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?**

A) 25%

B) 10%

C) 20%

D) 60%

E) 80%**C) 20%**

Explanation:

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 kg**D) 16 kg**

Explanation:

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 kg**E) 50 kg**

Explanation:

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 these**E) None of these**

Explanation:

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)%**

Explanation:

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?**

A) 1,77,366

B) 1,66,733

C) 1,76,736

D) 1,80,766

E) 1,69,766**A) 1,77,366**

Explanation:

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?**

A) 1925

B) 1175

C) 1275

D) 1100

E) 1800**B) 1175**

Explanation:

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?**

A) 2/5

B) 1/4

C) 2/3

D) 1/2

E) 1/3**C) 2/3**

Explanation:

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?**

A) 3457

B) 3491

C) 3578

D) 3591

E) 3500**D) 3591**

Explanation:

Number of smokers after 3 months will be = 4000 * (1 + (5/100)) (1 – (5/100)) (1 – (10/100))

= 3591**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?**

A) 4500

B) 2000

C) 3000

D) Cannot be determined

E) None of these**B) 2000**

Explanation:

Let number of girls = x, then no of boys = (5000-x). then

10% of (1000-x) + 15% of x = (5600-5000)

Solve, x = 2000

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