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)
What is the best way to attack this question?
Thanks guys
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maihuna
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assuming d = 100, time taken in x% distance = x/40
" remaining dist = (100-x)/60
total time t = 3x + 2*100 - 2x)/120*100 = (x+200)/120
Avg = d/t = 100/(x+200)*120
" remaining dist = (100-x)/60
total time t = 3x + 2*100 - 2x)/120*100 = (x+200)/120
Avg = d/t = 100/(x+200)*120
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GMATGuruNY
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We can plug in a value for the distance and a value for x.jscpba 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)
What is the best way to attack this question?
Thanks guys
Let distance = 100 miles.
Let x = 40.
Distance for 40% of the trip = .4*100 = 40 miles.
Since the rate for this portion is 40mph, time = d/r = 40/40 = 1 hour.
Distance for remainder of the trip = 100-40 = 60 miles.
Since the rate for this portion is 60mph, time = d/r = 60/60 = 1 hour.
Total time = 1+1 = 2 hours.
Average speed for the whole trip = (total distance)/(total time) = 100/2 = 50. This is our target.
Now we plug x = 40 into all the answer choices to see which yields our target of 50.
Only answer choice E works:
(12,000)/(x+200) = 12000/(40+200) = 50.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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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I was also having some issues with this problem and could not quite understand why I had to choose 100 as distance to make this problem work. It seemed weird to me that total average speed was also 100 (R1 = 40 mph + R2 = 60 mph) I know I am supposed to save time on solving these types of problems, and know that Francine traveled X percent of total distance but what if I had selected total distance to be 50 rather than 100? It seems that answer choice E only works if distance = 100 and I think this is what is confusing me. Any help would be greatly appreciated. Thanks!
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Brent@GMATPrepNow
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Mitch is demonstrating one of two approaches to solving these kinds of questions (variables in the answer choices). I call his method the Input-Output approach, and it useful for avoiding some cumbersome algebra.rsanz27 wrote:I was also having some issues with this problem and could not quite understand why I had to choose 100 as distance to make this problem work. It seemed weird to me that total average speed was also 100 (R1 = 40 mph + R2 = 60 mph) I know I am supposed to save time on solving these types of problems, and know that Francine traveled X percent of total distance but what if I had selected total distance to be 50 rather than 100? It seems that answer choice E only works if distance = 100 and I think this is what is confusing me. Any help would be greatly appreciated. Thanks!
If you use the same approach, but use 50 miles instead of 100 miles, you should reach the same conclusion.
Cheers,
Brent
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Brent@GMATPrepNow
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Here's the algebraic approach.jscpba 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)
I always 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
Brent Hanneson - Creator of GMATPrepNow.com
