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GMAT Quant: Arithmetic with Inequalities

Here are four GMAT Data Sufficiency questions involving inequalities.

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Full solutions will come at the end of this article.

 

The arithmetic of equations: a review

First of all, let’s review what should be more familiar — the arithmetic of equations.   Suppose A = B and P = Q.  The soundbyte is: you can combine them in almost any way imaginable to get a new valid equation.  You can add them, in either order (A + P = B + Q) or (A + Q = B + P).   You can subtract then, either one from the other, in either order (four subtraction equations: e.g. A – P = B – Q).  You can multiply them in either order (A* P = B * Q) or (A * Q = B * P).  You can divide either one by either other (assuming you are not dividing by zero), in either order (four division equations: e.g. A/P = B/Q).  You can raise either one to the power of the other, in either order (four exponentiation equation: e.g. A^P = B^Q).

(Note: for the GMAT, you are responsible for know what happens when you raise a base, say 3, to the power of integers or to the 1/2.  Technically, you can raise any positive base to any power — say, 3 to the pi — but that’s more complicated than the GMAT expects you to know.  For negative bases, things get dicier.  For the GMAT, just worry about negative bases to integer powers, (–2)^3 or (–2)^(–1).   Non-integral powers of negatives leads into complex numbers — again, beyond what the GMAT expects you to know).

Finally, the above may seem simple, and if A & B & P & Q are just individual numbers, then those equations are pretty much “duh”-simple.  Things get considerably more interesting if some or all of those letters are not individual numbers but algebraic expressions.  Even with four different algebra expressions, all this still holds for equations.

 

Adding inequalities

Everything gets trickier with inequalities.  First of all, an equation, such as A = B, is inherently symmetrical and “two-sided”, but an inequality is more a one-sided, unidirectional thing.   With any arithmetic of inequalities, we must consider the direction of the inequality.   Furthermore, equations are very intuitive, but some of the arithmetic of inequalities is quite unintuitive.

Adding inequalities is not so bad: you can add inequalities with the same direction of inequality.  Thus, if A > B, and P > Q, then it must be true that A + P > B + Q.  That always works, and it is in many ways what you’d expect.

 

Subtracting inequalities

This is the one that’s much trickier.   If A > B, and P > Q, then naïvely one might expect that (A – P) would be greater than (B – Q), but that’s not necessarily the case.  For example, suppose we have 10 > 5 and 2 > 1 — then we could subtract them in the same direction of inequality, and we’d get 8 > 4, which still works.  BUT, suppose 10 > 5, and 100 > 1, both true — now, if we subtract in the same direction of inequality, we find that (–90) is not greater than 4.   If we subtract two inequalities in the same direction of inequality, we may get another true statement, but we are not guaranteed that this will lead to a true statements, so it’s a very unwise move in problem solving.

Here’s the real head-scratcher: we can’t subtract inequalities with the same direction of inequality, but we CAN subtract inequalities with the opposite direction of inequality — in other words, if A > B, and P > Q, then it must be true that (A – Q) > (B – P).  That may be perplexing symbolically, so think about it in real world terms.  Suppose Ann has more money than Bob.  Suppose I take less from the richer person, Ann, and more from the poorer person, Bob.  When I am done, Ann still will have more than Bob — in fact, a greater difference than existed between them before!

 

Multiplication and division

Everything gets much hairier with multiplying or dividing inequalities when you consider — one or both sides could be negative.  Hmmm.  If we multiply or divide both sides of inequality by a single negative number, that’s perfectly legal, but we must remember to reverse the direction of inequality.   What happens if we were to multiply or divide inequalities and negatives were involved?  For example, if we know that x > –6 and y > –7, then what can we say about the product xy?  As it turns out, we could pick an x and a y that would satisfy the original inequalities and make the product xy equal absolutely any number on the number line.  This mindboggling bit of math is just to demonstrate why the GMAT is not going to touch multiplying or dividing general inequalities with a ten-foot pole!

