offers hundreds of GMAT video lessons and practice questions. Go there now.
Sign up or log in to Magoosh GMAT.

GMAT Quant: Practice Problems with Percents

First, here are eight practice problems exploring typical GMAT percent themes.

1. The original price of a suit is $200.  The price increased 30%, and after this increase, the store published a 30% off coupon for a one-day sale.  Given that the consumers who used the coupon on sale day were getting 30% off the increased price, how much did these consumers pay for the suit?

  1. $182
  2. $191
  3. $200
  4. $209
  5. $219

2. The profits of QRS company rose 10% from March to April, then dropped 20% from April to May, then rose 50% from May to June.   What was the percent increase for the whole quarter, from March to June?

  1. 15%
  2. 32%
  3. 40%
  4. 62%
  5. 80%

3. Bert and Rebecca were looking at the price of a condominium.  The price of the condominium was 80% more than Bert had in savings, and separately, the same price was also 20% more than Rebecca had in savings.  What is the ratio of what Bert has in savings to what Rebecca has in savings.

  1. 1:4
  2. 4:1
  3. 2:3
  4. 3:2
  5. 3:4

4. Company KW is being sold, and both Company A and Company B were considering the purchase.  The price of Company KW is 50% more than Company A has in assets, and this same price is also 100% more than Company B has in assets.  If Companies A and B were to merge and combine their assets, the price of Company KW would be approximately what percent of these combined assets?

  1. 66%
  2. 75%
  3. 86%
  4. 116%
  5. 150%

5. There are 300 seniors at Morse High School, and 40% of them have cars.  Of the remaining grades (freshmen, sophomores, and juniors), only 10% of them have cars.   If 15% of all the students at Morse have cars, how many students are in those other three lower grades?

  1. 600
  2. 900
  3. 1200
  4. 1350
  5. 1500

6. A scientific research study examined a large number of young foxes, that is, foxes between 1 year and 2 years old.   The study found that 80% of the young foxes caught a rabbit at least once, and 60% caught a songbird at least once   If 10% of the young foxes never caught either a rabbit or a songbird, then what percentage of young foxes were successful in catching at least one rabbit and at least one songbird?

  1. 40%
  2. 50%
  3. 60%
  4. 80%
  5. 90%

7. A book store bought copies of a new book by a popular author, in anticipation of robust sales.  The store bought 400 copies from their supplier, each copy at wholesale price W.  The store sold the first 150 copies in the first week at 80% more than W, and then over the next month, sold a 100 more at 20% more than W.  Finally, to clear shelf space, the store sold the remaining copies to a bargain retailer at 40% less than W.  What was the bookstore’s net percent profit or loss on the entire lot of 400 books?

  1. 30% loss
  2. 10% loss
  3. 10% profit
  4. 20% profit
  5. 60% profit

8.  At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30.  These two ticket types were the only source of revenue for this concert.  If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R?


Solutions will be given at the end of this article.


Thoughts on percents

First of all, here are some previous blogs on percents:

(1) Fundamentals of percents, including using multipliers for percent increase and decreases.

(2) Solution of a percent with variables problem, from the OG

(3) Another solution of percent with variables problem, from the OG

(4) Scale factors & percent change

(5) Solution & Mixing problems

(6) Ratios & Proportions

Especially in those first three, you can find some useful hints for the foregoing problems.


The BIG percent mistake

Folks who are rusty at math and/or returning to it after a long absence may not appreciate something known well to anyone who writes math problems: certain mistakes are as predictable as clockwork.  Anyone who writes problems knows — in such-and-such kind of problem, the vast majority of test-takers will make such-and-such very predictable mistake.  Of course, on any multiple choice standardized test, that predictable mistake consistently will be among the answer choices: it’s as if the test-maker sets up a huge butterfly net, and the unwitting test-takers run into this trap like lemmings running to the sea.   I know this sound cruel, but the purpose of a good standardized test question is to distinguish those who know their stuff from those who don’t, and highly predictable errors are great ways to draw such a distinction very clearly.

