A person sells bananas at Rs 3 per piece and incurs a loss of Rs 20, but if he sells at Rs 3.25 per piece he gains Rs 30. Find the total number of bananas sold.

Introduction:
This problem uses linear profit equations with the same unknown quantity. When the selling price per unit changes while the lot size stays fixed, the change in total profit is directly proportional to the change in price multiplied by the number of units. We translate the two scenarios into equations and solve for the number of bananas.

Given Data / Assumptions:

• Selling price case 1: Rs 3 per banana → overall loss Rs 20
• Selling price case 2: Rs 3.25 per banana → overall gain Rs 30
• Let n be the number of bananas and c be the cost price per banana (rupees)

Concept / Approach:
If SP is p, total profit = n * (p − c). Loss is negative profit; gain is positive profit. Set up two equations and eliminate c to get n. This is a standard “two price points, one quantity” technique in profit and loss.

Step-by-Step Solution:
For Rs 3: n * (3 − c) = −20 ⇒ n * (c − 3) = 20For Rs 3.25: n * (3.25 − c) = 30Add the two (c − 3) + (3.25 − c) = 20/n + 30/n ⇒ 0.25 = 50/nHence n = 50 / 0.25 = 200

Verification / Alternative check:
With n = 200, the price increase of Rs 0.25 raises revenue by 200 * 0.25 = Rs 50, which moves profit from −20 to +30 (a difference of Rs 50). Consistent.

Why Other Options Are Wrong:
100/120: Would yield revenue change 25 or 30, not enough to shift −20 to +30.2400: Would imply a profit swing of 600, far larger than observed.

Common Pitfalls:
Mixing up unit profit with lot profit; forgetting that loss is negative in the first equation; trying to determine the cost price first (it cancels out neatly).

Final Answer:
200