d Word Problem 2 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 2

Problem 2: A man sells two houses for Rs 3990 each. On one he gains 5% and on the other he loses 5%. Find his gain or loss.


Here gain means profit.

Case 1 ( Profit)

Sale Price (SP)=3990

Cost Price (CP)=x

Profit (P)=?

We use profit formula here;


But given he earns profit 5% (of cost price) i-e

P=5% Of CP

P=\frac{5}{100}\times x



Put  values in above formula;

\Rightarrow \frac{x}{20}=3990-x

Here we calculate value of ‘x’ i-e the cost price

\Rightarrow \frac{x}{20}+x=3990

\Rightarrow \frac{21x}{20}=3990

\Rightarrow x=\frac{20}{21}\times3990=Rs. 3800

So cost price of house is 3800 rupees


Profit Or Gain = SP - CP=3990-3800=190 Rs

So he gained Rs 190 for one house.

Case 2 (Loss)

Let cost price for second house = y Rs

CP = Y


Here we use Loss (L) formula


Loss is given to be 5% (of cost price) i-e

L= 5% of CP


L= 5% of y

L=\frac{5}{100}\times y


Put values in above formula


\Rightarrow \frac{y}{20}-y=-3990

\Rightarrow -\frac{19y}{20}=-3990

\Rightarrow y=\frac{20}{19}\times3990

\Rightarrow y=4200

So cost price for second house is Rs 4200


Loss= CP-SP =4200-3990=210 Rs

So the loss for second house is Rs 210


Net Gain or Gain

Net gain or profit is calculated by subtracting profit from loss or loss from profit (which ever is bigger)

Here loss is bigger than profit ,


Net Loss= Loss-Gain=210-190=20 Rs

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