# Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 46

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. The length of a rectangle is 20% more than its breadth. What will be the ratio of the area of this rectangle to the area of a square whose side is equal to the breadth of the rectangle?
A) 5 : 6
B) 6 : 5
C) 3 : 2
D) 2 : 3
B) 6 : 5

2. Mr. Duggal invested Rs 20,000 with rate of interest at 20 % per annum. The interest was compounded half yearly for first one year and in the next year it was compounded yearly. What will be the total interest earned at the end of two years?
A) 9500
B) 5040
C) 8200
D) 9040
D) 9040
Explanation:
Amount after 1st year = 20,000 [(1+10/100)]2 = Rs 24,200
Amount after 2nd year = 24,200 [(1+20/100)]1 = Rs 29,040
So CI = 29040 – 20000 = 9040

3. If a man walks 20 km at 5 km/hr, he will be late by 40 minutes. If he walks at 8 km/h, how early from the fixed time will he reach?
A) 15 min
B) 25 min
C) 50 min
D) 70 min
D C) 50 min

Explanation:
Let difference in times of late and early timings be ‘t’ min sot t/60 hrs
When we are given 2 speeds, and difference in timings, the formula to be used is
Distance = Product of speeds/Difference of speeds * Difference in time
So here, 20 = 8*5/(8-5) * t/60
Solve, t = 90 minutes
Man is 40 minutes late, so 50 minutes early makes difference equal to 90 minutes.

4. In covering a certain distance, the speeds of A and B are in the ratio of 3 : 4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is
A) 1 hr
B) 1 ½ hr
C) 2 hrs
D) 2 ½ hrs
C) 2 hrs
Explanation:
Let time taken by A is t hrs, then by B is (t – 30/60) hrs
Speeds ratio = 3 : 4
Distance is same. So
3 * t = 4 * (t – ½)
Solve, t = 2 hrs

5. The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each students gets same number of pens and same number of pencils is
A) 91
B) 910
C) 1001
D) 1911
A) 91
Explanation:
LCM (1001, 910) = 91

6. Oranges are bought at 7 for Rs 3. At what rate per hundred must they be sold to gain 33%?
A) Rs 56
B) Rs 57
C) Rs 58
D) Rs 60
B) Rs 57
Explanation:
7 for Rs 3, so for 1 Re oranges are 7/3
So for Re 1 to gain 33%, number of oranges sold = 7/3 * 100/(100+33) = 100/57
So per hundred oranges = 57 Rs

7. By mistake, instead of dividing Rs 117 among A, B and C in the ratio ½ : 1/3 : ¼, it was divided in the ratio of 2 : 3 : 4. Who gains the most and by how much?
A. A, Rs 25
B. B, Rs 3
C. C, Rs 20
D. C, Rs 25
D. C, Rs 25
Explanation:
Ratio in which divided = 2 : 3 : 4
Actual ratio = ½ : 1/3 : ¼ or 6 : 4 : 3
Both ratios show that A and B have lost and C has gained the most.
C got = 4/(2+3+4) * 117 = 52
Actually C would have got = 3/(6+4+3) * 117 = 27
C gained 52 – 27 = 25

8. The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is
A) 8.5 cm2
B) 8.75 cm2
C) 7.75 cm2
D) 7.50 cm2
B) 8.75 cm2

9. 5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5 women and 3 men do it?
A) 4
B) 5
C) 6
D) 8
B) 5
Explanation:
5 m in 6 days, so 1 m in 5*6 days, so 3 men in 5*6/3 = 10 days
10 w in 5 days, so 1 women in 10*5 days, so 5 women in 10*5/5 = 10 days
So 3 men and 5 women in 10*10/(10+10) = 5

10. The selling price of an article is marked at 20% above the cost price. At the time of selling, he allows a discount of 8%. His profit percent is:
A. 20%
B. 10.4%
C. 8.6%
D. 12%
B. 10.4%
Explanation:
Use M.P. = (100 + p%)/(100 – d%) * C.P.
MP is 20% above CP.
120/100 C.P. = (100 + p%)/(100 – 8) * C.P.
Find p%

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