Percentage - Key Formulas
Basic Percentage Formula
To calculate percentage:
To find the value when percentage is known:
Percentage Change
To calculate percentage increase or decrease:
If the result is positive, it's an increase; if negative, it's a decrease.
Successive Percentage Change
For successive percentage changes of x% and y%:
Example: 10% increase followed by 10% decrease is not 0%
Percentage Point
The difference between two percentages:
Example: Change from 10% to 15% is a 5 percentage point increase
Percentage Problems
Practice aptitude questions on percentage calculations
Question 1:
The difference between two numbers is 1550. If 8% of one number is 10% of the other number, then find the two numbers.
Let two numbers be x and y.
It is given that, 8% of x = 10% of y
Therefore,
8/100 × x = 10/100 × y
x = (10/8) × y = (5/4) × y
Difference between two numbers (x – y) = 1550
Substituting the value of x, we get
(5/4) × y – y = 1550
(5/4 - 1) × y = 1550
(1/4) × y = 1550
y = 1550 × 4 = 6200
x = (5/4) × 6200 = 7750
The two numbers whose difference is 1550 are 6200 and 7750.
Question 2:
Two numbers P and Q are such that, the sum of 2% of P and 2% of Q is two-third of the sum of 2% of P and 6% of Q. Find the ratio of P and Q.
The sum of 2% of P and 2% of Q is two-third of the sum of 2% of P and 6% of Q.
This means that,
2% of P + 2% of Q = (2/3) × (2% of P + 6% of Q)
(2/100) × P + (2/100) × Q = (2/3) × [(2/100) × P + (6/100) × Q]
(1/50) × P + (1/50) × Q = (1/75) × P + (1/25) × Q
(1/50 - 1/75) × P = (1/25 - 1/50) × Q
(1/150) × P = (1/50) × Q
P/Q = 150/50 = 3/1
Therefore, the ratio of P and Q is 3 : 1.
Question 3:
50% of a number is 18 less than two-third of that number. Find the number.
Let the number be x.
It is given that, 50% of a number is 18 less than two-third of that number. This means that,
(2/3) × x - 50% of x = 18
(2/3) × x - (50/100) × x = 18
(2/3 - 1/2) × x = 18
(4/6 - 3/6) × x = 18
(1/6) × x = 18
x = 18 × 6 = 108
The number is 108.
Question 4:
When 35 is subtracted from a number, it reduces to its 80%. Find the four-fifth of that number.
We are given, 35 when subtracted from a number, reduces to its 80%.
Therefore,
Let the number be x.
x - 35 = 80% × x
x - 35 = (80/100) × x
x - (80/100) × x = 35
(1 - 0.8) × x = 35
0.2 × x = 35
x = 35 / 0.2 = 175
Now, we are asked to find the four-fifth of that number i.e x
(4/5) × x = (4/5) × 175 = 140
Hence, the four-fifth of the number is 140.
Question 5:
The value of a lathe machine depreciates at the rate of 10% per annum. If the cost of machine at present is Rs. 160,000, then what will be its worth after 2 years?
Present value of machine = Rs. 160,000
Rate of depreciation = 10% per annum
The value of machine after n years = P × (1 - R/100)^n
= 160,000 × (1 - 10/100)^2
= 160,000 × (0.9)^2
= 160,000 × 0.81
= Rs. 129,600
After 2 years, the cost of machine = Rs. 129,600
Question 6:
The value of a Xerox machine depreciates at the rate of 10% per annum. If the cost of machine at present is Rs. 75,000 then what was the value of machine before 2 years?
Cost of Xerox machine at present = Rs. 75,000
Rate of depreciation = 10%
The value of machine n years ago = P / (1 - R/100)^n
= 75,000 / (1 - 10/100)^2
= 75,000 / (0.9)^2
= 75,000 / 0.81
= Rs. 92,592.60
Therefore, the value of machine before 2 years ≈ Rs. 92,600
Question 7:
The current birth rate per thousand is 30, whereas corresponding death rate is 10 per thousand. Find the net growth rate in terms of population increase in percent.
We are given that,
1) Current birth rate per thousand is 30
2) Corresponding death rate is 10 per thousand
Hence, net growth on 1000 = Current birth rate - death rate
= 30 - 10 = 20
We are asked to find, net growth rate in terms of population increase in percent
(which means net growth on 100)
Net worth on 100 = (Net worth on 1000 / 1000) × 100
Net worth on 100 = (20 / 1000) × 100 = 2%
Therefore, the net growth rate is 2%.
Question 8:
The total population of a city is 6500. The number of males and females increases by 5% and 10% respectively and consequently the population becomes 7000. Find the number of males in the village.
We are given that,
1) Total population of city = 6500
2) Increase in male and female population = 5% & 10% respectively.
3) Final population of city = 7000
Let's assume that number of males = x
Number of females = 6500 - x
Therefore, after increase in 5% male and 10% female, the population becomes 7000
x + 5% of x + (6500 - x) + 10% of (6500 - x) = 7000
x + (5/100) × x + 6500 - x + (10/100) × (6500 - x) = 7000
(5/100) × x + 6500 + (10/100) × (6500 - x) = 7000
(5/100) × x + 6500 + (10/100) × 6500 - (10/100) × x = 7000
(5/100) × x - (10/100) × x + 6500 + 650 = 7000
(-5/100) × x + 7150 = 7000
(-5/100) × x = -150
x = 3000
Number of males = 3000
Number of females = 6500 - 3000 = 3500
Question 9:
The present population of a country is 10 crores. If it rises to 17.28 crores during next 3 years, then find uniform rate of growth in population.
