## 24 Oct 2016

### IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions

IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions Set-22:
Dear Readers, Important Practice Aptitude Questions for IBPS Clerk and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

1). 24   14   26   ?   28   16   30
a)    13
b)    17
c)    15
d)    11
e)    20

2). 5   6   16   57   ?   1245   7506
a)    240
b)    248
c)    244
d)    250
e)    260

3). 2   9   ?   105   436   2195   13182
a)    32
b)    45
c)    40
d)    30
e)    25

4).  11   6   7   12   26   67.5   ?
a)    205.5
b)    195.5
c)    216.5
d)    200.5
e)    222.5

5). 3   ?   113   673   3361   13441   40321
a)    13
b)    14
c)    15
d)    16
e)    17

Directions (Q. 6-10): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer
a)    if p > q
b)    if p ≥ q
c)    if p < q
d)    if p ≤ q
e)    if p = q or no relation can be established between p and q.

6).
I. p^2 - 26p + 168 = 0
II. q^2 - 25q + 156 = 0

7).
I. 6p - 5q = -47
II. 5p + 3q = 11

8).
I. 2.3p - 20.01 = 0
II. 2.9q - p = 0

9).
I. p = 1764
II. q^2 = 1764

10).
I. p^2 - 13p + 42 = 0
II. q^2 + q - 42 = 0

1).c)  2).b)  3).d)  4).a)  5).e)  6).e)  7).c)  8).a)  9).b) 10). b)

SOLUTION:
1). The series consists of two series 1 and 2:
Series 1 : 24   26   28   30
+2   +2   +2
Series 2 : 14   15   16
+1    +1

2). The series is
5 × 1 + 12 = 6; 6 × 2 + 22 = 16; 16 × 3 + 32 = 57; 57
× 4 + 42 = 244; 244 × 5 + 52 = 1245;
1245 × 6 + 62 = 7506.

3). The series is
(2 + 7) × 1 = 9; (9 + 6) × 2 = 30;
(30 + 5) × 3 = 105; (105 + 4) × 4 = 436;
(436 + 3) × 5 = 2195; (2195 + 2) × 6 = 13182

4). The series is:
11 × 0.5 + 0.5 = 6; 6 × 1 + 1 = 7; 7 × 1.5 + 1.5 = 12;
12 × 2 + 2 = 26; 26 × 2.5 + 2.5 = 67.5; 67.5 × 3 +
3 = 205.5.

5). The series is:
3 × 8 – 7 = 17; 17 × 7 – 6 = 113; 113 × 6 – 5 = 673;
673 × 5 – 4 = 3361;
3361 × 4 – 3 = 13441; 13441 × 3 – 2 = 40321

6).
I. p^2 - 26p + 168 = 0
P^2 - 12p - 14p + 168 = 0
p (p - 12) - 14(p - 12) = 0
(p - 12) (p - 14) = 0
p = 12, 14
II. q^2 - 25q + 156 = 0
q^2 - 13q - 12q + 156 = 0
q(q - 13) - 12(q - 13) = 0
(q - 12) (q - 13) = 0
q = 12, 13
Hence, no relation can be established between p and q

7).
Equation (I) × 3 = 18p - 15q = -141
Equation (II) × 5 = 25p + 15q = 55
43p = -86
P = - 86/43 = 2
5p + 3q = 11
3q = 11 - 5p
3q = 11 + 10
3q = 21
q = 7
p < q

8).
I. 2.3p - 20.01 = 0
P = 20.01 / 2.3 = 8.7
II. 2.9q - p = 0
or, p = 2.9q
q = 8.7 /2.9 = 8.7
p > q

9).
I. p = 1764
p = 42
II. q^2 = 1764
q = + 42
p ≥ q

10).
I. p^2 - l3q + 42 = 0
p^2 - 6p - 7p + 42 = 0
p(p - 6) - 7(p - 6) = 0
(p - 6) (p - 7) = 0
p = 6, 7
II. q^2 + q - 42 = 0
q^2 + 7q - 6p - 42 = 0
q(q + 7) - 6(q + 7) = 0
(q - 6)(q + 7) = 0
q = 6, - 7
p ≥ q