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# Percentage - MCQ 5

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Percentage - MCQ 5

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This mock test of Percentage - MCQ 5 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Percentage - MCQ 5 (mcq) to study with solutions a complete question bank. The solved questions answers in this Percentage - MCQ 5 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Percentage - MCQ 5 exercise for a better result in the exam. You can find other Percentage - MCQ 5 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### If x is 20% more than y, then by what percent y is smaller than x.

Solution:

Answer – a) 50/3 % Solution: x = 120y/100 or x = 6y/5 y = 5x/6. Percentage by which y is smaller Than x is [(x – 5x/6)/x]*100 = 50/3 %

QUESTION: 2

### In an alloy, there is 15% of brass, to get 90 kg of brass, how much alloy is needed ?

Solution:

Answer – c) 600 kg Solution: Let X kg of alloy is needed. So, 15/100 of X = 90. So X =600 kg

QUESTION: 3

### 25 litre of solution contains alcohol and water in the ratio 2:3. How much alcohol must be added to the solution to make a solution containing 60% of alcohol ?

Solution:

Answer – c) 12.5 ltr Solution: Initially alcohol 2/5 * 25 = 10 ltr and water is 15 ltr.
To make a solution of 60% alcohol (10+x)/25+x = 60/100. X = 12.5

QUESTION: 4

In an examination if a person get 20% of the marks then it is fail by 30 marks.Another person who gets 30% marks gets 30 marks more than the passing marks. Find out the total marks and the passing marks.

Solution:

Answer – a) 600 and 150 Solution: 20% of X = P – 30 (X = Maximum marks and P = passing marks) 30% of X = P + 30. Solve for X and P.

QUESTION: 5

A company has produced 900 pieces of transistor out of which 15% are defective and out of remaining 20 % were not sold. Find out the number of sold transistor.

Solution:

Answer – c) 612 Solution: No of transistor sold = 900*(85/100)*(80/100) = 612

QUESTION: 6

In an election the votes between the winner and loser candidate are in the ratio 5:1. If total number of eligible voters are 1000, out of which 12% did not cast their vote and among the remaining vote 10% declared invalid. What is the number of votes the winner candidate get ?

Solution:

Answer – d) 660 Solution: Ratio b/w winner and loser 5:1 Total no of votes casted actually = 1000*(88/100)*(90/100) = 792 5x + x = 792, X =132 Votes of winner candidate = 5*132 = 660

QUESTION: 7

If the price of a commodity is increased by 30%, by how much % a consumer must reduce his consumption so to keep the expenditure same ?

Solution:

Answer – c) 300/13 Solution: If commodity price is increased then reduction in consumption will be [(increase in price)/100 + increase in price]*100. (30/130)*100 = 300/13%

QUESTION: 8

1000 sweets need to be distributed equally among the school students in such a way that each student gets sweet equal to 10% of total students. Then the number of sweets, each student gets.

Solution:

Answer – a) 10 Solution: No of students = T. Each student gets 10% of T.
So , T students get T^2/10 sweets.
T^2/10 = 1000. So T = 100. So each student gets 10 sweets

QUESTION: 9

Rishi salary is first increased by 20% and then decreased by 25%. How much percent the salary increased/decreased ?

Solution:

Answer – b) 10% Solution: Take 100 as rishi salary.
Increased by 20% percent = 120.
Then decreased by 25%, i.e = (75/100)*120 = 90.
So percentage decrease is 10%.

QUESTION: 10

The income of a person is 10000 and its expenditure is 6000 and thus saves 4000rs. In the next year his income is increased by 10% and its expenditure increased by 20%. Now his saving is what percent lower than the previous saving.

Solution:

Answer – a) 5% Solution: Initially I-E = S (I = Income, E = expenditure, S = saving) 10000-6000 = 4000(saving) Now, I = 11000 and E = 7200. So saving = I – E = 3800. [(4000-3800)/4000]*100 = 5%

QUESTION: 11

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?

Solution:

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%

QUESTION: 12

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?

Solution:

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

QUESTION: 13

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?

Solution:

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

QUESTION: 14

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.

Solution:

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

QUESTION: 15

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?

Solution:

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

QUESTION: 16

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?

Solution:

A) 1,77,366
Explanation: New population = 1,60,000 [(1 + (3/100)] [(1 + (2.5/100)] [(1 + (5/100)]

QUESTION: 17

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?

Solution:

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

QUESTION: 18

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?

Solution:

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)

QUESTION: 19

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?

Solution:

D) 3591
Explanation: Number of smokers after 3 months will be = 4000 * (1 + (5/100)) (1 – (5/100)) (1 – (10/100))
= 3591

QUESTION: 20

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?

Solution:

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