Brian takes a weekend trip to visit a friend. What is his

[AAPL]

Veritas Prep

Brian takes a weekend trip to visit a friend. What is his average rate for the there-and-back trip?

1) Brian took the same route for both segments.
2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.

OA C

[Jay@ManhattanReview]

Veritas Prep

Brian takes a weekend trip to visit a friend. What is his average rate for the there-and-back trip?

1) Brian took the same route for both segments.
2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.

OA C

We have to find out the average rate for the there-and-back trip.

Let's take each statement one by one.

1) Brian took the same route for both segments.

No information about speeds. Insufficient.

2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.

If the distances for the first segment and that for the second segment are equal, we can get the answer; however, if they are not equal, we cannot get the unique answer. Insufficient.

(1) and (2) together

Say the distance for each segment is x miles.

Time take to cover the first segment = x/80 hours;
Time take to cover the second segment = x/60 hours

Average speed = Total distance / Total time = 2x / (x/80 + x/60) = 2 / (1/80 + 1/60); x gets cancelled.

We will get a unique value of Average speed. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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[fskilnik@GMATH]

Veritas Prep

Brian takes a weekend trip to visit a friend. What is his average rate for the there-and-back trip?

1) Brian took the same route for both segments.
2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.

$$? = {{{\rm{dist}}\,{\rm{there}} + {\rm{dist}}\,{\rm{back}}} \over {{\rm{time}}\,{\rm{there}} + {\rm{time}}\,{\rm{back}}}}\,\,\,\,\,\,\,\,\left[ {{{{\rm{miles}}} \over {\rm{h}}}} \right]$$

Each statement alone has a trivial bifurcation, hence they will be omitted.


Let´s use UNITS CONTROL, one of the most powerful tools covered in our course!


$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,{\rm{dist}}\,{\rm{there}} = {\rm{dist}}\,{\rm{back}}\,\,{\rm{ = d}}\,{\rm{miles}} \hfill \cr
\,{\rm{time}}\,\,{\rm{there}} + {\rm{time}}\,{\rm{back}}\,\,\,{\rm{ = }}\,\,{\rm{d}}\,\,\,{\rm{miles}}\,\, \cdot \,\,\left( {{{1\,\,{\rm{h}}} \over {80\,\,{\rm{miles}}}}} \right)\,\,\,\,\, + \,\,\,\,\,\,\,{\rm{d}}\,\,\,{\rm{miles}}\,\, \cdot \,\,\left( {{{1\,\,{\rm{h}}} \over {50\,\,{\rm{miles}}}}} \right) \hfill \cr} \right.$$
$$? = {{2d} \over {\,\,d\left( {{1 \over {80}} + {1 \over {50}}} \right)\,\,}} = {2 \over {\,\,{1 \over {80}} + {1 \over {50}}\,\,}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{C}} \right)$$


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

Regards,
Fabio.