EgmatQuantExpert wrote:
Machines X and Y can work at their respective constant rates to manufacture a certain production unit. If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?
(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.
(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.
\(?\,\,:\,\,{T_X}\,,\,\,{T_Y}\,\,{\rm{relationship}}\,\,\,\,\,\,\left( {? = {T_X}\mathop \to \limits^{\Delta \% } {T_Y} = {{{T_Y} - {T_X}} \over {{T_X}}} = {{{T_Y}} \over {{T_X}}} - 1} \right)\)
Important: the ratio of time taken (for any given job) is the inverse of the ratio of the work done (for any given time).
\(\left( 1 \right)\,\,{{{T_{X \cup Y}}} \over {{T_Y}}} = {2 \over 3}\,\,\,\,\, \Rightarrow \,\,\,\,\,{{{W_{X \cup Y}}} \over {{W_Y}}} = {3 \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,{W_{X \cup Y}} = 3k \hfill \cr
\,{W_Y} = 2k \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,{W_X} = k\)
\({{{W_Y}} \over {{W_X}}} = {2 \over 1}\,\,\,\,\, \Rightarrow \,\,\,\,{{{T_Y}} \over {{T_X}}} = {1 \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)
\(\left( 2 \right)\,\,{T_Y} = 3{\rm{h}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,{{\rm{T}}_{\rm{X}}} = 3{\rm{h}}\,\,\,\, \Rightarrow \Delta \% = 0 \hfill \cr
\,{\rm{Take}}\,\,{{\rm{T}}_{\rm{X}}} = 4{\rm{h}}\,\,\,\, \Rightarrow \Delta \% \ne 0 \hfill \cr} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik ::
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