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GMAT Quant: Must Be True Problems

First, try these challenging GMAT practice problems. Give yourself a strict time-limit of six minutes for the set.

1) If A is a number, which of the following must be true for any A?

2) If F and G are integers, with F < G, which of the following must be true?

3) If J is an integer, which of the following must be true?

4) If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

Attacking “must be true” problems

For many math problems, one has to focus on finding the one right answer. That approach, with these problems, is highly problematic. If you are super-talented with math, fluent in Algebra, then perhaps you can scan the list of choices and immediately spot the correct answer. If you are talented enough to do that, then good for you!

For most folks, that approach is simply not feasible. The much more efficient solution involves elimination. You see, if any statement “must be true”, then it will be true for any value you pick. It takes incredible skill (or fantastic luck) to pick a single value that eliminates four answer choices and leaves only one. If that happens by chance, great, but don’t focus on that. Focus on speed and efficiency. Pick one value, eliminate what you can, then pick another, eliminate more, etc. Whittle down the choices until you are left with one. That’s the most efficient approach for these problems.

You have to pick values to plug in. At the start of picking, pick very easy values — with a few easy choices, you should be able to eliminate two or three answer choices, making your overall job much easier.

What values to pick?

If you picked the same values for all four questions above, that’s a problem. Those questions were specifically written so that the allowable numbers would be different in different questions.
When you hear “A is a number”, of what do you think? Remember that the general word “number”, like the general term “human being”, includes all types. Just as the individuals under the term “human being” constitute a bewildering variety, so do the citizens of the realm of “numbers.” Numbers include positive & negative, wholes & fractions & decimals, square-roots, pi, etc. etc. It’s a realm of perfect democracy — for example, -(pi)/5 is just as much a number as is 8. If, when a GMAT problem says “number”, your mind defaults to {1, 2, 3, 4, ..}, the “counting numbers”, then with all due respect, that’s really a third-grade way of thinking about the word “numbers”, and the GMAT will viciously punish that kind of thinking. You always must be aware of all possibilities for any category.

numbers = everything, positive/negative, zero, fractions, decimals, everything on the continuous infinity of the number line.

integers = positive and negative whole numbers = { … -3, -2, -1, 0, 1, 2, 3, …}

positive integers = the counting numbers, 1, 2, 3, 4, ..}; this list does NOT include zero, because zero is neither positive nor negative.

Remember that “odd integers” includes both positive and negatives, and “even integers” includes positives & zero & negatives. Prime numbers are, by definition, a subset of positive integers. There are no negative primes, and neither zero nor one is a prime number. It’s a very good idea to have the first few prime numbers memorized:

Primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, …)

Notice that 2 is the only even prime number (all other even numbers are divisible by 2, and hence are not prime). After 2, the primes are an irregularly spaced set of odd numbers that continues to infinity. In fact, let’s talk a moment about their pattern

The pattern of prime numbers

In this section, I am going to dip into very advanced math, far far beyond the GMAT realm. In modern mathematics, most work around the “pattern of prime numbers” revolves around the Riemann Hypothesis, the single hardest un-answered question in all of mathematics. The great mathematician Bernhard Riemann proposed this grand question as a conjecture in an 1859 paper, and it has utterly baffled the most brilliant mathematical minds on the planet every since. Forget about answering the question — you need close to a Ph.D. in mathematics just to understand what the question is asking and what it has to do with prime numbers! (A brief graph of the central player of the Riemann Hypothesis, the Riemann Zeta Function, is shown below.)

All of which is a long-winded way of saying: there is no easy algebraic formula that always produces prime numbers. A simple pattern for prime numbers absolutely does not exist. Thus, in a “must be true” question, any answer of the form “[simple algebra expression] is a prime number” cannot possibly be true all the time. That absolutely has to be a wrong answer.


If you had any big “aha!” while reading this article, you may want to give those problems another glance before reading the solutions below. If you have any further questions or observations, please let us know in the comments section below.



Practice problem explanations

1) First of all, notice that A is a “number” — could be positive or negative or zero, whole or fraction or negative. Let’s first try a very easy choice: A = 0. Then, for the five answer choices, we get:
(A) undefined
(B) false
(C) true
(D) true
(E) undefined

Remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers. We are down to (C) & (D). Try another relatively simple choice, A = -2.
(C) 1 = 1, true
(D) +2 ≠ -2, false

Remember that if A is negative, a negative squared is a positive, and the square-root sign always has a positive output. Therefore, the left side of the (D) equation is always positive, and can’t be true if A is negative.
Answer = (C).

