d Word problem 4_ | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word problem 4_

Problem 4: A manufacturer sales a product to a whole sale dealer at 20% profit. The whole sale dealer sales the product to a retailer at 15% profit. The retailer sales the product to a customer in Rs 897 with 30% profit. What is the making cost of the manufacturer?



Let’s suppose cost price of manufacturer is ‘x’

i-e CP = x

It is given that manufacturer’s profit is 20% of cost price;

i-e P= \frac{20}{100} of CP

\Rightarrow P= \frac{20}{100} \times x


\Rightarrow P= \frac{20}{100} \ x

Now using formula for sale price;



\Rightarrow SP=\frac{20x+100x}{100}

\Rightarrow SP=\frac{120x}{100}

Whole Sale Dealer

Now this will become  CP for whole sale dealer ;

\Rightarrow CP=\frac{120x}{100}

now wsd will earn 15% profit on his this CP;


\Rightarrow P=15% of\frac{120x}{100}

\Rightarrow P=\frac{15}{100} \times(\frac{120x}{100})

\Rightarrow P=\frac{18x}{100}

Again using formula for SP (for whole sale dealer);





Now this SP of whole sale dealer will become CP of retailer;

\Rightarrow CP=\frac{138x}{100}

Retailer earns 30% profit on this CP;

\Rightarrow P= \frac{30}{100} of (\frac{138x}{100})

\Rightarrow P= \frac{30}{100} \times (\frac{138x}{100})

\Rightarrow P= \frac{207x}{500}

Now again for sale price use;


\Rightarrow SP= \frac{207x}{500}+ \frac{138x}{100}

But given in question;

Retailer's SP= Rs 897

\Rightarrow 897= \frac{207x}{500}+ \frac{138x}{100}

\Rightarrow 897= \frac{897x}{500}

\Rightarrow x= \frac{897\times 500}{897}

\Rightarrow x= RS 500

So CP of manufacturer is 500 Rs.

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