Present population of country = 10 crores
After 3 years, population of country = 17.28 crores
To find: Rate of growth R%
After 3 years, the population is 17.28 crores. Therefore,
P × (1 + R/100)^n = 17.28
10 × (1 + R/100)^3 = 17.28
(1 + R/100)^3 = 17.28/10 = 1.728
(1 + R/100)^3 = (12/10)^3
(1 + R/100) = 12/10 = 1.2
R/100 = 0.2
R = 20%
Rate of growth in population = 20%
Question 10:
The population of different trees in a field increased by 10% in first year, increased by 8% in second year and decreased by 10% in third year. If at present the number of trees is 26730, then find the number of trees in the beginning.
We are given, number of trees:
1) First year = increased by 10%
2) Second year = increased by 8%
3) Third year = decreased by 10%
Let the initial number of trees be x.
After first year: x × (1 + 10/100) = 1.1x
After second year: 1.1x × (1 + 8/100) = 1.1x × 1.08 = 1.188x
After third year: 1.188x × (1 - 10/100) = 1.188x × 0.9 = 1.0692x
We know that the present number of trees is 26730.
Therefore, 1.0692x = 26730
x = 26730/1.0692 = 25000
The number of trees in the beginning was 25000.
Question 11:
The price of diesel increases by 50%. Find by how much percent a truck owner must reduce his consumption in order to maintain the same budget?
If the price of goods increases by R%, then the reduction in consumption so as not to increase the expenditure can be calculated using the formula:
(R/(100 + R)) × 100%
In this case, R = 50%
= (50/(100 + 50)) × 100%
= (50/150) × 100%
= 33.33%
The truck owner must reduce his consumption in order to maintain the same budget by 33.33%.
Question 12:
The price of rice falls by 15%. By what percentage a person can increase the consumption of rice so that his overall budget does not change?
If the price of goods decreases by R%, then the increase in consumption so as not to decrease the expenditure can be calculated using the formula:
(R/(100 - R)) × 100%
In this case, R = 15%
= (15/(100 - 15)) × 100%
= (15/85) × 100%
= 17.64%
Therefore, the person can increase his consumption by 17.64%.
Question 13:
In an examination, P scored 30% marks and failed by 15 marks. Q scored 40% marks and obtained 35 marks more than those required to pass. Find the pass percentage.
We have to calculate the pass percentage.
In case of P: He scores 30% out of total marks, but fails by 15 marks. Hence, the simple equation formed is (30% of x) + 15 = pass marks
In case of Q: He scores 40% out of total marks, but gets 35 marks more than required to pass. Hence, the simple equation formed is (40% of x) - 35 = pass marks
1) First calculate total marks.
Let total marks be x.
(30% of x) + 15 = (40% of x) - 35
(30/100) × x + 15 = (40/100) × x - 35
0.3x + 15 = 0.4x - 35
0.3x - 0.4x = -35 - 15
-0.1x = -50
x = 500
Total marks = 500
2) As we know the total marks, we can calculate the passing marks.
Passing Marks = (30/100) × 500 + 15 = 150 + 15 = 165
Therefore,
Passing Percentage = (165/500) × 100 = 33%
Required pass percentage = 33%
Question 14:
In a science examination, the average obtained by entire class was 80 marks. If 10% of students scored 92 marks and 20% of students scored 90 marks, then what was the average of remaining students?
Here, we do not know the number of students in the class. So let the number of students be 100 and the required average is y.
1) 10% of students scored 92 marks = 10 students scored 92 marks
2) 20% of students scored 90 marks = 20 students scored 90 marks
3) Therefore, from 100 students, the remaining students are 70
4) Average obtained by 100 students = 80 marks
Considering the given parameters, form the equation.
(10 × 92) + (20 × 90) + (70 × y) = (100 × 80)
920 + 1800 + 70y = 8000
70y = 8000 - 2720
70y = 5280
y = 75.43
The average of remaining students = 75.43
Question 15:
A student attempts x number of questions. He answers 15 correctly out of first 20 questions and of the remaining questions, he answers 1/3 correctly. If all questions have same credit and the student gets 50% marks, then find the value of x.
Given:
1) Student attempts x questions.
2) Out of 20 questions he answers 15 correctly and of (x - 20) questions he answered 1/3 correctly.
3) The student gets 50% marks.
Therefore,
15 + (1/3) × (x - 20) = 50% of x
15 + (1/3) × (x - 20) = (50/100) × x
15 + (1/3) × (x - 20) = x/2
15 + (x/3 - 20/3) = x/2
15 + x/3 - 20/3 = x/2
45/3 + x/3 - 20/3 = x/2
(45 + x - 20)/3 = x/2
(25 + x)/3 = x/2
2(25 + x) = 3x
50 + 2x = 3x
50 = 3x - 2x
50 = x
Hence, the number of questions attempted by the student = 50