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.

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.

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.

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.

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.