2) For this problem, both F & G are integers — they could be positive or negative or zero, but they must be whole numbers. Here’s a suggestion — one good choice would be to pick a positive number with smaller absolute value and a negative number with a larger absolute value: say, F = -4 and G = 1. Let’s true those in the answers:
(A) 16 < 1 false
(B) -64 < 1 true
(C) 4 < 1 false
(D) 4 < 1 false
(E) 1-16 = -15 > 0 false

With one magical choice, we were able to eliminate four answer choices. Keep in mind, the combination small positive/big negative is a good one, and so is pairing a negative number with 0 (e.g. F = -4, G = 0), because zero is greater than any negative number.
Answer = (B)

3) Here, J is an integer, so it can be positive or negative or zero, but not a fraction or a decimal. Zero is always a special case among integers, so let’s start there. J = 0: then,
(A) undefined error
(B) 0 > 0 false
(C) 0 > 0 false
(D) 1 > 0 true
(E) -1 > 0 false
Here, the special case was enough to eliminate four answers. Again, remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers.
Answer = (D)

4) Here, p and q are odd prime numbers. First of all, eliminate (A), because there’s no simple algebraic rule that generates prime numbers. The easiest choices for numbers are p = 3 and q = 5. Try this:
(B) 3 + 5 = 8 is divisible by 4 — true
(C) 5 – 3 = 2 is not divisible by 4 — false
(D) 5 + 3 + 1 = 9 = (5^2) – (4^2) — true
(E) (3^2) + (5^2) = 34

This last number requires comment. Every odd number is the difference between two perfect squares, because adjacent squares (n + 1)^2 - n^2 = 2n + 1. When we subtract two even squares or two odd squares, the number is always divisible by 4. Thus, 34 cannot be the difference of any pair of squares. Thus, on the basis of this choice, we can eliminate (E). We are left with (B) & (D).

Try p = 3 and q = 7.
(B) 3 + 7 = 10 is not divisible by 4 — false
(D) 3 + 7 + 1 = (6^2) – (5^2) — true

In fact, since p & q are both odd, (p + q + 1) must be odd, and as stated above, any odd number is the difference of two perfect squares. That’s why (D) must be true.
Answer = (D)

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14 Responses to GMAT Quant: Must Be True Problems

  1. Shail August 5, 2015 at 11:19 am #

    Hi mike,

    In option D why dint you consider p=3 and q=5 (p<q)both are odd prime numbers and the difference of two perfect square say 3^2- 2^2.
    Upon substituting it in the equation,i get
    9=5 which is not correct.

    Again using other pair of prime number, say p=13 and q=17 (p<q) and the perfect square is say 4^2 and 5^2 (different integers)

    31=9 again its incorrect.

    And you have chosen D as the correct ans. Please explain.

  2. Ifrah February 15, 2015 at 12:52 am #

    Hey Mike!!
    Can you help me with this?
    if a^2* b^3 *c^4 *d^5 > 0 and d < 0 then which of the following is true?
    a) a<0
    b) b<0
    c) c<0
    d) ab<0

  3. maxime December 16, 2014 at 3:30 pm #

    Hi Mike,

    What about option 4a:

    If P&Q are prime, then 2P+Q=prime

    Since no factor will divide 2P+Q into an integer, the result must be a prime number

    • Mike MᶜGarry
      Mike December 18, 2014 at 10:02 am #

      Dear Maxime,
      I’m happy to respond.. 🙂 It’s important to understand: no simple rule, even a rule involving prime number inputs, can generate prime numbers. The pattern of prime numbers is beyond anything algebra can begin to handle. That’s a very deep idea that can save you some time.
      In this case, suppose p = 5 and q = 17 — then 2P + Q = 27 is not divisible by 5 or 17, but it’s not a prime number: it’s divisible by 3. In fact, 27 = 3^3.
      Does all this make sense?
      Mike 🙂

  4. Sriram November 20, 2014 at 4:06 am #

    I think I need some more clarity for 4)E ; why have you not considered the difference of squares of non-consecutive odd and even numbers, that is, how can we be sure that some number(34 in this case) cannot be the difference of the squares of say 2p and 2q+1, where p is not equal to q?

