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by datonman » Wed Nov 04, 2015 9:25 am

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Answers

A

B

C

D

E

Stats

Difficulty—

During a trip, Francine traveled 'x' percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of 'x', what was Francine's average speed for the entire trip?

A.)180-x/2
B.)x + 60/4
C.)300-x/5
D.)600/115-x
E.)12,000/x+200
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by Brent@GMATPrepNow » Wed Nov 04, 2015 10:14 am
NOTE: I added some brackets to avoid ambiguity.
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)
Here's an algebraic approach.
I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40

time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60

Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60
Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)
= E

Cheers,
Brent
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by [email protected] » Wed Nov 04, 2015 10:26 am
Hi datonman,

This question can also be solved by TESTing VALUES.

We're told that X% of a distance was spent traveling at 40 miles/hour while the rest of the distance was spent traveling at 60 miles/hour. We're asked for the AVERAGE SPEED for the entire trip.

IF....
Total Distance = 100 miles
X = 40%

40 miles was traveled at 40 miles/hour --> 1 hour of travel
60 miles was traveled at 60 miles/hour --> 1 hour of travel

Total Distance = (Av. Sp)(Total Time)
100 miles = (Av. Sp.)(2 hours)
100/2 = Av. Sp
50 miles/hour = Av. Sp.

So we're looking for an answer that equals 50 when X = 40....

Answer A: (180-40)/2 = 70 NOT a match
Answer B: (40+60)/4 = 25 NOT a match
Answer C: (300-40)/5 = 52 NOT a match
Answer D: 600/(115-40) = 8 NOT a match
Answer E: 120000/(240) = 50 This IS a MATCH

Final Answer: E

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Rich
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by Brent@GMATPrepNow » Wed Nov 04, 2015 10:29 am
Rich and I have demonstrated the two methods (Algebraic and Input-Output) for solving a question type I call Variables in the Answer Choices.
If you'd like more information on these approaches, we have some free videos:
- Variables in the Answer Choices - https://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - https://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - https://www.gmatprepnow.com/module/gmat- ... /video/935

Cheers,
Brent
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by Jeff@TargetTestPrep » Thu Jun 28, 2018 5:12 pm
datonman wrote:During a trip, Francine traveled 'x' percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of 'x', what was Francine's average speed for the entire trip?

A.)180-x/2
B.)x + 60/4
C.)300-x/5
D.)600/115-x
E.)12,000/x+200
Since average speed = (total distance)/(total time), we have (note: in the parenthesis is what we do to the numerator and denominator to obtain the next line)

Average speed = d/[(d * x/100)/40 + (d * (100 - x)/100)/60] (Divide by d)

Average speed = 1/[(x/100)/40 + ((100 - x)/100)/60] (Multiply by 100)

Average speed = 100/[x/40 + (100 - x)/60] (Multiply by 120)

Average speed = 12000/[3x + 2(100 - x)]

Average speed = 12000/[3x + 200 - 2x]

Average speed = 12000/[x + 200]

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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