If the price of an eraser is reduced by `25%`, a person can buy three more erasers for Rs. `2` . How many erasers can be bought for Rs. `2` as the original price ?

Options

A. A)`10`
B. B)`8`
C. C)`9`
D. D)`12`

Answer

To solve the problem step by step, let's denote the original price of an eraser as \( P \).

### Step 1: Understand the price reduction
The price of the eraser is reduced by 25%. Therefore, the new price \( P' \) can be calculated as:
\[ P' = P - 0.25P = 0.75P \]

### Step 2: Establish the relationship between the number of erasers
Let \( x \) be the number of erasers that can be bought at the original price \( P \) for Rs. 2. Thus, we have:
\[ x = \frac{2}{P} \]

When the price is reduced, the number of erasers that can be bought for Rs. 2 becomes \( x + 3 \):
\[ x + 3 = \frac{2}{P'} \]

### Step 3: Substitute the new price
Substituting \( P' \) into the equation gives:
\[ x + 3 = \frac{2}{0.75P} \]

### Step 4: Simplify the equation
We can rewrite the equation as:
\[ x + 3 = \frac{2}{\frac{3}{4}P} = \frac{2 \cdot 4}{3P} = \frac{8}{3P} \]

### Step 5: Set up the equation
Now we have two equations:
1. \( x = \frac{2}{P} \)
2. \( x + 3 = \frac{8}{3P} \)

### Step 6: Substitute \( x \) from the first equation into the second
Substituting \( x \) into the second equation:
\[ \frac{2}{P} + 3 = \frac{8}{3P} \]

### Step 7: Clear the fractions
To eliminate the fractions, multiply through by \( 3P \):
\[ 3P \left(\frac{2}{P}\right) + 3 \cdot 3P = 8 \]
This simplifies to:
\[ 6 + 9P = 8 \]

### Step 8: Solve for \( P \)
Rearranging gives:
\[ 9P = 8 - 6 \]
\[ 9P = 2 \]
\[ P = \frac{2}{9} \]

### Step 9: Find the original number of erasers
Now we can substitute \( P \) back into the equation for \( x \):
\[ x = \frac{2}{P} = \frac{2}{\frac{2}{9}} = 9 \]

### Conclusion
Thus, the number of erasers that can be bought for Rs. 2 at the original price is \( \boxed{9} \).

---