Best Answer

Let speed of stream = y kmph & speed of boat = x kmph

speed downstram((x + y)kmph) = 8 * 60 / 40 kmph

= 12 kmph

speed upstream((x - y) kmph) = 8/1 kmph

= 8 kmph

x + y = 12

x - y = 8

--------------

2x = 20,

2y = 4

x = 10,

y = 2

=>Speed of stream = 2 kmph, Speed of boat = 10 kmp

Q: A sailor goes 8km downstream in 40 minutes and returns in 1 hours Determine the speed of the sailor in still water and speed of the current?

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I presume they meet only back at the bus station. The bus that takes 20 minutes for its route returns to the bus station after 20 minutes, 40 minutes, 60 minutes, and so on - multiples of 20 minutes. The bus that takes 55 minutes for its route returns to the bus station after 55 minutes, 110 minutes, 165 minutes, and so on - multiples of 55 minutes. The will both be back at the bus station at the same time when a multiple of 20 equals a multiple of 55 - at a common multiple of 20 and 55. The first time this happens is the lowest common multiple of 20 and 55. The lcm(20, 55) is 220, so after 220 minutes they will first both be back at the bus station together. There are 60 minutes in an hour ⇒ 220 minutes = 220 ÷ 60 hours = 3 hours 40 minutes. Which means that the time they first meet back at the bust station is 6am plus 3 hours 40 minutes which is 9:40 am.

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