Obviously, as a test-taker, it is very much to your advantage to learn to spot these very predicable mistake patterns.  Just avoiding these will put you well ahead of the pack.   In many different articles on this blog, I discuss these predictable mistakes.

The big predictable mistake with percents has a few variations on the same pattern.

(a) if something increases by P%, then decreases by Q%, the net change is not (P – Q)%

(b) if something increases by P%, then increases by Q%, that’s not an increase of (P + Q)%

(c) [special case] if something increase by P%, then decrease by P%, we do not return to the original value.

(d) [more general case] if something increase by P%, then decreases by Q%, then increase by R%, the total change in percent is not (P – Q + R)%

What all of these have in common is a deep root error — you cannot figure out the total percent change of a series of individual percent changes by adding or subtracting the individual percents.   That’s the mistake.  People see “start at $100, 40% increase, followed by a 40% decrease“, and scores upon score of people, as predictably as the sunrise, will believe and insist without a shadow of a doubt that the end result must be $100.  This large crowd will be in unison and they will be wrong.

What’s the correct thing to do?  The correct way to treat any series of percent changes is to express each change as a multiplier, and then multiply all the multipliers.  That first blog above talks about creating the multipliers for percent increases & decreases.  Once we have that, we can figure out the real change.  For example, the multiplier for a 40% increase would be: 1 + 0.40 = 1.4, and the  multiplier for a 40% decrease would be 1 – 0.40 = 0.6; now, 1.4*0.6 =  = 0.84, so the final price in fact would be $84, which is a 16% decrease.



If the foregoing discussion gave you any insights, you may want to re-read the practice problems before reading the solution.  Here’s another practice problem from the Magoosh Product:


If you have any observations or questions, please let us know in the comments section.


Practice problem explanations

1) Given the foregoing discussion, it may be obvious now the trap-mistake answer is (C).   Even if you can’t remember the correct thing to do, at the very least, learn to spot the trap!

The multiplier for a 30% increases is 1 + 0.30 = 1.3, and the multiplier for a 30% decrease is 1 – 0.30 = 0.70, so the combined change is 1.3*0.7 = 0.91, 91% percent of the original, or a 9% decreases.  Now, multiply $200*0.91 = $182.  Answer = (A).

2) Given the foregoing discussion, it may be obvious now the trap-mistake answer is (C), which results from simply adding and subtracting the percents.   We need multipliers.

multiplier for a 10% increases = 1 + 0.10 = 1.1

multiplier for a 20% decreases = 1 – 0.20 = 0.8

multiplier for a 50% increases = 1 + 0.50 = 1.5

Now, multiply these.  First, multiply (0.8) and (1.5), using the doubling & halving trick.   Half of 0.80 is 0.40, and twice 1.5 is 3

(0.8)*(1.5) = (0.4)*(3) = 1.2

Now, multiply this by 1.1

1.2*1.1 = 1.32

Thus, the three percent changes combined produce a 32% increase.  Answer = (B).

3) The trap answer here would be to take the ratio of 80% and 20% — those don’t represent actually amounts that other person has, just the differences between amounts owned and the cost of the condo.  Think of this in terms of multipliers.  Use the variables:

B = amount Bert has in savings

R = amount Rebecca has in savings

P = price of the condominium

Then in terms of multipliers, the information given tells us that P = 1.8*B, and P = 1.2*R.  Set these equal.

1.8*B = 1.2*R


Answer = (C)

4) There are a few ways to solve this.  This is plug-in approach.  Suppose Company A has $100 in assets.  (Yes, unrealistic, but a convenient choice.)  Then Company KW is being sold for 50% more = $150.   Now, this $150 is 100% more than what company B has in assets —- i.e., $150 is double what company B has in assets, so company B has $75 in assets.  Now, suppose companies A & B pool their resources — together, they have $175 in assets.

