If it takes Jacob x hours to complete a project and it

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Veritas Prep

If it takes Jacob x hours to complete a project and it takes Mike y hours to complete the same project, how many hours will it take them to complete the project if they are working together?

$$\text{A. } \frac{xy}{x+y}$$
$$\text{B. } \frac{x+y}{xy}$$
$$\text{C. }x+y$$
$$\text{D. } xy$$
$$\text{E. }x-y$$

OA A.

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by Brent@GMATPrepNow » Sat Aug 25, 2018 6:02 am
AAPL wrote:Veritas Prep

If it takes Jacob x hours to complete a project and it takes Mike y hours to complete the same project, how many hours will it take them to complete the project if they are working together?

$$\text{A. } \frac{xy}{x+y}$$
$$\text{B. } \frac{x+y}{xy}$$
$$\text{C. }x+y$$
$$\text{D. } xy$$
$$\text{E. }x-y$$

OA A.
For work questions, there are two useful rules:

Rule #1: If a person can complete an entire job in k hours, then in one hour, the person can complete 1/k of the job
Example: If it takes Sue 5 hours to complete a job, then in one hour, she can complete 1/5 of the job. In other words, her work rate is 1/5 of the job per hour

Rule #2: If a person completes a/b of the job in one hour, then it will take b/a hours to complete the entire job
Example: If Sam can complete 1/8 of the job in one hour, then it will take him 8/1 hours to complete the job.
Likewise, if Joe can complete 2/3 of the job in one hour, then it will take him 3/2 hours to complete the job.

Let's use these rules to solve the question. . . .

It takes Jacob x hours to complete a project and it takes Mike y hours to complete the same project
So, applying Rule #1....
Jacob completes 1/x of the job in ONE HOUR
Mike completes 1/y of the job in ONE HOUR
So, in ONE HOUR, the two workers complete 1/x + 1/y of the job

1/x + 1/y = y/xy + x/xy
= (x + y)/xy

In other words, in ONE HOUR, the two workers complete (x + y)/xy of the job
Applying Rule #2, the total time to COMPLETE the job = xy/(x + y)

Answer: A

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Brent
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by Jeff@TargetTestPrep » Tue Sep 04, 2018 3:32 am
AAPL wrote:Veritas Prep

If it takes Jacob x hours to complete a project and it takes Mike y hours to complete the same project, how many hours will it take them to complete the project if they are working together?

$$\text{A. } \frac{xy}{x+y}$$
$$\text{B. } \frac{x+y}{xy}$$
$$\text{C. }x+y$$
$$\text{D. } xy$$
$$\text{E. }x-y$$
Jacob's rate is 1/x, and Mike's rate is 1/y. We can create the following combined rate expression:

1/x + 1/y = y/xy + x/xy = (x + y)/xy

Since time is the inverse of rate, it will take them xy/(x+y) hours to complete the project.

Answer: A

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by fskilnik@GMATH » Tue Sep 04, 2018 6:44 pm
AAPL wrote:Veritas Prep

If it takes Jacob x hours to complete a project and it takes Mike y hours to complete the same project, how many hours will it take them to complete the project if they are working together?

\[\text{A. } \frac{xy}{x+y} \,\,\,\,\, \text{B. } \frac{x+y}{xy} \,\,\,\,\, \text{C. }x+y \,\,\,\,\, \text{D. } xy \,\,\,\,\, \text{E. }x-y \]
Let´s use UNITS CONTROL, one of the most powerful tools of our method!

\[J\,\,:\,\,\,\frac{{x\,\,\,{\text{h}}}}{{1\,\,{\text{job}}}}\,\,\, \to \,\,\,\frac{1}{x}\,\,\,\frac{{{\text{job}}}}{{\text{h}}}\,\,\,\,\,\,\,\,;\,\,\,\,\,\,M:\,\,\frac{{y\,\,{\text{h}}}}{{1\,\,{\text{job}}}}\,\,\,\, \to \,\,\,\frac{1}{y}\,\,\,\frac{{{\text{job}}}}{{\text{h}}}\]
\[?\,\,\,:\,\,{T_{\,J\, \cup \,M}}\,\,\,\,\left( {1\,\,\,{\text{job}}} \right)\]

\[J \cup M\,\,:\,\,\,\,\,\,{T_{\,J\, \cup \,M}}\,\,{\text{h}}\,\, \cdot \left( {\frac{1}{x} + \frac{1}{y}} \right)\,\,\frac{{{\text{job}}}}{{\text{h}}}\,\,\, = \,\,\,1\,\,\,{\text{job}}\]
\[{T_{\,J\, \cup \,M}}\left( {\frac{{y + x}}{{xy}}} \right)\,\,\, = \,\,\,1\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \frac{{xy}}{{x + y}}\,\,\]

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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