jgk wrote:

A group of 3 small pumps and 1 large pump is filling a tank. Each of the 3 small pumps works at 2/3 of the rate of the large pump. If all 4 pumps work at the same time, they will fill the tank in what fraction of the time that it would have taken the large pump had it operated alone?

A. 1/6

B. 1/3

C. 2/3

D. 3/4

E. 4/3

Rates in tanks/hour Let large pump fill rate = \(\frac{1}{6}\)

Each small pump's rate is \(\frac{2}{3}\) of larger pump's rate:

\((\frac{1}{6}*\frac{2}{3})=(\frac{2}{18})=\frac{1}{9}\)

Three small pumps' combined work rate:

\((3)(\frac{1}{9})=(\frac{3}{9})=\frac{1}{3}\)

All

four pumps' combined work rate is

\((\frac{1}{6} +\frac{1}{3})=(\frac{3}{6})=\frac{1}{2}\)

Times takenWhen Work is 1, rate and time are inversely proportional. Flip the rate to get the time.*

Time taken by large pump alone:

Rate is \(\frac{1T}{6hrs}\). Time = 6 hours

Time taken by all four:

Rate is \(\frac{1T}{2hrs}\). Time = 2 hours

Fraction?Time taken together as a fraction of time taken by large pump alone?

\(\frac{2hrs}{6hrs}=\frac{1}{3}\)

Answer B

*

That is, work, W, is 1 tank to be filled. So for large pump alone, e.g., time, t = \(\frac{W}{r}\). Rate is \(\frac{1}{6}\). See below; time taken is the work rate, inverted.\(\frac{1T}{(\frac{1T}{6hrs})}= 1T * \frac{6hrs}{1T} = 6 hrs\)
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