Question

a number was increased by10% and then decreased by 10%. find the net increase and decrease percentage.

Answer 1

Given,

A number was increased by10% and then decreased by 10%

To find,

The net increase and decrease percentage.

Main solution:

Let the no. be x

Firstly, it is increased by 10%

So,

new no. = x +10x/100

$= x + \frac{x}{10}$

$=\frac{11x}{10}$

Now this no. is decreased by 10%

So ,

new no. = 11x/10 -[(10/100)×(11x/10)]

$= \frac{11x}{10} + \frac{11x}{100}$

$= \frac{110x + 11x}{100}$

$= \frac{121x}{100}$

$= 1.21x$

So, we can see that there is an increase in no.

Increase in no. = 1.21 x - x = 0.21x

Increase % =( increase /original value)×100

$= \frac{0.21x}{x} \times 100$

$= 0.21 \times 100$

$= 21\%$

Answer :- 21% increase