 # Percentages Concepts

The percentage concept is required in several quantitative aptitude topics to solve the related questions. Also, questions directly related to percentages holds good weightage and thus, the this topic is considered as one of the most important topics for the CAT. Learn the percentages for CAT easily along with detailed explanations and solved examples here.

### Definition of Percent:

“Per cent” means for every hundred. If you get 25 marks in an exam where the full marks is 50, your percentage marks will be what you will get in a total of hundred marks. (25/50× 100)

Using unitary method: In 50 you get 25; So in 1 you get 25/50 and in 100 you get 2550× 100 = 50 percent. Thus, what you get out of hundred becomes the percentage.

For a fraction, if the denominator is hundred, it is called a percentage and the numerator of the fraction is called rate per cent.

### Percentage for Comparison:

Percentage helps us to compare between different fractions when the denominator or the total number is different in each case. It is one of the simplest tools for the comparison of data.

Take, for example, this table below which shows the marks obtained by a student in 3 different subjects

 Subject Marks Obtained History 60 Math 25 English 45

From this data alone, we cannot compare the marks obtained for the various subjects.. Now suppose we have the data of the total marks obtained as follows

 Subject Marks Obtained Total Marks Marks Obtained/total marks * 100 History 60 100 60% Math 25 25 100% English 45 90 50%

Now as all the three subjects are represented on a scale of 100, it is easy to compare the marks for the three subjects and decide which subject has the student scored the maximum in.

### Presentations of Percentages:

#### 1) a% of b

a% of b ( a percent of b) = b% of a

a% of b is represented mathematically as (a×b)/100.

Example: 24% of 25 = [24×25/100]= 6 25% of 24 = [(25×24)/100]= 6.

#### 2) What percentage of a is b

This is represented as (b/a× 100)

Eg. What percentage of 75 is 25? (25/75)×100 = 33.33%

### Conversions Related to Percentages:

#### 1) To convert a fraction into a percentage

Multiply and divide by 100.

Keep the denominator as 100, the numerator you obtain is the required answer.

Take for example

21/22=(21/22 x100)/100=(2100/22)/100 = 95 6/11%.

#### 2) To find the fraction equivalent of a percentage

Divide by 100(after removing the percentage sign) e.g. ) 11 3/8%=91/8×100=91/800

eg) 25% = 25/100=1/4.

### Illustration:

Question: Recently I went to buy a laptop for myself. The dealer said that he has laptops of two companies: HP and Lenovo. He was selling the HP laptop for Rs. 42000 and told me that he will offer me the same at 7/8 of that price while the Lenovo laptop was for Rs. 46000 and he were offering it at ?4/5?^th of that price. I decided to buy the laptop on which I was getting a better percentage discount. Which one should I buy?

Solution:

HP: 7/8 means (7×100)/8 = = 87.5% means a discount of 100 – 87.5 = 12.5%

LENOVO: 4/5 means 4×100/5 = 400/5= 80 % means a discount of 100 – 80 = 20%. So I should buy the LENOVO laptop.

### The Concept Of Change:

There are two types of change

#### 1. Absolute value change

It is the actual change in the quantity. For example, if there are 10 rabbits in the first year and 15 rabbits in the second year, the absolute change in the number of rabbits is 5

#### 2. Percentage change

This can be obtained by calculating the absolute change and dividing it by the initial number of rabbits present e.g. ) percentage change = (Absolute value change)/(Original quantity)×100= 5/10=1/2×100= 50%.

### Percentage Change and Percentage Point Change

If the pass percentage of a class was 75 % in 1991 and 85% in 1992, we can calculate the percentage point change and the percentage change as follows

Percentage point change= Final percentage – Initial percentage= 85 %– 75% = 10 percentage points Percentage change= (Final percentage – Initial percentage)/(Initial percentage)×100 = (85 – 75)/75 x 100 = 40/3%.

• #### To Increase a Number By a Given Percent

Use the formula: (100 + rate)/100

Example: Increase 40 by 20%

Here the rate = 20 Multiply the number by the above formula to get the answer 40 x ((100 + 20))/100=120×40/100 = 48. (We are effectively computing(40+20/100×40)

It is easier to remember that given the base value=x, the final value after increase can be found as:

20% increase= (x+0.2x= 1.2x); 30% increase= (x+0.3x=1.3x); 5% increase= (x+0.05x=1.05x).

