# Quantitative Aptitude Questions for IBPS Mains Exam Set – 33

Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS and RRB exams. We have included Some questions that are repeatedly asked in exams.

1. Two unbiased die are thrown simultaneously. Find the probability that the sum of the numbers on both the die is a composite number.
A. 5/12
B. 7/12
C. 11/12
D. 7/36
B. 7/12
Explanation:
Sample space is
{(1,1) (1,2)………… (1,6)
(2,1) (2,2)………….. (2,6)
.
.
(6,1) (6,2)…………… (6,6)}
So total pairs are 36.
The maximum sum can be 6+6 = 12.
Upto 12, lets take prime numbers – 2,3,5,7,11
Pairs for these – {(1,1)(1,2)(2,1)(1,4)(4,1)(2,3)(3,2)(1,6)(6,1)(2,5)(5,2)(3,4)(4,3)(5,6)(6,5)}
These are total 15
Probability for sum as prime numbers = 15/36
Other that prime numbers are called composite numbers.
So, probability for composite numbers = 1 – 15/36 = 21/36 = 7/12

2. A started a business by investing Rs 8,000. After 6 months B joined him with some money. After a year the total profit was equally shared between the two. Find the amount invested by B.
A. Rs 10,000
B. Rs 12,000
C. Rs 16,000
D. Rs 18,000
C. Rs 16,000
Explanation:
8000 is invested for 12 months, B’s amount (x) is invested for 6 months
So ratio of profits = 8000 * 12 : x * 6 = 16000 : x
Since the profits are equally divided, so 16000/x = 1/1, this gives x = 16000

3. The diameter of a cylinder is increased by 50%. To maintain the same volume, how much percent should the height be decreased?
A. 55.56%
B. 65.56%
C. 65%
D. 52.25%
A. 55.56%
Explanation:
Diameter increases by 50%, so radius also increases by 50%
Vol of cylinder = πr2h
There is multiplication of two r’s, so by successive increase formula ( x+y+(xy)/100)
x = y = 50
r2 increases by 50+50+(50*50)/100 = 125%
Now r2 increases by 125%, to maintain same volume height should decrease by 125/(125+100) * 100 = 500/9 %
* Note – When one quantity increases by p%, other decreases by p/(100+p) * 100

4. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is
A. 70 litres
B. 72 litres
C. 75 litres
D. 80 litres
D. 80 litres
Explanation:
Mixture contains 10% water.
The water which will be added will be 100% water
The new mixture contains 50% milk, this gives 50% water
So by allegation rule
Mixture           Water
10                    100
.             50
50                      40
This gives 50 : 40 = 5 : 4
Let x litres of water to be added to 100 litres mixture.
So 100/x = 5/4
x = 80

5. A vessel is full of milk. 6 litres are drawn from vessel and filled with water. This operation is performed two more times. The ratio of the quantities of milk and water in vessel is 8 : 117. How much milk did the vessel contain originally?
A. 15 litres
B. 10 litres
C. 20 litres
D. 30 litres
B. 10 litres
Explanation:
When the original quantity is x litres, and y litres of mixture is poured out and again filled with water with operation being performed n times.
The quantity left is = x [1 – y/x]n
Here the final ratio is 8 : 117
So,
8/8+117 = x [1 – 6/x]3 / x
8/125 = [1 – 6/x]3
2/5 = [1 – 6/x]
X = 10

6. A scored 73 marks in subject P, 56% in subject Q, and some marks in subject R. Maximum marks in each of the subject is 150. And the overall percentage got by A is 54%. Find A’s marks in subject R.
A. 80
B. 84
C. 83
D. 86
D. 86
Explanation:
Let total marls in 3 subjects is x
Maximum marks = 3*150 = 450
So x/450 * 100 = 54
Solve, x = 243
Now let marks in subject Q = y
So, y/150 * 100 = 56
Solve, y = 84
Marks in P = 73, Q = 84, and total marks = 243
So marks in subject R = 243 – (73+84) = 86

7. In a family of 5 members, with youngest member’s age ‘x’, the average of their ages is 19. ‘x’ years ago, the average of the family was 17.5. Find ‘x’.
A. 5
B. 6
C. 7
D. 8
A. 5
Let the present total age of 4 members excluding youngest member = y
So, (x+y)/5 = 19
x+y = 95
now ‘x’ years ago, the youngest member was not born, so total members were 4 only. And also ‘x’ years ago, the total age of these 4 members was = y – 4*x
so (y-4x)/4 = 17.5
y-4x = 70
we have 2 equations, y-4x = 70 and x+y = 95. Solve both, we get x = 5

8. A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random. What is the probability that they are not of the same color?
A. 3/44
B. 3/55
C. 52/55
D. 41/44
D. 41/44
Explanation:
3 marbles are to be drawn from total 12 balls, so total ways are 12C3
= 12*11*10 / 3*2*1 = 220
Case 1 : none is green
Ways are 5C3 = 10
Case 2 : none is yellow
Ways are 4C3 = 4
Case 3 : none is white
Ways are 3C3 = 1
So probability that all are of same color = 10+4+1 / 220 = 15/220 = 3/44
So probability that none is of same color = 1 – 3/44 = 41/44

9. A circular cylinder can hold 61.6 cm3 of water. If the height of the cylinder is 40 cm and the outer diameter is 1.6 cm, then the thickness of the material of the cylinder is
A. 0.1 cm
B. 0.2 cm
C. 0.3 cm
D. 0.4 cm
A. 0.1 cm
Explanation:
So 22/7 * r2 * 40 = 61.6
Solve, r = 0.7
Outer diameter = 1.6, so outer radius = 1.6/2 = 0.8
So thickness = 0.8 – 0.7 = 0.1

10. A man reaches at 10 AM from P to Q when his average speed is 15 km/hr and reaches at 12 noon when the average speed is 10 km/hr. What should be the average speed to reach at 9 AM?
A. 18 km/hr
B. 20 km/hr
C. 22 km/hr
D. 25 km/hr
B. 20 km/hr
Explanation:
Let distance from P to Q is ‘d’ and time taken to reach at 9 AM is ‘t’
Then with 10 AM, time becomes ‘t+1’, so d/(t+1) = 15
With 12 noon, time becomes ‘t+3’, so d/(t+3) = 10
Solve the equations, t = 3 hrs, d = 60 km
So speed to reach at 9 AM = 60/3 = 20 km/hr

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