Let focus, though, on a special case that could, in very advance problems, appear on the GMAT.  Suppose we know that all four numbers or expressions are positive: A > B > 0 and P > Q > 0.  Then, as with addition, we can multiply inequalities with the same direction: A*P > B*Q must be true.   And, as with subtraction, we can divide inequalities with the opposite direction: A/Q > B/P.  Again, remember the caveat: everything must be positive for these patterns to work.  If anything can be negative, things get much more complicated, so complicated that the GMAT won’t ask about them.

 

Absolute value inequalities

Absolute value inequality is a sizeable topic, with some mind-bending twists and turns, but chances are very good the GMAT is not going to probe this topic all that deeply.   In fact, probably most of the absolute value inequalities on the GMAT will be of the form: |x – p|, where p is some given fixed number.

Here, we must remember a few key mathematical facts.   First of all, subtraction gives us distance on the number line.   Technically, subtraction gives signed distance.  What do I mean by that?  Well, 5 – 2 = +3, indicating that it’s a signed distance of positive 3, i.e. three units to the right, from 2 to 5; by contrast, 2 – 5 = –3 indicating that it’s a signed distance of negative 3, i.e. three units to the left, from 5 to 2.

For ordinary distance, distance in the geometric sense of the word, we don’t care about sign or direction — the distance between two points is just a positive number and is the same, whether from A to B or from B to A.   That’s where absolute value comes in.  The expression |p – q| is the distance between numbers p & q on the number line.  That is a HUGE idea.

Thus, |x – 5| is the distance between variable point x and fixed point 5.  The expression |x – 5| < 2 indicates the set of all points x that have a distance to the point 5 of less than two.  Immediately, just thinking about this logic, and without any further calculations, we can see that |x – 5| < 2 is entirely equivalent to 3 < x < 7.   Recognize that this is exactly the kind of math the GMAT adores: with simple logic, you can jump to conclusions without having to do any calculations.  The GMAT goes crazy for math of this sort.   I can’t emphasize enough how important this particular set of logically interconnected ideas is.   If the GMAT asks you anything at all about absolute value inequalities, it is highly likely it would be something of this genre, and unlikely that it would be anything else in this extensive topic.

 

Summary

If you had some insights reading this article, I encourage you to take another look at the practice questions at the top before reading the solutions below.   Here’s a similar question from inside Magoosh.

5) http://gmat.magoosh.com/questions/960

If you have anything to add, or any questions, please let us know in the comment section below.

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Solutions to practice problems

1) In this problem, the tempting incorrect answer would be (A), or somehow would involve statement #1 as sufficient, but it’s not.  This gets into an idea discussed in this blog —- we can’t subtract inequalities with the same direction of inequality, but we can subtract inequalities with the opposite direction of inequality.

Statement #1: So the prompt inequality and this statement’s inequality would be true for a + b = 15, c + d = 7, b = 6, d = 2

prompt inequality: 15 > 7  — true

statement #1 inequality: 6 > 2 — true

prompt question: 9 > 5 — a > c, an answer of “yes”

But, both inequalities would still be true for a + b = 20, c + d = 18, b = 15, d = 3

prompt inequality: 20 > 18  — true

statement #1 inequality: 15 > 3 — true

prompt question: 5 < 15 — a < c, an answer of “no”

We can make different choices consistent with all the given statements that would produce either a “yes” or  “no” answer to the prompt question.  This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question.  This statement, alone and by itself, is insufficient.

Statement #2: Here, we are allowed legitimately to subtract the inequalities, because the directions of inequality are in opposite order.  When we subtract (b < d) from (a + b > c + d), we get a > c, a definitive “yes” answer.

Another way to think about it: (a + b) = LARGER, and (c + d) = SMALLER, so of course, the former is greater than the latter.  Now, say that b = tiny, and d = bigger, so of course, b < d.   Now, think about the two differences we will compare:

(i) a = (a + b) – b = LARGER – tiny

(ii) c = (c + d) – d = SMALLER – bigger

Obviously, it start with something LARGER, and subtract something tiny, the result will be greater than starting with something SMALLER and subtracting something bigger.  Therefore, a > c, a definitive answer to the prompt question.

