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# John has to hammer 100 railroad spikes for a new line his company is b

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Math Expert
Joined: 02 Sep 2009
Posts: 51218
John has to hammer 100 railroad spikes for a new line his company is b  [#permalink]

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21 Mar 2017, 03:56
00:00

Difficulty:

35% (medium)

Question Stats:

75% (01:53) correct 25% (02:25) wrong based on 163 sessions

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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

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Posts: 331
Re: John has to hammer 100 railroad spikes for a new line his company is b  [#permalink]

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21 Mar 2017, 10:42
1
1
Time John takes to hammer 8 spikes = 1 hour
Time John takes to hammer 1 spike = 1/8 hour
Time John takes to hammer 50 spikes = 50 * 1/8 hours = 25/4 hours

Time Paul takes to hammer 1 spike = 1/8 hour

Time Paul and John takes to hammer 1 spike = 1/2*8 hours = 1/16 hours
Time Paul and John takes to hammer 50 spikes = 50 * 1/16 hours = 25/8 hours

hours are saved with Paul’s help = Time taken by John to hammer 50 spikes - Time taken by Paul and John to hammer 50 spikes

= 25/4 - 25/8 = (50-25)/8 = 25/8 hours

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Re: John has to hammer 100 railroad spikes for a new line his company is b  [#permalink]

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15 May 2018, 15:11
1
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Bunuel wrote:
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

John can complete 100 spikes in 100/8 = 25/2 hours if he works all alone by himself.

After he is halfway done, he has worked 50/8 = 25/4 hours. Since Paul joins in at that point and their combined rate is 16 spikes per hour, the remaining 50 spikes are completed in 50/16 = 25/8 hours. So the actual total time to complete the 100 spikes is 25/4 + 25/8 = 50/4 + 25/8 = 75/8 hours.

Therefore, the time saved is 25/2 - 75/8 = 100/8 - 75/8 = 25/8 hours.

Alternate Solution:

John can complete the entire job by himself in 100/8 = 25/2 hours.

He completes the first half of the job by himself. For the second half of the job, since Paul works at the same rate as John, they each have completed 1/2 x 1/2 = 1/4 of the total job. We see, then, that Paul has completed 1/4 of the entire job, thus saving John (1/4) x (25/2) = 25/8 hours.

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Re: John has to hammer 100 railroad spikes for a new line his company is b  [#permalink]

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27 Sep 2018, 07:49
Bunuel wrote:
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.
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Re: John has to hammer 100 railroad spikes for a new line his company is b &nbs [#permalink] 27 Sep 2018, 07:49
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