"> Word Problem 3 | Job Exam Rare Mathematics (JEMrare)

Word Problem 3

Problem 3 : A sells a house to B at profit of 5% and B sells it to C at profit of 10%. If C pays Rs 231 for it, what did it cost to A ?



Solution:

Here we have to find the cost price which A paid for it

We take A, B and C’s cases individually:

A’s Case:

Suppose cost price (CP) for A is x. Given here that A earns 5% profit i-e profit is 5% of cost price.

Since A is earning profit, We use Profit formula here which is;

P=SP-CP

Putting values of P (profit) and CP (cost price)

5% Of x=SP-x

\Rightarrow \frac{5}{100}x= SP-x

From there we can get sale price SP for A

\Rightarrow \frac{5}{100}x+x= SP

Or

\Rightarrow SP=\frac{105x}{100}

Now this sale price will become cost price for B

 

 

B’s Case

B also earned profit, we again use here profit formula.

P=SP-CP

Given in question that B earns 10% profit of his cost price so,

\Rightarrow 10% of \frac{105x}{100}=SP-\frac{105x}{100}

\Rightarrow \frac{10}{100}\times \frac{105x}{100}=SP-\frac{105x}{100}

Taking -ve term to other side;

\Rightarrow \frac{10}{100}\times \frac{105x}{100}+\frac{105x}{100}=SP

Or

SP= \frac{10}{100}\times \frac{105x}{100}+\frac{105x}{100}

= (\frac{10}{100}+1)\times \frac{105x}{100}

=\frac{11\times105x}{10\times100}

After  simplification, we get sale price for B as;

SP=\frac{231x}{200}

Now this sale price of B will become cost price of C

C’s Case

We are given that C pays Rs 231 to B i-e cost price of C is Rs. 231. But we found C’s cost price to be \frac{231x}{200}

Both are cost price for C , so these should be equal;

i-e

\frac{231x}{200}=231

x=231 \times \frac{200}{231}

x=200

So, A’s cost was Rs 200/-