If Rebeca drives to work at x mph she will be one minute lat

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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.


Answer: [spoiler]____(C)__[/spoiler]
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by fskilnik@GMATH » Thu Feb 14, 2019 5:35 pm

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fskilnik@GMATH 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.

Regards,
Fabio.
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Rebeca

by GMATGuruNY » Thu Feb 14, 2019 7:31 pm

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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.

The correct answer is C.
Last edited by GMATGuruNY on Fri Feb 15, 2019 4:33 am, edited 1 time in total.
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by fskilnik@GMATH » Fri Feb 15, 2019 4:01 am

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Hi Mitch!

Thanks for joining.

We all know that Data Sufficiency is related to the uniqueness (and viability) of potential answers, but "easier math" is a nice contribution.

Just one important detail. Rebeca (my mother!) has only one "c", please... ;)

Regards,
Fabio.
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by GMATGuruNY » Fri Feb 15, 2019 4:48 am

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fskilnik@GMATH wrote:Just one important detail. Rebeca (my mother!) has only one "c", please... ;).
In deference to your mother, the offensive second "c" has been removed from my solution.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by fskilnik@GMATH » Fri Feb 15, 2019 5:00 am

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GMATGuruNY wrote: In deference to your mother, the offensive second "c" has been removed from my solution.
LoL... I was only joking, but I thank you for that.
You are not only an excellent teacher/professional... you are a gentleman!
Have a great weekend!
Regards,
Fabio.
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