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September 20, 2014

September 20, 2014

Posted by **Louis** on Saturday, October 22, 2011 at 3:43am.

- Math -
**Reiny**, Saturday, October 22, 2011 at 9:06amtime needed by 2nd faucet alone --- t hrs

rate for 2nd faucet alone = 1/t

time needed by 1st faucet alone --- t+2 hrs

rate of 1st faucet along = 1/(t+2)

combined rate = 1/t + 1/(t+2)

= (t+2 + t)/(t(t+2) = (2t+2)/(t(t+2))

then 1/[ 2t+2)/(t(t+2)) ] = 4/3

t(t+2)/(2t+2) = 4/3

3t^2 + 6t = 8t + 8

3t^2 - 2t - 8 = 0

(t-2)(3t+4) = 0

t = 2 or t = a negative time

the 2nd faucet would take 2 hours alone

the 1st facucet would take 4 hours alone

- Math -
**Anonymous**, Monday, August 20, 2012 at 10:32am1 + 1

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