Some expert can give a better answer, but for me this is a simple realization problem. You're given rate for J, whatever that rate is. The question asks for both machines working together. Since you don't have the rate for the other machine, you have no clue what time it will take. That being said, examine the answer choices:

1) says they start at 7am. This alone doesn't solve anything so A is out.
2) says then end at 723. Since this alone is useless, B is out. Now examine them together: if they start at 7 and end at 723, they take 23 mins; hence, you now know how long the 2 machines take to do the task together. Since you know the rate of just J and now for J and P, you know the difference in time, which is the question; hence, answer is C.

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.