Question

A number decreased by 18% becomes 410. find the number

Answer 1

Question

1. A number decreased by 18% becomes 410. Find the number.

Required answer:-

• 500 is the required number

Solution

Let the number be 100.

Since,                decrease in number = 18% 0f 100 = 18

⁂ After decrease, the number becomes = 100 - 18 = 82

Applying unitary method,

             When the decreased number = 82, the original number = 100

                 $\sf{=> When \: the \: decreased \: number = 1, \: the \: original \: number = \dfrac{100}{82}$

and,                     when the decreased number = 410,

                     $\sf{the \: original \: number = \dfrac{100}{82} \times 410=500}$

Alternative method (Algebraic method) :

 Let the original number be x

      $\sf{x-18\% \: of \: x=410 =>x-\dfrac{18x}{100}=410$

$\sf{i.e. \: \: \: \dfrac{100x-18x}{100}=410 => \dfrac{82x}{100}=410 \: i.e. \: x=410 \times \dfrac{100}{82}=500$

Direct method :-

If a number is decreased by x%,

 $\sf{the \: new \: number = [\dfrac{100-x}{100}] \times the \: original \: number}$

and, if a number is increased by x%,

$\sf{ the \: new \: number = [\dfrac{100+x}{100}] \times the \: original \: number}$

Here, the decrease in number = 18% and the new (decreased) number is 410.

$\sf{ The \: new \: number=\dfrac{100-8}{100} \times the \: original \: number}$

$\sf{=> 410=\dfrac{82}{100} \times the\: original \: number}$

$\sf{=> The \: original \: number = \dfrac{410 \times 100}{82}=500$