CP of 1 article be ` 1 Then, CP of 15 article = ` 15 = SP of 12 article And CP of 12 article = ` 12
1 – 60 y 1 y 60 18 – 15 54 5 and t = 54 5 18
Profit % =
57. 1; Let the salary of Divya be ` x
750 = – 4 = 46 seconds 15 49. 5; Total number of person = 8 Host can sit on a particular seat in one way Hence, remaining can sit in 7 places in 7P7 = 7! ways The number of required arrangement = 7! × 1 = 7! 50. 2; Vowels (E U A I) = 4 Required number of ways = 4C3 × 4C2
=
3 × 100 = 25 % 12
90 % of 9% of x = 12960
x = 12960 ×
100 100 × = ` 1,60,000 90 9
58. 4; Let the amount of water added be x litres. 5 The original mixture contains 620 litres of milk 8
4! 4! × = 4 × 6 = 24 ways. 3! 2! 2!
2
3 and 620 litres of water.. 8
64. 5; Required ratio = = 12 × 680 : 19 × 520 = 3 × 68 : 19 × 13 = 204 : 247
77.5 5 5 = 77.5 3 x 4 20 × 77.5 = 15 × 77.5 + 5x 5 × 77.5 = 5x x = 77.5 litre
Now,
59. 4; (A + B) did
65. 1; Ratio of transport is 13 : 15, in (Telangana and Maharashtra)
9 9 of the work and C did 1 – 11 11
66. 5; From (i) mystery of better life 53 97 79 21 From (ii) life is a mystery 24 27 79 53 From (iii) she is playing better 24 21 12 22 From (i) and (ii) life mystery – 79 53
2 work 11
=
(A + B)’s share : C’s share =
So, C’s share =
60. 4;
From (ii) and (iii) is – 24
9 2 : =9:2 11 11
But, we cannot find the code for ‘life’ from combination of all three.
2 × 6600 = ` 1200 11
67. 5; 111 – 21 ÷ 7 × 2 + 4 – 127 = ? By changing the symbols, we get, 111 × 21 + 7 ÷ 2 – 4 × 127
5x 16 9 6x 16 10
= 111 × 21 +
50x + 160 = 54x + 144 4x = 16 x = 4 The numbers are 20 and 24
= 2331 +
7 4669 – 1016 3653 – 508 = 2 2 2
B ODYG UARD
19 620 = ` 117.8 crore = 100
FDI in other sector in Telangana =
7 – 4 × 127 2
68. 4;
61. 1;FDI in entertainment sector in Kerala
69. 5; From I. Berth is father of Adil Berth > Berth’s wife ...... (i)
11 650 = ` 71.5 crore 100
Frank > Donna > Erik .....(ii) Adil’s mother > Frank ....(iii)
Required percentage more =
12 19 680 : 520 100 100
Combining (i), (ii) and (iii); we get
117.8 – 71.5 100 65% more. 71.5
Berth > Chandani > Frank > Donna > Erik Clearly Chandni is Adil’s mother and Berth’s wife.
14 750 100 70% 62. 3; Required percentage = 100 26 580 100
But Adil’s score is not mentioned. Hence I is not sufficient alone.
63. 1; Required answer
Berth > Chandani > Frank > Donna > Erik > Adil
=
From II. Nephew of Erik has lowest score. From I and II together Bertha is the winner.
18 680 650 750 620 580 520 100
Hence, both I and II are required to answer.
= ` 684 crore 70. 1; Given expression A< B > C > D = E > T
3
Now, option 1) B E follows
possibility I exists. Thus, conclusion I follows, and conclusion II does not follow.
Option 2) we cannot compare C and A.
81. 4; Some sweets are drink (I) + Some drinks are not solid (O) = I + O = No conclusion. Hence I does not follow.
Option 3) C > T is true. Hence (C T) does not follow. Option 4) B > T is true. Hence (B T) does not follow.
Again, All liquids are sweet (A) + Some sweets are drink (I) = A + I = I = No conclusion. Hence, neither conclusion I nor II follows.
(71–75) :
71. 4
(82–86) :
Participant
Country
Dance Style
Hari
Bulgaria
Bush
Arjun
Belize
Horo
Nilesh
Brazil
Chhau
Gaurav
Algeria
Rumba
Pawan
Cuba
Brukdown
Step I: 35 quant hear cute 50 65 98 giant 71 82 oliver melody silent 19
Amit
Fiji
Samba
Step II: hear cute 50 65 98 giant 71 82 oliver melody silent 19 quant 35
Deepak
Australia
Gavotte
72. 3
73. 3
74. 2
Words are arranged in reverse alphabetical order and the numbers are arranged in ascending order. One word and one number is arranged in each step. The arrangement is done from right to left. Input: 35 quant hear cute 50 65 98 silent giant 71 82 19 oliver melody
Step III: hear cute 65 98 giant 71 82 melody silent 19 quant 35 oliver 50 Step IV: hear cute 98 giant 71 82 silent 19 quant 35 oliver 50 melody 65
75. 4
76. 1; There is no negative statement. Hence, I follows and II does not follow.
Step V: cute 98 giant 82 silent 19 quant 35 oliver 50 melody 65 hear 71
77. 5; Some questions are answer (I) conversion some answers are questions (I) + All questions are puzzle (A) = I + A = I = Some answers are puzzles (I) Conversion Some puzzles are answers. Hence, conclusion I follows.
Step VI: cute 98 silent 19 quant 35 oliver 50 melody 65 hear 71 gaint 82 Step VII: silent 19 quant 35 oliver 50 melody 65 hear 71 gaint 82 cute 98 82. 3
Again, All questions are puzzles (A) + No puzzle is a problem (E) = A + E = E = No question is a problem. Hence, conclusion II follows.
83. 4
84. 3
85. 5
86. 4
(87–91):
78. 5; Some answers are puzzle (I) + No puzzle is problem (E) = O= Some answers are not problems. Hence, conclusion II follows. Also, the possibility in I exists . All problems being answer is a possibility. Therefore I also follows.
Name
Classroom
Test Centre
State
Position
Abhishek
Cuttack
Bangalore
Bihar
4
Deepak
Patna
Chennai
UP
1
79. 2; All hours are minute (A) + Some minutes are time (I) = A + I = No conclusion.
Dipika
Delhi
Dehradun
Uttrakhand
3
Vipin
Lucknow
Kolkata
Odisha
2
Hence I does not follow. Again, No second is an hour (E) + All hours are minute (A) = E + A = O* = Some minutes are not second. But the possibility in II exists. Thus, conclusion II follows.
87. 2 (92–93)
80. 1; No second is an hour (E) + All hours is minute (A) = E + A = O* = Some minutes are not second. But the
92. 3;
4
88. 5
89. 1
90. 2
91. 4
(–)
W (+)
(+) Male (–) Female
(–)
T
(+)
V
S C
Hence, C is grandchild of W, or W is grandmother of C 93. 5; Family Tree Z (+)
(–)
B
Y
C
U A
Since, the gender of ‘C’ is not specific. Hence, it is not possible to determine the relation between C and U. 94. 4 95. 5 (96–100):
Candidate Ruhi
96. 4
(i)
(ii)/(a)
(iii)/(b)
(iv)
×
Rohit
Joseph
Karthik
Osheen
97. 3
98. 3
99. 3
100. 1
5
SBIPO-PT-003 1. 1
2. 4
3. 2
4. 5
5. 1
6. 5
7. 3
8. 3
9. 2
10. 2
35. 1;
(11–15): D C A B F E 11. 1
12. 3
13. 2
14. 5
15. 4
16. 1
17. 2
18. 3
19. 5
20. 1
? = 656 + 114 = 770
36.4;
21. 1; Replace ‘detail’ with ‘detailed’ 22. 2; Add ‘the’ before ‘energy’ 23. 1; Replace ‘national’ with ‘nationally’
? = 26 + 3 = 29
24. 5; No error
37.3;
25. 1; Replace ‘provide’ with ‘provides’ 26. 1
27. 2
28. 3
29. 1
30. 5
31.2; I. x2 + 5x – 4x – 20 = 0 ? = 99 × 4 + 12= 408
(x – 4) (x + 5) = 0 x = 4, – 5 II. y2 + 8y + 5y + 40 = 0
38. 2; ? = 46.81
(y + 8) (y + 5) = 0 y = –8, – 5 x>y 32.5; I. x2 – 6x – 5x + 30 = 0
40. 5; ? =
8. 5 0.42 × 160 – × 750 100 100
= 13.6 – 3.15 = 10.45
x = 5, 6 II. y2 – 8y – 5y + 40 = 0
4200 60 30 70 30 = = 100 21 21 42. 3; ? 960 × 780 ÷ 24 = 31200 43. 1; Let the cost price of mobile be ` x and laptop be ` (78000 – x)
41. 4; ?
(y – 8) (y – 5) = 0 y = 8, 5 No relationship between ‘x’ and ‘y’ exits. 33. 3; I. (x + 5)2 = 0 x = –5
5y – 3
2
=0 =
3
x×
5
25 15 5 – (78000 – x) × = 78000 × 100 100 100
25x + 15x – 78000 × 15 = 78000 × 5 40x = 78000 × 20 x = ` 39000 44.3; Suppose the man works overtime for x hours.
x
(5x + 3) (2x – 7) = 0 x=
= 159 27 + 53 × 27 – 159 27 = 1431
(x – 5) (x – 6) = 0
II.
39. 1;? = 3 27 53 27 – 159 27
Now, working hours in 5 week = 5 × 6 × 8 = 240 hr.
–3 7 , 5 2
240 × 40 + x × 45 = 10, 500
2
II. y + 17y – 4y – 68 = 0 (y – 4) (y + 17) = 0 y = 4, –17 No relationship between x and y exists.
9600 + 45x = 10,500 x=
10,500 – 9600 = 20 hr.. 45
total hours of work = 240 + 20 = 260 hr.. 1
45. 4; Let the number of boys and girls be x each 3(x – 10) = x
1
50. 2; P’s share Q’s share =
2x = 30
5 1 100
x = 15 Total number of students = 2x = 30 46. 2; Number of questions attempted correctly = 70% of 10 + 50% of 30 + 60% of 45 = 7 + 15 + 27 = 49
Q’s share =
x 45 x × 100 = 6.25
5 1 100
8
400 × 12615 = ` 6000 841
t 51. 4; Pipes should be closed after = y 1 – minutes x
100x = 281.25 + 6.25x
[ x = A, y = B, t = Full]
281.25 = 3 kg. 93.75
16 8 = 27 1 – = 27 × = 9 min 24 24
45 3 = 12 tins 4
52. 5; Average speed for the total journey (8 4)(8 – 4) = 6 km/hr.. 8 53. 5; Required answer = 2 × 3! × 3! = 2 × 3 × 2 × 3 × 2 = 72
=
5 48. 1; Sugar contained = × 8 = 0.4 100
Now, % sugar in remaining solution =
1
441 P’s share = 841 × 12615 = ` 6615
60 × 85 = 51 100 Required answer = 51 – 49 = 2 47. 5; Total paint required = 45 + wastage = 45 + x
Total number of tins =
:
100 2 = (21)2 : (20)2 = 441 : 400 =1: 105
Passing grade =
x=
6
8
54. 1; Required probability =
0.4 × 100 = 5.71% 7
C3
52
C3
[honour card of red = 2 × 4 (Ace, King ,Queen, Jack)
49. 3; For distance,
=
8! 52! ÷ 3! 5! 3! 49!
