## 15 Jul 2017

### Crack IBPS Exam 2017 - Quantitative Aptitude Scoring Part (Day-31)

Crack IBPS Exam 2017 - Quantitative Aptitude Scoring Part (Day-31):
Dear Readers, Nowadays most of the aspirants are facing huge trouble to increase the overall marks. To score high you need to practice more and more standard questions daily. “Practice does not make perfect, Only Perfect Practice makes perfect”.

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Direction (1-10): What value should come in place of question mark (?) in the following questions?
1). (0.04)² ÷ (0.008) * (0.2)⁶ = (0.2)^?
7
6
5
4
3
(0.04)² ÷ (0.008) * (0.2)⁶ = (0.2)^?
= 0.2^1 * 0.2⁶ = (0.2)^7
2. 13.141 + 31.417 – 27.118 + 15.247-14.214 = ?
18.465
18.473
18.688
18.645
18.697
13.141 + 31.417 – 27.118 + 15.247-14.214 = ?
59.805 – 41.332 = 18.473
3. 36 * 14 – 49 * 784÷112
151
171
141
161
131
36*14 – 49*784÷112
504 – 343 = 161
4. √676 * 12 – 864 ÷ 36 = ? + 61
234
227
232
244
256
√676 * 12 – 864÷36 = ? + 61
x = 227
5). 468 ÷ 39*15 + 166 = (?)³ + 130
5
6
4
7
2
12 * 15 + 166 = (?)³ + 130
(?)³ = 216
? = 6
6. 185% of 600 + 25% of 240 = ?% of 1000
136
128
117
121
None of these
185% of 600 + 25% of 240 = ?% of 1000
1110 + 60 = x% of 1000
x = 1170/10 = 117
7. 2/3 of 7/5 of 75% of 640 = ?
396
448
454
382
424
2/3 of 7/5 of 75% of 640 = 2/3 * 7/5 * 75/100 * 640 = 64 * 7 = 448
8. 15.4 * 14.5 * x = 2798.75
15.2
15.3
12.9
12.5
19.2
15.4 * 14.5 * x = 2798.75
x = 2798.75/223.3 = 12.5
9. 7344 + (5.4)² + √? = 7437.16
4156
4401
4096
4553
4335
7344 + (5.4)² + √? = 7437.16
√? = 7408 – 7344 = 64 => ? = 4096
10. 567 ÷ (10.8 * 2.5) ÷ 3 = ?
6
7
8
4
5
567 ÷ (10.8 * 2.5) ÷ 3 = ?
? = 567 ÷ (27) ÷ 3 = 7
What approximate value should come in place of question mark (?) in the following questions?
11. 21.003 × 39.998 – 209.91 = 126 × ?
5
4
3
2
6
21 × 40 – 210 = 126 × ?
630 = 126 × ?
? = 5 (approx)
12. 1440.0003 ÷ 23.999 × 2.5 × 3 = ?
450
500
420
360
520
1440 ÷ 24 × 2.5 × 3 = ?
60 × 2.5 × 3 = 450 (approx)
13. 3995.009 – 290.999 – 129.989 × 2 = ?
3410
3445
3435
3465
3530
3995 – 291 – 130 × 2 = ?
? = 3444
= 3445 (approx)
14. (15)² + (19.99)² + (24.001)² = ?
1250
1200
1300
1120
1160
(15)² + (20)² + (24)² = 225 + 400 + 576 = 1200 (approx)
15. (99999 ÷ 999 ÷ 9) × 99.99 = ?
1250
1000
1050
1220
1110
(99999 ÷ 999 ÷ 9) × 99.99 = 11.12 × 100 = 1110 (approx.)
16. 6575 / 17.98 × 42.03 / 6.87 = ?
2190
2280
2090
2150
None of these
6575 / 18 × 42 / 7 = (6575 / 18) × (42 / 7) = 365 × 6 = 2190
17. 12.002 × 15.005 – 8.895 × 6.965 = ?
130
117
105
110
None of these
12 × 15 – 9 × 7 = 180 – 63 = 117
18. 12.664 × 22.009 × 17.932 = ?
5100
5200
5148
5199
None of these
13 × 22 × 18 = 5148
19. 16.978 + 27.007 + 36.984 – 12.969 – 9.003 = ?
50
51
52
59
65
17 + 27 + 37 – 13 – 9 = 81 – 22 = 59
20. 18% of 602 + 27.8% of 450 = ?
234
260
225
220
250
[(18 × 600) / 1000] + [(28 × 450) / 100] = 108 + 126 = 234
Direction (21 – 25): What will come in place of question mark (?) in the following number series?
21. 698, 554, 454, 390, 354, ?
338
348
388
398
None of these
The series is -12^2 -10^2, -8^2,…..
22. 5, 7, 10, 36, 136, ?
789
890
690
698
None of these
The series is * 1 + 2, * 2 – 4, * 3 + 6,…..
23. 2, 11, 46, ?, 286
141
136
156
190
None of these
The series is
2 * 5 + 1 = 11;
11 * 4 + 2 = 46;
46 * 3 + 3 = 141, ….
.
24. 3, 10, 32, ?, 460, 2315
112
132
115
119
None of these
The series is,
3 * 1 + 7 * 1 = 10
10 * 2 + 6 * 2 = 32
32 * 3 + 5 * 3 = 111,…..
25. 6072, ?, 200, 48, 14, 5
1045
1010
1005
1052
None of these
The series is, -12 / 6, -10 / 5, -8 / 4, ……

Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
a) If X > Y
b) If X < Y
c) If X ≥ Y
d) If X ≤ Y
e) If X = Y or cannot be established
26) I. X² – 3X + 2= 0
II. 9Y² – 9Y + 2 = 0

a)
b)
c)
d)
e)
(x-1)(x-2) = 0
X=1,2
(3y-1)(3y-2) = 0
Y =1/2,1/3
Hence, x > y
27). I. X² – 6X + 8= 0
II.Y² – 8Y + 15 = 0

a)
b)
c)
d)
e)
(x-4)(x-2) = 0
X=4,2
(y-5)(y-3) = 0
Y=5,3
Therefore, the relation cannot be established
28). I. 2X² – 3X – 2= 0
II. 6Y² + 7Y + 2 = 0

a)
b)
c)
d)
e)
(x-2)(2x+1) = 0
X = 2,-1/2
(3y+2)(2y+1) = 0
y = -2/3,-1/2
Therefore X ≥ Y
29). I. 2X² + 7X + 6= 0
II. 2Y² + Y – 3 = 0

a)
b)
c)
d)
e)
(x+2)(2x+3) = 0
X= -2,-3/2
(2y+3)(y-1) = 0
Y=1,-3/2
Therefore, y ≥ x
30). I. 6X² + 11X + 3= 0
II. 3Y² – 2Y – 1 = 0

a)
b)
c)
d)
e)