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Stark and Roggers are both baking pies. stark can make M pies per hour

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Stark and Roggers are both baking pies. stark can make M pies per hour  [#permalink]

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New post Updated on: 04 Oct 2018, 20:10
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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.

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Originally posted by Abhi077 on 04 Oct 2018, 11:40.
Last edited by Bunuel on 04 Oct 2018, 20:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Stark and Roggers are both baking pies. stark can make M pies per hour  [#permalink]

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New post 04 Oct 2018, 20:03
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

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Stark and Roggers are both baking pies. stark can make M pies per hour  [#permalink]

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New post 05 Oct 2018, 14:06
Abhi077 wrote:
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.

\(S:\,\,\,{{M\,\,{\rm{pies}}} \over {1\,\,{\rm{h}}}}\,\,\,\,\,;\,\,\,\,\,\,y\,{\rm{h}}\,\,\,\left( {S\,\,{\rm{alone}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{UNITS}}\,\,{\rm{CONTROL}}} \,\,\,\,\,\,? = yM\,\,\,\,\left[ {{\rm{pies}}} \right]\)

\(R:\,\,\,{{J\,\,{\rm{pies}}} \over {1\,\,{\rm{h}}}}\,\,\,\,\,;\,\,\,\,\,\,x\,{\rm{h}}\,\,\,\left( {S \cup R} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{UNITS}}\,\,{\rm{CONTROL}}} \,\,\,\,\,\,yM + \left( {M + J} \right)x = 100\,\,\,\,\left( * \right)\,\,\,\,\left[ {{\rm{pies}}} \right]\,\)

\(\left( 1 \right)\,\,\,M + J = 25\,\,\,{\rm{and}}\,\,\,\left( * \right)\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {5,15,20,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{75}}\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,5,15,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{50}}\,\,\,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,x = 2\,\,\,{\rm{and}}\,\,\,\left( * \right)\,\,\,\,\,\left\{ \matrix{
\,\left( {{\rm{Re}}} \right){\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,5,15,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{50}}\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {M,y,J,x} \right) = \left( {10,4,20,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{40}}\,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,M + J = 25 \hfill \cr
\,x = 2 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,yM\,\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)


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

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Fabio.
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Stark and Roggers are both baking pies. stark can make M pies per hour  [#permalink]

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New post 10 Oct 2018, 08:01
Abhi077 wrote:
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.


From the question stem we can create this equation M + (M+J) * x = 100

We need to find the variables in that equation in order to solve it.

statement 1 provides us with M+J but that is not sufficient to find out the value of M.

Statement 2) provides us that x = 2

when we combine it we have (M+J) * x as a known variable then we can find M.

1 + 2 provides a unique answer, hence sufficient.

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Stark and Roggers are both baking pies. stark can make M pies per hour   [#permalink] 10 Oct 2018, 08:01
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Stark and Roggers are both baking pies. stark can make M pies per hour

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