=
876 14 = 52 51 50 5525
55. 4; Required probability = Distance travelled in 8 days by 1st man, D1 = 12 × 8 = 96 km. and by 2nd man, D2 = 20 × 8 = 160 km For remaining distance, let both take t days to reach the final destination. x – 96 = 2 × 12 × t x – 96 = 24 t .......... (i) and x – 160 = 20 t ...... (ii) From (i) and (ii); we get, x – 96 – x + 160 = 24 t – 20 t 64 = 4 t t = 16 days. Total number of days = 16 + 8 = 24 days.
=
12
C3
52
C3
=
12! 52! ÷ 3! 9! 3! 49!
12 11 10 11 = 52 51 50 1105
56. 4; There are 10 letters in the word COMPONENTS and there are two O’s and two N’s Number of consonants = 7, number of vowels = 3 Relative order of vowels and consonants remain unchanged, means vowel will occupy vowels place and consonant will occupy consonant place. Required number of ways =
= 7 × 6 × 5 × 4 × 3 × 3 = 7560 2
7! 3! × 2! 2!
4
57. 1; Required probability =
C2 5 C2 6 C 2 15
65. 1; From statement A, Required answer =
C2
66. 3; min fin bin gin trains are always late .......... (i) gin din cin him drivers were always punished .. (ii) bin cin vin rin drivers stopped all trains ......... (iii) din kin fin vin all passengers were late ......... (iv) From Eq. (i) and (iv), fin late Eq. (ii) and (iii), cin drivers Eq. (ii) and (iv), din were Hence, drivers were late cin din fin 67. 3; V = W < X < Y > Z Conclusions (I) Z > X (x) (II) Y > V (x) (III) W < Y ( ) (IV) W = Y (x) Only III is true. 68. 2; K > L < O = M < N Conclusions (I) N > O ( ) (II) M > N (x) (III) K < N (x) (IV) L = N (x) Only I is true. 69. 5
5! 6! 15! 4! ÷ = 2! 2! 2! 3! 2! 4! 2! 13!
=
6 10 15 31 = 15 7 105
58. 4; Number of students in college C5 In 2014 =
15 × 30,000 = 4500 100
In 2013 =
12 × 25000 = 3000 100
4500 – 3000 × 100 = 50% 3000
Required % increase = 59. 2;
2013
2014
% increase
C1 17 × 250 = 4250
16 × 300 = 4800
12.94%
C2 20 × 250 = 5000
22 × 300 = 6600
32%
C3 21 × 250 = 5250
20 × 300 = 6000
14.28%
C4 10 × 250 = 2500
8 × 300 = 2400
decrease
C5 12 × 250 = 3000
15 × 300 = 4500
50%
C6 15 × 250 = 3750
13 × 300 = 3900
4%
C7 5 × 250 = 1250
6 × 300 = 1800
44%
Hence, minimum increase is in the college C6
Brajesh is at distance of 10m away and in West direction from his starting point S. 70. 2 In all groups, first and third letters are equidistant from middle letter as they appear in English alphabets.
60. 3; Hence, decrease is in the college C4.. 61. 4; Required % =
17 20 250 100 16 22 300
81%
62. 1; Required ratio = 15 × 250 : 13 × 300 = 25 : 26 71. 2; 72. 5;
63. 5; Combining the both statements A and B will not give the perimeter of semicircle. 64. 3; From A,
From B,
40 × 80 = 32. 100
30 40 N1 = × 90 N1 = 120 100 100 20 36 N2 = × 50 N2 = 90 100 100
73. 4
combining both statements A and B; N1 120 4 = = =4:3 N2 90 3
3
they are dependent. Thus, eventually we get one equation in two variables, which have infinite solution set. 88. 3; From statement I, If Karan turns 90° towards his right, he will face South. So, Rahul is facing North. From statement II,
74. 1
In the end, Priya faces North. So, Rahul also faces North. 89. 2; From statement I,
75. 5;
(76-80): Here words are arranged in alphabetical order, while numbers in descending order from the right side. Input: app is ox 32 87 ever shoe 72 69 96 66 Step I: app ever is ox 32 87 shoe 72 69 66 96 Step II: app ever is ox shoe 32 72 69 66 96 87 Step III: app ever is ox shoe 32 69 66 96 87 72 Step IV: app ever is ox shoe 32 66 96 87 72 69 Step V: app ever is ox shoe 32 96 87 72 69 66 Step VI: app ever is ox shoe 96 87 72 69 66 32
From statement II,
Hence, Y is grandmother of N.
76. 5; app ever is ox shoe 32 96 87 72 69 66 77. 3 (81-85):
81. 1
78. 5
82. 2
79. 1
83. 4
90. 1; Check option by pick method. By options 1) W = X < Y < A = B > C > D (91–95):
80. 3
84. 3
85. 5;
86. 5;From I, ‘a’ is a prime number. But a unique value of a is not known, a can be 2, 3, 5, 7......... Hence, statement I alone is not sufficient. From II, again a unique value of a is not known, a can be 2, 4,6...Hence statement II alone is not sufficient. Combining I and II, that ‘a’ is even and prime. Hence, I and II together are sufficient to answer the question.
Person
Movie
Subject
P
Swat
Geology
Q
Troy
Computer
R
Rush
Topology
S
Heat
Statistics
T
Avengers
Spanish
V
Constantine Psychology Congo
X
91. 4
87. 4; From I and II, we get 2 equation in two variables. But 4
92. 5
93. 4
94. 3
Polity
95. 1
96. 2
97.1
98. 2; Second highest number = 746;
746 = 373 2
Third highest number = 519. Difference = 519 – 373 = 146 99. 4; 519 915; 328 823; 746 647; 495 594; 837 738; lowest number = 594 100. 5; 519 951; 328 832; 746 674; 495 549; 837 783 Highest number = 951 or 519
5
SBIPO-PT-004 II. 7y2 + 21y + 3y + 9 = 0
1. 1 2. 5 3. 4 4. 2 5. 5 6. 2; Replace ‘organization’ with ‘organizations’ 7. 5; No error 8. 4; Replace ‘in’ with ‘under’ 9. 3; Replace ‘from’ with ‘for’ 10. 3; Replace ‘can’ with ‘could’ 11. 2; same 12. 4; opposite 13. 3; opposite 14. 4; same 15. 2; opposite (16–20): E D A C B 16. 1 17. 2 18. 3 19. 5 20. 4 21. 5 22. 1 23. 2 24. 3 25. 4 26. 3 27. 1 28. 2 29. 3 30. 4 5 2 5 1 2 5
3 y = – ,–3 7
16
239 – (12)2 = 95 96 95 + (13)2 = 264 264 – (14)2 = 68 68 + (15)2 = 293 Hence there should be 95 in place of 96.
5 (2x + 5) (x + 4) = 0 x = – ,– 4 2 II. 2y2 – 10y + 7y – 35 = 0
41. 5; The series is: 128
5 320 2
7 1120 2 9 1120 5040 2 11 5040 27720 2 320
7 (2y + 7) (y – 5) = 0 y = – ,5 2 Relationship between x and y does not exist. 35. 5; I. 12x2 – 20x – 21x + 35 = 0 7 , 4 3 (4y – 5) (y – 3) = 0
x=
13 180180 180190 2 Hence, there should be 180180 in place of 180190. 42. 1; He sells 920 grams of rice and gains 80 grams 27720
5 ,3 4 Relationship between x and y does not exist 36.5; I. 4(x2 – 1) = 60 x2 – 1 = 15 x = 4 II. 3(y2 + 1) = 51 y2 + 1 = 17 y = 4 Hence x = y 37. 2; I. 28x2 – 21x + 12x – 9 = 0 (7x + 3) (4x – 3) = 0
y=
80 100 = 8.69% 920 43. 2; Let the number of days he was absent be x days. 180 (40 – x) – 20 x = 5200 7200 – 180x – 20x = 5200 7200 – 200x = 5200 Gain % =
3 3 x= – , 7 4
SBI PO (PRE) Mock-Test-9
5 625 630 . 2
Hence there should 625 in place 7630 39. 4: The series is: 21 × 2 – 1 = 41 48 41 × 4 – 2 = 162 162 × 6 – 3 = 969 969 × 8 – 4 = 7748 Hence, there should 41 in place of 48. 40. 3; The series is: 118 + (11)2 = 239
? 824 1236 ? 150 32.2; 100 33. 1; 392 × 8.4 = 3292.8 34. 5; I. 2x2 + 8x + 5x + 20 = 0
II. 4y2 – 12y – 5y + 15 = 0
5 5 5 40 , 40 100 , 100 250 , 2 2 2
250
25
(4x –7) (3x – 5) = 0
x y
38. 1; The series is:
5
31.4; ?
(7y + 3) (y + 3) = 0
x= 1
2000 = 10 days 20
44. 2; Required number = LCM of (25, 30, 35, 40) – 7 = 4200 – 7 = 4193 45. 3; Time stopage rest per hour =
Difference of speed 60 minute Speed without stoppage
6x 7 11 18x + 21 = 11x + 77 x7 3 7x = 56 x = 8 Manorama’s present age = 6x = 48 years 4 40 1.6 kg 100
N1 1 100 4 4:7 N 2 5 35 7
53. 1; Required probability =
n – m! m! (n – 1)!
(Where n = number of person, m = particular person)
1.6 x 12 40 x 100
9 – 2! 2! 7 ! 2! 2 1 = = (9 – 1)! = 8! 8 4
160 100x 480 12x 88x = 320 x = 3.64 kg
48. 3;
Anshu, Bharti and Charita’s age are 9, 27, 21 years 52. 3; Required number of ways = 7! 6! = 7 × 6! × 6! = 7 × (6!)2
Let x kg of pure salt be added. Then,
N2 35 N1 5 100
51. 2; Let the age of Anshu be x years Bharti’s age = 3x Charita’s age = x + 12 x + 3x + x + 12 = 57 or, 5x + 12 = 57 x= 9
54 – 36 60 30 min per hour Now, 36 46. 5; Let Manorama son’s age be x years Manorama’s age = 6x
47. 4; Amount of salt in 40 kg of solution =
2xy 54. 5; Required probability = (x y) (x y 1)
80x 94y 86 xy
=
80x + 94y = 86x + 86y 8y = 6x
2 8 3 24 = 11 10 55
(x = yellow, y = green)
(y 1)!x! x boys 55. 2; Required probability = (x y)! y girls
x 8 4 y 6 3 4:3
49. 1; Radius of formed cone = 4 cm Slant height = 5 cm
=
7! 7! 7 6 5 4 3 2 7 = 13! 13 12 11 10 9 8 1716
n(n – 1)(n – 2)(n – 3) 24 (Where n = number of sides in polygon)
56. 3; Required answer = 5 cm 3 cm
=
12(12 – 1)(12 – 2)(12 – 3) 24
=
11 10 9 99 5 495 2
4 cm
22 4 5 = 62.8 cm2 Surface area of formed cone = πrl = 7
50. 1; N2 – 35% of N1 =
N2 –
57. 2; From statement B, Principal = x ; interest = x
4 N 5 2
Then,
4 35 N2 N1 5 100
x 4 r 100 x r= = 25% 100 4
58. 4; From statement A, area of the square =
2
1 (diagonal)2 2
SBI PO (PRE) Mock-Test-9
=
1 (16) 2 2
2009
2010
2011
2012
57.8%
57.05%
59.52%
60%
128 cm 2
2013
2014
62.07% 67.05%
2
64 2 From statement B, area of the square = 256 cm 4
Hence, highest is for the year 2014. (66–70):
59. 3; From statement A and B;
D
Cost of 1 apple = `45
Cost of 1 orange = 2 × 15 = ` 30 Required answer = 3 × 45 + 2 × 30 = 135 + 60 =` 195
H
So, we cannot conclude anything. 61. 4; Required ratio
8740 37 = 437 : 466 9320 466
vegetables are fruits. Hence, conclusion II does not follow. 72. 4; Some branches are plants (I) + No plant is the tree (E) I + E = O = Some branches are not trees. Hence, conclusion I does not follow. There is negative statement thus the possibility can not be exist. Hence, conclusion II follows. 73. 5; No plant is a tree. conversion No tree is a plant.
