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Working together, two servicemen can clean a swimming pool in 6 hours. If both of them worked together for the first 3 hours, and the first stopped for one reason or another, it will require the second 4 hours more to rub clean pool. How long would it take each of them separately to do the job?

Give the complete solution. Thank You.

Sagot :

Mlcparra16idn
Let the time needed for the first one to do the job alone be x and for the second be y.

This would mean that their rates would be 1/x and 1/y.

1 / (1/x+1/y) = 6
1 / (x+y/xy) = 6
xy/(x+y) = 6
xy = 6x + 6y

3 ((x+y)/xy) + 4 (1/y) =  1
(3x + 3y + 4x)/xy = 1
7x + 3y = xy

6x + 6y = 7x + 3y
3y = x

xy = 6x + 6y
xy = 6(x+y)
xy = 6(3y+y)
xy = 24y
x = 24

y = 8

Therefore it would take 24 hours for the first person to work alone and 8 hours for the second.
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