Question 1

If the price of petrol is increased by 20%, by what percentage should the consumption be decreased by the consumer, if the expenditure on petrol remains unchanged?

Accepted Answer

Required decrease in consumption = [a/(100 + a)] x 100% = [20/(100 + 20)] x 100% = (20/120) x 100% = 162/3%

Question 2

In a company , there are 724 employees. 25% employees left the job. Find the number of rest of employees.

Accepted Answer

Number of employees leaving job = 25% of 724 = (1/4) x 724 = 181 ∴ Number of rest of the employees = 724 - 181 = 543

Question 3

An agent sells goods of value of ₹ 15000. The commission which he receives at the ration of 12 1/ 2%, is

Accepted Answer

Agent receives commission of the rate = 121/2% Total value of goods sold = ₹ 15000 ∴ Agent's commission = 121/2% of 15000 = [(25/2) x 15000] /100 = ₹ 1875

Question 4

The price of certain article is ₹ 15000. But due to slump in the market, its price decreased by 8%. Find the new price of the article.

Accepted Answer

Given that, P = ₹ 15000, and Q = 8 According to the formula, Required new price = [(100 - Q)/100] x P = [(100 - 8)/100] x 15000 = (92/100) x 15000 = ₹ 13800

Question 5

A man losses 20% of his money. After sending 25 % of the remainder, he has ₹ 480 left. What is the amount of money he originally had?

Accepted Answer

Let a man have = ₹ N. Remainder money, after losses 20% of his money = N - 20% of N = N - (N x 20)/100 = (5N - N)/5 = 4N/5 Remainder money, after spending 25% of his remainder money = 4N/5 - 25% of (4N/5) = 4N/5 - (4N/5) x (25/100) = 4N/5 - N/5 = (4N - N)/5 = 3N/5 According to the question, 3N/5 = 480 ⇒ x = 480 x 5/3 = ₹ 800

Question 6

12 of 26% of 5/78 of 38% of 7/152 of 10000 =?

Accepted Answer

Ans = 12 x (26/100) x (5/78) x (38/100) x (7/152) x 10000 = 12 x (5/3) x (7/4) = 35

Question 7

In an examination, the marks obtained by Shantanu is 40% less than the marks obtained by Kamal, then marks obtained by Kamal is how much per cent more than the marks obtained by Shantanu?

Accepted Answer

Let kamal's marks be 100 ∴ Shantanu's marks = 60 Kamal has 40 marks more than Shantanu ∴ Required percentage = (40/60) x 100% = 662/3 %

Question 8

Because of scarcity of rainfall, the price of a land decrease by 12% and its production also decrease by 4%. What is the total effect on revenue?

Accepted Answer

Net effect = [- 12 - 4 + {(-12) (-4)/100}]% = (-16 + 0.48)% = -15.52%

Question 9

Mangulal a shopkeeper, marks the prices of his goods at 20% higher than the original price. After that he allows a discount of 5%. What profit or loss did he get?

Accepted Answer

Given that, a = 20% and b = 5% According to the formula Required percentage = [20 - 5 - {20 x (5/100)}% = 14%

Question 10

The prices of sugar is incerased by 25%. If a family wants to keep its expenses on sugar unaltered, then the family will have to reduce the consumption of sugar by

Accepted Answer

Let original price be ₹ 100 Then, increased price = ₹ 125 ∴ Reduction in consumption = [(125 - 100)/125 ] x 100% = (25/125) x 100 % = 20%

Question 11

If the prices of tea falls down by 6% by how much per cent must a householder increased its consumption, so as not to decreased expenditure?

Accepted Answer

Let original price be ₹ 100 Then, reduced price = ₹ 94 ∴ Increase in consumption = [(100 - 94)/94] x 100 % = (6/94) x 100 % = 618/47%

Question 12

A student has to score 40% marks to get through. if he gets 40 marks and fails by 40 marks, then find the maximum marks set for the examination.?

Accepted Answer

Let maximum marks be N. According to the question, 40N/100 = 40 + 40 ⇒ 40N/100 = 80 ∴ N = 200

Question 13

A candidate scores 20% and fails by 50 marks, while another candidate who score 40% marks, gests 30 marks more than the minimun required marks to pass the examination. Find the maximum marks for the examination.

Accepted Answer

Let the maximum marks be N. According to the question, (20N/100) + 50 = (40N/100) - 30 ⇒ 20N/100 = 80 ∴ N = (80 x 100)/20 = 400

Question 14

The population of a town is 126800. It increase by 15% in the 1st year and decrease by 20% in the 2nd year. What is the population of the town at the end of 2 yr.

Accepted Answer

Given, R = 15% and R2 = 20% Required population = P(1 + R1/100) (1 - R2/100) = 126800(1 + 15/100) (1- 20/100) = 126800 (1 + 3/20 ) (1- 1/5) = 126800 x (23/20) x (4/5) = 116656

Question 15

The population of a town is 705600. If it increases at the rate of 5% per annum, then what will be its population 2yr hence?

Accepted Answer

Given that, P = 705600, R = 5% and n = 2 According to the formula, Population after n yr = p(1 + R/100)n ∴ Population after 2 yr. = 705600 x (1+ 5/100)2 = 705600 x (105/100)2 = 705600 x (21/20 x 21/20) = 777924

Question 16

The population of a town is 1058400. if it increase at the rate of 5% per annum, then find the population of the town 2 yr ago.

Accepted Answer

Given that, p = 1058400, R = 5% and n = 2 According the formula, Population n yr ago = P/(1 + R/100) n ∴ Population 2yr ago = 1058400/(1+ 5/100)2 = 1058400 x 20/21 x 20/21 = 960000

Question 17

The population of a city is 250000. It is increasing at the rate of 2% every year. The growth in the population after 2 yr is?

Accepted Answer

Population after 2 yr = P (1 + R/100)2 = 250000 (1 + 2/100)2 ⇒ 250000 x (51/50) x (51/50) = 260100 ∴ Growth = 260100 - 250000 = 10100

Question 18

In a school, 10% of boys are equal to the one - fourth of the girls. What is the ration of boys and girls in that school?

Accepted Answer

Let the number of boys = B and number of girls = G Then, 10% of B = 1/4 of G ⇒ B/10 = G/4 ⇒ B/G = 10/4 = 5/2 ⇒ B : G = 5 : 2

Question 19

Income of Suman is first increased by 7% and then it is decreased by 7%. What is the change in her income?

Accepted Answer

In such case, there is always decrease. Given that, a = (common increase or decreased) = 7% According to the formula, Decreased Percentage = a2/100 % = 72/100 % = 0.49%

Question 20

48% of the 1st number is 60% of the 2nd number. What is the ratio of the 1st number to the 2 nd number?

Accepted Answer

Let 1st number be M and 2nd number be M. According to the question, 48% of M = 60% N ⇒ (M x 48)/100 = (N x 60)/100 ⇒ M/N = (60/100) x (100/48) = 5/4 ∴ M : N = 5 : 4

Percentage Questions