62. 5; Percentage of students getting selected from institute ‘O’ in 700 100 50.72% 1380
780 100 54.54% 1430 Required percentage increase
2010 =
=
Again, No plant is a tree (E) + All leaves are trees conversion .
54.54 – 50.72 100 7.5% 50.72
Some trees are leaves (I) E + I = O*. Some leaves are not plants. Hence, conclusion II follows. 74. 3; No bread is rice (E) + No rice is pulses (E) = E + E = No conclusion. But there is a complimentary statement. Hence, either conclusion I or II follows. 75. 5; No bread is rice (E)) conversion . No rice is bread.
63. 2; Percentage of students getting selected in the year 2013 J
K
52.14% 62.07%
L 58.67%
M
N
O
58.86% 58.78% 58.82%
Hence, conclusion I follows. Again, No rice is pulses (E) + Some pulses are fruits (I)= E + I = O* = Some fruits are not rice. Hence, conclusion II follows.
Hence, the highest percentage is for Institute K. 64. 3; Required Percentage =
800 1000 770 940 820 820 100 (1560 1680 1380 1560 1420 1460)
76. 2 77. 2; Fourth to the left of tenth from the right end (10 + 4) = 14 th from the right end i.e. –4 78. 3 79. 4
5150 100 57% 9060 65. 1; Percentage of students getting placement from Institute K in the given years:
=
SBI PO (PRE) Mock-Test-9
B F
66. 3 67. 2 68. 1 69. 4 70. 5 71. 1; All leaves are fruits (A) + All fruits are vegetables (A) = A + A = A = All leaves are vegetables. Hence, conclusion I follows. Again, All fruits are vegetables (A) conversion Some
1360 1460 1420 1500 1480 1520 = 1540 1610 1560 1520 1580 1510
2009 =
C
A
60. 5; Hence, the number of students is not given.
=
E
G
Now, losf of 4 apple = 4 × 45 = 2 × 3 × orange
3
80. 3;
N
AC PL E T Q S B ZYH C MU I
C
Tenth from the left end
81. 1; Given statements: M > T ..... ...(i) T R ..........(ii) K > M .........(iii) Combining all, K > M > T R M > R is true K R is not true 82. 2; Given statements: W < F .......(i) N > H .............(ii) F N .............(iii) Combining all,W < F N > H H < W is not true. F > H is true. 83. 1; Given statements H = D ......... (i) H T ............(ii) P N > D .....(iii) Combining all, P N > D = H T T D is true. P H is not true. 84. 5; Given statement K R .................(i) M < W ...............(ii) R > M = D ........ (iii) Combining all, KR > D = M < W K > D is true M < K is true 85. 4; Given statements J N ......... (i) N D.... .... ..(ii) M > D ........ (iii) Combining all, J ND < M J = D is not true. M > N is not true. 86. 3; +2
+2
P K I Similarly, F +2
A I +2
+2
+2
J
B
A
E
W B
A
S
(89–90): E > B > C > D > A, E > B > D > C > A 64
64
89. 3 90. 2 (91–95): H J I B G row-I facing south row II facing north F ACE D 91. 4 92. 2 93. 1 94. 2 95. 3 96. 5; G + D 7 + 4 = 11, H + D 8 + 4 = 12, F + G 6 + 7 = 13, H + F 8 + 6 = 14, J + F 10 + 6 = 16 Hence, JF does not belong to the group. Become the sum of the letter is perfect square. 97. 3; P R O V I D E D 98. 1; From I, WYZX Only I is sufficient to answer the question. 99. 5; From I and II write with pen ––– na pa sa this pen is black ––– ra sa ka ta Hence, pen –– sa Both are sufficient to answer the question.
+2
100. 4; From I, From II,
V
Q
T H +2
D
or,,
N I G H T +2
C
D
Q has two children.
brother
(–)
M
+2
H
H C K V J
D
Q is either uncle or aunt of H Hence, both are not sufficient to answer the question.
87. 2; –1
–1 –1 –1 S V U T W –1 –1 –1 –1 Q U T S R –1 –1 –1 –1 P T S R Q 88. 5; More the two posibility make. But we take only two we can’t determine
4
SBI PO (PRE) Mock-Test-9
SBIPO-PT-005 1. 3; Replace ‘can’ with ‘could’
x=
2. 1; Replace ‘will’ with ‘had’
4 5 , 3 2
Relationship between x and y cannot be established.
3. 2; Replace ‘to’ with ‘of’ 4. 5; No error
37.3; I. 7x2 + 2 7 x 8 + 8 = 0
5. 3; Replace ‘like’ with ‘as if’ 6. 1
7. 3
8. 5
9. 4
10. 2
11. 2
12. 1
13. 3
14. 5
15. 1
16. 5
17. 4
18. 3
19. 5
20. 3
2
=
0
7x
7 x+ 8 =0 x=
8
– 8 7
II. y2 – 2 × 10y + (10)2 = 0
(21–25): E F C A D B 21. 1
22. 5
23. 3
24. 4
25. 2
26. 5
27. 5
28. 2
29. 1
30. 4
(y – 10)2 = 0 Hence, x y
y = 10
38. 3; I. x2 + 3x – 2x – 6 = 0
31.5;
(x –2) (x + 3) = 0 ? = 240 ÷ 5 = 48
x =2, –3
32.4; The series is 112–1, 122–1, 132–1, 142 –1, 152–1,.........
II. 2y2 – 12y – 7y + 42 = 0 (y – 6) (2y – 7) = 0
? =16 – 1 = 255 2
7 2 Hence, x < y 39. 1;? = 76 ×75.25 = 5719 40. 4; ? = 2332.08 41. 5; ? 60000 + 3000 ÷ 4000 × 6000
y = 6,
33. 2; ? =25 + 7 = 32
49 34. 1; ? = 107 + 42 = 149
= 60000 +
35.1; Multiplying equation (I) by 3 and equation (II) by 2 and adding then we get.
250 220 42. 3; ? × 740 + × 600 100 100
15 x + 12y + 14x –12y = 219 + 100 29x = 319 x = 11
= 1850 + 1320 = 3170 43. 2; Let the CP of an article be ` x.
Again putting the value of x in equation (ii), we get y=
50 – 7x 50 – 77 = = 4.5 –6 –6
First SP =
34 136 x= x 25 100
New CP =
76 19 x= x 100 25
New SP =
145 19 551 × x= x 100 25 500
x>y 36.5; I. 6x2 – 8x + 21x – 28 = 0
(2x + 7)(3x – 4) = 0 x=–
3 × 6000 = 64500 4
7 4 , 2 3
Now,
II. 6y2 – 15 y + 8y – 20 = 0
(2y – 5)(3y + 4) = 0
1
34x 551x – = 258 25 500
680x – 551x = 258 500
Now, r = 25 ×
129 x = 258 x = `1000 500
4 27 208 Amount = 1200 1 100 = 100 × 1200
44.1; Let the numbers be 4x, 5x and 6x, then, 16x2 + 25x2 + 36x2 = 2772 77 x2 =2772 x = 6
= ` 2496 49. 1; Let x litres of water be added.
Sum of the numbers = (4x + 5x + 6x) = 15x = 90 45. 4; Required time =
Milk =
l 2 l 2 t1t 2 t 2 l1 – t 1 l 2
Now, =
125 200 4.5 8
50. 4; Let the number of girls be x Then, x × 8x = 6272
46. 5; Reduction in the consumption
or, 8x2 = 6272
28 = × 100 = 21.875% 128
or, x = 28 x = 28 (Number of girls can’t be negative)
47. 3; Suppose C alone will complete the work in x days 2 1 Work completed by A in 2 days = = 10 5
51. 4;
1 4 = 5 5
Total work completed by B and C together = 8×
= 28 3 P + Q + R = 93
Q R – 2 6
= 20 2 Q + R = 52 P’s present age = 93 – 52 = 41 years. 52. 4; Total salary = 5 × 38000 + 7 × 12000 + 10,000
5 4
= ` 2,84,000 53. 5; The word MODERN has 6 letters, in which 2 vowels and 4 consonants. Now, 2 vowels can be placed at any of the two place, out of the three marks. Required number of ways = 4! × 3! = 72 54. 1; Required number of ways = 3! × 5! × 4! × 3! = 6 × 5 × 4 × 3 × 2 × 4 ×3 × 2 × 3 × 2 = 6 × 20 × 6 × 4 × 6 × 6 = 103680
= 10 days. 1 1 1 + = 15 x 10
1 1 3– 2 1 1 = – = = 10 15 30 30 x
x = 30 days. r4 = 2400 48. 2; 1200 1 100
25 + r =
P Q R – 3 3
and
Since, the remaining work will be completed by B and C together in 8 days.
4 60 = 3 24 x
84 = 4x x = 21 litres
325 4.5 8 = = 117 seconds. 1000 – 900
Then,
5 2 × 84 = 60 litres; water = × 84 = 24 litres 7 7
180 = 96 + 4x
8 125 – 4.5 200
Remaining work = 1 –
108 = 27% 100
55. 1; Required number of diagonals =
24 25 = 50 r = 25% 12
5 5 – 3 =5 2
56. 3; Required probability = nCr × pr × qn–r p = probability of happening, 2
66. 5; No artist is a writer (E) + All writers are honest (A) = E + A = O* = Some honest are not artists. Hence, I does not follow. Again, All writers are honest (A) + No honest is an engineer (E) = A + E = E = No writer is an engineer +No artist is a writer (E) = E + E = No conclusion but the possibility in II exists. Hence II follows. 67. 4;Some plants are trees (I) + All tress are woods (A) = I + A = I = Some plants are woods. Thus the possibility in I exists. Hence conclusion I follows. Again, All tress are woods (A) + All woods are shrubs (A) = A + A = A . All trees are shrubs implication Some trees are shrubs. Hence conclusion II follows. 68. 3; Some honests are poor (I) + No poor is rich (E) = I + E = O = Some honests are not rich. Hence, conclusion I does not follow. Again, from statement first, second and third conclusion II may follow. Hence conclusion II does not follow. 69. 4; No window is a wall (E) + All walls are buildings (A) = E + A = O* = Some buildings are not windows. Hence the possibility in I exists. Hence, conclusion I follows. Again, some doors are windows (I) + No window is a wall (E) = I + E = O = Some doors are not walls. Hence, the possibility in II exists. Hence, conclusion II follows. 70. 5; All laptops are computers (A) + Some computers are tablets (I) = A + I = No conclusion. Hence, conclusion I does not follow. Again, from statement first the possibility in II exists. Hence conclusion II follows. (71-75):
q = probability of not happening; p + q = 1 1 3 1 = 4C 3 × 2 2
=4×
57. 4; Required probability =
1 1 1 × = 8 2 4
48
C4
52
C4
=
48! 52! ÷ 44! 4! 48! 4!
