# Two pipes can empty a pool in 3 1/2 hours. if the smaller pipe can empty it in 14 hours alone, how long can the larger pipe empty the same pool alone?

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P = S x T

smaller pipe 's speed is S

P=1, T =14

1 = S (14)

S = 1/14 pool per hour

bigger pipe's speed is S

let x the time of the bigger pipe to empty a pool , P=1

1= S x

S = 1/x pool per hour

together speed is (1/14) + (1/x) = (x/14x) + (14/14x) = (x+14)/14x pool per hour

back to the formula P = S x T

where P = 1 , T = 7/2 (together time 3 1/2 hour)

1 = [(x+14)/14x] (7/2)

manipulate the equation,

1 = (7x +98)/28x

28x = 7x + 98

28x -7x = 98

21x = 98

x = 98/21

x = 14/3 or 4 2/3

it takes 4 2/3 Hours to empty the pool by the bigger pipe alone