# Quantitative Aptitude Practice Questions for IBPS Clerk – Set 4

Quantitative Aptitude Practice Questions for IBPS Clerk – Set 4
Directions (1 – 5): What will come in place of question mark (?) in the given series:
1. 19    26     46     91     173     ?
a) 243
b) 294
c) 304
d) 345
e) None of these
2. 110  125  230  545  ?  2525
a) 1238
b) 1240
c) 1413
d) 1150
e) None of these
3. 420     413     392     357     ?     245
a) 313
b) 308
c) 284
d) 272
e) None of these
4. 43    52     81     74     135     ?
a) 96
b) 104
c) 106
d) 168
e) None of these
5. 10      14      25      55      140      ?
a) 386
b) 120
c) 388
d) 396
e) None of these
Directions (6 – 10): What value will come in place of question mark (?) in the following questions: (You are not expected to calculate the exact value)
6. 124% of 4896 + 48.96 of 124 = ?
a) 12152
b) 10152
c) 11152
d) 11052
e) 13152
7. 539. 79 + √1089.13 – 564.81 = ?
a) 16
b) 8
c) 128
d) 13
e) 64
8. 33.69 × 18.73 + 29.84 = ?2
a) 25
b) 16
c) 24
d) 26
e) 28
9. 137.45 × 128.86 + 3037.62 = ? × 149.04
a) 129
b) 139
c) 149
d) 229
e) 239
10. 7803.0031 ÷ ?2 = 19682.67
a) 13
b) 21
c) 17
d) 15
e) 19
Directions (11 – 15): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
a) if x > y
b) if x < y
c) if x ≥ y
d) if x ≤ y
e) if x = y or relationship between x and y can’t be established

Solutions:
1. C)
Series I        19       26         46          91           173                     ?
Series II  :        7                  20               45                 82                 ?
Series III :             13              25          37           ?
Series IV  :                         12               12           12
0            0
Clearly, the pattern in series III is +12.
So, the missing term in series III = 37 + 12 = 49;
missing term in series II = 82 + 49 = 131;
missing term in series I = 173 + 131 = 304.
2. A)
Series Pattern                                 Given Series
110                                                     110
110 + 15 (=1×3×5) = 125               125
125 + 105 (=3×5×7) = 230            230
230 + 315 (=5×7×9) = 545            545
545 + 693 (=7×9×11) = 1238        1238
1238 + 1287 (=9×11×13) = 2525 2525
3. B)
Series Pattern                                      Given Series
420                                                          420
420 – 7 (=1×7) = 413                           413
413 – 21 (=3×7) = 392                         392
392 – 35 (=5×7) = 357                         357
357 – 49 (=7×7) = 308                         308
308 – 63 (=9×7) 245                         245
4. A)
5. C) 14 – 10 = 4
25 – 14 = 11 = 4 × 3 – 1
55 – 25 = 30 = 11 × 3 – 3
140 – 55 = 85 = 30 × 3 – 5
X = 140 + 85 × 3 – 7
= 140 + 248 = 388
6. A)   7. B)   8. D)   9. B)   10. C)
When x = 6, x > y and when x = −2, x < y
Thus, the relationship between x and y can't be established.
Hence, option E is correct.
12. E) Since both equations are of the form ax2 ± bx − c = 0, both equations have one positive and one negative root.
Hence, the relation between x and y can't be established.
When x = 21/5, x > y
When x = −3 and y = 3, x < y
For either value of x, x > y
When x = 8.5, x > y
When x = 2, x < y
Hence, the relationship between x and y cannot be established.