A man rides one third of the distance from A to B at the rate of "a" km/hr and the remainder at "2b" km/hr. if he had traveled at the uniform rate of "3c" km/hr, he could have ridden from A to B and back again in the same time.
Prove that 2/c = (1/a) + (1/b)
Answer:
Let the distance between A to B be y kms.
so, time taken for the journey = (y/3) / a + (2y/3) /2b ---(1)
In the same time he can cover a distance of "2y" at '3c" km/hr
time taken = 2y/ 3c ------------(2)
Equating (1) and (2), we get
(y/3) / a + (2y/3) /2b = 2y/ 3c
or, y / 3a + 2y / 6b = 2y/ 3c
or, 1/ 3a + 1 / 3b = 2 / 3c
or, 1/a + 1/b = 2/c
Prove that 2/c = (1/a) + (1/b)
Answer:
Let the distance between A to B be y kms.
so, time taken for the journey = (y/3) / a + (2y/3) /2b ---(1)
In the same time he can cover a distance of "2y" at '3c" km/hr
time taken = 2y/ 3c ------------(2)
Equating (1) and (2), we get
(y/3) / a + (2y/3) /2b = 2y/ 3c
or, y / 3a + 2y / 6b = 2y/ 3c
or, 1/ 3a + 1 / 3b = 2 / 3c
or, 1/a + 1/b = 2/c
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