=
48 47 46 45 38916 = 52 51 50 49 54145
58. 1; From statement A, Average speed of bus = 3 ×
120 = 120 km/hr 3
59. 5;Data in both the statements are not sufficient to answer. 60. 3; From both the statements, radius = 196 = 14 cm Area of the circle =
22 × 14 × 14 = 616 cm2 7
61. 4; Suppose, the expenditure on production of rice and wheat in 2012 was ` 100 each. 136 Required ratio = 128 = 17 : 16
62. 3;The sales of wheat in 2014 was 136% of the expenditure on wheat in that year. Hence, the required production = 24.5 ×
100 136
= ` 18.01 lakh. 63. 2; Required profit =
30 × 18 = `5.4 lakh. 100
64. 5; Data inadequate 65. 1; Suppose the amount of profit earned through the sale of rice in 2010 and 2011 was `1000 and `900 respectively. This means that 32% of the expenditure on rice in 2010 is equal to `1000. Similarly 18% of expenditure on rice in 2011 is equal to `900. 1000 100 32 = 5:8 Required ratio = 900 100 18
3
71. 3 72. 2 73. 1 74. 5 75. 3 (76-80): no risk of damage do me ra su ........ (i) improve policy risk of insurance holders ti la ra su ju ve ..... (ii) no risk policy su me la ...... (iii) damage of insurance holders do ti ve ra .... (iv) from (i), (ii) and (iv), of ra ........(v) from (i), (iv) and (v) damage do........ (vi) from (i), (ii), (iii) and (v), risk su ......... (vii) from (ii), (iii) and (vii), policy la ......... (viii) from (i), (v), (vi) and (iii), no me ........ (ix) from (ii) and (iv), insurance/holders ti/vi ........ (x) from (iii), (v), (vii), (ix) and (x), improve ju 76. 2 77. 1 78. 3 79. 5 80. 5 81. 2; D = E < F ......... (i) G > H > E .......... (ii) combing (i) and (ii) G > H> D = E < E we can’t compare F and H. But G > D. Hence II is true. 82. 5; M < N < O, P>N>Q
Combining both,
R is the wife of M. 87. 5; D is mother of E
M < N < O or Q I; L> H 84. 2; Can’t compare P and H. Hence I is not true. I < M or M > I, Hence II is true. 85. 5; A > B > C > D >E ; C F=G;D>M>N Gender of H is not given. 88. 5; It is given that L is the father of P. L#M δ N?P#Q θ R
A > N, Hence I is true.
B > G, Hence II is true. 86. 2; T is nephew of N (Given)
4
From II; Rahul is 40th from the top and Nisha is 14th from the bottom. From III; Shivani is in the middle of Rahul and Nisha From II and III; Nisha’s rank from the top =55 – 14 + 1= 41 + 1 = 42 Rahul is 40th from top. Shivani’s rank is 41st from top. Shivani’s rank from bottom = 55 – 41 + 1 = 14 + 1 = 15th Either I or II and III are sufficient. 98. 5; From I;
Either ‘ θ ’ or ‘ δ ’ 89. 5; Except Yuan all are internationally accepted currencies. 90. 4; Lake Superior is a fresh water body while rest are salt water body. (91–95):
Name
Competition Name
Day
Team
A
Hockey
Monday/Saturday
White
B
Dancing
Tuesday
Black
C
Singing
Friday
Blue
D
Painting
Monday/Wednesday/ Thursday/Saturday
Green
E
Tennis
Monday/Wednesday/ Thursday/Saturday
Black
F
Cooking
Monday/Wednesday/ Thursday/Saturday
White
From II; Y M From III;
From I and III;
91. 4 92. 3 93. 4 94. 1 95. 2 96. 5 Given that __ > T > ___ > S > __ From I; Q is not the heaviest. From II; T > R > S i.e. R is in the middle. From III; __ > __ > P > S > __ Using only II; ___ > T > R > S > ____ i.e. R is in the middle. From III; __ T > P > S > __ i.e. P is in the middle. Either II or III alone is sufficient. 97. 3; From I; Sahil’s rank from the botton 25th Sahil’s rank from the top = 55 – 25 + 1 = 30 + 1 = 31 Shivani’s rank is 31 – 10 = 21st So, Shivani’s rank from the bottom = 55 – 25 + 1 = 34 + 1 = 35th
99. 4
5
All together are not sufficient. 100. 1
SBIPO-PT-006 (1–5): D B A C F E 1. 2
2. 1
3. 3
4. 5
5. 4
6. 1
7. 3
8. 2
9. 5
10. 4
11. 2
12. 1
13. 3
14. 3
15. 1
? = 141 × 4 + 7.5 = 571.5 38. 2; 474 + 12
9
21. 1
22. 3
23. 2
24. 4
25. 1
26. 3
27. 3
28. 5
29. 5
30. 3
73
85
77
+4
+8 +4
31.3; ? = 5184 + 512 – 81 = 5615 7 6231 14.5 740 30 900 – 31 100 100
= 27 – 5 =22 34. 3; ? 2800 28 6 – 126 – 245
1 3 y – ,– 4 4 x < y 42. 4;
= 600 – 126 – 245 = 600 – 371 = 229 1
?
1 2
1018
I. x2 – 15x – 11x + 165 = 0 or, (x – 11) (x – 15) = 0
229 51 51 212 1018 916 1018 53 ? ?
or, x = 11, 15 II. y2 – 21y – 15y + 315 = 0
102 ?
?
1 1 ? 2 4
or, (y – 15) (y – 21) = 0 or, y = 15, 21
36. 4;
xy
17
21 2
+2
37 2
+4
73
137
2
2
+6
43. 1; I. 3x2 – 21x – 11x + 77 = 0 or, (3x – 11) (x – 7) = 0
+8
? 73 64 137
37.3;
or, x =
571.5 141 44.5 18.5 ×1 + 7.5 ×2 + 7.5 × 3 + 7.5 × 4 + 7.5
11
SBI PO (PRE) Mock-Test-11
+22
+14 +6
121
99 +8
7 9 (5x + 7) (4x + 9) = 0 x = – ,– 5 4 2 II. 16y + 12y + 4y + 3 = 0 (4y +1) (4y +3) = 0
33. 5; ? (24)3 ÷ (8)3 – 5
51
× 5 –5
? 99 22 121 41. 3; I. 20x2 + 45x + 28x + 63 = 0
= 1407 + 107. 3 – 270 = 1244.3
5275
40. 4;
20. 4; Replace ‘wave’ with ‘waves’
132651 3
1056
×5–2 × 5 –3 × 5 –4 ? = 1056 × 5 – 5 = 5275
19. 4; Add ‘a’ before ‘flat line’
44944
212
43
18. 4; Replace ‘intense’ with ‘intensity’
52441 53
+ 48
39. 1;
17. 2; Replace ‘in’ with ‘over’
35. 1;
+ 36
? = 546 + 48 = 594
16. 4; Replace ‘often’ with ‘oftenly’
32.4; ? =
+ 24
594
546
510
486
11 ,7 3
II. 2y2 + 16y + 7y + 56 = 0 or, (2y + 7) ( y + 8) = 0
1
y= –
7 ,–8 2
50. 5; Speed of train =
x > y 44. 5; I. 63x2 + 35x + 36x + 20 = 0 (7x +4) (9x + 5) = 0
Speed of bike =
450 75 km/hr 6
75 37.5 km/hr 2
Time taken by bike =
4 5 x = – ,– 7 9 II. 45y2 + 27y + 25y + 15 = 0 (9y + 5) (5y + 3) = 0
300 8 hr 37.5
51. 1; Sonali’s percentage =
5 3 y = – ,– 9 5
35 39 40 37 32 100 60 5
183 100 61% 300
52. 4; Ratio of investment in three years of Rakesh and Kunal
No relationship between x and y exists 45. 3;
5 = 50,000 1 70,000 2 : 80000 2 = 1, 90, 000 : 2,00,000 = 19 : 20
I. x2 + 9x + 7x + 63 = 0
(x + 7) (x + 9) = 0 x = –7, –9 II. 6y2 – 24y – 13y + 52 = 0
Rakesh share of profit =
(6y – 13) ( y – 4 ) = 0 y =
13 ,4 6
= ` 3,42,000 53. 4; Using Alligation method,
x
4 9
6 13
70 100 x 350 x = 350 500 100 70
5 11
47. 3; Ratio of their uses = 8 : 10 : 12 = 4 : 5 : 6 Rent paid by R =
48. 2;
6 2250 = ` 900 15
6 –5 = 1 13 11 143
Part of tank filled in 1 hour
1 5 –4 11 9 = 99 1 1 : 9 : 13 143 99
1 1 1 65– 3 1 – = 20 24 40 120 15
Required ratio =
Thus, required time for the tank to be filled = 15 hrs.
Required quantity =
49. 2; 15 A = 10 B
9 22 9 litre 22
54. 2; Let the number be x
3A = 2B
Then, x2 – (68)2 = 7697
Now, B – 10 = 3A – 22 or, B – 10 = 2B – 22 or, (Put 3A = 2B)
or, x2 = 7697 + 4624
or, B = 12
or, x = 111
A =
19 7,02,000 39
55. 5; According to the question,
2 12 8 3
11x 13 3x 13 48 2
Required sum = 20
2
SBI PO (PRE) Mock-Test-11
61. 2; Average production for five years for
14x 26 48 7x + 13 = 48 7x = 35 x = 5 2 Son’s present age = 3 × 5 + 13 = 28 years
Nokia
56. 1; Statement A alone is sufficient to answer the
35 50 30 55 45 lakh 43 lakh 5
Samsung
question. We know that whenever any odd number is divided by any odd number, it gives an odd number.
25 40 40 45 55 lakh 41 lakh 5
From Statement B, we get A is either even or odd as
40 45 55 40 40 lakh 44 lakh 5
the sum of an even and an odd number is odd. If B is
The average production of microman is maximum.
Micromax
odd then A is even and if B is even, A is odd. 62. 3; Required percentage increase =
57. 4; From Statement A, Required ratio = 62 : 38 = 31 : 19 From Statement B, Required ratio = 62 : 38 = 31 : 19
63. 1; Percentage change (rise/fall) in the production of Samsung mobile in comparison to the previous year:
Either A or B alone are sufficient to answer the question. 58. 4; From Statement A, time of travel =
80 1 3 hr 25 5
From Statement B, Let the initial speed = S And initial time = t then, S =
or,
80 t
.... (i)
80 1 t– S5 2
C1
3 8 C1 8 3 Probability of not getting 1 purple ball P E = 1 – 8 8
60. 5; Odds in favour = PE : P E
8
C1
5
:1 –
8
C1 C1
For 2013
45 – 40 100 12.5% 40
For 2014
55 – 45 100 22.22% 45
40 100 160% 25
For 2011
45 100 112.5% 40
For 2012
55 100 137.5% 40
For 2013
40 100 88.89% 45
65. 4; Required ratio = 130 : 140 = 13 : 14
5 :1– 5 : 3 8 8
SBI PO (PRE) Mock-Test-11
40 – 40 100 0% 40
40 100 72.73% 55 Hence, maximum is in the year 2010.
