Hello Aspirants. Welcome to Online Quantitative Aptitude Section Section with explanation in AffairsCloud.com. Here we are creating question samples from Percentage with explanation, which is common for competitive exams. We have included some questions that are repeatedly asked in bank exams !!!
- 40% of the students like Mathematics, 50% like English and 10% like both Mathematics and English. What % of the students like neither English nor Mathematics?
E) 80%C) 20%
n(M or E) = n(M) + n(E) – n(M and E)
n(M or E) = 40+50-10 = 80
so % of the students who like neither English nor Mathematics = 100 – 80 = 20%
- A watermelon weighing 20 kg contains 96% of water by weight. It is put in sun for some time and some water evaporates so that now it contains only 95% of water by weight. The new weight of watermelon would be?
A) 17 kg
B) 15 kg
C) 18.5 kg
D) 16 kg
E) 18 kgD) 16 kg
Let new weight be x kg
Since the pulp is not being evaporated, the quantity of pulp should remain same in both cases. This gives
(100-96)% of 20 = (100-95)% of x
Solve, x = 16 kg
- If the price of wheat is reduced by 2%. How many kilograms of wheat a person can buy with the same money which was earlier sufficient to buy 49 kg of wheat?
A) 58 kg
B) 60 kg
C) 52 kg
D) 55 kg
E) 50 kgE) 50 kg
Let the original price = 100 Rs per kg
Then money required to buy 49 kg = 49*100 = Rs 4900
New price per kg is (100-98)% of Rs 100 = 98
So quantity of wheat bought in 4900 Rs is 4900/98 = 50 kg
- Monthly salary of A is 30% more than B’s monthly salary and B’s monthly salary is 20% less than C’s. If the difference between the monthly salaries of A and C is Rs 800, then find the annual salary of B.
A) Rs 14,500
B) Rs 16,800
C) Rs 15,000
D) Rs 16,000
E) None of theseE) None of these
Let C’s monthly salary = Rs 100, then B’s = (100-20)% of 100 = 80, and A’s monthly = (100+30)% * 80 = 104
If difference between A and C’s monthly salary is Rs 4 then B’s monthly salary is Rs 80
So if difference is Rs 800, B’s monthly salary is (80/4) * 800 = 16,000
So annual salary = 12*16,000
- Mixture 1 contains 20% of water and mixture 2 contains 35% of water. 10 parts from 1st mixture and 4 parts from 2nd mixture is taken and put in a glass. What is the percentage of water in the new mixture of glass?
A) 17 (5/7)%
B) 24 (2/7)%
C) 28 (1/5)%
D) 24 (1/7)%
E) 18 (2/7)%B) 24 (2/7)%
Water in new mixture from 1st mixture = (20/100) * 10 = 2 parts
Water in new mixture from 2nd mixture = (35/100) * 4 = 7/5 parts
Required % =[ [2+ (7/5)]/(10+4)] * 100
- 3 years ago the population of a town was 1,60,000. In the three respective years the population increased by 3%, 2.5% and 5% respectively. What is the population of town after 3 years?
E) 1,69,766A) 1,77,366
New population = 1,60,000 [(1 + (3/100)] [(1 + (2.5/100)] [(1 + (5/100)]
- There are 2500 students who appeared for an examination. Out of these, 35% students failed in 1 subject and 42% in other subject and 15% of students failed in both the subjects. How many of the students passed in either of the 2 subjects but not in both?
E) 1800B) 1175
Failed in 1st subject = (35/100) * 2500 = 875
Failed in 1st subject = (42/100) * 2500 = 1050
Failed in both = (15/100) * 2500 = 375
So failed in 1st subject only = 875 – 375 = 500
failed in 2nd subject only = 1050 – 375 = 675
passed in 1st only + passed In 2nd only = 675+500
- A bucket is filled with water such that the weight of bucket alone is 25% its weight when it is filled with water. Now some of the water is removed from the bucket and now the weight of bucket along with remaining water is 50% of the original total weight. What part of the water was removed from the bucket?
E) 1/3C) 2/3
Let original weight of bucket when it is filled with water = x
Then weight of bucket = (25/100) * x = x/4
Original weight of water = x – (x/4) = 3x/4
Now when some water removed, new weight of bucket with remaining water = (50/100) * x = x/2
So new weight of water = new weight of bucket with remaining water – weight of bucket = [(x/2) – (x/4)] = x/4
So part of water removed = [(3x/4) – (x/4)]/(3x/4)
- In a survey done by a committee, it was found that 4000 people have smoking habit. After a month this number rose by 5%. However due to continuous advices given by the committee to the people, the number reduced by 5% in the next month and further by 10% in the next month. What is the total number of smokers after 3 months?
E) 3500D) 3591
Number of smokers after 3 months will be = 4000 * (1 + (5/100)) (1 – (5/100)) (1 – (10/100))
- There are 5000 students in a school. The next year it was found that the number of boys and girls increased by 10% and 15% respectively making the total number of students in school as 5600. Find the number of girls originally in the school?
D) Cannot be determined
E) None of theseB) 2000
Let number of girls = x, then no of boys = (5000-x). then
10% of (1000-x) + 15% of x = (5600-5000)
Solve, x = 2000