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GMAT Official Guide Practice Problem: Francine’s Trip

OG 12th Edition, Problem Solving Practice Problem #149

Here is the problem that appears in the OG:

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) 12000/(x + 200)


This problem is a double-whammy, because it’s both a problem about average velocity, and a problem with variables in the answer choice.  I am going to explain the principles of average velocity problems, then solve the problem in two ways: the traditional algebraic approach and the plug-in-numbers approach.


What you need to remember about average velocity

This is a problem-type you are almost guaranteed to see on the GMAT: some object travels this distance at this speed for the first leg of the trip, then that distance at that speed for the second leg of the trip; what’s the average velocity?

First of all, you can never simply average the numerical values of the two speeds given to get the average speed.  The only formula in the world for average speed is:

Average Speed = {Total Distance}/{Total Time}

What you always need to do is find the distance & time of the first leg (using D = RT), then the distance and time of the second leg, then add the two distances for the total distance, add the two times for the total time, and divide them to find the average velocity.  Almost always, if you have to use a variable for the total distance, that variable will cancel in the final division.  If you can remember that logic, you can solve GMAT Average Velocity questions.


Approach #1: Algebraic

This is the approach of which your Algebra Two teacher would have approved: a purely symbolic approach.  Use the variable D for the total distance.

In the first leg of the trip, she went x% of D at 40 mph.  Write the percent as a fraction, then we have:

d_1 = x/100 * D

and we can calculate the time of the first leg via

t_1 = d_1 / v = {{x/100}*D}/40 = {x*D/4000}

For the second leg, she went the remainder — (100 - x)% — of D at 60 miles per hour.  Change this percent to a fraction, and multiply to find the distance of this leg:

d_2 = {(100-x)/100}*D

We can calculate the time of the second leg via

t_2 = d_2 / v = {{(100-x)/100}*D}/60 = (100-x)*D/6000

Now,we can add the times of the two legs of the trip to get the total time:

 t_T = t_1 + t_2

t_T = x*D/4000 + (100-x)*D/6000

The LCD is 12000

t_T = {x*D/4000} * {3/3} + {(100-x)*D/6000}*{2/2} = {3x}*D/12000 + 2*(100-x)*D/12000 

t_T = {3x*D + 200*D - 2x*D}/12000 = {x*D + 200*D}/12000 = (x+200)*D/12000

Finally, we simply divide the total distance, D, by this expression for the total time, to find the average velocity.

v_average = D/{(x+200)*D/12000} = {D/1}*{12000/{D*(x + 200)}} = 12000/{x + 200}

The D’s canceled, as predicted, and we are left with an algebraic expression for the average velocity.  This, of course, is answer E.


Approach #2: Plugging in numbers

Your Algebra #2 teacher probably wouldn’t have approved of this method, and maybe even made you feel bad about trying it back in high school, but it is a perfectly legitimate approach to solving problems with variables in the answer choices.

First, we will pick a number for x, but we will be careful: one spectacularly bad choice would be x = 50%, because then the distances of the two legs would be equal, and the ratio of one distance to the other would be one.  We don’t know, but the ratio of the two distances could be an important factor somewhere in the calculation, so we don’t want anything that could be an important factor equaling one.  Then, from numbers alone, you wouldn’t know whether that factor had been multiplied or not.  As a general rule, when you have to pick a number to represent a percent, steer clear of 50%.

Here, I’m going to suggest the somewhat less obvious x = 20%.  Francine goes 20% (one-fifth) of her trip at 40 mph and 80% (four-fifths) of her trip at 60 mph.  We are going to be dividing distances by 40 and 60 to get times, so I want to pick a number for distance that will be divisible by 40, by 60, by 5, etc.  I am going to pick D = 600 miles.  Then, Francine goes 20% of 600 mi, or 120 miles, at 40 mph in the first leg, and 80% of 600 mi, or 480 mi, at 60 mph in the second leg.

t_1 = d_1 / v = 120/40 = 3 hours

t_2 = d_2 / v = 480/60 = 8 hours

Total Time = 3 hr + 8 hr = 11 hr

v_average = D_T / t_T = {600 mi} / {11 hr}

Now, that’s a recognizable fraction, 600/11.  When we plug x = 20 into the answer choices, one of the answer choices should give that the average velocity is 600/11.


(A) (180 - x)/2 = (180 - 20)/2 = 160/2 = 80  fail!

(B) (x + 60)/4 = (20 + 60)/4 = 80/4 = 20 fail!

(C) (300 - x)/5 = (300 - 20)/5 = 280/5 = 56 fail!

(D) 600/(115 - x) =  600/(115 - 20) = 600/95 — whatever that is, it’s not 600/11

(E) 12000/(x + 200) = 12000/(20 + 200) =12000/220 = 1200/22 = 600/11 bingo!


Once again, Answer = E.


Here, you will notice that the plug-in-number approach is quicker, more efficient, and it hones in on the correct answer just as well as the purely algebraic approach.  Forget what your Algebra Two teacher wanted you to do.  Do whatever method feels most comfortable, most natural, to you.


Here’s another question on average speed, just for more practice:


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6 Responses to GMAT Official Guide Practice Problem: Francine’s Trip

  1. Ramy April 8, 2016 at 5:47 pm #

    Hi Mike,

    For the second approach, I understood everything until you mentioned to plug x = 20 back into the answer choices. Can you please clarify this part?

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert April 19, 2016 at 10:48 am #

      What’s basically happening here is that if you choose one value for x, you’ll always get the same final answer for x, provided you go with speed information given.

      In other words, if Francine traveled at 40mph for x percent of the trip and 60mph for the rest of the trip, a value of x will always lead to a final calculation of 600/11 as Francine’s average speed. Similarly, if you placed the value of x at 10% of the time for 40mph, you’d always get a value of 400/7. x=10 will lead to 400/7 with the given information no matter what the distance is, and x = 20 will lead to 600/11 regardless of distance as well. (Try it— use distances other than Mike’s chosen D value of 600, and you’ll see what I mean.)

      So if you can plug in x for time spent at 40mph, provided the 60mph alternate speed holds true for the other part of the trip, the correct answer formula will equal 600/11 with an x value of 20, even without any specific distance included int he formula, because distance doesn’t matter.

  2. Sriram July 10, 2014 at 1:29 pm #

    Thought of sharing a different perspective on this problem, which ironically leads to a 50% value of x. Let me explain. We know the ratio of the speeds, they’re 2:3, so the ratio of time would be 3:2. Now, from the speeds, we can compute the distance travelled as 120 each(3*40 and 2*60) – assuming the travel times are 3 and 2 hrs. This gives us a 1:1 ratio on the distance. Finally, the weighted average using a 2:3 time ratio on speed gives us 48km/hr. From the answer choice, you can see 120,000/(200 + 50) = 48.

    • Mike MᶜGarry
      Mike July 10, 2014 at 2:12 pm #

      Dear Sriram,
      Yes, that’s a very insightful way to solve the problem. Of course, I didn’t demonstrate an approach like that in the solutions, because many students don’t every know where to begin and want to see something more step-by-step. If you can look at a problem like this and have insights about quick ways to solve it, that puts you way ahead of the game. Best of luck to you!
      Mike 🙂

  3. Sachin November 10, 2012 at 9:49 pm #

    if you take x=0, it becomes more easier 🙂
    Lovely, loved the 2nd approach! Thanks a lot!

    • Mike MᶜGarry
      Mike November 10, 2012 at 10:00 pm #

      Sachin: I’m glad you like it. Yes, checking x = 0 can sometimes be a highly efficient way to eliminate some answers. Best of luck to you.
      Mike 🙂

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