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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.
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.