A dealer bought two tables for Rs 3120.


A dealer bought two tables for Rs 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then, he found that each table was sold for the same price. Find the cost price of each table.


Los $s$ on the first table $=15 \%$

Therefore, $S . P=C . P\left(\frac{100-l \text { oss } \%}{100}\right)$

S. P $=\frac{85 x}{100}=$ Rs. $0.85 x$

Gain on the second table $=36 \%$

Therefore, S.P $=C . P\left(\frac{100+g \text { ain } \%}{100}\right)$

$S . P=$ Rs. $1.36(3120-x)$

Since both tables have the same S.P,

So, $0.85 \mathrm{x}=1.36(3120-x)$

$0.85 x=4243.20-1.36 x$

$2.21 x=4243.20$


$x=$ Rs. 1920

So, the cost price of $t h e$ first table $i s$ Rs. 1920 .

Cost price of the second table $=$ Rs. $(3120-1920)=$ Rs. 1200

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now