Question

A man bought 2 boxes for Rs.1300. He sold one box at a profit at 20% and another box at 12%. If the selling price of both boxes are the same, then find the cost price of each box.

Answer 1

Hint: we should know about the definition of profit and loss. If the difference of selling price and cost price is positive, then the profit is said to be obtained. If the difference of selling price and cost price is negative, then the loss is said to be obtained. We know that if the C.P of an item is x, then the profit percentage is equal to \[\dfrac{profit}{C.P}\times 100\]. We know that if C.P of an item is x, then the loss percentage is equal to \[\dfrac{loss}{C.P}\times 100\]

Complete step by step answer:
From the question, it was given that a man bought 2 boxes for Rs.1300. He sold one box at a profit at 20% and another box at 12%. It was given that the selling boxes of both boxes are the same.
Let us assume the selling price of each box is equal to P.
Let us assume the cost price of the first box is equal to \[{{P}_{1}}\].
Let us assume the cost price of the second box is equal to \[{{P}_{2}}\].
We know that the sum of the cost price of both boxes is equal to Rs.1300.
\[\Rightarrow {{P}_{1}}+{{P}_{2}}=1300....(1)\]
We were given that the first box given a profit percentage is equal to 20%.
We know that
\[\text{profit percentage=}\dfrac{profit}{C.P}\times 100\]
By using this concept, we get
\[\begin{align}
  & \Rightarrow 20=\dfrac{\operatorname{Profit}\text{of box 1}}{{{P}_{1}}}\times 100 \\
 & \Rightarrow \dfrac{20}{100}=\dfrac{\operatorname{Profit}\text{ of first box}}{{{P}_{1}}} \\
 & \Rightarrow \operatorname{Profit}\text{ of first box}=\dfrac{{{P}_{1}}}{5}....(2) \\
\end{align}\]
We were also given that the second box given a loss percentage is equal to 12%.
\[\text{loss percentage=}\dfrac{loss}{C.P}\times 100\]
By using this concept, we get
\[\begin{align}
  & \Rightarrow 12=\dfrac{\text{Loss of box 2}}{{{P}_{2}}}\times 100 \\
 & \Rightarrow \dfrac{12}{100}=\dfrac{\operatorname{Profit}\text{ of first box}}{{{P}_{2}}} \\
 & \Rightarrow \operatorname{Profit}\text{ of first box}=\dfrac{3{{P}_{2}}}{25}....(2) \\
\end{align}\]
We know that the difference between selling price and cost price gives us either profit or loss.
So, for box 1 we get
\[\begin{align}
  & \operatorname{Profit}\text{ of box 1= S}\text{.P-C}\text{.P} \\
 & \Rightarrow \dfrac{{{P}_{1}}}{5}=P-{{P}_{1}} \\
 & \Rightarrow P=\dfrac{6{{P}_{1}}}{5}....(3) \\
\end{align}\]
So, for box 2 we get
\[\begin{align}
  & \text{Loss of box 1= S}\text{.P-C}\text{.P} \\
 & \Rightarrow \dfrac{-3{{P}_{2}}}{25}=P-{{P}_{2}} \\
 & \Rightarrow P=\dfrac{22{{P}_{2}}}{25}....(4) \\
\end{align}\]
From equation (3) and equation (4), we get
\[\begin{align}
  & \Rightarrow \dfrac{6{{P}_{1}}}{5}=\dfrac{22{{P}_{2}}}{25} \\
 & \Rightarrow 15{{P}_{1}}=11{{P}_{2}} \\
 & \Rightarrow {{P}_{1}}=\dfrac{11}{15}{{P}_{2}}.....(5) \\
\end{align}\]
Now we will substitute equation (5) in equation (1), we get
\[\begin{align}
  & \Rightarrow \dfrac{11}{15}{{P}_{2}}+{{P}_{2}}=1300 \\
 & \Rightarrow {{P}_{2}}\left( \dfrac{26}{15} \right)=1300 \\
 & \Rightarrow {{P}_{2}}=750.....(6) \\
\end{align}\]
Now we will substitute equation (6) in equation (5), we get
\[\begin{align}
  & \Rightarrow {{P}_{1}}=\dfrac{11}{15}(750) \\
 & \Rightarrow {{P}_{1}}=550....(7) \\
\end{align}\]
So, it is clear that the cost price of the first box is equal to Rs.750 and the cost price of the second price is equal to Rs.550.

Note: Students should be careful about the calculation part during solving this problem. If a small mistake is done, then it will lead to a wrong result. Students should take a particular variable to represent a function and this assumption should be followed until the end of the problem. The assumption should not be changed at the middle of the problem or end of the problem. This will give wrong results. Students should have a clear view at these points.