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06 Aug 2026, 11:41
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Two workers, Ava and Ben, are hired to mow a field at a community park. If Ava and Ben work at their respective constant rates, by what percent is the time Ben would take to mow the field alone less than the time Ava would take to mow the field alone?
(1) Ben, working alone at his constant rate, would take 2 hours less to mow the field than Ava, working alone at her constant rate, would take.
(2) Ava and Ben, working simultaneously and independently at their respective constant rates, can mow the field in \(\frac{3}{5}\) of the time that Ben, working alone at his constant rate, would take to mow the field.
(1) Ben, working alone at his constant rate, would take 2 hours less to mow the field than Ava, working alone at her constant rate, would take.
(2) Ava and Ben, working simultaneously and independently at their respective constant rates, can mow the field in \(\frac{3}{5}\) of the time that Ben, working alone at his constant rate, would take to mow the field.
06 Aug 2026, 11:41
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Official Solution:
Two workers, Ava and Ben, are hired to mow a field at a community park. If Ava and Ben work at their respective constant rates, by what percent is the time Ben would take to mow the field alone less than the time Ava would take to mow the field alone?
Let \(A\) be Ava’s rate and \(B\) be Ben’s rate, measured in fields per hour.
Ava’s time to mow the field alone is \(\frac{1}{A}\).
Ben’s time to mow the field alone is \(\frac{1}{B}\).
The question asks by what percent \(\frac{1}{B}\) is less than \(\frac{1}{A}\).
(1) Ben, working alone at his constant rate, would take 2 hours less to mow the field than Ava, working alone at her constant rate, would take.
This gives only the difference between their times, not the percent difference.
For example, if Ava takes 6 hours and Ben takes 4 hours, then the percent by which Ben’s time is less is:
\(\frac{2}{6} * 100 = 33\frac{1}{3}\%\)
But if Ava takes 10 hours and Ben takes 8 hours, then the percent by which Ben’s time is less is:
\(\frac{2}{10} * 100 = 20\%\)
So the percent can be different.
Not sufficient.
(2) Ava and Ben, working simultaneously and independently at their respective constant rates, can mow the field in \(\frac{3}{5}\) of the time that Ben, working alone at his constant rate, would take to mow the field.
Since Ava’s rate is \(A\) and Ben’s rate is \(B\), their combined rate is \(A + B\) fields per hour. So the time they take together is:
\(\frac{1}{A + B}\)
Ben’s time to mow the field alone is:
\(\frac{1}{B}\)
So statement (2) gives:
\(\frac{1}{A + B} = \frac{3}{5} * \frac{1}{B}\)
\(5B = 3A + 3B\)
\(2B = 3A\)
\(\frac{B}{A} = \frac{3}{2}\)
Thus:
Ben’s rate : Ava’s rate \(= 3:2\)
Since time is inversely proportional to rate:
Ben’s time : Ava’s time \(= 2:3\)
Therefore, Ben’s time is \(\frac{1}{3}\) less than Ava’s time, or \(33\frac{1}{3}\%\) less.
Sufficient.
Answer: B
Two workers, Ava and Ben, are hired to mow a field at a community park. If Ava and Ben work at their respective constant rates, by what percent is the time Ben would take to mow the field alone less than the time Ava would take to mow the field alone?
Let \(A\) be Ava’s rate and \(B\) be Ben’s rate, measured in fields per hour.
Ava’s time to mow the field alone is \(\frac{1}{A}\).
Ben’s time to mow the field alone is \(\frac{1}{B}\).
The question asks by what percent \(\frac{1}{B}\) is less than \(\frac{1}{A}\).
(1) Ben, working alone at his constant rate, would take 2 hours less to mow the field than Ava, working alone at her constant rate, would take.
This gives only the difference between their times, not the percent difference.
For example, if Ava takes 6 hours and Ben takes 4 hours, then the percent by which Ben’s time is less is:
\(\frac{2}{6} * 100 = 33\frac{1}{3}\%\)
But if Ava takes 10 hours and Ben takes 8 hours, then the percent by which Ben’s time is less is:
\(\frac{2}{10} * 100 = 20\%\)
So the percent can be different.
Not sufficient.
(2) Ava and Ben, working simultaneously and independently at their respective constant rates, can mow the field in \(\frac{3}{5}\) of the time that Ben, working alone at his constant rate, would take to mow the field.
Since Ava’s rate is \(A\) and Ben’s rate is \(B\), their combined rate is \(A + B\) fields per hour. So the time they take together is:
\(\frac{1}{A + B}\)
Ben’s time to mow the field alone is:
\(\frac{1}{B}\)
So statement (2) gives:
\(\frac{1}{A + B} = \frac{3}{5} * \frac{1}{B}\)
\(5B = 3A + 3B\)
\(2B = 3A\)
\(\frac{B}{A} = \frac{3}{2}\)
Thus:
Ben’s rate : Ava’s rate \(= 3:2\)
Since time is inversely proportional to rate:
Ben’s time : Ava’s time \(= 2:3\)
Therefore, Ben’s time is \(\frac{1}{3}\) less than Ava’s time, or \(33\frac{1}{3}\%\) less.
Sufficient.
Answer: B
