SOLUTION: two water taps together can fill a tank in 9 3/8 hours.the tap of larger diamer taker 10 hours less than the smaller one to fill the tank separately find the time in which each tap

Question 1054900: two water taps together can fill a tank in 9 3/8 hours.the tap of larger diamer taker 10 hours less than the smaller one to fill the tank separately find the time in which each tap can separately fill the tank.
Found 2 solutions by jorel555, josmiceli:
Answer by jorel555(1290) (Show Source): You can put this solution on YOUR website!
Let n be the larger tap. Then the smaller tap would be n+10. So:
1/n+1/(n+10)=1/(75/8)=8/75
75(n+10)+75n=8n�+80n
8n�-70n-750=0
4n�-35n-375=0
(4n+25)(n-15)=0
n=15,-25/4
Throwing out the negative result, we get the bigger tap filling the tank in 15 hours, and the smaller tap in 25 hrs.. ☺☺☺☺
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Rate of filling with both taps open:
[ 1 tank filled ] / [ 9.375 hrs ]
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Let = time in hrs for smaller tap to fill tank
= time in hrs for larger tap to fill tank
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Add the rates of filling of the 2 taps to get
the rate of filling with both taps open
----------------------------------------------

Multiply both sides by

Use quadratic formula to solve for

---------------------

hrs
and

-----------------
The smaller tap takes 25 hrs by itself
The larger tap takes 15 hrs by itself
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check:

OK

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