Math, asked by nitinwaidande5121, 16 days ago

# A and B are two pipes connected to a tank. A is the filler pipe and B is the ejector. The capacity of the tank is 6000L.The ejector pipe can eject water at the rate 4L/min higher than the filler pipe. The filler pipe can fill the tank 5 min later than the ejector pipe can eject. Find the flow rate of the ejector pipe in L/min(approx.).

0

Step-by-step explanation:

## solution:-

ejector pipe   flow rate    =  E  L/Min

Filler Pipe flow rate  =  E - 4  L/Min

Capacity of Tank  = 6000 L

Time Taken by Ejector pipe =  6000/E    Min

Time Taken by filler pipe = 6000/(E - 4)    Min

The filler pipe can fill the tank 5 min later than the ejector pipe can eject.

=>  6000/(E - 4)     = 6000/E    + 5

=> 6000E   = 6000(E - 4)   + 5E(E - 4)

=> 6000E = 6000E - 24000 + 5(E² - 4E)

=> 5(E² - 4E)  - 24000  = 0

=> E² - 4E - 4800  = 0

=>  E  = (4 ± √(4² - 4*(-4800)) ) /2

=> E =   2 +   √4801  ( ignoring - ve value)

=> E  = 2 + 69.3

=> E = 71.3

flow rate of the ejector pipe = 71.3 L/min

## Ihopeitwillhelpyoutounderstand❤

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