MathRevolution wrote:
[
Math Revolution GMAT math practice question]
Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?
\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)
\(1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{?\,\,{\rm{toys}}} \over {1\,\,{\rm{hour}}}}\)
\(\left. \matrix{
2\,\,{\rm{mach}}\,\,\, \to \,\,\,{{360\,\,{\rm{toys}}} \over {h + 3\,\,{\rm{hours}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{180\,\,{\rm{toys}}} \over {h + 3\,\,{\rm{hours}}}}\,\,\,\, \hfill \cr
3\,\,{\rm{mach}}\,\,\, \to \,\,\,{{360\,\,{\rm{toys}}} \over {h\,\,{\rm{hours}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{120\,\,{\rm{toys}}} \over {h\,\,{\rm{hours}}}} \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{180} \over {h + 3}} = {{120} \over h}\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,h = 6\)
\(1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{120\,\,{\rm{toys}}} \over {h = 6\,\,{\rm{hours}}}} = {{20\,\,{\rm{toys}}} \over {1\,\,{\rm{hour}}}}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 20\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
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