## 20 Sep 2016

### Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions [Answers Updated]

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

Directions (Q. 1-5): Evaluate

1). 7% of 77 + 77% of 777 + 777% of 7777
a)    58090.54
b)    60457.25
c)    68709.16
d)    71000.01
e)    None of these

2). 19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7 ÷ 2 + 14 – 3.5 – 9 × 8.5) × 4
a)    201
b)    241
c)    -241
d)    -281
e)    281

3). {(0.013)3 + 0.000000343} / {(0.013)2 – 0.000091 + 0.000049} =
a)    0.020
b)    0.002
c)    0.023
d)    0.021
e)    None of these

4). 552 + 392 – 452 – 112
a)    1600
b)    1900
c)    2100
d)    2500
e)    None of these

5). If [(156)2 ÷ 8 × 36] ÷ y = 117 × 24, what is the value of y?
a)    24
b)    39
c)    44
d)    32
e)    47

Directions (Q. 6-10): In the following questions two equations numbered I and II are given. You have to solve the equations and select appropriate option.
a)    x ≥ y
b)    x < y
c)    x ≤ y
d)    x > y
e)    x = y or the relationship cannot be established

6). I. √(2025)x + √(4900) = 0
II. (81)1/4y + 3√(343) = 0

7). I. 15x2 – 8x + 1 = 0
II. y2 + 9.68 + 5.64 = 16.95 – 1.59

8). I. (113 - 35) / 6 = x3
II. 4y3 = (500 / 4) + 5y3

9). I. 3x2 + 8x + 4 = 0
II. 4y2 – 19y + 12 = 0

10). I. (4/√x) + (7/√x) = √x
II. y2 – [(11)5/2 / √y] = 0

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

Solution:
1). 7% of 77 + 77% of 777 + 777% of 7777
= 0.07 × 77 + 0.77 × 777 + 7.77 × 7777
= 61030.97

2). 240×1 ÷ 180 = 4/3
12÷ 4 × (240 × 1 ÷ 180) = 3 × (4/3) = 4
101 + 98 ÷ 7 – 113 = 101 + 14 – 113 = 2
19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) = 19 + 4 × 6 ÷ 2
= 19 + 4 × 3 = 31
7÷2 + 14 – 3.5 – 9 × 8.5 = 3.5 + 14 – 3.5 – 76.5 = -62.5
(7÷2 + 14 – 3.5 – 9 × 8.5) × 4 = - 62.5 × 4 = -250
19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7÷2 + 14 – 3.5 – 9 × 8.5) × 4
= 31 – (-250) = 281

3). (0.013)3 + 0.000000343 = 0.00000254
(0.013)3 – 0.000091 + 0.000049 = 0.000127
Therefore,
{(0.013)3 + 0.000000343} / {(0.013)2 – 0.000091 + 0.000049} = 0.00000254 / 0.000127
= 2.54 /127 = 0.02

4). 552 – 452 = (55 + 45) × (55 - 45) = 1000
392 – 112 = (39 + 11) × (39 - 11) = 1400
Sum = 1000 + 1400 = 2400

5). (156 × 156 × 36) / 8 = 117 × y × 24
y = (156 × 156 × 36) / (117 × 8 × 24)
now, 156 = 13 × 21, 36 = 12×3, 117 = 13×9
y = (13 × 12 × 13 × 12 × 12 × 3) / (13 × 9 × 8 × 12 × 2)
= (13 × 12 × 12 × 3) / (9 × 8 × 2) à 12 × 12 = 144 = 9 × 8 × 2
So, y = 13 × 3 = 39

6). I. 45x + 70 = 0 or x = (-70/45) = (-14/9)
II. 3y + 7 = 0 or y = -(7/3)

7). I. 15x2 – 5x – 3x + 1 = 5x(3x - 1) –1 (3x - 1) = (5x - 1) (3x -1)
So, x = 0.2, 1/3
II. y2 = 0.04
y = ± 0.2
Hence, x≥y

8). I. 6x3 = 113 – 35 = 1296
x3 = 216 or x = 6
II. y3 = -125
y = -5
Hence, x>y

9). I. 3x2 + 8x + 4 = 0
3x2 + 6x + 2x +4 = 0
3x(x + 2) + 2 (x + 2) = 0
(3x + 2) (x + 2)= 0
x = -2 or -(2/3)
II. 4y2 – 19y + 12 = 0
4y2 – 16y – 3y +12 = 0
4y(y -4) – 3(y - 4) = 0
(y-4) (4y - 3) = 0
y = 4 or 3/4
Hence, x<y

10). I. (4+7) / √x = √x or x = 11
II. (y)2 + ½ = (11)5/2
y = 11