Notice, first of all, combined they have more in assets than the cost of KW, the price of KW would be a percent less than 100%.  Even if nothing else, we could eliminate (D) & (E), and practice solution behavior.

Part = price of KW = $150

Whole = combined assets = $175


Answer = (C)

5) Let x = the number of students other than seniors (freshmen + sophomores + junior).  We know 40% of the 300 seniors have cars.  Well, 10% of 300 is 30, so 40% is 4 times this —- 4*30 = 120 seniors have cars.  We know 10% of the other students have cars, so that would be 0.1*x.  The total number of students with cars is 120 + 0.1x.  That’s the PART.

The total number of students = 300 + x.  That’s the WHOLE.

PART/ WHOLE x 100% = 15%, which means that PART/ WHOLE = 0.15, which means PART = 0.15*(WHOLE).  That can be our equation.

(120 + 0.1x) = 0.15(300 + x)

120 + 0.1x = 45 + 0.15x

75 + 0.1x =  0.15x

75 = 0.15x – 0.10x = 0.05x

150 = 0.10x

1500 = x

Answer = (E)

6) This is less about percents and more about probability, particularly the probability OR-rule.   Let R = the event that a young fox catches at least one rabbit, and let S = the event that a young fox catches at least one songbird.   Using algebraic probability notation, we know P(R) = 0.8 and P(S) = 0.6.   We know P((not R) and (not S)) = 0.1, and the complement of [(not R) and (not S)] would be [R or S], so by the complement rule, P(R or S) = 1 – 0.1 = 0.9.  The question is asking for P(R and S).  The OR rule tells us

P(R or S) = P(R) + P(S) – P(R and S)

0.9 = 0.6 + 0.8 – P(R and S)

0.9 = 1.4 – P(R and S)

0.9 + P(R and S) = 1.4

P(R and S) = 0.5

Answer = (B)

7) First of all, amount paid = 400*W.  That was the bookstore’s total expenditures.  The total revenue came in three stages

150 copies @ 80% more than W = 150*1.8W = 300*.9W = 270W

100 copies @ 20% more than W = 100*1.2W = 120W

150 copies @ 40% less than W = 150*0.6W = 300*0.3W = 90W

Notice, all the percent changes were converted to multipliers.   Also, notice the use of the doubling & halving trick in the first and third lines.

Total revenue = 270W + 120W + 90W = 480W

Well, the store took in more revenue than they spent, so they made a profit, not a loss.   Notice that 10% more than 400 would be 440, so 480 would be a 20% increase.

Answer = (D)

8) There are a few different methods of solution.  I will show a numerical approach.   Let’s try some simple cases — suppose they sold nothing but balcony tickets: then B = 100 and R = 100.  If we plug in R = 100 …


Right away, with one choice, we know that (A) and (C) are out.  At this point, even if we could do nothing else, we could still use solution behavior.

Now, suppose the revenue they took in was half and half, so that R = 50%.  Well, in order for the revenue from balcony tickets to equal the revenue from orchestra tickets, they would have to be sold in a ratio of 5:3, which means that 5/8 of the tickets sold would be balcony tickets.  Thus, if we plug in R = 50, we should get B = (5/8)*100  — notice, we don’t actually need to calculate that out: the fraction is fine.  In the calculations below, notice the use of the doubling & halving trick in the denominator, to get the factor of 100.


Backsolving gets us to answer (D) very efficient.

Now, here’s an algebraic approach, which is longer, but some folks want to see this anyway for the algebra practice:

Let N = the total number of tickets sold.  Then:



AAAA big fraction

In the big fraction, in each term: cancel the N’s, cancel a factor of 10, and cancel the “divided by 100″:

AAAA smaller fraction

Multiply both sides by the denominator.

