# Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 37

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 exams. We have included Some questions that are repeatedly asked in exams.

1. A man travel 400 km by train at 80 km/hr, 600 km by ship at 30 km/hr, 500 km by aeroplane at 250 km/hr and 150 km by car at 50 km/hr. What is the average speed for the entire distance?
A. 30 km/hr
B. 45 km/hr
C. 55 km/hr
D. 60 km/hr
C. 55 km/hr
Explanation:
Total time = 400/80 + 600/30 + 500/250 + 150/50 = 30 hrs
Total distance = 400 + 600 + 500 + 150 = 1650 km
So average speed = 1650/30 = 55

2. There are 48 students in a class. Rs 24 are distributed among them so that each boy gets 60 p and each girl gets 40 p. Find the number of boys and girls in the class?
A. 16 girls, 32 boys
B. 24 girls, 24 boys
C. 22 girls, 26 boys
D. 32 girls, 16 boys
B. 24 girls, 24 boys
Explanation:
Rs 24 distributed among 48 students, so average money each student gets = 24/48 = 1/2 Re or 50 p.
By method of allegation:
Girl                                    Boy
40                                     60
.                        50
(60-50)=10                  (50-40)=10
Ratio is 10 : 10 or 1: 1
So there are equal number of boys and girls.

3. An alloy contains copper, zinc and nickel in the ratio of 5 : 3 : 2. The quantity of nickel (in kg) that must be added to 100 kg of this alloy to have the new ratio 5 : 3 : 3?
A. 8
B. 10
C. 12
D. 15
B. 10
Explanation:
Copper=(5/10)*100=50
Zinc=(3/10)*100=30
Nickel=(2/10)*100=20

To make ratio = 5 : 3 : 3 we have to add 10kg nickel
so as to make total nickel=20+10=30
and now the ratio = 50 : 30 : 30=5 : 3 : 3

4. In a rectangle if length increases by 20% and breadth decreases by 25%, then what is the percentage change in the perimeter of the rectangle?
A. 20%
B. 45%
C. -5%
D. Cannot be determined
D. Cannot be determined

Explanation:
Perimeter = 2(l+b)
New peri = 2(1.2l + 0.75b)
Change in perimeter = 2 [(1.2l – l) – (0.75b – b)]
= 2[0.2l – 0.25b]
As the values of l and b are not known, change cannot be determined.

5. A boat takes 10 min to reach to a place upstream. It can come back to the same place in 5 min down the stream. If the speed of the stream is 2 m/sec, then what is the speed of the boat?
A. 6 m/s
B. 4 m/s
C. 5 m/s
D. 8 m/s
A. 6 m/s
Explanation:
Convert time to seconds.
Use the formula:
B = [tu + td] / [tu – td] * R
B = [600+300] / [600-300] * 2
B = 6 m/sec

6. Two die are tossed simultaneously. Find the probability that the multiplication of the two numbers on the die is a prime number.
A. 1/6
B. 1/2
C. 13/24
D. 2/5
A. 1/6
Explanation:
Maximum product can be 6*6 = 36
Prime numbers = 2, 3, 5, 7, …
But after 5, the pairs for multiplication are not available.
So the required 6 pairs are – {(1,2)(2,1)(1,3)(3,1)(1,5)(5,1)}
Prob = 6/36 = 1/6

7. 12 men can complete a work in 9 days. After 6 men worked for 6 days, 6 more men joined. How many days will be required to complete the remaining work?
A. 5
B. 6
C. 7
D. 4
B. 6
12 m = 9d
So 1 m = 12*9 days
So 6 m = 12*9/6 = 18 days.
They worked for 6 days. So the work completed in first 6 days is 1/18 * 6 = 1/3
Remaining work = 1 – 1/3 = 2/3
Now 2 more men employed, total 12 m are there. They can complete one work in 9 days.
So they will complete 2/3 work in 9 * 2/3 = 6 days.

8. Ruchika buys 2 lots of mangoes. The first lot is at Rs 30 per dozen and the second lot is at Rs 50 per dozen. She sold all at Rs 40 per dozen. Find her profit or loss percent if she spent same amount of money on each of the lots.
A. loss 6 2/3%
B. profit 6 2/3%
C. profit 5%
D. loss 5%[
B. 6 2/3%
Explanation:
Let she spends Rs 150 on each lot (LCM of 50 and 30)
So first type = 150/30 = 5 dozen
Second type = 150/50 = 3 dozen
In total she spent 150+150 = 300 Rs for 8 dozen
She sells all at Rs 40 per dozen. She earns 8*40 = Rs 320
Profit% = (320-300)/300 * 100 = 6 2/3%

9. A man covered a certain distance at some speed. Had he moved 3 km/hr faster, he would have taken 40 min less. If he had moved 2 km/hr slower, he would have taken 40 min more. The distance in km is
A. 35
B. 36
C. 37
D. 40
D. 40
Explanation:
For these types of ques, when u r not given actual speed, but given faster and slower speeds, use:
Let faster speed = s1 = 3, slower speed = s2 = 2
Average speed of journey = 2* (s1*s2)/s1-s2 = 2* (3*2) /(3-2) = 12 km/hr
Now distance = (s+s1) * (s-s2) / (s+s1) – (s-s2) * difference in times
Here difference in times is 40 mins more – 40 mins early = 80 min = 80/60 hrs
So, distance = (12+3) * (12-2) / (12+3) – (12-2) * 80/60 = 40 km

10. 10 men and 15 women can complete a work in 6 days. It takes 100 days for one man to complete the same work. How many days will be required for 45 women to complete the same work?
A. 1 2/3 days
B. 5 days
C. 3 1/3 days
D. 3 days
B. 5 days
Explanation:
1 man takes 100 days, so 10 men take 100/10 = 10 days.
Let 15 women takes x days
So, 1/10 + 1/x = 1/6
Solve, x = 15
15 women take 15 days, so 45 women takes = 15*15/45 = 5 days

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