GMAT Math: Strange Symbols

The Quant section sometimes features GMAT algebra problems with strange symbols. These symbols should not fluster you too much as long as you remember that they do not represent standard mathematical notation. Instead, the symbols pertain only to the problem and are defined by the GMAT (or whatever prep material you happen to be using).

Let’s have a look at a simple example:

$$Q prime = 3Q – 3$$. What is the value of $$(2 prime) prime$$?

(A) 2
(B) 3
(C) 5
(D) 6
(E) 9

Explanation

To approach strange symbols think of the Q’ as a recipe. To the right of the equals sign are the steps (or the recipe) you have to follow.

Another way of looking at it, whatever we see in place of Q’ we want to plug it into the ‘Q’ in $$3Q – 3$$. Therefore $$2 prime = 3(2) – 3 = 3$$. Because the question has two apostrophe signs, we want to repeat this procedure to get, $$3 prime = 6$$. Answer (D).

This is a basic problem, one that if you saw it on the GMAT, would not bode well. So let’s try a problem that will make you sweat a little more.

$$A&&B = sqrt{b} – a$$. What the value of p in $$16&&p = 9$$?
(A) -5
(B) 9
(C) 13
(D) 25
(E) 625

Explanation

Be careful not to fall the trap that switches the order of b and a. Our equation should read: $$sqrt{p} – 16 = 9$$. Solving for p:

$$sqrt{p} = 25$$
$$p = 625$$.

For those who are looking to score a Q51, here are two brutally difficult questions. If you think you know the answer, go ahead and post it below with an explanation.

Brutal Question #1

$$x@y = 2sqrt{x} + y^2$$. What is the difference between the least and the greatest possible values of x + y, if x@y is an integer less than 15?

(A) 9
(B) 10
(C) 49
(D) 50
(E) 52

Brutal Question #2

[[x]] is equal to the lesser of the two integer values closest to non-integer x. What is the absolute value of $$[[-pi]] + [[-sqrt{37}]]$$?

(A) $$[[9.4]]$$

(B) $$[[4 pi]]$$

(C) $$[[sqrt{99}]]$$

(D) $$[[sqrt{120}]]$$

(E) $$[[sqrt{143}]]$$

Brutal Question #1 – E
Brutal Question #2 – E

16 Responses to GMAT Math: Strange Symbols

1. lecroissant November 12, 2017 at 1:03 am #

Thanks for this section! I was wondering if we want to work out the Brutal problems in a reasonable fastest amount of time (i.e. not a math genius but have had enough training), are there any shortcut methods you could use?

• Magoosh Test Prep Expert November 16, 2017 at 10:44 am #

There aren’t really shortcuts specific to “strange symbol” GMAT Quant problems. Instead, the best thing you can do is disregard the symbol itself and focus on the operation it stands for. From there, find the best possible shortcuts based on the operation. For a good breakdown of efficient solves for the “brutal” problems in this post, see the comments left by the awesomely-named commenter “Brutalized.” On March 24, 2017 in the comments field for this post, Brutalized gives some very good explanations of best processes for the harder questions here.

2. Brutalized March 24, 2017 at 12:28 am #

For Brutal Question #1:

To satisfy x@y the range of x and y are
x: {0,1,4,9,16,25,36,49}
since the x values are doubles in the original operation, the max we can take is the square of 7 as when doubled, it will give us 14. Any values higher than that will not suffice the requirement of being an integer less than 15 when operated on and any values lower will not suffice either.
y: (-3,-2,-1,0,1,2,3}

MAX values: X (49) + Y (0)
MIN values: X(0) + Y (-3)

Difference between max and min gives us -52, which is the required result.

For Brutal Question #2

The operation tells us that [[x]] is the lesser of two integers closest to x’s original value.
For example, [[1.5]] would be 1, or [[-1.5]] would -2.

We have to find the value of [[-pi]]+[[-sqrt(37)]]

Let’s approximate the values:
pi ~ 3.14
sqrt(36) = 6, so sqrt(37) is slightly larger than that.

Since the parity is negative, the operation is as follows:

[[-3.14]] + [[-6.xx]]

Therefore, it’s -4 – 7 and since we remove the parity and only take the magnitude, it’s just 11.