    • Mike MᶜGarry
      Mike November 20, 2014 at 10:42 am #

      Dear Sriram,
      My friend, think about what algebra tells us. Suppose the two numbers, whether odd or even, are 4 apart; then
      (n + 4)^2 – n^2 = 8n + 16, which must be divisible by 8.
      Suppose the two numbers, whether odd or even, are 6 apart; then
      (n + 6)^2 – n^2 = 12n + 36, which must be divisible by 12.
      You can see where the pattern is going. All of these differences will have be divisible not only by 4 but by various multiples of 4. This leaves poor 34 out in the cold.
      Even if the algebraic approach is not natural for you, I would suggest sitting down with a calculator and looking for patterns yourself. Try to subtract pair of squares in all combinations. Look for patterns. Do your best to find a pair of squares you could subtract to get 34. I think if you try this, you will start to see the patterns for yourself, and you will understand more deeply. Looking for the patterns among numbers yourself is far more valuable than the empty speculation about what they “might” do. Playing with numbers, open ended exploration, is precisely how you build number-sense.
      Does all this make sense?
      Mike 🙂

      • Sriram November 20, 2014 at 1:16 pm #

        I think you’ve missed my point. I understand that the difference between the squares of any “pairs” of odd and even numbers must be divisible by 4; Answer choice E simply talks about the difference of squares of integers and your proof assumes that these integers are odd or even pairs. What I was trying to imply was that (2q+1)^2 -(2p)^2 will not be divisible by 4, where p is not equal to q.

        • Mike MᶜGarry
          Mike November 20, 2014 at 2:20 pm #

          Dear Sriram,
          My friend, think about basic number properties. If P is even, then so is P^2; if Q is odd, then so is Q^2. An (ODD – EVEN) or (EVEN – ODD) would have to be an odd number. If both P & Q are even, or both are odd, then the difference of squares (P^2 – Q^2) is divisible by 4. If one is even and one is odd, the difference of squares (P^2 – Q^2) must be an odd number. Either way, 34 is not a possible result. There is no possible way to subtract two squares of integers and get a result that is an even number not divisible by 4.
          Again, my friend, you are asking questions in the abstract, without exploring the patterns among numbers for yourself. You can not understand math purely by learning abstract rules. You have to do real calculations with real numbers, and look for the patterns yourself. Number sense comes from seeing for yourself. Math is not a spectator sport.
          Does all this make sense?
          Mike 🙂

  5. Adam August 11, 2013 at 5:04 am #

    In this expression: (D) 5 + 3 + 1 = 9 = 52 – 42 — true
    how did you get numbers 52 and 42?
    and finally, how did you get following result: (E) 32 + 52 = 34

    • Mike MᶜGarry
      Mike August 11, 2013 at 12:46 pm #

      Those were all mistakes in how the exponents showed up. I believe I have corrected those, so it should make sense now. Thank you for pointing this out.
      Mike 🙂

  6. GMATQuantlover July 2, 2013 at 7:20 am #

    Hi Mike,
    I really enjoyed the explanation thanks.. but can you elaborate more on this part cause I didn’t understand it

    “Every odd number is the difference between two perfect squares, because adjacent squares (n + 1)2 – n2 = 2n + 1.”

    in the explanation of the last question

    • Mike MᶜGarry
      Mike July 2, 2013 at 10:12 am #

      Dear GMATQuantlover,

      Think about it. The difference between 1^2 and 2^2 is 1 + 2:
      1^2 + 1 + 2 = 2^2
      The difference between 2^2 and 3^2 is 2 + 3:
      2^2 + 2 + 3 = 3^2
      The difference between 57^2 and 58^2 is 57 + 58:
      57^2 + 57 + 58 = 58^2
      That pattern holds for the squares of all positive integers. Since any odd number can be expressed as the sum of two consecutive integers, any odd number can be the difference between two squares. For example, consider 47 — 47 = 23 + 24, so 47 must be the difference between 23^2 and 24^2:
      23^2 + 47 = 24^2
      Check all this on a calculator, and check it with other numbers, to verify to yourself that it works. It is highly unlikely to be tested directly on the GMAT, but it’s just a handy math trick to have up your sleeve.
      Mike 🙂

      • Aniket June 8, 2015 at 5:56 am #


        In this case, considering two consecutive integers p and q, where q > p.
        q^2 – p^2 = q + p. Since p and q are consecutive integers, (p + q) will always be odd and hence the any odd number can be expressed as the difference between two squares.

        But, q^2 – p^2 = (q – p)*(q + p)
        So, (q – p)*(q + p) = q + p.
        So wont q – p = 0 and q = p. But we have considered p and q to be two consecutive integers!! Is this not a contradiction?

        Please correct me if I am wrong. Studying for GMAT, need to get my doubts cleared.
        Thanks in advance.

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