A shopkeeper sells a saree at $ 8 \% $ profit and a sweater at $ 10 \% $ discount, thereby, getting a sum Rs 1008 . If she had sold the saree at $ 10 \% $ profit and the sweater at $ 8 \% $ discount, she would have got Rs 1028 . Find the cost price of the saree and the list price (price before discount) of the sweater.

Given:

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs. 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs. 1028.

To do:

We have to find the cost price of the saree and the list price (price before discount) of the sweater.

Let the cost price of the saree and the list price of the sweater be $x$ and $y$ respectively.

Selling price of the saree when it is sold at a profit of 8%$=x+\frac{8}{100}x=1.08x$

Selling price of the sweater when it is sold at a discount of 10%$=y-\frac{10}{100}y=0.9y$

Selling price of the saree when it is sold at a profit of 10%$=x+\frac{10}{100}x=1.1x$

Selling price of the sweater when it is sold at a discount of 8%$=y-\frac{8}{100}y=0.92y$

According to the question,

$1.08x + 0.9y = 1008$.....(i)

$1.1x + 0.92y = 1028$.....(ii)

Multiplying equation (i) by 0.92 on both sides, we get,

$0.92(1.08x+0.9y)=0.92(1008)$

$0.9936x+(0.92)(0.9)y=927.36$.....(iii)

Multiplying equation (ii) by 0.9 on both sides, we get,

$0.9(1.1x+0.92y)=0.9(1028)$

$0.99x+(0.9)(0.92)y=925.2$.....(iv)

Subtracting equation (iv) from equation (iiii), we get,

$[0.9936x+(0.9)(0.92)y]-[0.99x+(0.9)(0.92)y]=927.36-925.2$

$0.9936x-0.99x+(0.9)(0.92)y-(0.9)(0.92)y=2.16$

$0.0036x=2.16$

$x=\frac{2.16}{0.0036}$

$x=\frac{21600}{36}$

$x=600$

Substituting $x=600$ in equation (ii), we get,

$1.08(600)+0.9y=1008$

$648+0.9y=1008$

$0.9y=1008-648$

$0.9y=360$

$y=\frac{360}{0.9}$

$y=\frac{3600}{9}$

$y=400$

The cost price of the saree and the list price (price before discount) of the sweater are Rs. 600 and Rs. 400 respectively.

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Updated on: 10-Oct-2022

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