Question

Mayank visits three shops one by one. At each shop, he spends two thirds of the money he had at that shop.If he is left with rs. 32 with him in the end after visiting all 3 of them. Find the amount of money he had in the beginning.

Answer 1

The amount he has initially is x
The amount he has spend in first shop 2x/3
The amount with him after visiting first shop is x/3.
The amount he spend in second shop is 2x/9
The amount with him after visiting second shop is x/3-2x/9=x/9
The amount spent in third shop is 2x/27
The remaining balance is x/9-2x/27=x/27=32
X=27*32=864

Answer 2

Answer:

He had 810 rupees in the beginning .

Step-by-step explanation:

Given,Mayank visits three shops one by one.

Let,he had total x rupees at the beginning.

It is also given he spends two thirds of the money which he had at that shop.

So,at the first shop,he spend

$= x \times \frac{2}{3} = \frac{2x}{3} \: rupees$

Remaining money

$= x - \frac{2x}{3} \\ = \frac{3x - 2x}{3} \\ = \frac{x}{3}$

Again he spend at second shop

$= \frac{2}{3} \: of \: \frac{x}{3} \\ = \frac{2x}{9} \: rupees$

Remaining money now

$= \frac{x}{3} - \frac{2x}{9} \\ = \frac{3x - 2x}{9} \\ = \frac{x}{9} \: rupees$

Again he spend at third shop

$= \frac{2}{3} \times \frac{x}{9} \\ = \frac{2x}{27} \: rupees$

Remaining money

$= \frac{x}{9} - \frac{2x}{27} \\ = \frac{3x - 2x}{27} \\ = \frac{x}{27} \: rupees$

It is also given after three shop spending he is left with rs. 32.

According to question,

$\frac{x}{27} = 32 \\ x = 27 \times 32 \\ x = 810$

This is a problem of Algebra.

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