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Abhi077
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Difficulty:
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Question Stats:
64% (02:05) correct
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Stark and Roggers are both baking pies. stark can make M pies per hour; Roggers can make J pies per hour. stark begins baking pies alone, but later Roggers joins him. If 100 pies are finished x hours after Roggers joins stark, how many pies has stark already made by the time Roggers begins to help?
(1) Working together stark and Roggers can bake 25 pies per hour.
(2) Once Roggers arrives, Stark and Roggers complete the remainder of the pies in 2 hours.
(1) Working together stark and Roggers can bake 25 pies per hour.
(2) Once Roggers arrives, Stark and Roggers complete the remainder of the pies in 2 hours.
04 Oct 2018, 21:03
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Stark and Roggers are both baking pies. stark can make M pies per hour; Roggers can make J pies per hour. stark begins baking pies alone, but later Roggers joins him. If 100 pies are finished x hours after Roggers joins stark, how many pies has stark already made by the time Roggers begins to help?
M*y+(M+J)*x=100, where y is time Stark works alone..
We are looking for value of My
1) Working together stark and Roggers can bake 25 pies per hour.
So M+J=25
My+25x=100
Insufficient
2) Once Roggers arrives, Stark and Roggers complete the remainder of the pies in 2 hours.
So x=2..
Insufficient
Combined..
My+25*2=100.....My=50
Sufficient
C
M*y+(M+J)*x=100, where y is time Stark works alone..
We are looking for value of My
1) Working together stark and Roggers can bake 25 pies per hour.
So M+J=25
My+25x=100
Insufficient
2) Once Roggers arrives, Stark and Roggers complete the remainder of the pies in 2 hours.
So x=2..
Insufficient
Combined..
My+25*2=100.....My=50
Sufficient
C
05 Oct 2018, 15:06
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Abhi077
\(S:\,\,\,{{M\,\,{\rm{pies}}} \over {1\,\,{\rm{h}}}}\,\,\,\,\,;\,\,\,\,\,\,y\,{\rm{h}}\,\,\,\left( {S\,\,{\rm{alone}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{UNITS}}\,\,{\rm{CONTROL}}} \,\,\,\,\,\,? = yM\,\,\,\,\left[ {{\rm{pies}}} \right]\)
\(R:\,\,\,{{J\,\,{\rm{pies}}} \over {1\,\,{\rm{h}}}}\,\,\,\,\,;\,\,\,\,\,\,x\,{\rm{h}}\,\,\,\left( {S \cup R} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{UNITS}}\,\,{\rm{CONTROL}}} \,\,\,\,\,\,yM + \left( {M + J} \right)x = 100\,\,\,\,\left( * \right)\,\,\,\,\left[ {{\rm{pies}}} \right]\,\)
\(\left( 1 \right)\,\,\,M + J = 25\,\,\,{\rm{and}}\,\,\,\left( * \right)\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {5,15,20,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{75}}\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,5,15,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{50}}\,\,\,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,\,x = 2\,\,\,{\rm{and}}\,\,\,\left( * \right)\,\,\,\,\,\left\{ \matrix{\\
\,\left( {{\rm{Re}}} \right){\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,5,15,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{50}}\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,4,20,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{40}}\,\, \hfill \cr} \right.\)
\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
\,M + J = 25 \hfill \cr \\
\,x = 2 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,yM\,\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
