Question

A number is first decreased by 10%. and then increased by
10%. The number so obtained is 50
less than original number.
The original number is
:​

Answer 1

ANSWER:

• The original number is 5000.

GIVEN:

• A number is first decreased by 10%.

• Then it is increased by 10%.

• The number so obtained is 50 less than original number.

TO FIND:

• The original number.

EXPLANATION:

Let the number be x.

Let y be the 10 % of x.

$\sf \dfrac{y}{x} \times 100 = 10\%$

$\sf \dfrac{y}{x} \times 10 = 1$

$\sf x = 10y$

$\sf y = \dfrac{x}{10}$

When x is decreased by 10 %, the resulting number will be x - y

Let z be the 10 % of x - y [ As x - y is increased by 10 % ]

$\sf \dfrac{z}{x - y} \times 100 = 10\%$

$\sf \dfrac{z}{x - y} \times 10 = 1$

$\sf x - y = 10z$

$\sf z = \dfrac{x - y}{10}$

When x - y is increased by 10 %, the resulting number will be x - y + z

x - y + z = x - 50 [ As the resulting number is 50 less than the original number ]

Substitute the value of y and z.

$\sf x - \dfrac{x}{10} + \dfrac{x - y}{10} = x - 50$

$\sf{Again \ substitute \ y = \dfrac{x}{10} }$

$\sf x - \dfrac{x}{10} + \dfrac{x - \dfrac{x}{10} }{10} = x - 50$

$\sf x - \dfrac{x}{10} + \dfrac{\dfrac{10x - x}{10} }{10} = x - 50$

$\sf x - \dfrac{x}{10} + \dfrac{9x}{100} = x - 50$

$\sf \dfrac{100x}{100} - \dfrac{10x}{100} + \dfrac{9x}{100} = x - 50$

$\sf \dfrac{100x - 10x + 9x}{100} = x - 50$

$\sf \dfrac{99x}{100} = x - 50$

$\sf 99x = 100x - 5000$

$\sf 99x - 100x = - 5000$

$\sf - x = - 5000$

$\sf x = 5000$

Hence the original number is 5000.

VERIFICATION:

$\sf We\ know\ that \ y = \dfrac{x}{10}$

Substitute x = 5000

$\sf y = \dfrac{5000}{10}$

$\sf y = 500$

$\sf We\ know\ that \ z = \dfrac{x - y}{10}$

Substitute x = 5000 and y = 500

$\sf z = \dfrac{5000 - 500}{10}$

$\sf z = \dfrac{4500}{10}$

$\sf z =450$

$\sf We\ know\ that \ x - y + z = x - 50$

Substitute x = 5000, y = 500, z = 450

$\sf5000 - 500+ 450 = 5000- 50$

$\sf5000 - 50 = 4950$

$\sf4950 = 4950$

HENCE PROVED.