Question

A man sold a chair and a table together for Rs. 1520 thereby making a profit of 25% on the chair and 10% on table. By selling them together for Rs. 1535 he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
(a) C.P. of chair $=Rs.800$ , C.P. of table $=Rs.500$
(b) C.P. of chair $=Rs.700$ , C.P. of table $=Rs.900$
(c) C.P. of chair $=Rs.600$ , C.P. of table $=Rs.700$
(d) None of these

Answer 1

Hint: First, we will assume the cost price of the chair as some variable x and that of table to by y. Then we will convert the given statement in the form of an equation. So, we will have two equation i.e. $x+25\%x+y+10\%y=1520$ and $x+10\%x+y+25\%y=1535$ . On solving this and substituting the values in the equation, we will find values of x and y which will be our required answer.

Complete step-by-step answer:
Here, we are told that if the chair is sold making 25% profit and the table is told at 10% profit then, the total selling price is Rs. 1520. We will assume the chair to be variable x and table to be y. So, in mathematical form we can write it as
$x+25\%x+y+10\%y=1520$
On further solving, we can write it as
$x+\dfrac{25}{100}x+y+\dfrac{10}{100}y=1520$
On taking LCM, we will get
$100x+25x+100y+10y=152000$
$125x+110y=152000$ …………………………..(1)
Now, similarly if the chair is sold making 10% profit and the table is told at 25% profit then, the total selling price is Rs. 1535. So, we can write this as
$x+10\%x+y+25\%y=1535$
On further solving, we can write it as
$x+\dfrac{10}{100}x+y+\dfrac{25}{100}y=1535$
On taking LCM, we will get
$100x+10x+100y+25y=153500$
$110x+125y=153500$ ……………………….(2)
On taking 5 common from both equations (1) and (2), we will get answer as
$25x+22y=30400$ …………………..(3)
$22x+25y=30700$ ……………………..(4)
Now, we will multiply equation (3) with 22 and (4) with 25 and then subtracting it, we will get as
$\begin{align}
  & \text{ }550x+484y=668800 \\
 & \underline{{}^{-}550x\overset{-}{\mathop{+}}\,625y={}^{-}767500} \\
 & \text{ }0x-141y=-98700 \\
\end{align}$
Thus, we get value of y as
$-141y=-98700$
$y=\dfrac{-98700}{-141}=700$
Now, we will substitute this value of y into (3), we will get as
$25x+22\left( 700 \right)=30400$
On further solving we will get as
$25x=30400-15400=15000$
$x=\dfrac{15000}{25}=600$
Thus, the cost price of chair i.e. x is Rs. 600 and that of table i.e. y is Rs. 700.
Hence, option (c) is the correct answer.

Note: Another method of solving this problem is by putting the values given in option in the equations $x+25\%x+y+10\%y=1520$ and $x+10\%x+y+25\%y=1535$ . If we take option (a): then the cost price of chair is Rs. 800 and table is Rs. 500. So, if we put in equation, we will get $800+\dfrac{25}{100}\cdot 800+500+\dfrac{10}{100}\cdot 500$
On solving we get answer as
$800+200+500+50=1550$ that is not equal to 1520. So, option (a) is not correct. In the same way by putting all the options and seeing which equation satisfies, will give us the same answer.