GMAT Solution and Mixing Problems

Attention, mad scientists out there!  Consider these two practice questions.


1) A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution.  How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

(A) 200

(B) 400

(C) 500

(D) 600

(E) 800


2) A chef mixes P ounces of 60% sugar solution with Q ounces of a 10% sugar solution to produce R ounces of a 25% sugar solution.  What is the value of P?

[Statement #1] Q = 455 mL

[Statement #2] R = 660 mL


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Mixtures, solutions, and concentration

First of all, keep in mind: this is relatively rare kind of GMAT Math problem.  There’s a good chance you could take five GMATs in row and never see a mixture problem.  Nevertheless, they do arise occasionally, and it’s good to be familiar with them.

Let’s think for a moment about “concentration” — not the mental quality, but in the chemical solution sense.  What does it mean to say we have a “400 units of a 6% phosphoric acid solution”?  Whatever those units are, the total amount of mixture, of solution, we have is 400 units, and of that, 6% is pure phosphoric acid.  Well, 6% of 400 is 24, so we know we have 24 units of pure phosphoric acid.  That is the amount of concentrate we have in our solution.  (Those “units” could be units of volume or units of mass, but that’s far more detail that you need to know on the GMAT).

I bet “phosphoric acid” sounds highly technical and abstruse: it’s actually one of the ingredients listed on every can of Coca Cola.


The two equations

The secret to any mixture or concentration problem is to use the two equations.  First is the amount of stuff, the total volume.  This is called the volume equation.  The basic idea is:

This makes sense when you think about it: the volume of the resultant solution had to come from the volumes of the two things we mixed.

The second equation is similar.  This concerns, specifically, the amount of concentrate, of whatever the chemical or substances is of which we have a solution.  In problem #1, the concentrate is phosphoric acid, and in problem #2, sugar.   The amount of concentrate that winds up in the resultant solution must come from somewhere.  It must come from the amount of concentrate in the two solutions mixed.

As in the section above, these amounts of concentrate will always be (the concentration percentage) times (the total volume of the solution concerned).

If you had trouble with the problems above before, go back to them, and see if you can set up both of these equations and solve.  Give them another try before reading the explanations below.

Also, here’s another practice question:



Practice problems explanations

1) We could backsolve from the numerical answer choices, but let’s use a straight algebra approach.  Let X equal the units of 12% phosphoric acid solution we use, and let Y be the units of 10% sulfuric acid solution that result.

The volume equation is:

400 + X = Y

In the first solution, we have 6% of 400, or 24 units of phosphoric acid.

In the second solution, we have 12% of X = 0.12*X of phosphoric acid.

In the resultant solution, we have 10% of Y = 0.10*Y of phosphoric acid.

The concentration equation is:

24 + 0.12*X = 0.10*Y

Multiply this by 100, to clear the decimals:

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2400 + 12X = 10Y

Everything is even, so divide by 2 to simplify:

1200 + 6X = 5Y

We want X, so let’s multiply the volume equation by -5 and add that to this equation we just got:

Answer = E

2) This one is very elegant.  We  have three variables — the amount of 60% sugar solution, the amount of 15% sugar solution, and the amount of the resultant 25% sugar solution.  Three variables.  We have two equations: the volume equation and the concentration equation.  Right now, three variables and two equations: we can’t solve.

Now, look at the statements.  Each statement gives us the value of one of the variables.   If we get the value of one variable, that’s no longer a variable, and thus we are down to two variables with two equations: that’s a situation in which we can find a full solution.  Thus, given the value of either Q or R, we enter a situation in which we can solve for everything, and thus we would know P.  Therefore, each statement, by itself, is sufficient.  Answer = D


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12 Responses to GMAT Solution and Mixing Problems

  1. John August 20, 2017 at 11:24 am #

    For the first problem .. easy number sense shortcut:

    400 units of 6% solution is 24(units solute)/400(units solution)
    400 units of 10% solution would be 40(units solute)/400(units solution)

    We would need 16 additional units of solute to have 10% solution @ 400 units.

    In terms of the difference between units of solute in the current solution (less than 10%) and units of solute in 10% solution.. each 100 units of 12% solution we add, we get 2 units of solute ‘closer’ to 10% solution.

    16 units of solute needed / 2 units gained per increment of 100 = 8 increments of 100 units (of 12% solution) needed.

    Thus the answer E.

    Basically it’s a number sense way to look at it .. essentially using growth/decay instead of algebra. Better for the right brain.

  2. Ryan June 5, 2016 at 6:24 pm #

    In order to make #2 work, it appears to me that you should change R to 650, instead of 660. I understand that it doesn’t really matter, in terms of getting the problem correct, but I always like to fully solve problems so that I can completely grasp what’s going on.

