Question

A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and returns at a speed 5c in the same time, then:

• A. \( \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \)
• B. \( a + b = c \)
• C. \( \frac{1}{a} + \frac{1}{b} = \frac{2}{c} \)
• D. None of these

Hint:

Always use the formula: Time = Distance ÷ Speed. If total time is equal for round trips, equate individual time expressions carefully.

Answer 1

The Correct Option is A

A man travels three-fifths of a distance \( AB \) at a speed \( 3a \), and the remaining two-fifths at a speed \( 2b \). Let total distance \( AB = d \).

Time from A to B:
First part: \( \frac{3d}{5} \) at speed \( 3a \) ⇒ Time = \( \frac{3d}{5 \cdot 3a} = \frac{d}{5a} \)
Second part: \( \frac{2d}{5} \) at speed \( 2b \) ⇒ Time = \( \frac{2d}{5 \cdot 2b} = \frac{d}{5b} \)
Total time from A to B = \( \frac{d}{5a} + \frac{d}{5b} \)

Time from B to A and back to B:
Total distance = \( 2d \) at speed \( 5c \) ⇒ Time = \( \frac{2d}{5c} \)

Given: Both times are equal:
\[ \frac{d}{5a} + \frac{d}{5b} = \frac{2d}{5c} \] Cancel \( d \) and multiply all terms by 5: \[ \frac{1}{a} + \frac{1}{b} = \frac{2}{c} \div 2 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \]

Answer: \( \boxed{\frac{1}{a} + \frac{1}{b} = \frac{1}{c}} \)