Question

If the price of the eraser is reduced by 25% , a person can buy 2 more erasers for a rupee. How many erasers are available for a rupee?
A.8
B.6
C.4
D.2

Answer 1

Hint: In this question, we have to find the number of erasers available for a rupee. Therefore, we can take it to be a variable x. Then, we can find the cost of each eraser in terms of x. Then, using the given information, we can construct algebraic equations and solve it to get the value of x which will be the required answer to this question.

Complete step-by-step answer:
Let the number of erasers available for a rupee be x. Then,
Cost of each eraser $\times $ x = 1 Rupee
$\Rightarrow $ Cost of each eraser= Rupees $\dfrac{1}{x}$ ………………… (1.1)
Now, from (1.1), $25%$ of the price of an eraser will be
25% of the price of each eraser= 25% of $\dfrac{1}{x}$ = $\dfrac{25}{100}\times x=0.25x...................(1.2)$
Now, if the price of the eraser is reduced by 25%, then from (1.1) and (1.2), we obtain
New price of the eraser= $\dfrac{1}{x}-0.25\times \dfrac{1}{x}=\left( 1-0.25 \right)\dfrac{1}{x}=\dfrac{0.75}{x}.......................(1.3)$
It is given that the person can buy two more erasers when the price of each eraser is reduced by 25%, therefore, he can now buy x+2 erasers in 1 Rupee. Thus, using (1.3), we can write
Total cost of (x+2) erasers= 1 Rupee
$\begin{align}
  & \Rightarrow (x+2)\times \dfrac{0.75}{x}=1\text{ } \\
 & \Rightarrow 0.75x+2\times 0.75=x \\
 & \Rightarrow 2\times 0.75=\left( 1-0.75 \right)x \\
 & \Rightarrow 1.5=0.25x \\
 & \Rightarrow x=\dfrac{1.5}{0.25}=6 \\
\end{align}$
Therefore, we obtain the answer to be 6 which is the original number of erasers the person could buy with 1 Rupee. This matches option (b) and hence (b) is the correct answer to the given question.

Note: We should note that while solving the equations, we have converted each number into its decimal representation. However, we can also write the numbers in their fractional form, for example $0.25=\dfrac{1}{4}$ and solve the equations. The answer will remain the same in any of the two methods.