A man travels from A to B, covering three-fifths of distance AB at speed 3a…
2024
A man travels from A to B, covering three-fifths of distance AB at speed 3a and the remaining two-fifths at speed 2b. If, in the same time, he completes the entire round trip from B to A and back to B, traveling both legs at a constant speed 5c, which relation follows?
Answer: C. 1/a + 1/b = 2/c — ConceptFor uniform motion, time equals distance divided by speed. For a journey split into parts, total time is the sum of the times for the parts; for a…
- A.
1/a + 1/b = 1/c
- B.
a + b = c
- C.
1/a + 1/b = 2/c
- D.
1/a + 1/b = 1/(2c)
Attempted by 2 students.
Show answer & explanation
Correct answer: C
Concept
For uniform motion, time equals distance divided by speed.
For a journey split into parts, total time is the sum of the times for the parts; for a round trip, the two equal one-way distances are added.
Application
Let AB = D. The first part is 3D/5 at speed 3a, so its time is (3D/5)/(3a) = D/(5a).
The remaining part is 2D/5 at speed 2b, so its time is (2D/5)/(2b) = D/(5b).
Hence the A-to-B time is D/(5a) + D/(5b) = (D/5)(1/a + 1/b).
The route B-to-A-to-B covers 2D at speed 5c, so its time is 2D/(5c).
Equating the times gives (D/5)(1/a + 1/b) = 2D/(5c). Cancelling D/5 yields 1/a + 1/b = 2/c.
Cross-check
Multiplying the final relation by abc gives bc + ac = 2ab, which is the same relation obtained after clearing denominators in the time equation.
Therefore, 1/a + 1/b = 2/c.