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Contradictory PS problem

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Contradictory PS problem

by dddanny2006 » Sat Mar 22, 2014 5:21 pm
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)


My approach,I use the weighed averages method
Since nothing is given about x ,I assume the answers will hold good for all different value's of x%.So,now lets assume x to be 50%.

40(0.5) + 60(0.5)=50

Average speed equals 50.

Lets check the answers if any of those on substitution of 50% give is 50 as the average.

C and E are close.I'd say C.But the answer is E

(300-50)/5 =50

Please tell me why Im wrong?I have seen the methodology that gives us E as the answer but I want to know why this method doesnt work here.

Thanks

Dan
Last edited by dddanny2006 on Sat Mar 22, 2014 5:36 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Sat Mar 22, 2014 5:29 pm
dddanny2006 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)
Here's the algebraic approach.
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
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by dddanny2006 » Sat Mar 22, 2014 5:43 pm
Thank you.I was wondering where I could have made mistakes.I've been doing alot of weighed average problems recently,and as a result use that technique most of the times.Can you please tell me where I might have gone wrong?

Thanks sir

PS:If you have time,could you please clear my doubt here too-

https://www.beatthegmat.com/by-1945-the- ... tml#713922

Thank you sir.


Brent@GMATPrepNow wrote:
dddanny2006 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)
Here's the algebraic approach.
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
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by Brent@GMATPrepNow » Sat Mar 22, 2014 8:12 pm
dddanny2006 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)


My approach,I use the weighed averages method
Since nothing is given about x ,I assume the answers will hold good for all different value's of x%.So,now lets assume x to be 50%.

40(0.5) + 60(0.5)=50

Average speed equals 50.

Lets check the answers if any of those on substitution of 50% give is 50 as the average.

C and E are close.I'd say C.But the answer is E

(300-50)/5 =50

Please tell me why Im wrong?I have seen the methodology that gives us E as the answer but I want to know why this method doesnt work here.

Thanks

Dan
The problem is highlighted above in green
This isn't a weighted question in the sense you are using it.
If we drive 1/2 a distance at speed X and the other 1/2 at speed Y, the average speed does not equal the average of X and Y.

To illustrate this, consider the following example:
Bob sets out on a 2-mile trip.
For the first mile, he drives at a speed of 0.000001 miles per hour (so it takes over 100 years to travel that first mile)
For the second mile, he drives at a speed of 100,000 miles per hour (so it takes less than 1 second to travel the second mile)

What's the average speed for the entire 2-mile trip? Is it (0.000001 + 100,000)/2?
No.
(0.000001 + 100,000)/2 ≈ 50,000 miles per hour. That's pretty fast for a 2-mile trip that took more than 100 years to complete.

IMPORTANT: Average speed = (total distance)/(total time)

So, in your example (in green), we need to determine the total distance and divide it by the total time. Try that and see what happens.

Cheers,
Brent
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by Gurpreet singh » Sat May 28, 2016 2:42 am
Plug in value

let total distance be 240 ie LCM of 40 & 60

Journey 1
X=50% so distance traveled is 120.

Distance(120) = speed(40)*time(t1)= solving for t1=3

Journey 2

Distance(120)=speed(60)*time(t2) =solving for t2=2

Avg speed= total distance(240)/total time(t1+t2)=240/5=48

Now in answer choices plugin 50 and the option which gives 48 as the result is the answer.

Answer is E=12000/(x+200)
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by [email protected] » Sat May 28, 2016 8:28 am
Hi All,

There's a full discussion of this question here:

https://www.beatthegmat.com/average-spee ... 81762.html

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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