Question 1

72% of the students of a certain class took Biology and 44% took Mathematics. If each student took Biology or Mathematics and 40 took both. The total number of students in the class was?

Accepted Answer

Let the total number of students be 100 Then, n(A ∩ B) = n(A) + n(B) - n(A ∪ B) = (72 + 44 - 100)% = 16% Now,16% of N = 40 ⇒ (16 x N) / 100 = 40 ∴ N = (100 x 40) / 16 = 250

Question 2

If the side of a square is increased by 30%, its area is increased by?

Accepted Answer

Let side of square = 100 cm Area = (100 x 100)cm2 = 10000 cm2 New area = (130 x 130)cm2 = 16900 cm2 Increase in area = (6900 / 10000) x 100% = 69%

Question 3

The radius of circle is increased by 1%. What is the increased percent in its area?

Accepted Answer

Let radius of circle = 100 m New radius = 101 m original area = [π x (100)2 ]m2 New area = [π x (101)2 ]m2 Increase% = [π x {(101)2 - (100)2} / { π x (100)2} ] x 100 % = 201 / 100% = 2.01%

Question 4

The length and breadth of square are increased by 30% and 20% respectively. The area of the rectangle so formed exceeds the area of the square by?

Accepted Answer

Let length of square = 100 m and breadth of square = 100 m Area = (100 x 100) m2 = 10000 m2 New length = 130 m and breadth = 120 m New area of rectangle = (130 x 120) m2 = 15600 m2 Increase % = (5600 / 10000) x 100 % = 56%

Question 5

The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area?

Accepted Answer

Let length of a rectangle = 100 m, Breadth = 100 m New length = 160 m, New breadth = B meters Then, 160 x B = 100 x 100 ⇒ B = (100 x 100) / 160 = 125/2 Decrease in breadth = (100 - 125/2)% = 371/2%

Question 6

The price of coffee is increased by 15% and a house wife reduced her consumption of coffee by 15% and hence her expenditure on coffee?

Accepted Answer

Consumption Decreases = (common %)2/100 = 225/100 = 2.25%.

Question 7

In vocational course in a college 15% seats increase annually. If there were 800 students in 1992, how many student will be there in 1994?

Accepted Answer

Required number = 800 x (1 + 15/100)2 = 800 x (23/20) x (23/20) = 1058

Question 8

A reduction of 21% in the price of wheat enables a person to buy 10.5 kg more for Rs. 100. What is the reduced price per kg?

Accepted Answer

Let original rate = Rs. P per kg New rate = 79% of Rs. P per kg = Rs. (79 x P)/100) per kg ∵ {(100/79P)/100} - {100/P} = 10.5 ⇒ 10000/79P - 100/P = 10.5 ⇒ 10000 - 7900 = 10.5 x 79P ∴ P = 2100/(10.5 x 79) ∴ P = 2100/(10.5 x 79) Reduced rate = Rs. (79/100) x 2100/(10.5 x 79) per kg = Rs. 2 per kg.

Question 9

For a sphere of radius 10 cms. the numerical value of the surface area is how many per cent of the numerical value of its volume?

Accepted Answer

Surface area sphere = 4π x (10)2 volume sphere=4π/3 x (10)3 = (4000π/3) cm3 ∴ Required percentage = {(400π x 3) / (4000π)} x 100 % = 30%

Question 10

5% income of A is equal to 15% income of B and 10% income of B is equal to 20% income of C. If income of C is Rs.2000, then total income of A, B and C is?

Accepted Answer

∵ 5% of A=15% of B and 10% of B = 20% of C ∴ A/20 = 3B/20 and B/10 = C/5 ⇒ B = 2C ⇒ A/20 = (3/20) x 2C = 3C/10 =(3/10) x 2000 = 600 ∴ A = (600 x 20) = 12000, B = (2 x 2000) = 4000 ∴ A + B + C = (12000 + 4000 + 2000) = 18000

Question 11

A city has a population of 300000 out of which 180000 are males. 50% of the population is literate. If 70% of the males are literate, the number of literate females is

Accepted Answer

Total population = 300000 Total number of males = 180000 Total Literates = 50% of total population = 150000 Number of literate males = 70% of males = 126000 ∴ Number of literate females = 150000 - 126000 = 24000

Question 12

If the price of mangoes is decreased by 40%, one could get 32 more mangoes for ₹ 120. Find the original price of each mango.

