# Quantitative Aptitude Questions for IBPS/RRB Exam Set – 18

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. Do see the solutions, you might get any shortcut.

1. What is the probability that when 3 cards are pulled from a pack of cards without replacement that we get 1 king, 1 queen and 1 jack?
A. 16/35139
B. 1/2179
C. 16/5525
D. 16/2179
C. 16/5525
Explanation:
Case 1: first chosen is king, second is queen and third jack:
Prob = 4/52 * 4/51 * 4/50
There will be 6 cases like this, so total prob = (4/52 * 4/51 * 4/50) * 6

2. In an examination consisting of 4 subjects, the marks obtained by Ram in 3 of them are 90%, 95% and 95% resp. Each subject is of equal marks. Under the given circumstances his average percentage marks for the examination cannot be
A. 94%
B. 90%
C. 93%
D. 96%
D. 96%
Explanation:
Let 100 be max marks of each subject. Check each option now.
From option D, (90+95+95+x)/4 = 96, it gives x = 104 which is not possible because max marks is 100.

3. The radius of a sphere is 14 cm. The cost of painting the surface of sphere is Rs 25 per square cm. If the radius of sphere is increased by 10%, then the cost of painting is increased by 20%. What is the percentage increase in the total cost of painting per square cm?
A. 45.2%
B. 54.27%
C. 20.3%
D. 18.2%
A. 45.2%
Explanation:
Radius increases by 10%, so surface area changes by 10+10+((10*10)/100) = 21%
Total cost of painting = Rate * Surface area of sphere
So change in cost of painting = 21+20+((21*20)/100) = 45.2%

4. In an election there were only two candidates A and B, B got 50% of the votes that A got. Had A got 200 votes less, there would have been a tie. How many people casted their votes in all?
A. 1000
B. 1500
C. 1200
D. 1800
C. 1200
Explanation:
Let total votes casted = x
then 50/100 * A + A = x
when A got 200 votes less, means these 200 votes goes to B, there is tie, so
A – 200 = 50/100 * A + 200
A = 800
x = 1200

5. A 3 : 2 milk and water solution is mixed with another 4 : 1 milk and water solution. If the volumes are 400 ml and 1,000 ml resp., then what is the ratio of milk to water in the resultant solution?
A. 9 : 5
B. 8 : 21
C. 5 : 26
D. 26 : 9
D. 26 : 9
Explanation:
In 400 lit, milk = 3/8 * 400 = 240 lit, water = 160 lit
In 1000 lit, milk = 4/5 * 400 = 800 lit, water = 200 lit
New ratio = (240+800)/(160+200)

6. In 40 litres mixture of milk and water, the ratio of milk to water is 7 : 1. In order to make the ratio of milk and water 3 : 1, the quantity of water should be added to the mixture will be
A. 6 3/4 litres
B. 6 2/3 litres
C. 6 litres
D. 6 1/3 litres
B. 6 2/3 litres
Explanation:
In 40 lit, milk = 7/8 * 40 = 35 lit, water = 5 lit
Let x lit of water to be added, so
35/(5+x) = 3/1

7. A man swims downstream 30 km and upstream 18 km, taking 3 hrs each time. What is the speed of current?
A. 6 km/hr
B. 4 km/hr
C. 2 km/hr
D. 10 km/hr
C. 2 km/hr
Explanation:
Downstream speed = 30/3 = 10, Upstream speed = 18/3 = 3
Speed of stream = (10 – 6)/2

8. A man can row 7 km/hr in still water. In a stream which is flowing at 3 km/hr, it takes him 7 hrs to row to a place and come back. How far is the place?
A. 25 km
B. 20 km
C. 10 hm
D. 15 km
B. 20 km
Explanation:
Distance = T (B^2 – R^2)/2*B
=7 (7^2 – 3^2)/2*7 = 20 km

9. If a sum of money at a certain rate of interest doubles in 5 years and at a different rate of simple interest becomes three times in 12 years, then the better rate of interest is in
A. case-1
B. case-2
C. equal in both cases
D. None of these
A. case-1
Explanation:
Case 1: SI = 2P – P = P, so P = P*r*5/100, so r = 20%
Case 1: SI = 3P – P = 2P, so 2P = P*r*12/100, so r = 16.66%

10. The population of a town increases 10% half-yearly. What is the population after one and half years, if the initial population is 1,000?
A. 1331
B. 1300
C. 1200
D. 1210
A. 1331
Explanation:
Solve it by the method of compounded semi-annually.

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