This statement allows us to give a definitive “yes” answer to the prompt, so this statement, alone and by itself, is sufficient.

Answer = B

2) Statement #1: if p = 2 and q = 1, then this statement’s equation is true, 4 > 1, and p > q, so the answer to the prompt is “yes”.

But, if p = –2 and q = –1, then it’s still true that the square of p is larger than the square of q, 4 > 1, but now it’s true that p < q, so the answer to the prompt question is “no.”

We can make different choices consistent with all the given statements that would produce either a “yes” or  “no” answer to the prompt question.  This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question.  This statement, alone and by itself, is insufficient.

Statement #2: if p = 2 and q = 1, then this statement’s equation is true, 8 > 1, and p > q, so the answer to the prompt is “yes”.

Switching to negatives won’t make a difference, because the cube of a negative is still negative.  The values p = –2 and q = –3 satisfy the statement inequality, and it’s still true that p > q

What about fractions?  If p = 1/2 and q = 1/3, then p cubed is still larger than q cubed, (1/8) > (1/27), and p is still greater than q.  No matter what numbers we pick, the inequality in statement #2 directly implies the prompt inequality.  The mathematical way to say this is: cubing, or taking a cube-root, preserves the order of any inequality.

This statement gives us a definitive answer of “yes” to the prompt question, so this statement, alone and by itself, is sufficient.

Answer = B

3)  Think about the distance interpretation of absolute value inequalities.   We want to know: is x further than two units away from 6 on the number line?

Statement #1:  the x-values allowed by this statement are x’s that are more than three units from 4.  Here’s a picture of these values, in green:

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Notice, the endpoints, 1 & 7, are not allowed, because they are exactly three units from 4, and exactly 3 is not greater than 3.  Most of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six.  Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer.  Different choices give different answers.   This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question.  This statement, alone and by itself, is insufficient.

Statement #2: the x-values allowed by this statement are x’s that are more than one unit from 8.  Here’s a picture of these values, in green:

awi_img3

Again, notice the endpoints are not included.  Many of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six.  In fact, the point 6 itself — which is a distance of zero units from 6 — is allowed by this statement!  Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer.  Different choices give different answers.   This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question.  This statement, alone and by itself, is insufficient.

Combined statements: Now, we combine the constraints of the individual statements.  Now, the allowed points must be both more than three units from 4 and more than one unit from 8.  Shown in green are the points that simultaneously satisfy both constraints:

awi_img4

Now, all the green points are more than two units away from 6, and it absolutely impossible to pick a value of x that simultaneously satisfies the constraints of both statements and is closer than two units to 6.   The combined statements allow for a definitive “yes” answer to the prompt.  Together, the statements are sufficient.

Answer = C

4) Statement #1: this gets at the fundamental meaning of the inequality.   If y is one more than x, it must be greater than x.  To add one to a number means to move it one unit to the right on the number line, so y must be one unit right of x, which means it is greater than x.  This allows us to determine a definite “yes” to the prompt question.   This statement, alone and by itself, is sufficient.

Statement #2: There are a couple ways to think about this one.  One is to treat different categories of numbers.

(i) if x is negative, then by squaring, y will be positive, and y > x.

(ii) if x = 0, y = 1, and y > x

(iii) if x is a fraction between 0 and 1, then its square will also be between 0 and 1, and adding one to this will produce a number greater than 1, between 1 and 2.  Therefore, y > x.

(iv) if x = 1, y = 2, and y > x

(v) if x > 1, then squaring x makes it bigger, and adding one makes that even bigger, so y > x

Thus, for all possible values of x, y > x.  Thus, the prompt gives a definitive “yes” answer.

