# Concept: Percentage

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CONTENTS

Concept of percentage is very important in any MBA Entrance Exam, because there are a lot of questions related to the use of percentages in Profit & Loss, Time & Work, Speed & Distance. Apart from this percentage plays an important role in handling Data Interpretation (DI) Calculation.

The word percent consists of two words ‘per’ and ‘cent’. Per means each and cent means hundred. Thus, percent means ‘on each hundred’ or in each hundred. The symbol % is used to represent percent. In brief it is written as p.c.

Percent or Percentage can be defined as :

A percentage is a fraction whose denominator is 100 and the numerator is a definite quantity which is called the rate percent.

To convert percentage into fraction/decimal, we need to divide the percentage by 100.

1% = $\frac{1}{100}$ | 2% = $\frac{1}{50}$ | 4% = $\frac{1}{25}$ | 8% = $\frac{2}{25}$ | 16% = $\frac{4}{25}$ |

5% = $\frac{1}{20}$ | 10% = $\frac{1}{10}$ | 20% = $\frac{1}{5}$ | 40% = $\frac{2}{5}$ | 60% = $\frac{3}{5}$ |

12.5% = $\frac{1}{8}$ | 37.5% = $\frac{3}{8}$ | 62.5% = $\frac{5}{8}$ | 87.5% = $\frac{7}{8}$ | 112.5% = $\frac{9}{8}$ |

To convert fraction/decimal into percentage, we need to multiply the fraction/decimal with 100.

$\frac{1}{1}$ = 1 = 100% | $\frac{1}{2}$ = 0.5 =50% | $\frac{1}{3}$ = 0.33 = 33.33% | $\frac{1}{4}$ = 0.25 = 25% |

$\frac{1}{5}$ = 0.2 = 20% | $\frac{1}{6}$ = 0.167 = 16.67% | $\frac{1}{7}$ = 0.1428 = 14.28% | $\frac{1}{8}$ = 0.125 = 12.5% |

$\frac{1}{9}$ = 0.111 = 11.11% | $\frac{1}{10}$ = 0.1 = 10% | $\frac{1}{11}$ = 0.0909 = 9.09% | $\frac{1}{12}$ = 0.833 = 8.33% |

- To convert (or express) any fraction a/b to rate percent – multiply it by 100 and put a % sign = $\frac{\mathrm{a}}{\mathrm{b}}$ × 100 %
- To convert (or express) any percentage P% into fraction – divide it by 100 = $\frac{\mathrm{P}}{100}$

**Example:**. Express the following in terms of percentage:

(a) 0.4 (b) 1.0
(c) 5/3 (d)
7x/y (e) 1.23

**Solution**:

(a) multiply the decimal fraction by 100.

∴ 0.4 = (0.4 × 100)% = 40%

(b) 1.0 = (1.0 × 100)% = 100%

(c) 5/3 = (5/3 × 100)% = 166 2/3%

(d) 7x/y = (7x/y × 100)% = 700x/y%

(e) 1.23 = (1.23 × 100)% = 123%

To convert a rate percent to a fraction divide it by 100 and delete the % sign.

**Example:** Express the following in terms of fraction:

(a) 22$\frac{1}{2}$% (b) 35%
(c) a^{2}/b% (d)
0.3% (e) 2/7%

**Solution:**

(a) Divide the given percentage by 100 to convert it into a fraction

∴ 22$\frac{1}{2}$% = (22$\frac{1}{2}$)/100 = 45/(2 × 100) = 9/40

(b) 35% = 35/100 = 7/20

(c) a^{2}/b% = a^{2}2/(b × 100) = a^{2}/(100b)

(d) 0.3% = 0.3/100 = 3/1000

(e) 2/7% = 2/(7 × 100) = 1/350

**Example:** Find

(a) 9% of 27 (b) 0.02% of 6500 (c)
7/2% of 80 (d) 125% of 64 (e) 10%
of 5% of 320

**Solution:**

(a) Multiply the number by P/100, if p% of the number is to be calculated.

∴ 9% of 27 = 9/100 × 27 = 243/100

(b) 0.02% of 650 = 0.02/100 × 6500 = 13/10

(c) 7/2% of 80 = 7/(2 × 100) × 80 = 14/5

(d) 125% of 64 = 125/100 × 64 = 80

(e) 10% of 5% of 320 = 10/100 × 5/100 × 320 = 1.6 = 8/5

**Example:** Find the following :

(a) 36 is what % of 144 (b) 7/8
is what % of 3/4

(c) What % of 80 is 16 (d) 0.625
is equal to what % of 1$\frac{7}{28}$

(e) 36 × 14 is what % of 1400 (f) R is what %
of N

**Solution:**

(a) Here we have to compare 36 with 144.

