# 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.

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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.

Solution:

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.

Updated on 10-Oct-2022 13:27:24