A totally different way to think about it: using coordinate geometry.  The graph of y = (x^2) is a parabola that passes through the origin.  The graph of y = (x^2) + 1 is this same parabola shift up one in the y-direction, passing through (0, 1) instead of the origin.  Now, compare this shifted parabola to the line y = x.  One of the special properties of the line y = x is that all points above this line, regardless of quadrant, have the property y > x.   Think about the graph:

awi_img5

The parabola is always above the line y = x, so every point on the parabola must satisfy y > x.

Either way, this allows us to determine a definite “yes” to the prompt question.   This statement, alone and by itself, is sufficient.

Both statements are individually sufficient.

Answer = D

 

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21 Responses to GMAT Quant: Arithmetic with Inequalities

  1. Elliot Holder October 18, 2017 at 3:23 pm #

    What above fractions? If p = 1/2 and q = 1/3, then p cubed is still larger than q cubed, (1/8) > (1/27), and p is still greater than q. No matter what numbers we pick, the inequality in statement #2 directly implies the prompt inequality. The mathematical way to say this is: cubing, or taking a cube-root, preserves the order of any inequality.

    Should it be What ABOUT fractions/

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert October 21, 2017 at 1:29 pm #

      Hi Elliot,

      Yes! Thanks for catching our typo! I have corrected the mistake 🙂

  2. pragya September 5, 2017 at 1:07 am #

    Hi mike!!

    First of all I love all your posts! thanks a ton for all the work you do!!

    Second, while solving the 4th sum – 1st statement y > x for all the values i put except for :

    -1/2 = -1/2 + 1 for which y=x

    Could you please explain where i am going wrong here

    (I might be making some silly mistake here, cause of all the studying!)

    I got the 2nd statement as sufficient.

    Thanks in advance!

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert September 10, 2017 at 5:45 pm #

      Hi Pragya,

      It looks like you did make a simple mistake here– but an easy one to make if your mind is already racing from lots of studying.

      If y is -1/2, then x would actually need to be -1.5 (aka, negative 1 and 1/2). Add 1 to -1.5, and you bring -1.5 up in value, which makes it a smaller negative number, -0.5, or -1/2.

  3. Praveen October 19, 2016 at 12:30 am #

    Hi,

    Can you please solve the below question using ‘subtracting inequalities’ method?

    Is ‘a’ positive?
    a – b > 0
    2a – b > 0

    This is what I did,
    Statement 1 and 2 are insufficient, now after I combine both the statements, I get,
    a>b and 2a>b. Which can be modified as a>b and -2a0 or a<0 . which means statement 1 and 2 are sufficient (Option 'C'). But the OA is Option 'E'

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert October 22, 2016 at 11:03 am #

      Hi Praveen,

      In order to subtract inequalities, we have to subtract the inequalities with different directions of inequalities. We have a – b > 0 and 2a – b > 0, since the signs are the same we have to subtract the opposite sides. So we get (a-b)-0 > 0-(2a-b). We have to distribute the subtraction sign in the right side of the equation: a-b>-2a+b. We can simplify this to 3a>2b, so even with the subtraction method we can’t determine if this a is positive. So neither statement A nor statement B is sufficient to solve this problem.

  4. Landry November 11, 2015 at 9:34 am #

    Hi Mike,

    in this video, you said, when one squares a fraction it get smaller. I guess this applies only to fractions with numerator < denominator. right?

  5. Diane L September 6, 2015 at 10:23 am #

    Hi Mike,

    For Absolute Value Inequalities, how about if | x+5 | < 2? Would the same logic apply? Could you explain this a bit?

    Thanks,
    Diane

    • Mike MᶜGarry
      Mike McGarry September 8, 2015 at 12:34 pm #

      Diane,
      I’m happy to respond. 🙂 Remember that for any linear expression of x, we can replace the expression with another variable, say, u = x + 5.
      Then, our inequality |u| < 2 easily reduces to -2 < u < 2, using the distance reasoning.
      Then, we substitute the expression of x for u:
      -2 < x + 5 < 2
      Then we subtract 5 from all three terms.
      -7 < x < -3
      A totally different way to think about that is with the distance formula. Remember that the expression {x + 5) is equivalent to [x – (-5)], the distance from negative five. We want points that are less than 2 from negative five on the number line, so they must go from negative three to negative seven, without including the endpoints. That's exactly what we go from the algebraic solution.
      Does all this make sense?
      Mike 🙂

  6. Samy November 25, 2014 at 5:44 pm #

    Hi Mike,

    This was an excellent blog post. One question — For Q3 above, I understand how you’ve combined the two inequalities in the latter part of your analysis graphically. Graphical presentation certainly helps. Unfortunately, I feel on standardized tests given time constraints, this may not be possible or the best course of action. Is there a way to combined absolute value inequalities (non-graphically)?