Value of percentage (%) = (Number to be compared)/(Number to be compared against) × 100

⇒ % = 36/144 × 100 = 25%

(b) Value of % =(7⁄8)/(3⁄4)×100 = 116$\frac{2}{3}$%

(c) Value of % = Result/(Original number) × 100 = 16/80 × 100 = 20%

(d) Value of % = 0.625/(1$\frac{7}{28}$) × 100 = 0.625/35 × 28 × 100 = 50%

(e) value of % = (36 × 14)/1400 × 100 = 36%

(f) value of % = R/N × 100 = 100R/N%

**Example:** 25% of a number is 20, what is 40% of that number? Also find the number

**Solution:**

25% → 20

⇒ 40%→ 20/25 × 40 = 32

Original number (N) = $\frac{20}{25}$ × 100 = 80

**Example:** A number A exceeds B by 25%. By what percent is B less than A?

**Solution:**

If B = 100, then A = 125

⇒ B = $\frac{100}{125}$ × 100 = 80% of A

∴ B is (100 - 80) = 20% less than A

When a number is increased by p%, it means p% of the number is added in itself.

**Example:** Increase 120 by 50%?

**Solution:**

We add 50% of 120 with 120.

∴ 120 + 120 × 50% = 120 + 60 = 180

Similarly, when a number is decreased by p%, it means p% of the number is subtracted from itself.

**Example:** Decrease 120 by 50%?

**Solution:**

We subtract 50% of 120 from 120.

∴ 120 - 120 × 50% = 120 - 60 = 120.

When a number 'I' is increased or decreased by P%.

Final number (F) = I ± P% of I = I(1 ± P%) = I$\left(1\pm \frac{P}{100}\right)$

Here, $\left(1\pm \frac{P}{100}\right)$ is called the multiplication factor.

Hence, when a number is to be increased or decreased by P%, we first calculate the multiplication factor and multiply if with the initial number to get the final number.

**Note**: Multiplication factor does not depend on the value of initial number. It only depends on
the percentage change.

- F = I$\left(1\pm \frac{P}{100}\right)$ = I × multiplication factor (mf)
- I = $\frac{\mathrm{F}}{\mathrm{mf}}$
- P% = $\frac{\mathrm{F\; -\; I}}{\mathrm{I}}$ × 100%
- When a number is multiplied with f (i.e. multiplication factor is f)

∴ f = 1 + $\frac{P}{100}$

∴ Percentage change = (f - 1) × 100%

**Example:** The daily wage is increased by 15%, and a person now gets Rs. 23 per day. What
was his daily wage before the increase?

**Solution:**

Multiplication factor = $\left(1+\frac{\mathrm{15}}{100}\right)$ = $\left(1+\frac{3}{20}\right)$ = $\frac{23}{20}$

∴ Original daily wage =$\frac{23}{\mathrm{23/20}}$ = 20

**Example:** From a man’s salary, 10% is deducted on tax, 20% of the rest is spent on
the education, and 25% of the rest is spent on food. After all these expenditures, he is left with Rs. 2,700
Find his salary.

**Solution:**

Let the salary be S.

10% of salary is deducted on tax.

∴ Remaining salary = $S\left(1-\frac{10}{100}\right)$

Now, 20% of this remaining salary is spent on education.

∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)$

Now, 25% of this remaining salary is spent on food.

∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ which is equal to 2700

⇒ $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = 2700

⇒ $S\times \frac{9}{10}\times \frac{4}{5}\times \frac{3}{4}$ = 2700

⇒ S = $\frac{2700\times 50}{27}$ = Rs. 5000

**Example** : A number is tripled. What is the percentage change.

**Solution**:

When a number is tripled, it is effectively multiplied with 3.

∴ % change = (3 - 1) × 100 = 200%

**Example** : A number is increased by 50%. By what percentage should it be decreased to get
back the original number.

**Solution**:

For 50% increase, multiplication factor = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$

∴ New number = Initial number × $\frac{3}{2}$

To get the original number, this new number must be multiplied with $\frac{2}{3}$

∴ % change = (2/3 - 1) × 100 = -33.33%

∴ The new number should be decreased by 33.33% to get back the original number.

The concept of successive percentage change deals with two or more percentage changes applied to quantity consecutively. In this case, the final change is not the simple addition of the two percentage changes (as the base changes after the first change).

**Example**: If a number I is successively increased by a% and then b%.

Number after a% increase = I$\left(1+\frac{a}{100}\right)$

Now, this number has to be again increased by b%.