R*(500 – 2B) = (3B)*100

500R – 2RB = 300B

Get all the B’s on one side

500R = 2RB + 300B

B*(2R + 300) = 500R


Answer = (D)

That’s the full-blown algebraic solution, although why anyone would want to slog through all that instead of using backsolving is beyond me!!

About the Author

Mike McGarry is a Content Developer for Magoosh with over 20 years of teaching experience and a BS in Physics and an MA in Religion, both from Harvard. He enjoys hitting foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Follow him on Google+!

22 Responses to GMAT Quant: Practice Problems with Percents

  1. Tango February 17, 2014 at 7:21 am #

    Can you please let me know what difficulty level do the above problems fall in?? I got all of them correct…. I just to see how much I should expect in the Math Section.

    • Mike
      Mike February 17, 2014 at 4:34 pm #

      Dear Tango,
      Well, it’s always hard to estimate difficulty, but none of them is especially easy, and the last in particular is challenging. I would say at least a few of them are 700 level. If solving all eight of these was no problem for you, then you certainly are in good shape for the GMAT Q section, at least on the topics of percents.
      Good job!
      Mike :-)

  2. barb February 1, 2014 at 1:46 am #

    Hey Mike. How can number 4 be done without the plug in approach? I am trying to make sense of the concepts of the percent change formulas conceptually.

    • Mike
      Mike February 1, 2014 at 1:29 pm #

      Dear Barb,
      Without plug-in, this problem becomes a sea of algebra. Here’s how I would do it algebraically. Let P be the price of KW, let A = A’s assets, and B = B’s assets. We know P = 1.5*A and P = 2B, which tells us that 1.5A = 2B or 0.75A = B. Therefore, combined assets A + B = A + 0,75A = 1.75*A, and so now the question is: P is what percent of 1.75*A? In other words, 1.5A is what percent of 1.75*A, or 1.5 is what percent of 1.75? We can double both — 3 is what percent of 3.5? Double again —- 6 is what percent of 7? Well, that’s close to 100%, just less than that. We know 1/7 is about 14%, so 6/7 would be about 100 – 14 = 86 percent.
      There’s no way to get to the answer without somehow wrestling a bit with the number. Personally, I think the plug in approach is easier than this algebraic approach, but you need to something like this — set the numbers in motion and follow their consequences. Without doing that in some form, I don’t think you can intuit the answer.
      Does this make sense?
      Mike :-)

  3. tiffany February 1, 2014 at 12:27 am #

    Hi Mike! Thank you so much for posting these. They are very helpful. Why is it that the answer for number 3 is 2:3. When I worked it out I got 3:2 since it is B:R. I am not sure if I am missing a concept here which is causing me to do this backwards?

    • Mike
      Mike February 1, 2014 at 1:17 pm #

      Dear Tiffany,
      Hmmm. I would have to see your steps to know where things went awry. I will say: just think about this conceptually. The price is 80% more than Bert has — in other words, it’s almost double what Bert has, and Bert has only a little more than half the price. By contrast, the same price is 20% more than Rebecca has — in other words, she has almost the whole price all by herself. Bert has barely more than 1/2 the price, and Rebecca has almost the whole price by herself, so Rebecca must have MORE than Bert. That means, in a B:R ratio, the R number must be bigger than the B number. Check all your work to make sure that, at each step, it makes sense that R > B. In fact, since we know the answer is 2:3, plug in B = 2 and R = 3 in each link of your work, and that will help you find the mistake.
      Does all this make sense?
      Mike :-)

  4. Titouprince January 8, 2014 at 9:32 am #

    I don’t understand why in the #7 the result is not the same as 1.8*1.2*0.6, just multiplying the multipliers?
    Thank you

    • Mike
      Mike January 8, 2014 at 10:10 am #

      Dear Titouprince,
      We would simply multiply the multipliers if a *single item* went up 80%, then up 20%, and then down 40%. Here, we are talking about different items — in fact, unequal batches of books. The 150 books that were sold at 80% more than W were sold and gone by the time they sold the next batch of 100 at 40% more than W. Multiplying together the multipliers is good when we want the cumulative effect of all the percent changes on a single thing. Here, the different percents are happening to different batches of books — nothing is accumulating in any one place. Does this make sense?
      Mike :-)

    • Titouprince January 8, 2014 at 10:40 am #

      Thanks a lot!

