violetsplash wrote:
Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?
(1) x > 0.8y
(2) y = x + 1
\(M\,\,\, - \,\,\,4\,\,{\text{min}}\,\,\, - \,\,\,x\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,M\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,5x\,\,{\text{widgets}}\)
\(N\,\,\, - \,\,\,5\,\,{\text{min}}\,\,\, - \,\,\,y\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,4y\,\,{\text{widgets}}\)
\(5x\mathop > \limits^? \,4y\)
\(\left( 1 \right)\,\,\,x > \frac{4}{5}y\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,5} \,\,\,\,\,5x > 4y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)
\(\left( 2 \right)\,\,\,\,y = x + 1\,\,\,\,\left\{ \begin{gathered}
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\, \hfill \\
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {5,6} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\, \hfill \\
\end{gathered} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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