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Contradictory PS problem

This topic has 3 expert replies and 2 member replies
dddanny2006 Master | Next Rank: 500 Posts Default Avatar
Joined
12 Jan 2012
Posted:
209 messages

Contradictory PS problem

Post Sat Mar 22, 2014 5:21 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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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    dddanny2006 Master | Next Rank: 500 Posts Default Avatar
    Joined
    12 Jan 2012
    Posted:
    209 messages
    Post 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-

    http://www.beatthegmat.com/by-1945-the-us-had-been-t274960.html#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

    Post 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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    Gurpreet singh Senior | Next Rank: 100 Posts Default Avatar
    Joined
    28 Apr 2016
    Posted:
    38 messages
    Thanked:
    1 times
    Post 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)

    Post Sat May 28, 2016 8:28 am
    Hi All,

    There's a full discussion of this question here:

    http://www.beatthegmat.com/average-speed-for-trip-t281762.html

    GMAT assassins aren't born, they're made,
    Rich

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