## 29 Sep 2016

### Mission IBPS PO 2016 - Practice Quantitative Aptitude Questions with Detailed Solutions

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): study the following table carefully to answer the questions that follow:
Production of Sugar (in tonnes) of three different States over the years

1). What is the average production of sugar of State A for all the years together?
a)    6.24 tonnes
b)    6.3 tonnes
c)    7.1 tonnes
d)    6.1 tonnes
e)    None of these

2). What is the approximate percentage increase in production of sugar in state B from 2006 to 2007?
a)    12
b)    18
c)    24
d)    10
e)    21

3). What is the difference between production of sugar of all the three states together in 2008 and 2005 respectively?
a)    6.9 tonnes
b)    4.3 tonnes
c)    6.1 tonnes
d)    5.1 tonnes
e)    None of these

4). What is the average production of sugar of all the three states in 2003 and 2004 together?
a)    4.1 tonnes
b)    4.7 tonnes
c)    5.1 tonnes
d)    4.8 tonnes
e)    None of these

5). What is the ratio between total production of sugar of all the three states in year 2006 and 2007 respectively?
a)    7 : 9
b)    6 : 7
c)    8 : 7
d)    7 : 8
e)    11: 12

6). A, B, and C starts from the same place and travel in the same direction at speeds 30km/hr, 40 km/hr and 60 km/hr respectively. B starts 2 hours after A, but B and C overtakes A at the same instant. How many hours after A did C start?
a)    1
b)    2
c)    4
d)    6
e)    8

7). A bag contains Rs 510 In the form of 50 paise, 25 paise and 20 paise coins in the ratio 2 : 3 : 4. Find the total number of coins.
a)    450
b)    1200
c)    1400
d)    1800
e)    2100

8). Rs. 1500 amounts to 1653.75 in 2 years compounded annually. At the same rate, how much will Rs. 1800 amount to in 3 years?
a)    2000
b)    2040
c)    2080
d)    2120
e)    2160

9). Ram has a mixture of 2 Acids X and Y in a can in the ratio 2 : 3. He removes 6 litres of Y, and replaces it with X, and the ratio between X and Y becomes 4 : 3. What was the volume of X in the can initially?
a)    12
b)    14
c)    16
d)    18
e)    Cannot be determined

10). A spherical metal shall has an outer radius of 11 cm, and inner radius of 4 cm. if the cost of metal is Rs. 12 per cm3, what is the cost of the shell?
a)    Rs. 6310
b)    Rs. 63712
c)    Rs. 89108
d)    Rs. 47234
e)    Rs. 12444

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

Solution:
Directions (Q. 1-5):
 State Year C B A 2003 4.3 3.1 3.9 2004 4.9 3.7 4.7 2005 5.6 4.4 5.8 2006 5.8 5.1 6.6 2007 6.7 6 7.3 2008 7.4 6.2 8.3

1). Total production of sugar of state A for all the years together = 3.9 + 4.7 + 5.8 + 6.6 + 7.3 + 8.3 = 36.6 tonnes
Average = 36.6/6 = 6.1 tonnes

2). Production of sugar in state Q in 2006 = 5.1 tonnes
Production of sugar in state Q in 2007 = 6 tonnes
Percentage increase = [(6-5.1)/5.1] × 100 = 17.64%

3). Total production of sugar of all the three states together in 2008 = 7.4 + 6.2 + 8.3 = 21.9 tonnes
Total production of sugar of all the three states together in 2005 = 5.6 + 4.4+ 5.8 = 15.8 tonnes
Difference = 21.9 – 15.8 = 6.1 tonnes

4). Total production of sugar of all the three states in 2003 and 2004 together = (4.3 + 3.1 + 3.9) + (4.9 + 3.7 + 4.7)= 24.6 tonnes
Average = 24.6/6 = 4.1 tonnes

5). Total production of sugar of all the three states in year 2006 = 5.8 + 5.1 + 6.6 = 17.5 tonnes
Total production of sugar of all the three states in year 2007 = 6.7 + 6 + 7.3 = 20 tonnes
Ratio = 17.5 : 20 = 7 : 8

6). Let the time taken by A when he was overtaken by B and C be T hours.
Time taken by B = T – 2 hours
Since, distance travelled is same by the time B overtakes A
30(T)= 40(T-2)
T = 8 hours
Distance travelled by A = 30×8 = 240 km
C takes 4 hours to cover a distance of 240 km
So, he starts 8-4 = 4 hours after A.

7). Let the number of 50 paise, 25 paise and 20 paise coins be 2n, 3n and 4n respectively
2n(0.5) + 3n(0.25) + 4n(0.2) = 510
n + 0.75n + 0.8n = 510
2.55n = 510
n = 200
total coins = 2n + 3n + 4n = 9n = 1800

8). 1653.75 = 1500 (1 + R)2
441/400 = (1 + R)2
Solving this gives us R = 0.05 (5%)
Required Amount = 1800 × 1.05 × 1.05 × 1.05 = 2083.725 = 2080

9). Let the initial volumes of X, Y be 2x, 3x.
We have (2x + 6)/(3x - 6) = 4/3
6x + 18 = 12x – 24
Or x = 7
Original volume of X = 2x = 14

10). Volume of Metal = (4/3) × 22/7 × (113 - 43)
= (4/3) × (22/7) × (1331 - 64)
= 4 × 22 × 181 / 3
Cost = 12× 4 × 22 × 181/3
= 4 × 4 × 22 × 181 = Rs. 63712