C1
For 2012
For 2014
5 3 Odds against = P E : PE 8 : 8 5 : 3
5
40 – 25 100 60% 25
64. 5; For 2010
59. 1; Probability of getting 1 purple ball PE
For 2011
Hence, maximum is for the year 2011.
..... (ii)
Solving (i) and (ii) we find the time Hence either statement A alone or B alone is sufficient to answer the question.
3
55 – 35 100 57% 35
3
30 55 45 40 45 55 : 3 3
79. 1; No phone is a mobile (E) + All calculators are mobiles Conversion Some mobiles are calculators (I) = E + I = O* = Some calculators are not phones. Hence, conclusion I follows. Some cameras are not mobiles (O) + Some mobiles are calculators (I) = O + I = No Conclusion. Hence, conclusion II does not follow.
(66–70): Persons
Subject
Period
A
History
VII
B
Maths
II
C
Economics
III
D
Hindi
I
E
English
VI
F
Sanskrit
V
G
Computer
IV
80. 4; There is negative statements, thus possibility can not be exist. Hence conclusion I does not follows. No planet is asteroids (E) + No asteroids are moon (E)= E + E = No conclusion. Hence, conclusion II does not follow. 81. 1; No planet is asteroids Conversion No asteroids is planet. Hence, conclusion I follows. All stars are planets (A) + No planet is asteroid (E) = A + E = E = No star is asteroid + No asteroids are moon. E = E + E = No conclusion. Hence conclusion II does not follow. 82. 2; The highest number is 956 Sum = 9 + 5 = 14 83. 1;
71. 1; Except 1 all are synonyms. 72. 5; E F F I C I E N C Y
(73–74). Given statements L Q > P .......... .(i) R < C > N = L .......(ii) Combining all: R < C > N = L Q > P 73. 1; C > Q is true. R P is not true. 74. 4; N P is not true. L < R is not true.
95 6
81 2
78 2 3 5 9 5 1 6
82 1 9 6 5 72 8 3 9 5 5 6 1 The lowest number is 395. 84. 4; 95 6
8 12 75. 5; Given statements : T > E C = H ...... (i) R M > H .......(ii) Combining all: T> EC=H < MR R > C is true. T > H is true. (76–77): Given statements Z > N ............(i) K N .............(ii) K = M D .....(iii) Combining all: Z> NK= MD 76. 4; Z > M is not true. N D is not true. 77. 5; N M is true. K D is true.
78 2 35 9 51 6
6 59 287 9 5 3 61 5 21 8 The second highest number is 659. 85. 3; The second highest number is 812. (86–89): In the rearrangement numbers are arranged in descending order and the words are arranged in English alphabetical order from left to right in alternate step. Input : 36 eager power 16 24 farm 46 sky ball 59 Step I : 59 36 eager power 16 24 farm 46 sky ball Step II : 59 ball 36 eager power 16 24 farm 46 sky Step III : 59 ball 46 36 eager power 16 24 farm sky Step IV : 59 ball 46 eager 36 power 16 24 farm sky Step V: 59 ball 46 eager 36 farm power 16 24 sky Step VI : 59 ball 46 eager 36 farm 24 power 16 sky. 86. 3 87. 3 88. 2 89. 4 (90–92):
78. 5; Some cameras are phones (I) + No phone is a mobile. (E) = I + E = O = Some cameras are not mobiles. Hence, conclusion I follows. No phone is a mobile Conversion No mobile is a phone. Thus, All mobiles are not phone. Hence, conclusion II follows.
(–)
(+)
B
D (+)
E
(–)
F
( +)
G
(–)
A C
4
SBI PO (PRE) Mock-Test-11
90. 2 91. 1 92. 5; We don’t know the gender of C. (93–97)
From both ;
M
M P
K
H
K
P
A
E C
F
J
N
Hence, both are not sufficient to answer the question. 99. 2; From I; We don’t know Y is in which direction of from W. In fact it could be in any direction. From II;
B 93. 4 94. 2 98. 4; From I;
95. 1
96. 5
97. 2
M
M P
K
L
L
G
D
N
J
N
P
K Z
3m
X
5m
Y
E
W S
L
Z is to the west of point Y.. 100. 5; From I ; Y > X > W From II; Z > U > V From both: Z > U > V > Y > X > W Hence, W is the youngest.
L
From II; L
P J
K
J
L
K
SBI PO (PRE) Mock-Test-11
P
5
SBIPO-PT-007 43 + 42 = 59
(1–5): D A E C B 1. 1
2. 5
3. 3
4. 2
5. 2
59 + 32 = 68 68 + 22 = 72
6. 1
7. 5
8. 2
9. 3
10. 4
11. 1
12. 2
13. 3
14. 4
15. 5
The series is:
16. 5
17. 1
18. 1
19. 5
20. 4
24 × 9 – 8 - 208
36. 4;
21. 2; Replace ‘more’ with ‘many’
208 × 8 –7 = 1657 1675
22. 3; Replace ‘try’ with ‘trying’
1657 × 7 – 6 = 11593
23. 1; Replace ‘were’ with ‘had’
11593 × 6 – 5 = 69553
24. 1; Replace ‘asked’ with ‘ordered’
37.4; The series is 21 × 1 + 6 = 27, 27 × 2 – 9 = 45, 45 × 3 – 12 = 123, 123 × 4 – 15 = 477, 477 × 5 – 18 = 2367 2667 × 6 – 21 = 14181 38. 5; I. 30x2 – 35x – 36x + 42 = 0
25. 5; No error (26–30): 26. 1
27. 3
28. 5
31.5; ? = 54 ÷ 58 ×(52)3 × 18.4 +
29. 1
30. 4
75 140 100
(5x – 6) (6x –7) = 0
= 5–4×56 ×18.4 + 105 = 25 × 18.4 + 105 = 400 + 105 = 565 32.2; ? = 251 + 1572 + 3300 – 2985 = 2138
x=
II. 12y2 – 28y – 9y + 21 = 0
(4y – 3) (3y –7) = 0
21 11 3 10 1 4 33. 1; ? 5 5 – 4 3 6 – 9
y=
21 11 3 60 3 – 8 – 5 5 4 18
21 11 55 21 264 – 275 – 5 5 24 5 120
21 20 504 – 5 – 11 11
6 7 , 5 6
3 7 , 4 3
No relationship between x and y exists. 39. 3; I. x2 – 6x – 9x + 54 = 0 (x – 6) (x – 9 ) = 0 x = 6, 9 II. y2 – 11y – 10y + 110 = 0 (y – 10) (y – 11) = 0 y = 10, 11 x < y
34. 3; The series is : 129 + 5 ÷ 2 = 67
40. 3; I.
2x 7
2
0
7
x= –
2
67 + 5 ÷ 3 = 24 25 24 + 5 ÷ 2 = 14.5
II.
14.5 + 5 ÷ 3 = 6.5
2
0
y
x
35. 1; The series is: –18 + 62 = 18 18 + 52 = 43 44
SBI PO (PRE) Mock-Test-12
3y – 8
1
8 3
46. 3; From A,
7 3 ,x – 10 2 II. 72y2 + 45y + 40y + 25 = 0 (9y + 5) (8y + 5) = 0
x= –
Time taken by train to cross a man =
5 5 y = – ,– 9 8 Hence, x < y
Speed of train =
3900 75 46 10 42. 5; ? 130
l 8
l 180 12
From statement B, speed of train = From, A and B
= 30 3450 10 = 348 43. 1; ?
length of train speed of train
l l 180 8 12
120 275 850 400 100 100
12l = 8l + 1440 4l = 1440
= 1020 + 1100 = 2120 44. 3;
l = 360 m 47. 1; n(s) = 3,6 ; 4,5 ; 4,6 ; 5,4 ; 5,5 ; 6,3 ; 6,4 a
Required probability =
a h
7 36 5
C 2 6 C2 4 C2
48. 5; Required probability =
b
From statement A, 2a + b = 24..................(i)
C2
10 15 6 31 = 105 105
=
1 4a 2 – b 2 6 .......(ii) 2 From (i) and (ii), we can find a & b
15
From statement B,
49. 4; Required number of ways =
b 4a 2 – b 2 4 45. 3; From statement A, Respective ratio of investment of Anamika, Divya and Charu = 16 : 14 : 13
8! 2!
= 8 × 7 ×6 × 5 × 4 × 3 = 20160
So, Area =
50. 5; Required number of ways = (4 women and 2 men) or (3 women and 3 men) or (2 women and 4 men) or (1 women and 5 men)
Let Charu joined for m months
or (6 men only)
From statement A and statement B Ratio of the profit will be = 16 × 12 : 14 × 12 : 13 × m
C 4
4
4 6
6
C
C 2 4 C3 6 C 3
4
2
6 C 4
6
( C1 C 5 ) C 6
and, 192x + 168x = 72000 360x = 72000 x = 200
C’s share = 85000 – 72000 13 × m × 200 = 13000
6! 4! 6! 4! 6! 4! 6! 1 2!4! 3! 3!3! 2!2! 2!4! 3! 5!
= 15 +
m = 5 months
Charu joined for 5 months.
