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A pool is fitted with two taps. When operating independently, tap b needs 6
hours more than tap a to fill the pool. Tap a needs 2 hours more than the time
taken when both taps are turned on at the same time. How long does it take
to fill the pool if two taps were turned on together at the same time?

 Mar 31, 2020
 #1
avatar+112282 
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Let  the time  for Tap A to fill  the pool  =  A hours

Let  the time  for Tap B to fill the pool  =  A + 6  hours

 

So  Tap A fills  1/A  of the pool in every hour

And Tap B fills  1/ {A + 6)  of the pool every hour

 

Add these  and we  get  that

 

1 / A  + 1 / ( A + 6)  =     [ A + 6 + A  ] /  [ A (A + 6)]  =  [ 2A + 6 ] / [ A^2 + 6A ]

 

The reciprocal of this is the  time it takes  both  taps working together to  fill the  pool = 

 

[ A^2  + 6A ]  / [2A + 6 ]    [ in hours ]

 

And  we  know  that

 

Time for both taps working together + 2 hours =    =  time for Tap A to fill the pool

 

So we have  that

 

[ A^2  + 6A ] / [2A + 6 } +  2  =   A

 

Simplify

 

A^2  + 6A  / [ 2A + 6 ]  =  A  -  2

 

A^2  + 6A   =  [ A - 2 ] [ 2A + 6]

 

A^2  + 6A =  2A^2  - 4A + 6A  - 12

 

A^2 + - 4A - 12  =  0       factor

 

(A - 6) ( A + 2)  =  0

 

Setting  the first factor to 0  and solving for A  gives us the positive  answer for A =  6 hours  to fill pool

 

And B takes  6 + 6 =  12   hours  to  fill the  pool

 

So....the time for both to fill the pool is

 

1/6  +  1/12

 

[12 + 6 ]  / [ 6 * 12 ]

 

18/ 72        take the reciprocal

 

72 / 18  =  4  hours  for  both to fill the  pool

 

 

cool cool cool

 Mar 31, 2020

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