      • Mike
        Mike January 8, 2014 at 10:45 am #

        You’re quite welcome! Best of luck to you!
        Mike :-)

  5. Dennis November 25, 2013 at 11:38 pm #

    Hi Chris, I think there is a spelling error in question 6:
    ” If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets”

    • Mike
      Mike November 26, 2013 at 8:49 am #

      This is Mike, the author of this blog. A spelling error in #8? I’m sorry, but I don’t follow. Which word do you believe is misspelled?
      Mike :-)

  6. satish October 29, 2013 at 12:20 am #

    Hi Mike,

    I am under the impression that solution given for problem 6 is right.Could please explain why it is wrong .This is based on the comment by Musa “P((not R) and (not S)) = 0.1, however this must be 0,20. So the right answer should be 0,60 nat 0,50 “

    • Mike
      Mike October 29, 2013 at 10:05 am #

      Dear Satish,
      The solution for #6 right now is indeed correct. When I first published this blog, at the end of September, problem #6 had a typo in it — one number in the question was different from what I intended, and consequently, the problem and solution did not match. A couple weeks ago, Musa made a comment pointing this out, and at that time, I corrected the typo, to make the problem & solution are perfectly consistent. This is the version you now see. Does this make sense?
      Mike :-)

  7. GMAT4o October 21, 2013 at 2:16 am #

    Hi Mike,

    Can you show algebraic approach to Prob 8.

    • Mike
      Mike October 21, 2013 at 10:29 am #

      Dear GMAT4o,
      Well, I added the algebraic solution to #8 as part of the solutions listed above. I certainly could not do all that in under 5 minutes, and I doubt anyone short of Will Hunting could. I hope it helps to see it for algebra practice, but solving with numerical plug-ins & backsolving is definitely more efficient.
      Mike :-)

      • GMAT40 October 21, 2013 at 6:19 pm #

        Thanks Mike,

        I used the back solving method but was somehow getting numbers and constants mixed up with algebraic approach. this as just to understand where was i going wrong.


        • Mike
          Mike October 22, 2013 at 10:38 am #

          You’re welcome. I’m glad you found it helpful. That 8th problem is a particularly tricky one. Best of luck to you.
          Mike :-)

  8. musa bilgehan October 10, 2013 at 7:08 am #

    Hi, I think the answer for the question 6 is wrong. In the solution iti is stated that: P((not R) and (not S)) = 0.1, however this must be 0,20. So the right answer should be 0,60 nat 0,50

    • Mike
      Mike October 10, 2013 at 10:20 am #

      Dear Musa,
      You’re perfectly correct. Thank you very much for pointing this out. :-) In the process of writing this blog, I decided to change the problem slightly, and thought I had consistently changed everything. I meant for the doubly-deprived fox to be 10%, not 20% of the population, so I just changed that in the question, to make it consistent with the solution. Good eye for detail — that skill will serve you well on GMAT Quant! :-) Thanks again!
      Mike :-)

      • Son April 16, 2014 at 1:50 pm #

        Hey Mike for Q 6 whats the right answer I am getting 0.6 too
        Is it right?

        0.7 + 0.5 +0.1 – intersection= 1
        Intersection equals = 0.3
        We want A or B event that give 0.7 -0.3 =0.4 and 0.5-0.3 =0.2
        Hence 0.6

        • Mike
          Mike April 16, 2014 at 3:11 pm #

          Dear Son,
          Notice that all the answers and explanations are given on the page. Please read the OE of #6, and let me know if you have any further questions.
          Mike :-)

Magoosh blog comment policy: To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!

Leave a Reply