SOLUTION: hello please help me to solve this,thankyou!! two hoses, when connected to a swimming pool, can fill the pool in 4 hours. If the larger hose alone is used, it can fill the pool in

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Question 1197446: hello please help me to solve this,thankyou!!
two hoses, when connected to a swimming pool, can fill the pool in 4 hours. If the larger hose alone is used, it can fill the pool in 6 hours less than the smaller hose how long will it take the smaller hose to fill the swimming pool alone?
Found 3 solutions by MathLover1, josgarithmetic, MathTherapy:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

We let the time taken when ....
the larger hose alone is used be
and the smaller hose as .
This would make their rates and respectively.
Take note that:

so,
of both hoses =
rate of both hoses =























This would mean that or but cannot be since that would make the time taken for the larger hose negative.
Therefore it will take the smaller hose hours to fill the swimming pool alone.

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
                       TIME       RATE

smallhose                x       1/x

bighose                  x-6     1/(x-6)

combined                  4      1/4

Combined-rate

-




------------ x can only be 12.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

hello please help me to solve this,thankyou!!
two hoses, when connected to a swimming pool, can fill the pool in 4 hours. If the larger hose alone is used, it can fill the pool in 6 hours less than the smaller hose how long will it take the smaller hose to fill the swimming pool alone?
Let time the smaller hose takes be S
Then time the larger hose takes = S - 6
Since both, when turned on, fills it in 4 hours, we get: 
                                                     4(S - 6) + 4S = S(S - 6) ------ Multiplying by LCD, 4S(S - 6)
                                                        
                                                                 0 = (S - 12)(S - 2)
                 
                 Time the smaller hose takes, working alone, or        or        S = 2 hours (ignore)
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