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If Rebeca drives to work at x mph she will be one minute late, but if

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If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post 14 Feb 2019, 07:30
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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.

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Re: If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post 14 Feb 2019, 11:49
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
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Re: If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post 14 Feb 2019, 18:45
fskilnik wrote:
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.

\(?\,\, = \,\,d\,\,\,\,\left( {{\rm{miles}}} \right)\)

\(\left. \matrix{
\left( {{\rm{late}}} \right)\,\,:\,\,\,d\,\,{\rm{miles}}\left( {{{60\,\,\min } \over {x\,\,{\rm{miles}}}}} \right)\,\,\, = \,\,\,{{60d} \over x}\,\,\min \,\,\, \hfill \cr
\left( {{\rm{early}}} \right)\,\,:\,\,\,d\,\,{\rm{miles}}\left( {{{60\,\,\min } \over {y\,\,{\rm{miles}}}}} \right)\,\,\, = \,\,\,{{60d} \over y}\,\,\min \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{stem}}} \,\,\,\,\,\,{{60d} \over x} - {{60d} \over y} = 2\,\,\,\,\left[ {\min } \right]\,\,\,\,\, \Rightarrow \,\,\,\,\,30d\left( {{1 \over x} - {1 \over y}} \right) = 1\,\,\,\,\,\,\,\left( * \right)\)


\(\left( 1 \right)\,\,\,y - x = 7\,\,\,\left( {y > x} \right)\,\,\,\,\left[ {mph} \right]\)

\(\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,8} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {1 - {1 \over 8}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? = {8 \over 7}\left( {{1 \over {30}}} \right) \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,9} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {{1 \over 2} - {1 \over 9}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? \ne {8 \over 7}\left( {{1 \over {30}}} \right) \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\)


\(\left( 2 \right)\,\,\,y = {{111} \over {100}}x\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,{{111} \over {100}}} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {1 - {{100} \over {111}}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? = {{111} \over {11}}\left( {{1 \over {30}}} \right) \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {100,111} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,30d\left( {{1 \over {100}} - {1 \over {111}}} \right) = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{viable}}!} \,\,\,\,\,? \ne {{111} \over {11}}\left( {{1 \over {30}}} \right) \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\)


\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,y - x = 7 \hfill \cr
\,y = {{111} \over {100}}x \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,{{111} \over {100}}x - x = 7\,\,\,\,\, \Rightarrow \,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\,\left( {y = x + 7\,\,{\rm{unique}}} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,d\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\)


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

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Fabio.
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If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post Updated on: 15 Feb 2019, 05:54
To make the math easier, we can replace the given values with rounder numbers, as follows:

fskilnik@GMATH wrote:
GMATH practice exercise (Quant Class 19)

If Rebeca drives to work at x mph she will be one hour late, but if she drives at y mph she will be one hour early. How far (in miles) does Rebeca drive to work?

(1) x and y differ by 10 miles per hour.
(2) y is 50% greater than x.


Since Rebeca arrives 1 hour late when traveling at the lower speed and 1 hour early when traveling at the higher speed, the time at the lower speed must be 2 hours greater than the time at the higher speed.

Statement 1:
Case 1: x = 10 mph and y = 20 mph
Since the rate ratio = 10:20 = 1:2, the time ratio = 2:1 = 4:2, implying 4 hours at the lower speed and 2 hours at the higher speed.
Since the trip takes 4 hours when traveling at the lower speed of 10 mph, the distance = 10*4 = 40 miles.

Case 2: x = 20 mph and y = 30 mph
Since the rate ratio = 20:30 = 2:3, the time ratio = 3:2 = 6:4, implying 6 hours at the lower speed and 4 hours at the higher speed.
Since the trip takes 6 hours when traveling at the lower speed of 20 mph, the distance = 20*6 = 120 miles.

Since the distance can be different values, INSUFFICIENT.

Statement 2:
Case 2 also satisfies Statement 2.
In Case 2, the distance = 120 miles.

Case 3: x = 2 mph and y = 3 mph
Since the rate ratio = 2:3, the time ratio = 3:2 = 6:4, implying 6 hours at the lower speed and 4 hours at the higher speed.
Since the trip takes 6 hours when traveling at the lower speed of 2 mph, the distance = 2*6 = 12 miles.

Since the distance can be different values, INSUFFICIENT.

Statements combined:
Only Case 2 satisfies both statements.
In Case 2, the distance = 120 miles.
SUFFICIENT.


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Originally posted by GMATGuruNY on 14 Feb 2019, 20:49.
Last edited by GMATGuruNY on 15 Feb 2019, 05:54, edited 1 time in total.
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Re: If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post 14 Feb 2019, 23:03
Took 3 minutes.

Rephrase---> x(t+1/60) = d and y(t-1/60)=d
(x-y)t = -1/60(x+y)
SO we need x y and t to find d the distance.

1. Insufficient as x-y = 7 does not solve the rephrased stem.
2. 1.1x=y gives a ratio. No info on time insuficient.
C----> 1.1x= y . Substituting in stem, we get t = 21/60.
This way we can find x and y as well. D can be found out. Sufficient.
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Re: If Rebeca drives to work at x mph she will be one minute late, but if  [#permalink]

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New post 15 Feb 2019, 05:10
Hi there, chetan2u and GMATGuruNY !

Thanks for your comments on improving the "calculations-related" issue.

We all know that Data Sufficiency is related to the uniqueness (and viability) of potential answers, but "friendly numbers" (as called by Chetan) are always welcomed.

Regards,
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Re: If Rebeca drives to work at x mph she will be one minute late, but if   [#permalink] 15 Feb 2019, 05:10
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