SOLUTION: I have a problem on my homework that has very little detail in the textbook. I need to know a formula for the following: Two pipes working together can fill a tank in 12hrs. Workin
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Question 377620: I have a problem on my homework that has very little detail in the textbook. I need to know a formula for the following: Two pipes working together can fill a tank in 12hrs. Working alone, the larger one can fill the tank in 18hrs less than the smaller one. How long does it take for the smaller tank working alone? Thank you so much! Found 2 solutions by edjones, ewatrrr:Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! Let x=time it take larger pipe to fill tank and y =time it take larger pipe to fill tank.
y=x+18
1/x + 1/y = 1/12
1/x + 1/(x+18) = 1/12
12(x+18)+12x=x(x+18) Multiply by 12x(x+18)
12x+216+12x=x^2+18x
x^2-6x-216=0
x=18 hr
y=36 hr
.
Ed
You can put this solution on YOUR website! Hi
the larger one can fill the tank in 18hrs less than the smaller one.
Let x represent the time for the smaller alone,
then (x-18hr)would be the time for the larger alone
Per hour is the Equalizer
1/x + 1/(x-18) = 1/12 Mulitiplying each term by 12x(x-18)so all denominators = 1
12(x-18) + 12x = x(x-18)
solving for x
12x - 12*18 +12x = x^2 - 18x
24x - 12*18 = x^2 - 18x
x^2 - 42x + 216 = 0
x = 6 Cannot use
x = 36hr, the time for the smaller alone. 18hr for the larger alone (36-18)
CHECKING our Anwer
1/36 + 1/18 = 2/72 + 4/72 = 6/72 = 1/12