∴ Final number = I$\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$

Similarly, when a number I is successively increased by a%, b% and c%

Final number = I$\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)\left(1+\frac{c}{100}\right)$

**Example** : When a number is successively changed by a% and b%, what is the overall percentage
change.

**Solution**:

When a number is successively changed by a% and b%, the overall
multiplication factor = $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$ = f

∴ percentage change = (f - 1) × 100% = $\left(1+\frac{a}{100}+\frac{b}{100}+\frac{a\times b}{{100}^{2}}-1\right)$ × 100% = a + b + $\frac{a\times b}{100}$

**Note**: a and b can be either positive or negative and the sequence of increase or decrease does not
affect the overall change.

- When a number I is successively increased by a% and b%

∴ Overall % change (P) = a + b + $\frac{ab}{100}$

- When a number I is successively increased by a% and then decreased by a%

∴ Overall % change (P) = - $\frac{{a}^{2}}{100}$%

- When a number I is successively increased by a% n times

∴ Final number (F) = I${\left(1+\frac{a}{100}\right)}^{n}$

- After increasing a number by a% successively n times, final number F is obtained

∴ Initial number (I) = $\frac{F}{{\left(1+\frac{a}{100}\right)}^{n}}$

- If P is the original value of asset, and the depreciation is r% per year

∴ Value after n years = P${\left(1-\frac{r}{100}\right)}^{n}$

**Example** : When a number is successively changed by 20% and 25%, what is the overall
percentage change.

**Solution**: **Method 1**: (Multiplication Factor)

Overall multiplication
factor = $\left(1+\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{6}{5}\times \frac{5}{4}$ = $\frac{3}{2}$

∴ % change = (3/2 - 1) × 100 = 50%

**Method 2**: (Formula)

Overall % change = 20 + 25 + $\frac{\mathrm{20}\times \mathrm{25}}{100}$ = 50%

**Example** : When a number is successively changed by -20% and 25%, what is the overall
percentage change.

**Solution**: **Method 1**: (Multiplication Factor)

Overall multiplication
factor = $\left(1-\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{4}{5}\times \frac{5}{4}$ = 1

∴ % change = (1 - 1) × 100 = 0% i.e. no change

**Method 2**: (Formula)

Overall % change = -20 + 25 + $\frac{\mathrm{-20}\times \mathrm{25}}{100}$ = 0%

**Example** : When a number is successively changed by -20% and -25%, what is the overall
percentage change.

**Solution**: **Method 1**: (Multiplication Factor)

Overall multiplication
factor = $\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = $\frac{4}{5}\times \frac{3}{4}$ = $\frac{3}{5}$

∴ % change = (3/5 - 1) × 100 = - 40%

**Method 2**: (Formula)

Overall % change = - 20 - 25 + $\frac{\mathrm{-20}\times \mathrm{-25}}{100}$ = - 40%

**Example** : When sugar price increase by 25%, by what percentage should a family reduced its
consumption so that monthly expenditure does not change.

**Solution**: **Method 1**:

Let the initial price of sugar = Rs. 100/kg
and quantity consumed = 10kg / month

∴ Total Expenditure = 100 × 10 = Rs. 1,000

Now, price increases by 25%, hence, new price = 125

New quantity consumed = 1,000/125 = 8 kg

∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%

**Method 2**: (Multiplication factor)

Expenditure = Price × Quantity

⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for
quantity

Overall multiplication factor for expenditure = 1 (no change)

Multiplication factor for price = $\left(1+\frac{25}{100}\right)$ = $\frac{5}{4}$

Let multiplication factor for quantity = f

∴ 1 = 5/4 × f

⇒ f = 4/5

∴ % change in quantity = (4/5 - 1) × 100 = -20%

**Example** : When sugar price increase by 50%, by what percentage should a family reduced its
consumption so that monthly expenditure goes up by only 20%.

**Solution**: **Method 1**:

Let the initial price of sugar = Rs. 100/kg
and quantity consumed = 10kg / month

∴ Total Expenditure = 100 × 10 = Rs. 1,000

Now, price increases by 50%, hence, new price = 150

Expenditure increases by 20%, hence, new expenditure = 1200

New quantity consumed = 1,200/150 = 8 kg

∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%

**Method 2**: (Multiplication factor)

Expenditure = Price × Quantity

⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for
quantity

Overall multiplication factor for expenditure = $\left(1+\frac{20}{100}\right)$ = $\frac{6}{5}$

Multiplication factor for price = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$

Let multiplication factor for quantity = f

∴ 6/5 = 3/2 × f

⇒ f = 4/5

∴ % change in quantity = (4/5 - 1) × 100 = -20%

**Example:** Groundnut oil is being sold at Rs. 27 per kg. During last month its cost was
Rs. 24 per kg. Find by how much % a family should reduce its consumption, so as to keep the expenditure the
same.

