Bunuel wrote:

A certain machine fills jars of molasses by drawing molasses at a constant rate from a supply tank that holds l liters of molasses when full. If the machine draws m liters of molasses from the supply take in s seconds and no additional molasses is added to the tank, how many minutes will it take the machine to empty a full supply tank?

A. l−60m/s

B. 60ls/m

C. 60lm/s

D. m/(60ls)

E. ls/(60m)

I picked numbers, centered around 60 secs/min. Tank capacity

\(l\) is a multiple of 60. Drain amount

\(\frac{m}{s}\) divides easily into 60.

Let

\(l = 600\) Let

\(m = 5\)Let

\(s = 1\)Rate is

\(\frac{5l}{1s}*\frac{60s}{1min}=\frac{300l}{1min}\)\(\frac{W}{r}=t\)

\(\frac{600L}{(\frac{300L}{1min})}= 600L* \frac{1min}{300L}= 2 min\)With

\(l = 600\),

\(m = 5\), and

\(s = 1\), find the answer choice where time taken = 2 minutes

Immediately eliminate answers A, B, C, and D.

For (A) and (C),

\(s = 1\): That tiny denominator would require a numerator of 2. But

\(l =600\), in the numerators, is huge. Not even close.

For (B) and (D):

\(m = 10\): That is too small a numerator to have

\(l = 600\) in the denominator. By POE, E remains.

E. ls/(60m)

\(\frac{(600*1)}{(60)(5)}=\frac{600}{300}=2\). That is a match.

Answer E

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