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert June 6, 2016 at 3:35 pm #

      If I understand you correctly, you’re saying you could more easily go through every step of the problem if you changed 660 to a more “round” number that’s more conducive of mental math and doubling/halving? If that’s what you’re saying, you are correct. If you want to go through the entire problem yourself to make sure the data really is sufficient and can get you to an answer at the end, modifying 660 to 650 (or another simpler number) does help.

      And I can understand why you’d want to do this even though it really doesn’t matter in terms of getting the correct answer. During GMAT practice, fully solving the problem can boost your mental math skills and allow you to check your answers to data sufficiency questions in the process. Of course, you’ll want to abandon this approach by test day. Fully solving data sufficiency problems burns through precious time on the exam itself. So as your studies progress, make sure you eventually stop relying on this technique.

  3. Wajid August 13, 2015 at 5:38 am #

    In response to Mike’s reply on July 5, 2014 at 11:39 am:

    The below video gives a simple explanation.

    I am sorry that I am posting even if i got no question. I am just bit enthused to share it.


  4. Wajid July 5, 2014 at 9:13 am #

    How about this:

    Let x be no. of bottles of solution, which is 6% concentrated (we know it is 400, but lets use that information later)

    Let y be no. of bottles of solution, which is 12% concentrated





    Therefore x/y=1/2

    so, for each bottle of x, we need 2 bottles of y.
    Therefore, for 400 bottles of x, we need 800 bottles of y.

    • Mike MᶜGarry
      Mike July 5, 2014 at 11:39 am #

      Dear Wajid,
      That is a perfectly fine solution for this particular problem. It’s not clear to me what logic you used to create the first equation, and whether that logic would generalize to other scenarios. Make sure you understand multiple ways of thinking about this: you always understand math more deeply when you understand multiple ways to interpret a problem.
      Does this make sense?
      Mike 🙂

      • Wajid July 6, 2014 at 4:09 am #

        I got it. I am thankful that I got good schooling for math.

        Let me make the explain the equation.
        The equation works for mixture problems.

        What I want to do here is, I want to get ratio.
        I want to find out,
        for each bottle of 6% concentrated solution
        how many bottles of 12% concentrated solution are required
        to make the resultant solution to be of 10 % concentrated.

        I assumed that
        I have X bottles of 6% concentration
        and Y bottles of 12% concentration.

        On mixing both the solutions, I got
        a total of X+Y bottles of solution, which is 10% concentrated solution


        Hope the explanation helps.


        I admire you man. I saw your SC videos.
        Your explanation is direct, powerful, sleek and by far, the best explanation ever given.

        With deep regards,

        • Mike MᶜGarry
          Mike July 6, 2014 at 12:01 pm #

          Dear Wajid,
          Thank you very much for your kind words. Best of luck to you!
          Mike 🙂

  5. leila November 14, 2013 at 7:38 am #

    Dear Mike, I don’t know if you still check this. I have just come across your posts and I am finding them very helpful. I have a question regarding the first practice problem explanation:

    You have written:

    “In the resultant solution, we have 10% of Y = 0.12*X of phosphoric acid.”

    How did you conclude that 10% of Y equals 0.12 X? shouldn’t 10% of Y equal 0.12 X + 24?
    How did you make the 24 disappear?

    Thank you.

    • Mike MᶜGarry
      Mike November 14, 2013 at 9:41 am #

      Dear Leila,
      Yes, we still check this blog. 🙂 I will apologize — there were a few typos in the lines you quoted, and these typos made those lines absolutely nonsensical. That’s why you found them confusing. I have corrected the typos, so now it should all be clear. Thank you for pointing this out. Best of luck to you.
      Mike 🙂

  6. Kevin October 1, 2012 at 7:44 am #

    For the second problem,

    if I have R = 660
    .6(P)+.10(Q)=165 –>
    6P+Q=1650, aren’t there multiple values for P and Q??

    Is 6P+Q much different than 2P+Q?

    If I see this problem with the clock ticking away, I will choose A instead of D.
    How do you know it’s sufficient without having to put more time on the clock to actually calculate all the variables for 6P and Q to make sure there’s only 1 solution here?

    Thanks for your help

    • Mike MᶜGarry
      Mike October 1, 2012 at 9:38 am #

      Kevin — For the second equation, you are only considering the “concentration” equation, .6(P)+.10(Q)=.25(R) — for one equation, given R, you would have multiple values of P & Q. Instead, you forgot the VOLUME equation P + Q = R. One equation with two unknowns leads to multiple values, but TWO equations with two unknowns leads to unique solutions, so long as the equations aren’t dependent (which they aren’t here.) That’s precisely why both statements are sufficient. Does this make sense?
      Mike 🙂

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