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John has to hammer 100 railroad spikes for a new line his

by AAPL » Thu Sep 27, 2018 4:24 am

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John has to hammer 100 railroad spikes for a new line his company is building. Working at a constant rate, he can hammer 8 spikes per hour. When he is halfway done, his coworker Paul begins working at the same constant rate. Compared to John completing all the work alone, how many hours are saved with Paul's help?

A. 5/2
B. 23/8
C. 25/8
D. 17/4
E. 25/4

OA is C.
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John has to hammer 100 railroad spikes for a new line his

by Jay@ManhattanReview » Thu Sep 27, 2018 4:44 am
AAPL wrote:Veritas Prep

John has to hammer 100 railroad spikes for a new line his company is building. Working at a constant rate, he can hammer 8 spikes per hour. When he is halfway done, his coworker Paul begins working at the same constant rate. Compared to John completing all the work alone, how many hours are saved with Paul's help?

A. 5/2
B. 23/8
C. 25/8
D. 17/4
E. 25/4

OA is C.
At the rate of 8 spikes per hour. John would have taken 100/8 = 25/2 hours working alone to complete the work.

Since John finished half the work, it means he put in (25/2)/2 = 25/4 hours.

With the joining of his friend whose rate is the same as that of John, they together would now take (25/4)/2 = 25/8 hours to complete the remaining work.

Total time take = 25/4 + 25/8 = 75/8 hours

Time saved = 25/2 - 75/8 = 25/8 hours

The correct answer: C

Hope this helps!

-Jay
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Last edited by Jay@ManhattanReview on Thu Sep 27, 2018 9:48 pm, edited 1 time in total.
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by fskilnik@GMATH » Thu Sep 27, 2018 7:44 am
AAPL wrote:Veritas Prep

John has to hammer 100 railroad spikes for a new line his company is building. Working at a constant rate, he can hammer 8 spikes per hour. When he is halfway done, his coworker Paul begins working at the same constant rate. Compared to John completing all the work alone, how many hours are saved with Paul's help?

A. 5/2
B. 23/8
C. 25/8
D. 17/4
E. 25/4
\[?\,\,\,:\,\,\,{\text{hours}}\,{\text{saved}}\,\,{\text{ = }}\,\,{\text{Time}}\,\left( {{\text{John}}\,,\,50\,\,{\text{spikes}}} \right) - {\text{Time}}\left( {{\text{John}} \cup {\text{Paul}}\,,\,50\,\,{\text{spikes}}} \right)\]
\[{\text{Time}}\,\left( {{\text{John}}\,,\,50\,\,{\text{spikes}}} \right)\,\, = \,\,50\,\,spikes\,\,\left( {\frac{{1\,\,{\text{h}}}}{{8\,\,{\text{spikes}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \frac{{25}}{4}\,\,{\text{h}}\]
\[{\text{Time}}\,\left( {{\text{John}} \cup {\text{Paul}}\,,\,50\,\,{\text{spikes}}} \right)\,\, = \,\,50\,\,spikes\,\,\left( {\frac{{1\,\,{\text{h}}}}{{16\,\,{\text{spikes}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \frac{{25}}{8}\,\,{\text{h}}\]
Obs.: arrows indicate licit converters.
\[{\text{?}}\,\,{\text{ = }}\,\,\frac{{25 \cdot \boxed2}}{{4 \cdot \boxed2}} - \frac{{25}}{8} = \frac{{25}}{8}\,\,\,\left[ {\text{h}} \right]\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Tue Oct 02, 2018 6:51 pm
AAPL wrote:Veritas Prep

John has to hammer 100 railroad spikes for a new line his company is building. Working at a constant rate, he can hammer 8 spikes per hour. When he is halfway done, his coworker Paul begins working at the same constant rate. Compared to John completing all the work alone, how many hours are saved with Paul's help?

A. 5/2
B. 23/8
C. 25/8
D. 17/4
E. 25/4
If John were to work alone, it would take him 100/8 hours or 200/16 hours.

For John to complete half the job, it takes him 50/8 hours. When his coworker Paul joins him, it takes them 50/16 hours to complete the remaining half of the job, or a total of 50/8 + 50/16 = 100/16 + 50/16 = 150/16 hours to complete the entire job.

Thus, the number of hours saved is 200/16 - 150/16 = 50/16 = 25/8 hours.

Answer: C

Scott Woodbury-Stewart
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