Question

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

Answer 1

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

Situation 1 : By selling a saree at 8 % profit and a sweater at 10 % discount, the shopkeeper gets Rs. 1008

$\Rightarrow \frac{108x}{100}+\frac{90y}{100}=1008$

$\Rightarrow \frac{27x}{25}+\frac{9y}{10}=\frac{1008}{1}$

Taking L.C.M. of the denominators and then solving it, we get,

$54x+45y=50400$ ...(1)

Situation 2 : By selling a saree at 10 % profit and a sweater at 8 % discount, the shopkeeper gets Rs. 1028.

$\Rightarrow \frac{110x}{100}+\frac{92y}{100}=1028$

$\Rightarrow \frac{11x}{10}+\frac{23y}{25}=\frac{1028}{1}$

Taking L.C.M. of the denominators and then solving it, we get,

$\Rightarrow 55x+46y=51400$ ..(2)

Now, multiplying the equation (1) by 55 and (2) by 54, we get.

$(54x+45y=50400)\times 55$

$=2970x+2475y=2772000$ ...(3)

$(55x+46y=51400)\times 54$

$=2970x+2484y=2775600$ ...(4)

Now, subtracting (3) from (4), we get.

$(2970x+2484y)-(2970x+2475y)=2775600-2772000$

$\Rightarrow 9y=3600$

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

$\Rightarrow y=400$

Putting the value of $y=400$ in (1), we get.

$54x+45y=50400$

$\Rightarrow 54x+(45\times 400)=50400$

$\Rightarrow 54x+18000=50400$

$\Rightarrow 54x=50400-18000$

$\Rightarrow 54x=32400$

$\Rightarrow x=\frac{32400}{54}$

$\Rightarrow x=600$

So, the cost price of a saree is Rs. $600$ and the list price of a sweater is Rs. $400$