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.
John has to hammer 100 railroad spikes for a new line his
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At the rate of 8 spikes per hour. John would have taken 100/8 = 25/2 hours working alone to complete the work.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.
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.
- fskilnik@GMATH
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\[?\,\,\,:\,\,\,{\text{hours}}\,{\text{saved}}\,\,{\text{ = }}\,\,{\text{Time}}\,\left( {{\text{John}}\,,\,50\,\,{\text{spikes}}} \right) - {\text{Time}}\left( {{\text{John}} \cup {\text{Paul}}\,,\,50\,\,{\text{spikes}}} \right)\]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
\[{\text{Time}}\,\left( {{\text{John}}\,,\,50\,\,{\text{spikes}}} \right)\,\, = \,\,50\,\,spikes\,\,\left( {\frac{{1\,\,{\text{h}}}}{{8\,\,{\text{spikes}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \frac{{25}}{4}\,\,{\text{h}}\]
\[{\text{Time}}\,\left( {{\text{John}} \cup {\text{Paul}}\,,\,50\,\,{\text{spikes}}} \right)\,\, = \,\,50\,\,spikes\,\,\left( {\frac{{1\,\,{\text{h}}}}{{16\,\,{\text{spikes}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \frac{{25}}{8}\,\,{\text{h}}\]
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.
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If John were to work alone, it would take him 100/8 hours or 200/16 hours.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
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
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