    I also read your blog-post titled “Understanding Absolute Values” that was great too, but I feel if the above question could be answered and addressed, then I would have covered most or wider range of questions/problem solving techniques/mastery that may be needed for the GRE.

    Thanks,

    Samy

    • Mike MᶜGarry
      Mike November 25, 2014 at 6:04 pm #

      Dear Samy,
      I’m very glad you enjoyed this blog post! 🙂 I will tell you: on this blog article, and on every article on this blog, I do not spend my time trying to think of unnecessarily complicated solutions. Every solution I show you is the most efficient. No, there is no straightforward algebraic way to combine these inequalities. The graphical solution is by far the quickest and most efficient solution. If you are not familiar enough to apply this solution with efficiency, then that’s exactly what you need to practice and understand in greater depth. Don’t allow your mathematical approach to be a prisoner of algebra. Part of success, on either the GMAT or GRE Quant section, is to be able to move fluidly back and forth between numerical, algebraic, and graphical perspectives.
      You may find this GMAT blog helpful:
      http://magoosh.com/gmat/2013/how-to-do-gmat-math-faster/
      Does all this make sense?
      Mike 🙂

  7. Sriram September 30, 2014 at 12:04 pm #

    Hi Mike,
    Just wondering, if A < B < 0 and P < Q <0 do you see any issues wrt multiplication and division?

    • Mike MᶜGarry
      Mike September 30, 2014 at 1:25 pm #

      Dear Sriram,
      This issue is extremely unlikely to arise on the GMAT, but I’m happy to address it. If A < B < 0 and P < Q < 0, then on multiplication, it's true that A*P > B*Q > 0, and it’s also true that (A/Q) > (B/P) > 0. Again, it is extremely unlikely that you would have to know this.
      Mike 🙂

    • Sriram September 30, 2014 at 1:27 pm #

      Realised my folly. You’re right about just positive numbers making the cut.

      • Mike MᶜGarry
        Mike September 30, 2014 at 1:29 pm #

        Dear Sriram,
        I’m glad you found that helpful. 🙂 Best of luck to you in the future!
        Mike 🙂

  8. Anuj November 7, 2013 at 10:42 am #

    Yes, it absolutely does. Thank you so much.

    AK

    • Mike MᶜGarry
      Mike November 7, 2013 at 10:48 am #

      You are quite welcome. Best of luck to you.
      Mike 🙂

  9. Anuj November 7, 2013 at 10:34 am #

    Hey Mike,

    I was under the assumption that the two statements could not contradict themselves. However in question one they do with bd.

    Could you clear this up for me?

    Thanks.

    AK

    • Mike MᶜGarry
      Mike November 7, 2013 at 10:41 am #

      Dear AK,
      You are perfectly correct. That question does not really reach the standards of a good GMAT DS. It’s purpose is more pedagogical, forcing students to focus on a very particular mathematical choice. Does that make sense?
      Mike 🙂

  10. prateek sharma November 5, 2013 at 5:37 am #

    hello Mike ..

    I was able to solve the above questions with ease .

    Are these questions similar to what asked in GMAT . Do they belong to 700+ level .

    Or I need to practice more difficult questions than these ?

    • Mike MᶜGarry
      Mike November 5, 2013 at 10:00 am #

      Dear Prateek,
      I don’t think the GMAT would get much harder than these, so if you found these easy, then you are probably in good shape on this topic.
      Mike 🙂


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