4 6 5 4 4 3 6 5 4 6 1 3 2 2 2
= 15 + 80 + 90 + 24 + 1 = 210 2
SBI PO (PRE) Mock-Test-12
51. 3; Let the wages of 1 man, 1 woman and 1 children be
Expenses on fruits be 17x Expenses on milk be 5x Total expenses = 36x Increased expenses
` x, ` y and ` z respectively. Then, 3x = 2y ; 2y = 3z
x : y : z = 2 : 3: 2
125 130 160 14x 17x 5x 100 100 100 = 17.5 x + 22.1x + 8x = 47.6x
=
5x + 4y + 6z = 28900
5x + 6x + 6x = 28900 17x = 28900 x = ` 1700; y =
3 1700 ` 2550 and z = ` 1700 2
Total increased percentage =
Required answer = 6x + 5y + 2z = 6 × 1700 + 5 × 2550 + 2 × 1700
56. 5;
= 10200 + 12750 + 3400 = ` 26350 52. 1; Relative speed of policeman = (33 – 15) ×
A 5 8
5 = 5 m/s 18
B 7 13 9 16
To catch the thief, the policeman has to gain 400 m. Time taken to gain 400 m=
400 = 80 sec 5
5 , 9 –7 16 13 = 16×13
Actual distance covered by policemen in 80 sec = 80 × 33 ×
57. 3; Correct average of guavas
Distance covered by thief = 733 – 400 = 333m =
53. 4; After successive discounts,
324 2130 2008 – 2880 690120 – 872 = = 2127.3 324 324
58. 1; Let x, y and z be the amounts invested in schemes A,
85 85 = ` 1734 100 100
B and C respectively. Then,
After single discount,
SP of watch = 2400
Required loss =
70 = ` 1680 100
x 8 y 10 z 12 2156 100 100 100
8x + 10x + 12z = 215600 4x + 5y + 6z = 107800
1734 – 1680 100 3% 1734
Now, z =
54. 2; Circumference of base = 2r 2r = 88
2×
4×
22 × r = 88 r = 14 cm 7
22 14 14 20 = 12320 cm3 = 7
125 5 180 9 x x and z y y 100 4 100 5
4 5 z 5 z 6 z 107800 5 9
16z 25z 6 z 107800 5 9
144z 125z 270z 539z 107800 107800 45 45
2 Volume of the cylinder = πr h
z 9000
55. 2; Let the expenses on medicine be 14x
SBI PO (PRE) Mock-Test-12
5 –9= 1 8 16 16
Required ratio = 5 : 13
5 733 m 18
SP of watch = 2400
11.6x 100 = 32.22% 36x
3
y=
44 – 24 100 83% 24 (66 – 70): P Table Tennis Germany’ Q Cricket Canada R Tennis USA S Golf India T Badminton France U Hockey Britain V Football Australia 66. 4 67. 2 68. 3 69. 5 70. 3 71. 3; S < V and V = M S < M, hence conclusion I is wrong Again, R < S and U > S R < S < U ; Hence, conclusion II is also wrong. 72. 1; R < S T R < T ; Hence conclusion I is wrong. R < S and S < V = M R Q and Q > N Hence, H > N Again, Q = L and Q < Z Q < Z is true. 74. 1; Relation between A and L can’t be determined. But from first statement H > L; hence, conclusion II holds true. 75. 2; L < T and R L R < T is true. Again, L R and R P L P Hence, conclusion II does not hold true. 76. 5; No copy is pen (E) + All pens are pencils (A) O* Some pencils are not copy. Hence I follows. Some books are copy (I) Conversion Some/few copy
5 z = ` 5000 9
=
Sum invested in scheme B is ` 5000
59. 5; Let Raman’s present age be x years x=
7 (x – 8) 5
5x = 7x – 56 2x = 56 x = 28 Raman’s brothers age at the time of his marriage = (28 – 8) – 6 = 14 years Present age of Raman’s brother = 14 + 8 = 22 years 60. 1; Let the number of houses be x.
Number of houses having one female only = 100 – 30 % of (100 – 45) % of x =
70 55 77 x= x 100 100 200
Required percentage =
77x 1 100 200 x
= 38.5% 61. 2; Average speed of bus A =
44 32 52 60 56 60 18 m/sec 6 5
= 182.4 m/sec 62. 3; Ratio of speed of bus A and bus B on day D5 = 56 : 40 = 7 : 5 Ratio of time taken to travel the same distance = 5 : 7 63. 5; Average speed of bus C =
32 24 44 44 52 44 40 km/h 6
is book. Hence II follows. 77. 1; From statement III No copy is a pen, it can be concluded that A copy can never be pen. Hence I follows. Again, Some books are copy (I) No copy is a pen (E) = I + E =O. = Some books are not pen is true but some pens are not book doesn’t hold true. 78. 1; Some papers are print I No print is picture (E) = I + E = O = Some papers are not pictures. Hence, possibility of some papers is not pictures is also true true.
Average speed of bus B =
48 44 24 52 40 56 44 km/h 6
Required percentage = 64. 1; Time taken =
40 100 91% 44
975 18.75 hour = 18 hour 45 min 52
65. 4; Required percentage increase
4
SBI PO (PRE) Mock-Test-12
94. 2; From I, From II (O, Q ) > N > P Q > R > (M, N, O, P) Only From II statement conclusion can be drawn. 95. 5; From I,
All page is picture Conversion Some pages are picture Hence, conclusion II doesn’t hold true. 79. 5; All page is picture All picture being page is a possibility is correct. Hence, conclusion I is follows. Again, Some papers are print All papers being print is a possibility correct. 80. 2; All boxes are cups = A + No bag is cup Conversion No cup is bag E
B
A + E E No box is a bag does hold. Hence possibility in I does not exist. From I, All bottles are boxes (A) + All boxes are cups(A) A+A A All bottels is cup (A) + No cup is a bag (E) E No bottle is a bag No bag is a bottle. Hence conclusion II follows. 81. 3; lowest no 129 sum of Ist and 3rd digit (1 + 9) = 10 82. 2; 879 , 547, 892, 219, 148 Third highest number = 547 83. 1; 987, 754, 289, 921, 814 Second lowest number = 754 84. 3; 798, 475, 928, 192, 481 Third lowest number = 481 (85 –88): Here, in the above rearrangement, words are arranged in reverse alphabetical order considering last letter of the word and number are arranged in decreasing order in each alternate step. Input: volume 20 resign 25 request betray 5 14 Step I: betray volume 20 resign 25 request 5 14 Step II: betray volume 20 resign request 5 14 25 Step III: betray request volume 20 resign 5 14 25 Step IV: betray request volume resign 5 14 25 20 Step V: betray request resign volume 5 14 25 20 Step VI: betray request resign volume 5 25 20 14 Step VII: betray request resign volume 25 20 14 5 85. 3 86. 5 87. 3 88. 5 (89–93):
A D
From II, E B or A
C
From I and II,
B E
A
F
D
C
On combining both the statement, we have C is third to the left of B. 96. 1; From I;
T Q
R
A or B
W
From II;
P 5m X
V
6m
S
P
W
U 89. 4
90. 4
91. 3
92. 2
SBI PO (PRE) Mock-Test-12
2m
R
T
5m
Q
4m
3m W
From statement I, x is north–east of W.
93. 3
5
R
(97–99): (+)
(–)
T
S (+)
(–)
R
Q (+)
P
(–)
U
(–)
W
97. 2 98. 3 99. 4 100. 4; except option ‘4’ the sum of their position in English alphabet is same, as AZ 1 + (26) (1) + (2 + 6) = 9
6
SBI PO (PRE) Mock-Test-12
SBIPO-PT-008 1. 2 2. 1 3. 3 4. 1 6. 1 7. 5 8. 2 9. 3 (11–15): B A D C E 11. 1 12. 2 13. 1 14. 3 16. 3; Replace ‘for’ with ‘since’ 17. 2; Replace ‘at’ with ‘in’ 18. 5; No error 19. 4; Add ‘the’ before ‘government’ 20. 4; Replace ‘were’ with ‘was’ 21. 2 22. 1 23. 3 24. 5 26. 2 27. 4 28. 4 29. 1
5. 4
40. 5; The series is
10. 2
448 – 8 220 2
15. 5
220 – 8 106 2 106 – 8 49 2
25. 2 30. 1
49 – 8 20.5 2 41. 1; 6x2 + 16x – 21x – 56 = 0 (2x – 7) (3x + 8) = 0
2820 5 11 48 1429 31.4; ? 50 = 282 + 528 + 1429 = 2239
32.1; ?
352 182 36 735
7 8 ,– 2 3 II. y2 + 4y + 7y + 28 = 0 (y + 7) (y +4) = 0 y = –7, –4 x>y 42. 1; I. 44x2 + 33x + 4x + 3= 0 (11x + 1) (4x + 3) = 0
x=
9 18 120
45 82 33. 5; ? 2400 280 16 100 100
1968 126 130.875 131
34. 3; ?
16
65 15 51200 = 50 16 26 24 100
1 3 ,– 11 4 II. y2 + y + 11y + 11 = 0 (y + 11) (y + 1) = 0 y = –11, –1 x y 43. 5; I. 3x2 + 12x + 5x + 20 = 0 (3x + 5) (x + 4) = 0
x–
2780 250 139 5000 ? = 19321 36. 2; ? % of 34900 = 242000 – 191744 = 50256
35. 2;
?
100 144 34900 37.4; The series is 62 + 1, 82 + 1, 102 + 1, 122 + 1, 142 + 1, ... 38. 4; The series is: 128 + 3 = 131 131 + 5 = 136 136 + 7 = 143
? 50256
5 x – ,–4 3 II. 2y2 + 6y + 7y + 21 = 0 (2y + 7) (y +3) = 0
7 y – , –3 2 No relationship between x and y exists. 44. 5; x2 – 2x – 7x + 14 = 0 (x –7) (x – 2) = 0 x = 7, 2 II. y2 + 5y – 7y – 35 = 0 (y – 7) (y + 5) = 0 y = 7, –5 No relationship between x and y exists.
