A man can buy a certain number of notebooks for ₹300. If the price of each notebook increases by ₹5, he can buy 10 fewer notebooks for the same ₹300. What is the original price (in rupees) of each notebook?

Introduction:
This question tests translating a real-life buying situation into an algebraic equation. The amount of money is fixed (₹300), so quantity equals total money divided by price per notebook. When price increases, the quantity decreases. The phrase “10 fewer notebooks” creates a relationship between the two quantities. Setting up the equation correctly is the main challenge; solving the resulting quadratic is straightforward.

Given Data / Assumptions:

• Total money available = ₹300
• Original price per notebook = p rupees
• New price per notebook = p + 5 rupees
• New quantity = original quantity - 10

Concept / Approach:
Quantity purchased = 300/price. So original quantity is 300/p and new quantity is 300/(p+5). The condition says 300/(p+5) = 300/p - 10. Solve this equation for p (take only the positive solution, since price must be positive).

Step-by-Step Solution:
Original quantity = 300/p New quantity = 300/(p + 5) Condition: 300/(p + 5) = 300/p - 10 Multiply by p(p+5): 300p = (p+5)(300 - 10p) Expand: 300p = 300p - 10p^2 + 1500 - 50p Bring terms together: 10p^2 + 50p - 1500 = 0 Divide by 10: p^2 + 5p - 150 = 0 Factor: (p + 15)(p - 10) = 0 Positive solution: p = 10

Verification / Alternative check:
If p = 10, original quantity = 300/10 = 30. New price = 15, new quantity = 300/15 = 20, which is 10 fewer. Works perfectly.

Why Other Options Are Wrong:
Any other price fails the exact “10 fewer” condition when you compute 300/p and 300/(p+5). Only ₹10 satisfies both quantities consistently.

Common Pitfalls:
Using 300 - 5 instead of p + 5, forgetting that quantity is division (not subtraction), or keeping the negative root (price cannot be negative).

Final Answer:
The original price of each notebook is ₹10.