# Percentage

29 April 2020

##### Practice Questions For Percentage

Percentage is one of the topics from Quantitative aptitude on which fixed questions are asked.

1) A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A) 45%

B) 45(5/11)%

C) 54(6/11)%

D) 54(6/11)%

Solution:

2) Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?

A) 42, 36

B) 42, 33

C) 44, 36

D) 44, 33

Solution :

3) In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A) 2700

B) 2900

C) 3000

D) 3100

Solution :

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4) Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

A) Rs. 250

B) Rs. 150

C) Rs. 300

D) Rs. 200

Solution :

5) A student has to obtain 33% of the total marks to pass. He got 125  marks and failed by 40 marks. The maximum marks are:

A. 500

B. 600

C. 800

D. 1000

Solution:

Given that the student got 125 marks and still he failed by 40 marks
=> The minimum pass mark = 125 + 40 = 165
Given that minimum pass mark = 33% of the total mark
=> total mark =33/100 =165
=> total mark = 16500/33 = 500

6) A man spends 35% of his income on food, 25% on children's  education and 80% of the remaining on house rent. What percent of his income he is left with?

A) 6%

B) 8%

C) 10%

D) 12%

Solution:

Let the total income be x. Then, income left = (100 -80)% of [100 - (35 +
25)] % of x
= 20% of 40% of x = 20/100 * 40/100 * 100) % of x = 8 % of x.

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7) A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition.

A) 23.4%

B) 54.8%

C) 64.5%

D) 87.8%

Solution:

Total number of fruits shopkeeper bought = 600 + 400 = 1000
Number of rotten oranges = 15% of 600
= 15/100 × 600
= 9000/100
= 90
Number of rotten bananas = 8% of 400
= 8/100 × 400
= 3200/100
= 32
Therefore, total number of rotten fruits = 90 + 32 = 122
Therefore Number of fruits in good condition = 1000 - 122 = 878
Therefore Percentage of fruits in good condition = (878/1000 × 100)%
= (87800/1000)%
= 87.8%

8) A reduction of 10% in the price of wheat enables a man to buy 50g of wheat   more for a rupee. How much wheat could originally be had for a  rupee?

A) 400g

B) 500g

C) 450g

D) 350g

Solution:

Original prices per kg= Rs/(100-p)y per kg
{(1 x 10)/(100 – 10)0.05} = Rs. 20/9 /kg
It means the man can buy 1 kg or 1000 gm in Rs 20/9 or he can buy
(1000 x 9)/20 = 450g in one rupee.

9) In a certain school, 20% of students are below 8 years of age. The  number of students above 8 years of age is 2/3 of the number of  students of 8 years of age which is 48. What is the total number of  students in the school?

A) 72

B) 80

C) 120

D) 100

Solution:

Let the number of students be x.
Then, Number of students above 8 years of age = (100 - 20)% of x =
80% of x.
80% of x = 48 + 2/3 of 48
80/100x = 80
x = 100.

10) If A's height is 40% less than that of B, how much percent B's  height is more than that of A?

A) 66.66%

B) 76.66%

C) 96.66%

D) 86.66%

Solution:

Excess of B's height over A's = [(40/(100 - 40)] x 100% = 66.66%.

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