143 + 9 = 152 151 152 + 11 = 163 39. 3; The series is 1399 – (14)2 = 1203 1203 – (12)2 = 1059 1059 – (10)2 = 959 989 959 – (8)2 = 895 895 – (6)2 = 859
SBI PO (PRE) Mock-Test-13
904 – 8 448 2
1
45. 3; From both the statement together we find the Arjun’s salary in 2013. 46. 3; From statement A ; 2(l + b) = 48 l + b = 24 From statement B; lb = 140. Combining A and B ; We get, l = 14, b = 10 Length of diagonal =
1 1 : 36 : 28 9 : 7 28 36 Difference of ratio = 9 – 7 = 2
A’s share : B’s share =
2 100 28.57% 7 50. 3; Let the marked price of bed sheet and pillow covers be 2x and 3x Let the discount on pillow cover be d% Then,
Required percentage =
142 102
= 196 100 = 296 2 74 m 47. 3; From A, Time taken to fill the cistern without leak = 8 hours
55% of 2x
1 Part of cistern filled without leak in 1 hour = 8 From Statement B, Time taken to fill the cistern in presence of leak = 10 hours 1 10 Combining both statements A and B,
Net filling in 1 hour =
8 8x 12 12x – x – 20 20 20 20
4x 20x 4 20 20
x
11 3d 3– 3 10 100
11 3d 3–3 10 100
30 In (37 × 11) min, distance covered = 37 11 37 = 330 km In last 29 min, distance covered
1 6
30 29 24.17 km 36 Required total distance cover in 436 minute = 330+ 24.17 km = 354.17 km
=
1 49. 5; A’s 1 day work = 28
B’s 1 day work =
60 = 30 min 1 min 37 min 50 In 37 min, distance covered = 30 km
4 1 24 6
Part of mixture replaced with wine =
110 3x(100 – d) 300 x x 100 100 100
60 80 min 4 min = 44 min = 120 Time taken by express train to cover 800 km = Time taken by it to cover 720 km + Time taken by it to cover 80 km 44 60 80 min = 720 min 80 120 = 396 min + 40 min = 436 min Time taken by local train in covering 30 km including stoppages
12 12x – l Quantity of water in the new mixture = 20 20
12x 8x 12 8 – x – 20 20 20 20
110 2 36 % 3 3 51. 4; Time taken by express train to cover 80 km including stoppage
8 8x – x l = 20 20
100
60 3x 2x 100
d
1 1 5–4 1 Work done by leak in 1 hour = – 8 10 40 40 Leak will empty the cistern in 40 hours. 48. 1; Let the container initially contains 20 l of liquid. Let ‘x’ part of liquid be replaced with wine Quantity of wine in the new mixture
100 – d of 3x
1 36
2
SBI PO (PRE) Mock-Test-13
52. 5; Data is insufficient to give the answer 53. 2; Let the principal be = ` 1000 30% of P = ` 300 Remaining sum = ` 700 20% of 700 = ` 140 Remaining sum = ` 560
10
57. 1; Required probability =
=
300 12 140 15 560 25 100 100 100 = 36 + 21 + 140 = ` 197 Now, we know that
Total interest =
Interest =
1000 R 1 100 R = 19.7% 54. 3; Part of tank filled by P1, P2 and P3 in 3 min 1 1 1 3 2 – 4 1 – 24 36 18 72 72
2010 = 1 1 67 Time taken to fill 1 – 24 36 72 th part of the tank = 3 × 67 = 201 min
Tap P1 fills
4x 135% 9 : 10 5x 120%
30 – 10 × 100 = 200% 10
70 – 30 × 100 = 133.33% 30 2012 = Decrease
2011 =
67 5 = 72 72
60 – 50 × 100 = 20% 50 2014 = Decrease Hence, maximum is in the year 2010 62. 3; Average production of wheat
2013 =
1 parts in 1 min. 24
5 2 1 1 = = Remaining part = 72 – 72 36 24
=
1 36 th part is filled by P2 in 1 min. Hence, required time to fill the whole tank = 201 + 2 = 203 min. 55. 4; Let the given distance be x km. Then, time taken to walk x km + time taken to ride x km = 86 min Time taken to walk 2x km + time taken to ride 2x km = 172 min Time taken to ride 2x km = 172 – 124 = 48 min
20 10 10 30 70 50 40 60 8
290 = 36.25 tonnes 36 tonnes 8 63. 5; Average production of rice
=
= 15 20 25 35 60 45 50 55 8 =
305 = 38.125 tonnes 8
Required percentage = 56. 3; Required probability =
=
11
C4
15
C4
38.125 ×100 105% 36.25
64. 5; Required percentage more =
11! 15! 11 9 8 10 22 = 4! 7! 4!11! 15 14 13 12 91
SBI PO (PRE) Mock-Test-13
C4
59. 5; Data inadequate 60. 3; Total number of letters of the word ‘DESIGN’ = 6 Consonants are D, S, G, N Vowel are E and I According to the question, none of the consonants are at either end hence; vowels will be at both ends. Required number of ways = 2P2 × 4P4 = 2 × 1 × 4 × 3 × 2 × 1 = 48 61. 2; Percentage increase in production of wheat 2008 = decrease 2009 = no change
P R T 100
Remaining part of the tank = 1 –
15
10! 5! 15! 10 10 4 3 2 20 = 9! 3! 2! 4!11! 15 14 13 12 273
58. 1; Required ratio =
or, 197 =
=
C1 5 C 3
16.67%
3
70 – 60 ×100 60
65. 2; Four (i.e. 2008, 2009, 2010, 2014) 66. 3; Given: P Q = N > S R = T Thus, we can’t compare P and T. Hence, conclusion I (P T) is not true. Again, N > T is true. Hence, conclusion II is true. 67. 4; Given: P N = Q M R S Thus, we cannot compare P and R. Hence, conclusion I (P > R) is not true. Again, P M is true. Hence, P = M may be true. Hence, conclusion II is also not true. 68. 1; Given: K = L M ........ (i) N O P.............(ii) M N Q..........(iii) K = L M N Q Thus, we can’t compare K and Q. Hence, conclusion I (K Q) is not true. From (i), (ii) and (iii), we get K = L M N O P Thus, L O is true. Hence, conclusion II is true. 69. 4; Given: N O < P = Q ......(i) Q > T = R ............(ii) Combining (i) and (ii), we get, N O < P = Q > T = R Thus, O < R is true. Hence, conclusion I (O = R) is not true. Again, P > R is true. Hence, conclusion II is true. 70. 2; Given: Z X = Y W > V > U Thus, Z > U is true. Again, Y > U or U < Y is true. Hence, both conclusion I and II are true. (71–73): good time for India mo jo ki la .......(i) time for money ki la ha.............(ii) earn more money ha re pa............(iii) manage time more ki pa ru.............(iv) from (i),(ii) and (iv), we get, time ki .....(v) from (i), (ii), (iv) and (v), we get, for la ........(vi) from (ii) and (iii), we get, money ha .....(vii) from (iii) and (iv), we get, more pa........(viii) from (iii), (vi) and (viii), we get, earn re ......(ix) from (iv), (viii) and (v), we get, manage ru .....(x) from (i), (v) and (vi), we get, good /India mo /jo 71. 1 72. 5 73. 3 74. 3; R O U T E P M 6 @ © 5 7 8 So, MORE 8 65
(75–80): (Jaipur) F B (Chennai)
(Delhi)D
C (Mumbai)
(Patna) G
(Lucknow) E
H (Goa) A (Bhopal)
75. 2 76. 2 77. 2 78. 3 79. 3 80. 4 (81–85): The machine rearranges the word in alphabetical order while numbers are arranged in descending order in each alternate step. In step I the word are arranged first then in next step largest number are arranged and so on in each alternate step. Input: above rule 92 followed by 73 65 47 the employees 18 in 12 company 45 Step I: above 92 rule followed by 73 65 47 the employees 18 in 12 company 45 Step II: above 92 by rule followed 73 65 47 the employees 18 in 12 company 45 Step III: above 92 by 73 rule followed 65 47 the employees 18 in 12 company 45 Step IV: above 92 by 73 company rule followed 65 47 the employees 18 in 12 45 Step V: above 92 by 73 company 65 rule followed 47 the employees 18 in 12 Step VI: above 92 by 73 company 65 employees rule followed 47 the 18 in 12 45 Step VII: above 92 by 73 company 65 employees 47 rule followed the 18 in 12 45 Step VIII: above 92 by 73 company 65 employees 47 followed rule the 18 in 12 45 Step IX: above 92 by 73 company 65 employees 47 followed 45 rule the 18 in 12 Step X: above 92 by 73 company 65 employees 47 followed 45 in rule the 18 12 Step XI: above 92 by 73 company 65 employees 47 followed 45 in 18 rule the 12 Step XII: above 92 by 73 company 65 employees 47 followed 45 in 18 rule 12 the 81. 3 82. 1 83. 5 84. 2 85. 4 86. 1; Some poems are novels (I) : Hence, possibility in I exists. Again, All books are poems (A) + No poem is good (E) = A + E= E = No book is good. Hence, II does not follows. 87. 4; All germs are infections (A) + All infections are bad (A) = A + A = A = All germs are bads. Hence, I does not follows. 4
SBI PO (PRE) Mock-Test-13
Similarly,
Again, No germs are harmful (E) conversion No harmful is germs (E) + All germs are infections (A) = E + A = O* = Some infections are not harmful. Hence, II does not follows. 88. 5; No metal is wood (E) + All woods are alloys (A) = E + A = O* = Some alloys are not metals. Hence, possibility in I exist. Again, Some papers are metals (I) + No metals are wood (E) = I + E = O = Some paper are no woods. Hence, II follows. 89. 5; All windows are doors (A) + Some doors are locks (I) = A + I = No conclusion. But possibility in I and II exists. Because there is no negative statement. 90. 1; All letters are words (A) + No words is a sentence (E) = A + E = E = No letter is a sentence. Hence, I follows. But II does not follow. 91. 3; do re me he is going ............(i) la me pa she is singing .........(ii) ta do ka he plays cricket.....(iii) From (i) and (ii), me is From (i) and (iii), do he From (i), re going Hence,‘re’ is code for going. 92. 3;
S E P TYPES
FOREST
+4
+5
+5
C Y
–4
P +2 R S +2 U –4 J +2 L +4 P +3 S –4 D +2 F 96. 2; % U T, © A E 97. 3; New arrangement: J 2 K 8 Q 7 U T I V 9 M D P 14 F A E 3 T W N L H
–4
2 8 7 9 1 4 3 34 17 2 2 99. 1; Seventh to the left of eleventh from the right end (11 + 7) = 18th from the right end i.e. %. 100. 2;
98. 5;
2 7 V 1
+6
U U H V T L
SBI PO (PRE) Mock-Test-13
Y T +4
After rearrange: 1 2 3 4 5 6 8 95. 4;
T S E R O F +3
+3
94. 3; 4 6 5 1 3 2 8
PH OTOGRAPH
+2
+2
T G S
93. 1; +1
+1
5
+1 +1 +1 +1
@ % 9 4
+2 +2 +2 +2
K T M ©
+3 +3 +3
7 V 1
SBIPO-PT-010 1. 4 2. 1 3. 2 4. 2 5. 1 6. 4 7. 1 8. 2 9. 1 10.3 11. 2 12. 4 13. 3 14. 5 15. 1 (16–20): C E A BD 16. 1 17. 3 18. 2 19. 4 20. 5 21. 2; Replace ‘develop’ with ‘developed’ 22. 3; Replace ‘costing’ with ‘cost’ 23. 2; Add ‘the’ before ‘initial’ 24. 1; Replace ‘Create’ with ‘Creating’ 25. 4; Add ‘at’ after ‘aims’ 26. 2 27. 4 28. 1 29. 5 30. 1 31.1; The series is: 6 × 11 – 10 = 56
5
x–
,–
2 2
2
3
II. y 2 – 6 y – 5 y 30 0
y– 5 y– 6 0 y 5, 6
y >x 36. 3; I. 55x2 + 33x + 20x + 12 = 0
11x 45x 3 0
56 × 9 – 8 = 496 496 × 7 – 6 = 3466
x–
3466 × 5 – 4 = 17326
4 3 ,– 11 5
II. 48y2 – 12y – 12y + 3 = 0
? = 17326 × 3 – 2 = 51976
(12y – 3) (4y – 1) 0
32. 5; The series is: 25 × 1.2 = 30
y
30 × 2.2 = 66 66 × 3.2 = 211.2
3 1 1 1 , , 12 4 4 4
y >x
211.2 × 4.2 = 887.04
37.2; I. x2 – 6x – 19x + 114 = 0
? = 887.04 × 5.2 = 4612.608
(x – 19) (x – 6) = 0
33. 2; The series is: (11)2 + 11, (12)2 + 12, (13)2 + 13,
x = 19, 6
121 + 11 = 132
II. y2 – 6y – 4y + 24 = 0
144 + 12 = 156
(y – 6) (y – 4) = 0
169 + 13 = 182 196 + 14 = 210
y = 6, 4
? = 225 + 15 = 240
xy
38. 5; I. 8x2 + 6x + 36x + 27 = 0
34. 3; The series is: (28 × 8) + 3 = 227
(2x + 9) (4x + 3) = 0
(28 ×7) + 4 = 200 (28 × 6) + 5 = 173
9 3 x = – ,– 2 4
(28 × 5) + 6 = 146
II. 15y2 + 18y + 25y +30 = 0
? = ( 28 × 4) + 7 = 119
(3y + 5) (5y + 6) = 0
35. 3; I.
2
6 x 4x 15x 2 10 0
2x 5
5 6 y – ,– 3 5
3x 2 2 0
Relationship between x and y cannot be established.
