AAPL wrote:Machine J runs at a constant rate and produces a lot consisting of 300 bottles in 4 hours. How much less time would it take to produce the lot of cans if both machines J and P were run simultaneously?
(1) Machines J and P start working simultaneously at 7 a.m.
(2) Machines J and P finish one lot by 7:23 a.m.
Please, can any expert assist me with this DS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
Sure, my pleasure!
Excellent opportunity to explore BIFURCATIONS with "extremal analyses", one of the most powerful tools of our method!
\[J\,\,\,\, \to \,\,\,\,\frac{{300\,\,\,{\text{bottles}}}}{{4\,\,\,h}}\,\,\,\,\,\,\,\,\left( {300\,\,\,{\text{bottles}}\,\, = \,\,\,1\,\,\,{\text{lot}}} \right)\]
\[?\,\,\,\,:\,\,\,\,{T_{\,J}} - {T_{\,P\, \cup \,J}}\,\,\,\,\,\,\left( {{\text{times}}\,\,{\text{for}}\,\,1\,\,{\text{lot}}} \right)\]
(1) Insufficient:
> If P takes (say) 10,000,000 h (!) to produce 1 lot, it will aggregate almost no efficiency to J (when working together). Hence the answer would be approximately 0h (? = 4-4).
> If P takes (say) 1 s (!) to produce 1 lot, it will aggregate an enormous efficiency to J (when working together). Hence the answer would be approximately 4h (? = 4-0).
(2) Insufficient: \[\left( 2 \right)\,\,\left\{ \begin{gathered}
\,\,{\text{if}}\,\,{\text{they}}\,\,{\text{start}}\,\,{\text{7}}\,\,{\text{a}}{\text{.m}}{\text{.}}\,\,\,\,\, \Rightarrow \,\,\,?\,\,\, = \,\,{T_{\,J}} - {T_{\,P\, \cup \,J}} = 4h - 23\min \,\, \hfill \\
\,\,{\text{if}}\,\,{\text{they}}\,\,{\text{start}}\,\,{\text{6}}\,\,{\text{a}}{\text{.m}}{\text{.}}\,\,\,\,\, \Rightarrow \,\,\,?\,\,\, = \,\,{T_{\,J}} - {T_{\,P\, \cup \,J}} = 4h - 1{\text{h}}23\min \hfill \\
\end{gathered} \right.\]
(1+2) Sufficient:
\[\left. \begin{gathered}
{T_{\,P\, \cup \,J}} = 23\min \,\,\, \hfill \\
{T_J} = 4{\text{h}} \hfill \\
\end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,? = \,\,4h - 23\min \]
Obs.: we assume 7 a.m. and 7:23 a.m. of the same day, otherwise the answer would easily be (E)!
The above follows the notations and rationale taught in the GMATH method.
