## Some important Questions Percentage

## Example 1

The length of a
rectangle increases by 14% and the breadth by 8%. What is the consequent
percentage increase in area? .

### Solution

Let length and
breadth of the rectangle be 1 and b. .

Area of
rectangle = 1b

Length is
increased by 14%. .

1 + 1 x

(1 + 0.141) =
1.141 .

breadth is
increased by 8%

= b + 8% b

= b + 0.08 b .

(1.08) b .

Area of
rectangle

= 1.141 x 1.08b
= (1.2312) 1b .

Increase in Area
is = 1.2312 1b - 1b .

= 0.2312 1b .

\ percentage increase = x 100

23.12% increase.
.

## Example 2

If the price of
rice rises by 20% by what percentage should a family reduce the consumption in
order to spend the same amount as before?

### Solution

(Price per kg) x (Number of kg consumed) = constant = Amount spent
by the family

Let the initial
price be al and consumption be b_{1}. .

The new price a_{2}
= a_{1 }= 1.2 a_{1}

Amount spent is
constant a_{1}b_{1} = a_{2}b_{2}

a_{1}b_{1}
= (1.2) a_{1} b_{2} .

= 83.33%

&⇒ 100 - 83.33 = 16.67%. .

\ Consumption has to be reduced by 16.67%. .

## Example 3

The population
of a city increases by 20% every year. If the present population is 4,32,000,
what was the population of the city two years ago? .

### Solution

Let the population two years ago be 100. .

Then the
population in the last year and the present year are 120 and 144. .

For two years

144 = 4,32,000

100 = ?

x 100

3000 x 100 =
3,00,000

Two years ago
the population was 3,00,000. .

## Example 4

Mohan’s salary
was first increased by 20%, then decreased by 20%. If his present salary is Rs.7200,
then what was his original salary? .

### Solution

Let Mohan’s salary be Rs.100. .

When increased
by 20%,

Mohan’s salary =
Rs.120 .

Again when
decreased by 20%, Mohan’s salary

= 120 — 24 = Rs.96.
.

But present
salary is Rs.7,200 .

\ for, 96 → 100

7200 → ?

Required salary is
x 100 i.e Rs.7500.

## Example 5

Ajay got 30% of
the maximum marks in a test and failed by 15 marks. Bala got 40% of the maximum
marks in it and got 5 marks more than the pass mark. Find the maximum marks and
also the pass marks. .

### Solution

Let the maximum marks be M. .

**Method 1:**

Ajay’s marks = M and Bala’s marks = M

Ajay failed by
15 marks and Bala got 5 marks more than the pass marks. .

\ Pass marks = M + 15 = M - 5

20 = M

&⇒ M = 200. .

Also pass marks
= 30% of 200 + 15 = 75

**Method 2:**

Let the pass
marks be P. .

Bala’s marks —
Ajay’s marks

= P + 5 - (P -
15) = 20

M – M = 20

M = 200