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

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Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 23 Nov 2014, 13:35
1
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A
B
C
D
E

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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.

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Re: Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 24 Nov 2014, 02:57
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aashu4uiit wrote:
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.



st.1 : clearly not sufficient. as nothing is mentioned about the rates.

st.2 : tells us about the rate. but nothing is mentioned about the segments. i.e. whether brian took the same route for both segments or he took the different route for each segment.

st.1 and st.2

we know that brain follows the same route for each segment. hence Distance traveled in each segment is constant. therefore, average speed can be calculated as

avg. speed = \(\frac{total distance}{total time}

= \frac{2D}{D/80 +D/50}
=\frac{2.80.50}{80+50}
=\frac{800}{13}
=61.5\)

hence answer should be C
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Re: Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 24 Nov 2014, 04:53
since , Average rate = total distance/total time

1) Brian took the same route for both segments.
This statement only talk about distances and no information as to how much Brian has covered been given.
INSUFFICIENT.

2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.
This statement provides the respective average speed taken during the each way of the journey.
But again there is no information as to how much distances were covered during each trip.
INSUFFICIENT.

If we combine both the statements Average rate = 2x/(x/80 +x/50). SUFFICIENT.

Answer C
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Re: Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 10 Dec 2014, 23:28
akumar5 wrote:
since , Average rate = total distance/total time

1) Brian took the same route for both segments.
This statement only talk about distances and no information as to how much Brian has covered been given.
INSUFFICIENT.

2) Brian averaged 80 mph for the first segment and 50 mph for the second segment.
This statement provides the respective average speed taken during the each way of the journey.
But again there is no information as to how much distances were covered during each trip.
INSUFFICIENT.

If we combine both the statements Average rate = 2x/(x/80 +x/50). SUFFICIENT.

Answer C


I am not sure if the question is poorly worded. How can we interpret segments to the actual journey to a place and from a place. There couple be multiple segments within the same route...
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Re: Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 06 Dec 2016, 03:34
manpreetsingh86 wrote:
aashu4uiit wrote:
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.



st.1 : clearly not sufficient. as nothing is mentioned about the rates.

st.2 : tells us about the rate. but nothing is mentioned about the segments. i.e. whether brian took the same route for both segments or he took the different route for each segment.

st.1 and st.2

we know that brain follows the same route for each segment. hence Distance traveled in each segment is constant. therefore, average speed can be calculated as

avg. speed = \(\frac{total distance}{total time}

= \frac{2D}{D/80 +D/50}
=\frac{2.80.50}{80+50}
=\frac{800}{13}
=61.5\)

hence answer should be C

Could you or somebody please elaborate why in the equation we just replaced D? I thought I wouldn't be able to solve it since I did not have a value for D (but did come up with the same equation).
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Re: Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 30 Dec 2016, 01:24
I marked answer as E as I assumed that weekend trip includes some stay at the trip place and we do not know stay time at place. Am I wrong to assume it as it is not mentioned in question or just use time and distance and stay time can be ignored for calculation.
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Brian takes a weekend trip to visit a friend. What is his average rate  [#permalink]

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New post 23 Jan 2019, 05:33
aashu4uiit wrote:
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.
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Brian takes a weekend trip to visit a friend. What is his average rate   [#permalink] 23 Jan 2019, 05:33
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