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Francine's trip

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Francine's trip

by alex.gellatly » Mon Sep 03, 2012 8:48 pm
I know this question has been posted before, but I'm looking for some other approaches.

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

(180-x)/2
(x+60)/4
(300-x)/5
(600)/(115-x)
12,000/(x+200)

Thanks
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by neelgandham » Mon Sep 03, 2012 10:02 pm
Algebric approach:
If S be the total distance travelled during the trip, then distance travelled in the first leg = (X/100)*S
Average speed during the first leg = 40mph
Time taken during the first leg = Distance travelled in the first leg/Average speed during the first leg = [(X/100)*S]/40 hours

Distance travelled in the second leg = Total distance - Distance travelled in the first leg
Distance travelled in the second leg = S - (X/100)*S = S*[(100-X)/100]
Average speed during the second leg = 60mph
Time taken during the second leg = Distance travelled in the second leg/Average speed during the second leg
Time taken during the second leg = S*[(100-X)/100]/60

Total time taken = S*[(100-X)/100]/60 + [(X/100)*S]/40]
= S*[(100-X)/6000 + (X/4000)]
= S*[2*(100-X)/12000 + 3X/12000]
= S*[(200 - 2X + 3X)/12000]
= S*[(200 + X)/12000]

Average speed of the trip = Total distance / Total time taken
= S/ S*[(200 + X)/12000]
= 12000/(200+X)

Answer E
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by neelgandham » Mon Sep 03, 2012 10:14 pm
The GMAT way:
Let the value of X be 0.So, the average speed of the trip is 60mph. Now, let us check the answer choices.

A) (180-x)/2 = 90mph, if x = 0. But if x = 0, the answer should be 60mph. So Answer choice A is incorrect.
B) (x+60)/4 = 15mph, if x = 0. But if x = 0, the answer should be 60mph. So Answer choice B is incorrect.
C) (300-x)/5 = 60mph - Hold on.
D) (600)/(115-x) = 5.something mph, if x = 0. But if x = 0, the answer should be 60mph. So Answer choice D is incorrect.
E) 12,000/(x+200) = 60mph - Hold on.

If x = 50 then
Time taken in the first leg = (S/2)/40 hours
Time taken in the first leg = (S/2)/60 hours
Average speed = S/{[(S/2)/40] + [(S/2)/60]} = 48mph.

Between C and E
C) (300-x)/5 = 50mph, if x = 0. But if x = 0, the answer should be 48mph. So Answer choice C is incorrect and anwer choice E should be the answer. Let us do a quick check
E) 12,000/(x+200) = 40mph. Bingo!
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by Anurag@Gurome » Tue Sep 04, 2012 12:07 am
alex.gellatly wrote:I know this question has been posted before, but I'm looking for some other approaches.

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

(180-x)/2
(x+60)/4
(300-x)/5
(600)/(115-x)
12,000/(x+200)

Thanks
Let us take the total distance = 1 mile
Then x% of 1 = x/100
So, Francine traveled x/100 mile at 40 mph. So, time taken = distance/speed = (x/100)/40 = x/4000
Remaining distance = 1 - x/100 = (100 - x)/100
Time taken to travel (100 - x)/100 mile at 60 mph = distance/speed = [(100 - x)/100]/60 = (100 - x)/6,000

Total time = x/4000 + (100 - x)/6000 = [3x + 2(100 - x)]/12000 = (x + 200)/12000

Therefore, total average speed = distance/time = 1/[(x + 200)/12000] = [spoiler]12000/(x + 200)[/spoiler]

The correct answer is E.
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by GMATGuruNY » Tue Sep 04, 2012 8:08 am
alex.gellatly wrote:I know this question has been posted before, but I'm looking for some other approaches.

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

(180-x)/2
(x+60)/4
(300-x)/5
(600)/(115-x)
12,000/(x+200)

Thanks
When the same distance is traveled at two different speeds, the average speed for the entire trip will be just a bit less than the average of the two speeds.
The reason is that it will take longer to travel the distance at the slower speed than at the faster speed.
The result: the average speed for the entire trip will be just a bit less than halfway between the two speeds.

Let x=50, so that 50% of the distance is traveled at 40mph and 50% is traveled at 60mph.
Halfway between 40 and 60 = 50.
Thus, the average speed for the entire trip must be just a bit less than 50.
Now we plug x=50 into the answers to see which yields an average speed just a bit less than 50.
Do only as much math as needed.

A: (180-x)/2 = (180-50)/2, which is not a bit less than 50.
B: (x+60)/4 = (50+60)/4, which is not a bit less than 50.
C: (300-x)/5 = (300-50)/5 = 50, which is not a bit LESS than 50.
D: (600)/(115-x) = 600/(115-50), which is not a bit less than 50.

The correct answer is E.

E: 12,000/(x+200) = 12,000/(50+200) = 48, which is a bit less than 50.
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by everything's eventual » Wed Sep 05, 2012 6:58 pm
I did something very similar to neelgandham's GMAT way

Let x = 50% and total distance be 100 miles.

S * T = D

1st part: 40 * 5/4 = 50

2nd part: 60 * 5/6 = 50


Average speed = Total distance / total time

= 100 / (50/24) = 48 MPH

Substitute x = 50 in all the given options and try to get 48.

(E) is the answer.
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