A man travels three-fifths of a distance AB at a speed of 3a, and the

1.	A man travels three-fifths of a distance AB at a speed of 3a, and the remaining at a speed of 2b. If he goes from B to A and returns at a speed of 5c in the same time, then(a) \(\frac1a+\frac1b=\frac1c\)(b) a + b = c (c) 3a + 2b = 5c(d) \(\frac1a+\frac1b=\frac2c\)
Answer» Time taken to cover AC = \(\frac{3x}{5\times3a}=\frac{x}{5a}hr\) Time taken to cover CB = \(\frac{2x}{5\times2b}=\frac{x}{5b}hr\) Time taken to cover BA and back AB = \(\frac{2x}{5c}\) Given, \(\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}\) ⇒ \(\frac1a+\frac1b=\frac2c\)

A man travels three-fifths of a distance AB at a speed of 3a, and the remaining at a speed of 2b. If he goes from B to A and returns at a speed of 5c in the same time, then(a) \(\frac1a+\frac1b=\frac1c\)(b) a + b = c (c) 3a + 2b = 5c(d) \(\frac1a+\frac1b=\frac2c\)

Answer»

Time taken to cover AC = \(\frac{3x}{5\times3a}=\frac{x}{5a}hr\)

Time taken to cover CB = \(\frac{2x}{5\times2b}=\frac{x}{5b}hr\)

Time taken to cover BA and back AB = \(\frac{2x}{5c}\)

Given, \(\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}\) ⇒ \(\frac1a+\frac1b=\frac2c\)

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A man travels three-fifths of a distance AB at a speed of 3a, and the remaining at a speed of 2b. If he goes from B to A and returns at a speed of 5c in the same time, then(a) \(\frac1a+\frac1b=\frac1c\)(b) a + b = c (c) 3a + 2b = 5c(d) \(\frac1a+\frac1b=\frac2c\)

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