If the price of rice is reduced by 20%, one can buy 2 kg more for Rs. 100. The reduced price of rice is

Options

A. Rs. 50 per kg.
B. Rs. 10 per kg.
C. Rs. 40 per kg.
D. Rs. 5 per kg.

Answer

To solve the problem, we need to find the reduced price of rice after a 20% reduction that allows us to buy 2 kg more for Rs. 100.

### Step-by-Step Solution:

1. **Let the original price of rice be \( X \) Rs. per kg.**
- This is our starting point to understand the price before any reduction.

2. **Calculate the reduced price after a 20% reduction.**
- A 20% reduction means the new price will be \( X - 0.2X = 0.8X \) Rs. per kg.

3. **Determine how much rice can be bought at the original price with Rs. 100.**
- At the original price \( X \), the quantity of rice that can be bought for Rs. 100 is \( \frac{100}{X} \) kg.

4. **Determine how much rice can be bought at the reduced price with Rs. 100.**
- At the reduced price \( 0.8X \), the quantity of rice that can be bought for Rs. 100 is \( \frac{100}{0.8X} \) kg.

5. **Set up the equation based on the information given in the problem.**
- According to the problem, the difference in quantity bought at the reduced price and the original price is 2 kg:
\[ \frac{100}{0.8X} - \frac{100}{X} = 2 \]

6. **Simplify the equation.**
- To simplify, find a common denominator:
\[ \frac{100X - 100(0.8X)}{0.8X^2} = 2 \]
- This simplifies to:
\[ \frac{100X - 80X}{0.8X^2} = 2 \]
- Which further simplifies to:
\[ \frac{20X}{0.8X^2} = 2 \]

7. **Cross-multiply to solve for \( X \).**
\[ 20X = 2 \times 0.8X^2 \]
\[ 20X = 1.6X^2 \]
- Rearranging gives:
\[ 1.6X^2 - 20X = 0 \]

8. **Factor out \( X \).**
\[ X(1.6X - 20) = 0 \]
- This gives us two solutions: \( X = 0 \) or \( 1.6X - 20 = 0 \).

9. **Solve for \( X \).**
\[ 1.6X = 20 \implies X = \frac{20}{1.6} = 12.5 \]

10. **Calculate the reduced price.**
- The reduced price is:
\[ 0.8X = 0.8 \times 12.5 = 10 \text{ Rs. per kg} \]

### Final Answer:
The reduced price of rice is **10 Rs. per kg**.