If machine J, working alone at its constant rate, takes 2 minutes to

Solution

Given:

• Machine J, working alone at its constant rate, takes 2 minutes to wrap 60 pieces of candy

To find:

• The time machine K will take, working alone at its constant rate, to wrap 120 pieces of candy

Analysing Statement 1

• As per the information given in statement 1, machine K, working alone at its constant rate, takes more than 5 minutes to wrap 60 pieces of candy
• More than 5 minutes gives us infinite number of possibilities, and hence, does not help us in getting any specific information

Hence, statement 1 is not sufficient to answer

Analysing Statement 2

• As per the information given in statement 2, machines J and K, working together at their respective constant rates, take 1 minute and 30 seconds to wrap 60 pieces of candy

o We know the combined efficiency of J and K as well as the individual efficiency of J
o Therefore, we can find out the individual efficiency of K also

Hence, statement 2 is sufficient to answer

Hence, the correct answer is option B.

Answer: B
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\(J\,\,\, - \,\,\,60\,\,{\text{candies}}\,\,\, - \,\,\,2\,\,{\text{min}}\)

\(K\,\,\, - \,\,60\,\,{\text{candies}}\,\,\left( * \right)\,\,\, - \,\,\,?\,\,\min\)

\(\left( * \right)\,\,{\text{Our}}\,\,{\text{FOCUS}}\,\,{\text{is}}\,\,{\text{half}}\,\,{\text{the}}\,\,{\text{official}}\,\,{\text{one}}!\)


\(\left( 1 \right)\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,? = 5.5\min \hfill \\\\
\,{\text{Take}}\,\,? = 6\min \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\)


\(\left( 2 \right)\,\,\frac{1}{{1\frac{1}{2}}} = \frac{1}{2} + \frac{1}{?}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{1}{?}\,\,\,{\text{unique}}\,\,\,\, \Rightarrow \,\,\,\,?\,\,\,{\text{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)

(This is the "work/rate shortcut".)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Missed "more than"

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