**Solution:**

Let the initial consumption of groundnut oil be = x and new expenditure = x_{1}

Expense (Initial) = 24x and Expense (Final) = 27x_{1}

∴ 24x = 27 × x_{1} [since expense has to remain the same]

x_{1 }= 8x/9

In the above fraction we can see that the new consumption is 8/9^{th} fraction of the old
consumption.

Thus, the reduction is 1/9^{th} of the original consumption.

Hence, percentage reduction in consumption is 1/9 × 100%.

**Example:** If the radius of the circle is increased by 20%, its area will increase by what %?

**Solution:**

We know that Area of circle = 𝜋 × r × r

overall multiplication factor for area = $\left(1+\frac{20}{100}\right)\times \left(1+\frac{20}{100}\right)$ = ${\left(1+\frac{20}{100}\right)}^{2}$ = ${\left(\frac{6}{5}\right)}^{2}$ = $\frac{36}{25}$

∴ % change in area (36/25 - 1) × 100 = 44%

Hence, increase in area = 44%

**Example:** If the sides of a square are increased by 30%. Find the resultant increase in
area.

**Solution:**

Let A ∝ a^{2} where ‘a’ = initial side

New area A_{1} ∝ (1.3a)^{2}

⇒ A_{1} ∝1.69a^{2}

Thus, it can be seen that new area is 69% more than the old area.

**Example:** A reduction of 25% in the price of Cadbury chocolates, enables a person to buy
5 kg more for Rs. 120. Find the original price of Cadbury chocolates per kg.

**Solution:**

Method 1 :

In this question the fixed expense = Rs. 120

If P is the initial price, reduced price = 0.75P

Now, new quantity (at the reduced price) – initial quantity = 5

120/0.75P - 120/P = 5

⇒ P = Rs. 8

Alternate Method:

Since the price reduces by 25%, the initial quantity can be bought for 120 × 0.75 = Rs. 90 at the reduced
price.

∴ Remaining amount = 120 - 90 = Rs. 30

Now, 5 kg extra can be bought for this Rs. 30

∴ Reduced price = 30/5 = Rs. 6 = 0.75P

∴ Initial price = 6/0.75 = Rs. 8

**Example:** When N is reduced by 4, it becomes 80% of itself. What is the value of N?

**Solution:**

The problem implies that

20% of N = 4 (since 100 – 80 = 20)

N = 4/20 × 100

∴ N = 20

**Example:** If 10% of an electricity bill is deducted, Rs. 45 is still to be paid. How
much was the bill?

**Solution:**

If 10% of the bill is reduced, still 90% has to be paid

Here the decreased bill = Rs. 45

⇒ 90% of Total Bill = 45

⇒ N = (45 × 100)/90

∴ N = 50

Hence, the bill was Rs. 50

**Example:** If the price of 1 kg of cornflakes is increased by 25%, the increase is Rs.
10. Find the new price of cornflakes per kg.

**Solution:**

Due to 25% increase, the price changes by Rs. 10

⇒ 25% of price = 10

⇒ price = (10 × 100)/25 = Rs. 50

∴ New price of cornflakes per kg is Rs. 50

**Example:** In an election between two candidates, a candidate who gets 62% of the total votes
polled is elected by a majority of 144 votes, Find the number of votes polled and votes secured by the
candidate. [All votes polled are valid votes]

**Solution:**

Let, the total number of votes be 100x

Winning candidate gets 62% = 62x votes

Losing candidate gets (100 - 62 =) 38% = 38x votes

Now, difference of votes is 144

∴ 62x - 38x = 144

⇒ 24x = 144

⇒ x = 6

Thus, total number of votes = 100x = 600 votes and winner gets = 62x = 372 votes

**Example:** The rate for admission to an exhibition was Rs. 5 and was later reduced by 20%. As
a result, the sale proceeds increased by 44%. What was the % increase in attendance?