Accepted Answer

Let the original price be ₹ N per kg. New price per kg = 60% of N = 3N/5 ∴ 120 / (3N/5) - 120/N = 32 ⇒ 600/3N - 120/N = 32 ⇒ 200/N - 120/N = 32 ⇒ 80/N = 32 ∴ N = 80/32 = 2.5 So, the original price is ₹ 2.50

Question 13

A student multiplied a number by 2/5 instead of 5/2. What is the percentage error in the calculation?

Accepted Answer

Let the number be N. Then, error = 5N/2 - 2N/5 = 21N/10 ∴ Error percentage = (21N/10) x (2/5N) x 100 % = 84%

Question 14

Anirudh secures 85%, 70%, 65% and 60% marks in test papers with 100, 120, 140 and 160, respectively as maximum marks. The percentage of his aggregate is?

Accepted Answer

Marks secured by Anirudh = [(85/100) x 100] + [(70/100) x 120] + [(65/100) x 140] + [(60/100) x 160] = (85 + 84 + 91 + 96) = 356 ∴ Required percentage = [356/(100 + 120 + 140 + 160)] x 100 % = (356/520) x 100% = 686/13%

Question 15

In an examination, 950 boys and 250 girls appeared. 90 % of the boys and 60 % of the girl passed the examination. The percentage of candidates failed is?

Accepted Answer

Number of failures = 10% of 950 + 40% of 250 = (10/100) x 950 + (40/100) x 250 = (95 + 100) = 195 Fail percentage = [195/(950 + 250)] x 100 = (195/1200) x 100 = 16.25%

Question 16

The population of a town was 150000 three years ago. If it is increased by 2%, 2.5% and 4%, respectively in last 3 years, then the present population is?

Accepted Answer

Percentage population = 150000 x (1 + 2/100) x {1 + (5/2)/100} x (1 + 4/100) = 150000 x (102/100) x (41/40) x (104/100) = 163098

Question 17

In a survey, it was found that 80% of those surveyed owned a car while 60% of those surveyed owned a mobile phone. If 55% owned both a car and a mobile phone, what percent of those surveyed owned a car or a mobile phone or both?

Accepted Answer

Percentage of car owners = 80% Percentage of mobile phone owners = 60% Percentage of people having both car and mobile phone = 55% Percentage of people having only car = 80 - 55 = 25% Percentage of people having only mobile phone = 60 - 55 = 5% Percentage of people having a car or a mobile phone or both = 55% + 25% + 5% = 85%

Question 18

In an examination, every candidate took Physics or Mathematics or both. 65.8% took Physics and 59.2 % took Mathematics. The total number of candidates was 2000. How many candidates took both Physics and Mathematics?

Accepted Answer

Let x% candidates take both the subjects. Percentage of candidates who opted Physics = 65.8% and percentage of candidates who opted Mathematics = 59.2% ∴ x = (65.8 + 59.2 - 100)% = (125 - 100)% = 25% Also, total number of candidates = 2000 ∴ Number of candidates who opted both the subjects = 25 x 2000/100 = 500

Question 19

Two numbers X and Y are respectively 20% and 28% less then a third number Z. By what percentage is the number Y less than that of the number X?

Accepted Answer

Here, Z = 100, X = 80 and Y = 72 Now according to the formula, Percentage change = (New amount - Original amount)/Original amount = [(80 - 72)/80] x 100% = (8/80) x 100% = 10% ∴ Y is less by X = 10%

Question 20

The population of a city increases at the rate of 5% per annum. There is additional annual increases of 5% due to influx of job seekers. The percent increase in population after 3 yr, is?

Accepted Answer

Total rate of increase in population per annum = 10% Initially, let the population be 1000 population after 3 yr = 1000(1 + 10/100)3 = (1000 x 11/10 x 11/10 x 11/10) = 1331 ∴ Increase in percentage = [(1331 - 1000)/1000] x 100% = (331/1000) x 100% = 33.1%

Percentage Questions