Question

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 difference between the cost price first and the second table.

Answer 1

Hint: The "by substitution" approach involves solving one of the equations (you choose which one) for one of the variables (you choose which one), then plugging it into the other equation, it is said to be "substituting" for the chosen variable and solving for the other. The first variable is then back-solved.

Complete step-by-step answer:
From the question, we understood that
The cost price of \[2\] tables= Rs. \[3120\]
Let Cost price of the first table be \[Rs.\;x\] .
Then, we can find that the cost price of the second table will be \[Rs.\;3120 - x\] .
Let's say the first table is sold for a profit and the second is sold for a loss.
The equation first table’s profit price is:
$ \Rightarrow $Selling price of first table = \[x + x \times \dfrac{{36}}{{100}}\]
$ \Rightarrow $Selling price of first table \[ = x + \dfrac{{9x}}{{25}}\]
$ \Rightarrow $Selling price of first table \[ = Rs.\dfrac{{34x}}{{25}}\]
Then the equation for the second table’s loss price is:
$ \Rightarrow $Selling price of second table = \[(3120 - x) \times \dfrac{{85}}{{100}}\]
$ \Rightarrow $ Selling price of second table \[ = Rs.\dfrac{{(85 \times 3120) - 85x}}{{100}}\]
So they can be equated (as given in the question) as follows;
 \[ \Rightarrow \dfrac{{34x}}{{25}} = \dfrac{{85 \times 3120 - 85x}}{{100}}\]
 \[ \Rightarrow 221x = 85 \times 3120\]
 \[ \Rightarrow x = \dfrac{{85 \times 3120}}{{221}}\]
 \[ \Rightarrow x = 1920\]
Cost price of first table \[ = Rs.\;1920\]
Cost price of second table \[ = Rs.\;(3120 - 1920) = Rs.\;1200\]
The difference between the cost price of first and second table is \[Rs.\;(1920 - 1200) = Rs.\;720\]
Hence, the final answer is found to be \[Rs.\;720\] .
So, the correct answer is “ \[Rs.\;720\] ”.

Note: Actually, the truth is that the possibility of error in these types of questions can be at the point where you have to calculate the cost of price of second after calculating the cost of price of first. So we should actually always apply the right formula and the right substitutions to get the correct solution.