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April 16, 2014

April 16, 2014

Posted by **jannaline inguito** on Wednesday, September 14, 2011 at 7:39am.

- math -
**Reiny**, Wednesday, September 14, 2011 at 10:06amlet the tank contain V units of volume

rate of first pipe = V/5 units/hr

rate of 2nd pipe = V/x units/hr

So the first pipe runs for 3 hrs

Volume = 3(V/5) = 3V/5

the second pipe runs for 4 hrs.

Volume filled by 2nd pipe = 4(V/x) = 4V/x

3V/5 + 4V/x = V

multiply by 5x

3Vx + 20V = 5xV

3x + 20 = 5x

x = 10

It would take 10 hrs using the 2nd pipe alone

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