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A water tank is filled by two pipes in 48 minutes. If the larger pipe alone can fill it in 40 minutes less time than the smaller pipe, find the time it takes for the smaller pipe to fill the tank alone.

 Apr 9, 2020
 #1
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Let  the  time  for  one pipe to fill the tank be  = T  (in minutes)

 

Let  the  time for the other pipe  to fill the T + 40    (in minutes)

 

So....in one  minute  the  first pipe fills  1/T  of the tank  and the second fills   1 / (T + 40)  of the tank

 

And both pipes fill1/48  of the tank in one minute

 

So....we  have  this equation

 

1 /T  +  1/ (T + 40)   = 1/48      simplify

 

(T + 40 + T ) / [ (T) (T + 40) ] =  1/48        cross-multiply

 

48 (2T + 40)  =  T(T +  40)

 

96T + 1920   = T^2  + 40T       rearrange as

 

T^2 - 56T  - 1920  =  0         factor  as

 

(T - 80)  (T + 24)  = 0

 

The first factor  set to  0 and solved   produces  a positive solution  for T  = 80  (min)

 

The  time  for the smaller pipe to fill the tank  =  T + 40  =  120  min

 

 

cool cool cool

 Apr 9, 2020

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