A dealer bought two tables for Rs 3120. He sold one of them at loss of

1.	A dealer bought two tables for Rs 3120. He sold one of them at loss of 15% and other at a gain of 36%. Then, he found that each table was should for the same price. Find the cost price of each table.
Answer» Cost price of 2 tables = Rs.3120 Let CP of first table = Rs. x Then, CP of second table would be = Rs. 3120 – x Let first table is sold at a gain and second at loss. Selling price of first table = \(x+x\times \frac{36}{100}\) = \(x+\frac{9x}{25}\) = Rs. \(\frac{34x}{25}\) Selling price of second table = (3120 - x) x \(\frac{85}{100}\) = Rs. \(\frac{85\times 3120-85x}{100}\) We have, = \(\frac{34x}{25}\) = \(\frac{85\times 3120-85x}{100}\) = 221x = 85 × 3120 = x = \(\frac{85\times 3120}{221}\) = 1920 Cost price of first table = Rs.1920 Cost price of second table = 3120 – 1920 = Rs.1200

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

Answer»

Cost price of 2 tables = Rs.3120

Let CP of first table = Rs. x

Then, CP of second table would be = Rs. 3120 – x

Let first table is sold at a gain and second at loss.

Selling price of first table = \(x+x\times \frac{36}{100}\) = \(x+\frac{9x}{25}\) = Rs. \(\frac{34x}{25}\)

Selling price of second table = (3120 - x) x \(\frac{85}{100}\) = Rs. \(\frac{85\times 3120-85x}{100}\)

We have,

= \(\frac{34x}{25}\) = \(\frac{85\times 3120-85x}{100}\)

= 221x = 85 × 3120

= x = \(\frac{85\times 3120}{221}\) = 1920

Cost price of first table = Rs.1920

Cost price of second table = 3120 – 1920 = Rs.1200