A man sold 12 candies for $10, incurring a loss of b%. He then sold 12 candies…

Concept: When the same item is sold at two different prices such that one sale is a loss of b% and the other is a profit of the SAME b%, both sales share one cost price C. This gives two equations, P1 = C × (1 − b/100) for the loss and P2 = C × (1 + b/100) for the profit; dividing one equation by the other eliminates C, leaving a single equation in b.

Applying this to the given prices ($10 and $12):

1. Let the cost price of the 12 candies be C.

2. Selling at $10 gives a loss of b%: 10 = C × (1 − b/100).

3. Selling at $12 gives a profit of b%: 12 = C × (1 + b/100).

4. Divide the second equation by the first to eliminate C: 12/10 = (1 + b/100) / (1 − b/100), i.e. 1.2 = (1 + b/100) / (1 − b/100).

5. Cross-multiply: 1.2 × (1 − b/100) = 1 + b/100, so 1.2 − 1.2b/100 = 1 + b/100.

6. Collect the b terms on one side: 1.2 − 1 = b/100 + 1.2b/100, so 0.2 = 2.2b/100.

7. Solve for b: b = (0.2 × 100) / 2.2 = 20 / 2.2 ≈ 9.09.

Cross-check:

With b ≈ 9.09%, the cost price works out to C = 10 / (1 − 9.09/100) ≈ $11. Checking the second sale: $11 × (1 + 9.09/100) ≈ $12, which matches exactly. This also matches the shortcut for this problem type: b = (P2 − P1) / (P2 + P1) × 100 = (12 − 10) / (12 + 10) × 100 = 2/22 × 100 ≈ 9.09%, confirming the result.

Thus, b ≈ 9.09%.