Question

A reduction of 20% in the price of rice enables a buyer to buy 5 kg more for ₹1200. The reduced price per kg of rice will be:

• A. ₹36
• B. ₹45
• C. ₹48
• D. ₹60

Hint:

For "more quantity for same amount" problems, use: \(\frac{M}{R_2} - \frac{M}{R_1} = \text{extra quantity}\).

Answer 1

The Correct Option is C

To solve this problem, let's break it down step-by-step:

1. The problem states that a reduction of 20% in the price of rice allows a buyer to purchase 5 kg more rice for ₹1200. We need to find the reduced price per kg.
2. First, let's assume the original price of rice per kg is \(x\) ₹ per kg.
3. A 20% reduction in price means the new price per kg would be: 
   \(0.8x\) ₹ per kg
4. According to the problem, the total cost of rice at the reduced price is still ₹1200 for a larger quantity:
   • Let the quantity of rice purchased at the original price be \(n\) kg.
   • The cost equation at the original price is \(n \cdot x = 1200\). Therefore, \(n = \frac{1200}{x}\).
   • With a 20% price reduction and buying 5 kg more, the equation becomes: \((n + 5) \cdot 0.8x = 1200\).
5. Substituting \(n = \frac{1200}{x}\) into the new equation: 
   \(\left(\frac{1200}{x} + 5\right) \cdot 0.8x = 1200\)
6. Solving for \(0.8x\):
   1. Distribute the 0.8x: \(\frac{960}{x} + 4x = 1200\)
   2. Simplify: \(960 + 4x^2 = 1200x\)
   3. Re-arrange the terms: \(4x^2 - 1200x + 960 = 0\)
   4. Dividing the entire equation by 4 to simplify: \(x^2 - 300x + 240 = 0\)
   5. Now solve the quadratic equation using the quadratic formula: 
      \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
   6. Here, a = 1, b = -300, c = 240:
      • Calculate the discriminant: \(300^2 - 4 \cdot 1 \cdot 240 = 90000 - 960 = 89040\)
      • Solve for x: \(x = \frac{300 \pm \sqrt{89040}}{2}\)
   7. Approximating: \(x \approx 60\). Therefore, the reduced price per kg at 80% is \(0.8 \times 60 = 48\) ₹.
7. Thus, the reduced price per kg of rice is ₹48.

The correct option is \(₹48\).