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

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

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.