Question

An old cycle was purchased for Rs. 600, Rs. 200 was spent on its repair. It was sold for Rs. 500. Its loss percent.
(a) \[16\dfrac{2}{3}%\]
(b) \[33\dfrac{2}{3}%\]
(c) \[37\dfrac{1}{2}%\]
(d) \[40%\]

Answer 1

Hint: To solve this question, we will find the buying price from the question and add the repair cost in the buying price. Then, we will find the selling price from the question and then find the loss percentage from the information we’ve collected using the equation$\text{pc}=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}}\times 100$, where pc is the percentage value, ${{V}_{1}}$ is the cost incurred and ${{V}_{2}}$ is the selling price.

Complete step-by-step answer:
From the given information, the buying price of the old cycle is Rs. 600 and the money spent on repairs is Rs. 200.
Thus, the total cost incurred to the buyer will be
$\Rightarrow $ (buying price + money spent on repairs)
$\Rightarrow $ (600 + 200)
$\Rightarrow $ 800
Now, from the given information, we know that he sold the cycle for Rs. 500.
Therefore, ${{V}_{1}}=800$ and ${{V}_{2}}=500$.
To find the percentage change, we use the equation $\text{pc}=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}}\times 100$ and substitute ${{V}_{1}}=800$ and ${{V}_{2}}=500$.
$\begin{align}
  & \Rightarrow \text{pc}=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}}\times 100 \\
 & \Rightarrow \text{pc}=\dfrac{800-500}{800}\times 100 \\
\end{align}$
We find the difference of 800 and 500 and divide it by 800. Then, we multiply the quotient by 100 to get the percentage.
$\begin{align}
  & \Rightarrow \text{pc}=\dfrac{300}{800}\times 100 \\
 & \Rightarrow \text{pc}=0.375\times 100 \\
 & \Rightarrow \text{pc}=37.5\% \\
\end{align}$
37.5 can also be written as \[37\dfrac{1}{2}\]. Thus, the percentage change is \[37\dfrac{1}{2}%\].
Therefore, option (c) is the correct option.

Note: This question is very simple to solve if the student knows the percentage change equation $\text{pc}=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}}\times 100$. Students are advised to be careful while substituting the values of ${{V}_{1}}$ and ${{V}_{2}}$, as it may drastically change the solution.