# Mission IBPS PO 2016 – Practice Quantitative Aptitude Questions

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): Each of these questions consists of a questions followed by information in three statements.
You have to study the question and the statements and decide that information in which of the statement(s) is/are necessary to answer the question.

1).What is the capacity of the cylindrical tank?
I. Radius of the base is half of its height.
II. Area of the base is 616 square meters.
III. Height of the cylinder is 28 meters.
a)   Only I and II
b)   Only II and III
c)   Only I and III
d)   All I, II and III
e)   Any two of the three
2).What is the speed of the train?
I. The train crosses a signal pole in 18 seconds
II. The train crosses a platform of equal length in 36 seconds.
III. Length of the train is 300 meters.
a)   I and III only
b)   II and III only
c)   I and II only
d)   III and either I or II only
e)   Any two of three

3).What is the staff strength of company X?
I. Male and female employees are in the ratio 2:3 respectively.
II. Of the officer employees 80% are male.
III. Total number of officer is 132.
a)   I and III only
b)   II and either III or I only
c)   All I, II and III
d)   Any two of three
e)   Cannot be determined

4).What is the two digit number?
I. Number obtained by interchanging the digits is more than the original number by 9.
II. Sum of the digits is 7.
III. Difference between the digits is 1.
a)   I and III only
b)   I and II only
c)   II and III only
d)    All I, II and III
e)   Cannot be determined

5).How many articles were sold?
I. Total profit earned was Rs. 1,596/-
II. Cost – price per article was Rs. 632/-
III. Selling price per articles was Rs. 765/-
a)   II and III only
b)   I and II only
c)   All I, II and III
d)   Any two of the three
e)   Question cannot be answered even with the information in all three statements

6).Simplify: ( √2 + √4 – √6)-1
a)   ( 1 + √2 + √3) / 4
b)   (1 – √2 + √3) / 4
c)   (1 + √2 – √3) / 4
d)   (1 – √2 – √3) / 8
e)   None of these

7).(15 + 8 – 15 ÷ (40 × 1 ÷ 8)) × 12 ÷ (8 – 4) + (6 ÷ 2 + 7 ×2 – 9)×1 ÷2
a)   48
b)   56
c)   121
d)   64
e)   None of these

8).(12 × 8 – 9 ÷ (45 × 1 ÷ 405)) × 6 ÷ (19 – 14) + (1 ÷ 2 + 7× 2.5 – 9 + 1.5) × 2
a)   17
b)   25
c)   31
d)   39
e)   None of these

9).(14 + 72 ÷ 4 × (300 × 1 ÷ 270)) × 6 ÷( 14 – 11) + ( 1 ÷ 2 + 12 × 3.5 – 9) × 4
a)   108
b)   128
c)   134
d)   187
e)   202

10).Evaluate: (0.943 × 0.943 – 0.943 × 0.057 + 0.057 × 0.057) / (0.943 × 0.943 × 0.943 + 0.057 × 0.057 × 0.057)
a)   0.886
b)   1. 1286
c)   0.32
d)   0.781
e)   None of these
1)e 2)d 3)e 4)b 5)c 6)a 7)d 8)d  9)e 10)e

Solution:
1).The capacity of the cylindrical tank is nothing but its volume
Volume of the tank = V =pr2h
Consider statement I and II:
Area of the base = 616 sq.m
pr2= 616
Hence, radius value can be calculated
From statement I, we can calculate the height of the cylinder
Hence, from statement I and II, volume of the tank can be calculated
Consider statement I and III:
Height of the cylinder is given and from statement I, radius of the cylinder can be calculated
Hence, from statements I and III, volume of the tank can be calculated
Consider statements II and III:
Radius of the cylinder can be calculated from statement II and height is given in statement III.
Hence, from these statements also, volume of the tank can be calculated
Hence, any two of the three statements is sufficient to find the capacity of the cylinder

2).Consider statement I and III:
Length of the train = 300 m
Speed of the train = 300/18 = 16.67 m/s
Consider statement II and III:
Length of train = length of platform = 300 m
Speed = 600/36 = 16.67 m/s
Consider statement I and II:
Since the length of the train is not known, we cannot determine the speed of the train
Hence, statement III is necessary and then either statement I or II is sufficient to determine the speed of the train

3).From the statements II and III:
Number of male officer employees = 0.8 ×132 = 105 and female officer employees = 27
But there is no information about non-officer employees number or ratio corresponding to officer employees. Hence, using all the statements also, we cannot determine the staff strength

4).Let the number be (10x +y)
From statement I:
(10y +x) – (10x + y) = 9
9y – 9x = 9
Y –x = 1
From statement II:
X + Y = 7
Hence, solving both, x =3 and y = 4
Hence, the two digit number is 34

5).Let the number of articles sold be N
Combining all the statements,
765N – 632N = 1596
133N = 1596
N = 12

6).( √2 + √4 – √6)-1 =( √2 + √4 + √6) / [(√2 + √4)2– (√6)2 ]
= (√2 + √4 + √6) / (2 + 4 + 2√8 – 6)à (√2 + √4 + √6) / 2√8
= (√2 + √4 + √6) / 4√2à (2 + √8 + √12) / 8
= 2(1√2 + √3) / 8à(1 + √2 + √3) / 4

7).40×1÷8 = 40/8 = 5
15 ÷ (40×1÷8) = 15/5 = 3
(15 + 8 – 15 ÷ (40×1÷8)) = 15 + 8 – 3 = 20
12÷(8-4) = 12/4 = 3
(15 + 8 – 15 ÷(40×1÷8)) × 12÷(8-4) = 20×3 = 60
(6 ÷ 2 + 7 × 2 – 9)× 1÷2 = (3 + 14 – 9) / 2 = 4
(15 + 8 – 15 ÷(40×1÷8)) × 12÷(8-4) + (6 ÷ 2 + 7 × 2 – 9)× 1÷2 = 60 + 4 = 64

8).12×8 = 96
9÷(45 ×1÷ 405) = 9 ÷ (1/9) = 9×9 = 81
(12 × 8 – 9 ÷(45×1÷405)) = 96 – 81 = 15
(12 × 8 – 9 ÷(45×1÷405)) × 6 ÷ (19-14) = 15 × 6 / 5 = 18
(1 ÷ 2 + 7 × 2.5 – 9 + 1.5) = 0.5 + 17.5 -7.5= 10.5
(1 ÷ 2 + 7 × 2.5 – 9 + 1.5) × 2 = 21
(12 × 8 – 9 ÷(45×1÷405)) × 6 ÷ (19-14) + (1 ÷ 2 + 7 × 2.5 – 9 + 1.5) × 2 = 18+21 = 39

9).300 × 1 ÷ 270 = 10/9
72 ÷ 4 = 18
72 ÷ 4 × (300 × 1 ÷ 270) = 18 × 10/9 = 20
(14 + 72 ÷ 4 × (300 × 1 ÷ 270)) = 14 + 20 = 34
6 ÷ (14 – 11) = 6/3 = 2
(1÷2 + 12 × 3.5 – 9)× 4 = (0.5 + 42 – 9)×4 = 134
(14 + 72 ÷ 4 × (300 × 1 ÷ 270)) × 6 ÷ (14 – 11) + (1÷2 + 12 × 3.5 – 9)× 4
= 34 × 2 + 134 = 68 + 134 = 202