SBI PO (PRE) Mock-Test-15
1
39. 1; ? 6200 × 960 ÷ 240 = 24800
40 2 60 3
1400 130 1435 40. 4; ? 100 = 1820 + 1435 = 3255
Now,
1 41. 3; ? = 100 – 561 = 56089 51
640 – x x 40 – 5 4 3
2560 – 4x – 5x 40 20 3
42. 2; ? =
43. 5; ? =
114 124 250 450 100 100 = 285 + 558 = 843
(2560 – 9x) × 3 = 800
5 6 25 18 43 21 35 105 105
7680 – 27 x = 800
44. 3; From statement A, Total CP of 80 kg = (5200 + 400) = ` 5600 From statement B, profit = 15% Combining both the statement A and B, we get SP =
640 – x x 40 – 100 80 60
x=
6880 254.81 km 27
49. 3; Let the money lent to Anshul be ` x Then, money lent to Rajat = ` (20000 – x)
115 5600 ` 6440 100
SI for Anshul = SP per kg of sugar =
6440 ` 80.5 80
SI for Rajat =
45. 4; Let edge of the cube be a metre. From statement A;
x 20 4 20x 100 25
20000 – x 24 4 20000 – x 24 100
25
According to the question,
a2 = 81 a = 9 m 20x 24 – 20000 – x 864 25 25
Volume = a3 = 93 = 729 m3
From statement B, a=9m 3
3
3
Volume = a = 9 = 729 m
44x – 19200 864 25
x ` 11400
46. 2; From statement B,
Amount lent to Rajat = ` (20000 – 11400) = ` 8600 50. 1; Speed of the person = 6 km/hr
1 Speed of boat in still water = 16 = 8 km/hr 2
47. 3; Let the original number be 10x + y
= 6
(10x + y) – (10y + x) = 36 9x – 9y = 36
5 5 m/s 18 3
Let the speed of train be x m/s
x –y = 4
20
and x + y = 8 x = 6; y = 2 The original number = 62
x–
48. 4; Let two trains meet at x km from station S1 Time taken by K to cover (640 – x) km – Time taken by J to cover x km.
280 5 x – 3
5 47 14 x m/s 3 3
Now the distance covered by train in 45 minutes.
2
SBI PO (PRE) Mock-Test-15
=
Then, according to the question,
47 60 45 42300 metres 3
x 1 150000 – x 3 2
Thus, time taken by person to cover this distance
2x = 450000 – 3x x = ` 90,000 Saving in Public Provided Fund
42300 25380 sec 5 6 18
= 1, 50,000 – 90,000 = ` 60,000 55. 2; Let the two-digit number be 10x + y Numbers obtained by interchanging the digits = 10y + x According to the question, 10x + y – (10y + x) = 54 ... (i) 9x – 9y = 54 x –y = 6
25380 1 7 = 7 hours 3 minutes 60 60 20
51. 4; Ratio of profit Anil, Deepak and Sohan = 24000 × 12 : 36000 × 7 + 36000 × 2 : 7 × 16000 + 4000 ×5
Again, y
= 288 : 324 : 132
... (ii) 3y = x From (i) and (ii) y = 3 and x = 9 Required number is 93 56. 1; Let the present age of Raju and his father be 2x and
= 72 : 81 : 33 = 24 : 27 : 11 Deepak’s share of profit =
52. 3; r1 264
27 26164 = ` 11,394 62
5x respectively.
7 42 cm 44
Hence, r2 352
7 56 cm 44
2x 4 5 5x 4 11
22x + 44 = 25x + 20 3x = 24 x = 8
Required difference π 562 – 422
Father’s age 5 years ago = 5 × 8 – 5 = 35 years 57. 5; SP = ` 15000
=
If Manoj offered a discount of 10% then the new SP = 15000 – 1500 = ` 13500
22 98 14 = 4312 cm2 7
53. 2; Let the price of sugar before increase be ` x
CP =
Consumption = 15 kg
13500 100 ` 12500 108
58. 4; 2M = 3W
Expenditure on sugar = 15x After increase
1M =
Expenditure on sugar = 120% of 15x = 18x Price of sugar = 128% of x =
32x per kg 25
3 5 W 1W W 2 2
M1D1 = M2D2 3 4
1 kg 16
5 D2 2
D 2 12
54. 1; Let Radha’s savings in National Saving Certificate be ` x.
SBI PO (PRE) Mock-Test-15
3 W 2
1M+ 1W=
18x 25 225 New consumption 32x 16
= 14
x 3
2 24 days 5 5
59. 3; Required number of ways 3
=
C 3
1
7
C 2 3 C 2 7 C 2 3 C3
6 14 × 100 24 8 = 43.75 %
64. 2; Required percentage =
3! 7! 3! 7! 1 2! 2!5! 2! 6!
3 7 3 3 7 1
65. 1; Required percentage more
= 63 + 21 + 1 = 85
60. 4; Let E = getting a king of club or queen of heart
=
210 – 56 21 10 – 4 14 100 × 100 = 56 4 14
=
154 × 100 = 275 % 56
n(E) = 2 2 1 Required probabilit y 52 26
(66–70):
19 14 61. 4; Required answer = × 4 lakh + × 3 lakh 100 100 42 76 lakh = 100 100
=
118 lakh = 1.18 lakh 100
62. 3; The number of people shopping online from Jabong in: Delhi =
6 × 14 lakh = 84000 100
Mumbai =
20 × 10 lakh = 2 lakh 100
Bengaluru =
Chennai =
8 × 8 lakh = 64000 100
28 × 4 lakh = 112000 100
Hence, maximum is in Mumbai 63. 5; Required average =
28 14 26 10 22 8 20 3 24 4 500
=
392 260 176 60 96 lakh 500
=
984 × 1,00,000 = 1,96,800 500
City
Activities
A
Jaipur
Acting
B
Bengaluru
Music
C
Patna
Sports
D
Delhi
Discussion
E
Lucknow
Dialogue
F
Bhopal
Painting
G
Mumbai
Adventure
H
Kolkata
Drama
66. 2 67. 3 68. 4 69. 5 70. 1 71. 4; Given expression, P > R , R < S T , P S From expression, P S and R < S T PS,TS Thus, we can’t compare P and T. Hence, conclusion I does not follows. Again, From expression , R < S T, it is obvious that S > R. Hence, conclusion II also does not follows. 72. 4; Given expression, N S < T and S = I, N I is true. Hence, N < I does not hold true. Again, From expression, N S < T > O and O E Thus, we can’t compare N and E. Hence, conclusion II also does not true. 73. 1; Given expression, N S < T > O and S = I We can conclude I < T. Hence, conclusion I holds true. Again, from expression, N S < T > O and O E T>O,EO Thus, we cannot compare T and E. Hence, conclusion II doesn’t hold true. 74. 5; Given expression, L = M < E and S = T > Z We can conclude Z < M and M < E
24 × 3 lakh = 72000 100
Hyderabad =
Person
lakh
4
SBI PO (PRE) Mock-Test-15
Now, Some bats are not black (O) + Some black is board (I) No conclusion. Hence, conclusion II does not hold true. (80–85): In the rearrangement words are arranged in ascending order from left to right as vowel comes first followed by consonant alternatively. Numbers are arranged according to the position of (last letter– 1) of the arrangement in the previous step from left to right first number at the extreme right then seems and so on in each alternate step. Input : steps 22 onion apt 19 tribe 4 new 18 13 Step I : apt steps 22, onion 19 tribe 4 new 18 13 Step II: apt steps 22 onion tribe 4 new 18 13 19 Step III: apt new steps 22 onion tribe 4 18 13 19 Step IV: apt new steps onion tribe 4 18 13 19 22 Step V: apt new onion steps tribe 4 18 13 19 22 Step VI: apt new onion steps tribe 4 18 19 22 13 Step VII: apt new onion steps tribe 4 19 22 13 18 Step VIII: apt new onion steps tribe 19 22 13 18 4 80. 2 81. 1 82. 5 83. 4 84. 1 85. 2 (86–90):
Z M< E Z O, S= T > Z M Y T S T Hence, conclusion II also holds true. 75. 2; Given expression, L = M < E and S = T > Z We can write, S = T > Z L as L M Thus, we can’t compare S and L. Hence, conclusion I does not follow. Again, from expression, Y S > O and S= T > Z M Thus, we can conclude Y S > Z Y > Z Hence, conclusion II holds true. 76. 1; Some tribes are teachers (I) conversion Some teachers are tribes (I) + All tribes are trainers (A) I + A I Some teachers are trainer. Hence, All teachers being trainers is a possibility holds true. Hence, I follows. Again, Some tribes are teachers (I) conversion Some teachers are tribes. Hence, Some tribes are not teachers doesn’t hold true. Hence, conclusion II does not follow. 77. 2; Some tribes are teachers (I) conversion Some
D
B
F
G
teachers are tribes (I) + All tribes are trainers (A) I + A I Some teachers are trainers. Again, No teacher is a typewriter (E) conversion No typewriter is a teacher. Now, No typewriter is teacher (E) + Some teachers are trainers (I) E + I O* Some trainers are not typewriters. Hence, I does not follow. Again, Some tribes are teachers (I) + No teacher is a typewriter (E) I + E O Some tribes are not typewriters. Hence, conclusion II follows.
H
C
E
A
86. 1 87. 3 88. 2 89. 3 90. 5 91. 1; From I, U < S < V < T Because it is given, S is only faster than U. Thus, S is second slowest. 92. 5; From I, From II, (+)
R
78. 1; Every Board is black (A) Every All the possibility All boards being black holds true. Thus conclusion I follows. Again, All colour is black conversion Some black is
(+)
Q
(–)
(+)
T
colour. Again, Some bats are not black (O) + Some black is colour (I) O + I No conclusion. Hence, conclusion II does not follow. 79. 1; Some bats are not black (O), We cannot draw any conclusion. Thus, all the possibilities in this case hold true. Hence, Some black is a bat is a possibility holds true. Conclusion I follows. Again, Every Board is black implication Some board
(+)
S
P
S
T
Thus, From I and II (+)
R
(+)
Q T
is black (I) conversion Some black is board.
(–)
S
P
Hence, From I and II together, we can conclude P has only
SBI PO (PRE) Mock-Test-15
5
one sister. 93. 4; From I, ‘He knows about you’ ‘ 3 12 7 4’ ... (i) ‘He knows everything’ 7 6 3 ... (ii) From (i) and (ii) ‘He knows’ either 7, 3 or 3, 7 We can’t find the code of ‘about’ From II, ‘It is about you’ 4 5 8 12 Hence, both statements I and II together are not sufficient to answer the question. 94. 3; except triangle all are 3 dimensional. 95. 2; Given
97. 2; (–)
I
Sister
Mother (+)
N
+2
–2
+2
98. 1; (–)
(+)
W
H
(+)
N
2
Q
Hence, W is grandmother of Q
S P E C T A C U L A R 19 16 5 3 20 1 3 21 12 1 18 +2
–2
S
–2
0 20 13 3 Similarly
–2
Brother
I is the maternal aunt of S
B R O A D 2 18 5 1 4 –2
(–)
Q
+2
+2
–2
17 18 3 5 18
–2
+2
–2
(–)
(+)
(–)
N
W
2) H
+2 –2
3 1 23 10 3 16 Q
96. 4; (–)
1) N
(+)
Sister
(+)
2) R
M
Brother
R
3)
H
(–)
(–)
T
N
(+)
(+)
T
N Hence, R is maternal uncle of N (+)
3) N
W
(–)
T
4) T
(+)
(–)
W
(+)
K
R Hence, sex of R is not determined. (+)
4) N
(–)
H
N
99. 2 100. 3
(+)
T
(+)
R
K
Hence, R is the nephew of N.
6
SBI PO (PRE) Mock-Test-15
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