**Solution:**

Sale proceeds = price × attendance

Multiplication factor for sales = $\left(1+\frac{44}{100}\right)=\frac{36}{25}$

Multiplication factor for price = $\left(1-\frac{20}{100}\right)=\frac{4}{5}$

Let the multiplication factor for attendance = f

⇒ 36/25 = 4/5 × f

⇒ f = 36/25 × 5/4 = 9/5

∴ % change in attendace = (9/5 - 1) × 100 = 80%

∴ increase in attendance = 80%

**Alternate Method**:

Let the initial attendace = 100 people

∴ Initial sale proceeds = 5 × 100 = Rs. 500

Now, new price = 5 × 0.8 = Rs. 4

and, new sales proceeds = 500 × 1.44 = Rs. 720

∴, new attendance = 720/4 = 180

∴, Increase in attendance = (180 - 100)/100 × 100 = 80%

* The information of price is redundant in this question.

**Example:** A piece of cotton cloth, 20m long, shrinks by 0.5% after washing. Find the
length of the cloth after washing.

**Solution:**

Let the length of the cloth after washing be x metro or Now it left 100 – 0.5 = 99.5%

∴ x = 20 × 99.5%

⇒ x = 19.9 meter

Hence, original length = 19.9 meter

**Example:** By mistake, a line was measured as 11.25 cm, which was 2.5% more than the
actual length. Find the actual length of the line?

**Solution:**

let actual length of line be x cm

∴ x(100 + 2.5)% = 11.25

⇒ x = (11.25 × 100)/102.5 = 10.9

Hence, actual length = 10.9 cm

**Example:** When the price of a pressure cooker was increased by 15%, the sale of pressure
cookers (quantity) decreased by 15%. What was the net effect on the sales?

**Solution:**

Let the original price be Rs. 100 also the number of item 100.

∴ Revenues = 100 × 100 = 10000

After change price = Rs. 115 and item = 85

Now, Revenue = 115 × 85 = 9775

∴ decrease % =(10000 - 9775)/10000 × 100

Hence, decreased by = 2.25%

Alternate Method :

1.15 × 0.85 = 0.9775

Thus, decrease = (0.9775 - 1) × 100 = 2.25%

**Example:**. A medical student has to secure 40% marks to pass. He gets 80 marks and fails by
60 marks. Find the maximum marks?

**Solution:**

Let the maximum number of marks be x

∴ x × 40% = 80 + 60 (he needs to pass)

⇒ x = 350

Hence, maximum marks = 350

**Example:** When 75% of a number is added to 75, the result is the number again. The
number is

**Solution:**

let the number be x

∴ 75% of x + 75 = x

⇒ 75 = x – 75% of x

⇒ x = 7500/25 = 300

Hence, the original number is 300

**Example:** What is the total number of candidates in an examination, if 31% fail, and the
number of those who pass exceeds the number of those who fail by 247?

**Solution:**

Let the total number of candidates be x

Pass % = 100 – 31 = 69

Now, 69% of x – 31% of x = 247

⇒ 38% of x = 247 ⇒ x = (247 × 100)/38 = 650

Hence, the total number = 650

**Example:** Rita loses 12$\frac{1}{2}$% of her money and after spending 70% of the remainder, has
Rs. 210 left. How much money did she have at first?

**Solution:**

Let Rita have Rs. x at first

30% of (87$\frac{1}{2}$% of x) = 210

or 3/10 × 7/8 × x = 210

⇒ x = Rs. 800. Hence, Rita had Rs. 800 at first.

**Example** 5% people of a village died by bombardment, 15% of the remainder left the
village on account of fear. If now the population is reduced to 3553. How much was it in the beginning?

**Solution:**

Let the population at beginning be x

∴ people left after bombardment = 95% of x

and number after leaving 15% people = 85% of (95% of x) = 3553 (given)

⇒ 85/100 × 95x/100 = 3553 ⇒ x = 4400

**Example:** In a factory, there are 40% technicians and 60% non-technicians. If the 60% of
the technicians and 40% of non-technicians are permanent employees, then find the percentage of worker who
are temporary?

**Solution:**

Let the number of employees be 100

∴ Number of technicians = 40

∴ Number of non-technicians = 60

Now, Number of temporary employee

= 40% of 40 + 60% of 60 = 16 + 36 = 52

Hence, 52% of the workers is temporary

**Example:** A salesman receives a salary of Rs. 1000 per month and commission of 8% on all
sales in excess of Rs. 2000. What should be the amount of his sales in a particular month if he were to earn Rs.
5000 in that month

**Solution:**

Let amount of his sales be x

∴ 1000 + 8% (x – 2000) = 5000

⇒ x = Rs. 52000

**Example:** During one year, the population of a town increased by 10% and during the next
year, it diminished by 10%. If at the end of the second year the population was 89100, what was it at the
beginning?

**Solution:**

Let the population at the beginning = x

net decrease % (10)^{2}/10% = 1%

∴ Population at the end = x × 99% = 89100

x = (89100 × 100)/99

x = 90,000