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Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

User Malexander
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1 Answer

7 votes

Answer:
(20)/(3)\text{ hours}

Explanation:

Let x be the speed of slower pump and 1.5x be the speed of faster pump to fill the swimming pool .

Then , According to the given question, we have the following equation:-


x+1.5x=(1)/(4)\\\\\rightarrow\ 2.5x=(1)/(4)\\\\\Rightarrow\ x=(1)/(10)=

Now, the time taken by faster pump to fill the pool is given by :-


t=(1)/(1.5x)=(10)/(1.5)=(20)/(3)\text{ hours}

Hence, the faster pump would take
(20)/(3)\text{ hours} to fill the pool if it had worked alone at its constant rate.

User Geoff Clayton
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