• #### To Decrease a Number By a Given Number

Use the formula (100 – Rate)/100

Example: Decrease 40 by 20%

Here the rate = 20. Multiply the number by the above formula to get the answer

40 x ((100- 20))/100=80×40/100=32. (We are effectively computing (40-20/100×40))

It is easier to remember that given the base value=x, the final value after decrease can be found as:

20% decrease= 0.8 x (0.8= 1-0.2); 30% decrease= 0.7x (0.7= 1-0.3); 5% decrease= 0.95x (0.95= 1-0.05).

To find the percentage increase or decrease of a given number, the concept of percentage change is used as explained above.

1) Percentage Increase = (Final Value – Initial Value)/(Initial Value) = (Total Increase )/(Initial Value) x 100%

2) Percentage Decrease = (Initial Value – Final Value)/(Initial Value) = (Total Decrease)/(Initial Value) x 100%

### Illustration:

Question: In the IPL match an analysis was done for the two openers of both teams: Delhi Daredevils (DD) and Deccan Chargers (DC). It was found that the two DC openers Gilchrist and Gibbs together scored 20% more than what the two DD openers Sehwag and Gambhir scored. Gambhir scored 30 and Gibbs scored 40. How much percentage less/more did Sehwag score than Gilchrist if Gilchrist scored 80?

Solution:

Say runs scored by Sehwag = a, Gambhir = b, Gilchrist = c and Gibbs = d.

Then (c + d) = 1.2 (a + b)

Also b = 30, d = 40, c = 80

80 + 40 = 1.2 (a + 30) => 120/1.2 = a + 30 => a = 70

So Sehwag = 70. So Sehwag scored (80 – 70)/80 x 100 = 10/80 x 100 = 12.5 % less than Gilchrist.

Note: Here we took 80 as the denominator because we need to calculate how much Sehwag scored less “than” Gilchrist. So, Gilchrist’s score has to be made the denominator as it is the base value that the difference is being compared to.

#### Basic Formulae related to Percentage Change:

a) If the price of a commodity increases by r%, then the reduction in consumption so as to not increase the expenditure is r/(100+r) x 100.This formula can also be used to compare incomes between two people.

e.g.) If P’s income is r% more than Q’s income, then Q’s income is less than P’s income by r/(100+r) x 100.

b) If the price of a commodity decreases by r%, then the increase in consumption so as to not decrease the expenditure is r/(100-r) x 100 . This formula can also be used to compare incomes between two people.

Example: If P’s income is r% less than Q’s income, then Q’s income is greater than P’s income by r/(100-r) x 100.

c) If there is a consecutive percentage change of a% and b%, the net change can be calculate as

Net change= a + b + ab/100

### Illustrations:

3. Anil is taking an examination which has two sections A and B. He starts with section A and when he proceeds to section B he realizes that the marks carried for each question has decreased by 40% as compared to section A. How many more questions (in percentage) should he solve in section B as compared to section A if he wants to score same marks in section A and section B?

Solution:-

Given that both sections carry equal marks. Solution: Here marks per question decreases by 40%, so using the formula (b), he should solve ((40/(100-40))×100)% = 66.66% more questions in section B if he wants to score same marks.

4. The petrol prices shot up by 20% due to Iraq War. Amit travels 2000 kms every month and his car gives a mileage of 20 kms per liter. By how many kilometers should he reduce his travel to maintain his expenditure to the previous level.

Solution:

At 20 kms per liter he bought 2000/20= 100 liters of petrol. Now the price the increased by 20%; using formula (a), the consumption has to be reduced by (20/(100+20))×100) % = 16.66%.

So, he should reduce his travel by (16.66×2000)/100= 333.33 kms to maintain the same expenditure level.

5. After receiving two successive pay hikes if Ashish’s salary becomes 15/8 of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as high (in %) as the first?

a) 15 %

b) 20%

c) 25%

d) 30%

Solution:

This question can be solved using two approaches:

Approach 1: (conventional)

Suppose first raise was of a % then the second raise = 2a % Now, using formula (c) net change: a + b + ab/100?a + 2a + 2axa/100

Now if initial salary was X, then % change = ((15 X)/8– X)/X×100=7/8× 100

So 3a + (2a^2)/100 = 700/8 solving we get a = 25%. So, the first raise was of 25%.

Approach 2: (Recommended)

In these types of questions it is quicker to go from the answer options. First calculate the net change= ( (15 X)/8– X )×100 = 700/8% = 87.5 %

Start with option (b) 20% First year raise 20% next year raise= 40% Net change (use formula (c)) 20 + 20 + 400/100 = 44% With option (d) Net change = 30 + 60 + 1800/100 = 108% This way, you will get option (c) 25% is the correct option.

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