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14 Feb 2019, 07:30
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C
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GMATH practice exercise (Quant Class 19)
If Rebeca drives to work at x mph she will be one minute late, but if she drives at y mph she will be one minute early. How far (in miles) does Rebeca drive to work?
(1) x and y differ by seven miles per hour.
(2) y is 11% greater than x.
If Rebeca drives to work at x mph she will be one minute late, but if she drives at y mph she will be one minute early. How far (in miles) does Rebeca drive to work?
(1) x and y differ by seven miles per hour.
(2) y is 11% greater than x.
14 Feb 2019, 11:49
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If Rebeca drives to work at x mph she will be one minute late, but if she drives at y mph she will be one minute early. How far (in miles) does Rebeca drive to work?
Let the distance be d, then exact time = \(\frac{d}{x}-\frac{1}{60}\) while travelling at x mph and \(\frac{d}{y}+\frac{1}{60}\) while travelling at y mph. So, \(\frac{d}{x}-\frac{1}{60}=\frac{d}{y}+\frac{1}{60}........\frac{d}{x}-\frac{d}{y}=\frac{1}{60}+\frac{1}{60}..........\frac{d(y-x)}{xy}=\frac{2}{60}......d=\frac{xy}{30(y-x)}\)
(1) x and y differ by seven miles per hour.
Thsi just tells us that y-x=7, thus d=\(\frac{xy}{30(y-x)}=\frac{xy}{30*7}\)
Insuff
(2) y is 11% greater than x.
y=1.11x, thus d=\(\frac{x*1.11x}{30*0.11x}=\frac{111x}{330}\)
Insuff
Combined we can find x and y and thus d..
Sufficient
C
Note.. You could have used some friendly terms... y-x=7=1.11x-x...0.11x=7...x=700/11?? A very unfriendly term
Could have taken the difference as 11 or y as 7% or 14% greater than x
Let the distance be d, then exact time = \(\frac{d}{x}-\frac{1}{60}\) while travelling at x mph and \(\frac{d}{y}+\frac{1}{60}\) while travelling at y mph. So, \(\frac{d}{x}-\frac{1}{60}=\frac{d}{y}+\frac{1}{60}........\frac{d}{x}-\frac{d}{y}=\frac{1}{60}+\frac{1}{60}..........\frac{d(y-x)}{xy}=\frac{2}{60}......d=\frac{xy}{30(y-x)}\)
(1) x and y differ by seven miles per hour.
Thsi just tells us that y-x=7, thus d=\(\frac{xy}{30(y-x)}=\frac{xy}{30*7}\)
Insuff
(2) y is 11% greater than x.
y=1.11x, thus d=\(\frac{x*1.11x}{30*0.11x}=\frac{111x}{330}\)
Insuff
Combined we can find x and y and thus d..
Sufficient
C
Note.. You could have used some friendly terms... y-x=7=1.11x-x...0.11x=7...x=700/11?? A very unfriendly term
Could have taken the difference as 11 or y as 7% or 14% greater than x
14 Feb 2019, 18:45
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fskilnik
\(?\,\, = \,\,d\,\,\,\,\left( {{\rm{miles}}} \right)\)
\(\left. \matrix{\\
\left( {{\rm{late}}} \right)\,\,:\,\,\,d\,\,{\rm{miles}}\left( {{{60\,\,\min } \over {x\,\,{\rm{miles}}}}} \right)\,\,\, = \,\,\,{{60d} \over x}\,\,\min \,\,\, \hfill \cr \\
\left( {{\rm{early}}} \right)\,\,:\,\,\,d\,\,{\rm{miles}}\left( {{{60\,\,\min } \over {y\,\,{\rm{miles}}}}} \right)\,\,\, = \,\,\,{{60d} \over y}\,\,\min \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{stem}}} \,\,\,\,\,\,{{60d} \over x} - {{60d} \over y} = 2\,\,\,\,\left[ {\min } \right]\,\,\,\,\, \Rightarrow \,\,\,\,\,30d\left( {{1 \over x} - {1 \over y}} \right) = 1\,\,\,\,\,\,\,\left( * \right)\)
\(\left( 1 \right)\,\,\,y - x = 7\,\,\,\left( {y > x} \right)\,\,\,\,\left[ {mph} \right]\)
\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,8} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {1 - {1 \over 8}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? = {8 \over 7}\left( {{1 \over {30}}} \right) \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,9} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {{1 \over 2} - {1 \over 9}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? \ne {8 \over 7}\left( {{1 \over {30}}} \right) \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\)
\(\left( 2 \right)\,\,\,y = {{111} \over {100}}x\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,{{111} \over {100}}} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {1 - {{100} \over {111}}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? = {{111} \over {11}}\left( {{1 \over {30}}} \right) \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {100,111} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {{1 \over {100}} - {1 \over {111}}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? \ne {{111} \over {11}}\left( {{1 \over {30}}} \right) \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\)
\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
\,y - x = 7 \hfill \cr \\
\,y = {{111} \over {100}}x \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,{{111} \over {100}}x - x = 7\,\,\,\,\, \Rightarrow \,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\,\left( {y = x + 7\,\,{\rm{unique}}} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,d\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
