Two dealers in sequence — An article passes through two hands and finally sells at 40% profit over the original cost. If the first dealer's profit is 20%, what is the second dealer's profit percentage?

Introduction / Context:
Successive profits compound multiplicatively when measured on their own respective cost bases. Here the final selling price is 40% above the original cost, and the first dealer takes 20% profit. The second dealer’s profit is then computed with respect to the first dealer’s selling price (the second dealer’s cost).

Given Data / Assumptions:

• Original cost = C.
• First dealer sells at 1.20C.
• Final price to the customer = 1.40C.

Concept / Approach:
The second dealer’s profit% is relative to his cost, which is 1.20C. Compute (1.40C − 1.20C) / 1.20C * 100.

Step-by-Step Solution:

Verification / Alternative check:
Factor view: (1.20) * (1 + x) = 1.40 ⇒ 1 + x = 1.40/1.20 = 1.1666… ⇒ x = 16 2/3%.

Why Other Options Are Wrong:
15 1/3%, 13 2/3%, and 12 1/2% do not satisfy the exact ratio 1.40/1.20. 20% misreads the base as the original cost instead of the second dealer’s cost.

Common Pitfalls:
Subtracting percentages directly instead of using the correct base for the second dealer (his own cost).

Final Answer:
16 2/3%