We need to find a value that is higher than eleven but lower than twelve.

sqrt(120) is just shy of 121 which is the perfect square of 11, so that’s out.
sqrt(137) is between 121 and 144, which is the perfect square of 12, so it’s lower end is at 11.

Therefore [[sqrt(137)]] is the required result.

• Magoosh Test Prep Expert March 24, 2017 at 7:46 am #

Hi there,

Great job on both problems! However, for Question 2 did you mean [[sqrt(143)]]. Your process is still correct, but there is no [[sqrt(137)]] answer option here. 😀

3. Adeniyi November 3, 2016 at 3:33 am #

Hello Chris,

Since x@y < 15, then x@y has the values {14,13,12,11,10,9,8,7,6,5,4,3,2,1,0}. This

means that x@y = 2sqrt{x} + y^2 = {14,13,12,11,10,9,8,7,6,5,4,3,2,1,0}.

For all values of x@y, make y=0 and 2sqrt{x} will only have the integer values of

x={0,1,4,9,16,25,36,49}

Next For all values of x@y, make x=0 and y^2 will only have the integer values of y={-3,-2,-1,0,1,2,3}.

So I agree with ArijitC38 about the values

x={0,1,4,9,16,25,36,49}
y={-3,-2,-1,0,1,2,3}

But from here, the X+Y (MAX) should be 49 (X) + 3 (Y) =52 and not 49 (X) + Y (0) =49 which will bring the final answer to 55. How is the X+Y (MAX) = 49. Please explain

• Magoosh Test Prep Expert November 3, 2016 at 6:29 pm #

The error that you’re making is that you’re combining two separate sets. What I mean by this is that to get x={0,1,4,9,16,25,36,49}, you SET y=0. So, the greatest set you can have here is y=0 and x=49. Then, the smallest set you can have here is y=0 and x=0.

Then, to get the set y={-3,-2,-1,0,1,2,3}, you SET x=0. So, the greatest set you can have here is y=3 and x=0. Then, the smallest set you can have here is y=-3 and x=0.

Please note that you can’t just take x=49 from one set and y=3 from another set when you got these values by fixing the other variable to be a certain number.

4. ArijitC38 March 26, 2016 at 9:58 am #

Solution to Brutal Question 1
Since the expression x@y is an integer less than 15 :-

Then the possible values for :-
x={0,1,4,9,16,25,36,49}
y={-3,-2,-1,0,1,2,3}
as any other value would be an outlier.

Now, in the context of the problem
X+Y (MAX) = 49 (X) + Y (0) =49.
X+Y (MIN) = 0 (X) + (-3 ) (Y) = -3.

Therefore, MAX – MIN = 49 – (-3) = 49+3=52 .

Regards,
Arijit.

• Shweta April 13, 2016 at 9:37 am #

Hi,
If an integer is less than 15, it has to be limited to 14 numbers. Please explain the logic behind choosing values for X and Y.

5. A Slayton November 23, 2015 at 12:52 pm #

Is there a fast way of accounting for the unique sums? Thank you!

6. nelson January 10, 2015 at 9:55 pm #

Hello and thanks,
But it says unique sums, so the negative for y goes away with square (get repeated), or missing something?
Thanks, Nelson

7. Farah July 21, 2014 at 5:13 pm #

Answer to Brutal Question #2 is E.

Since,
-π = -3.14..
and the non-integer -3.14 is between the integers -4 and -3. But we want the lesser;
so [[-π]] = -4

-√37 = -6.08..
and the non-integer -6.08 is between integers -7 and -6. But we want the lesser;
so [[-√37]] = -7

Therefore, -4 + (-7) = -4 – 7 = -11, but we want the absolute value;
so |-11| = 11

Answer choice E is [[√143]] = [[11.95..]] = 11, since the non-integer 11.95 is between integers 11 and 12, we pick the smaller number (11) which agrees with our answer.

8. Jeffrey Jose June 8, 2014 at 11:45 am #

The answer to second brutal question is E

The expression evaluates to | (-4) + (-7) | = | -11 | = 11

The answer choice (E) comes to 11.

9. Sebastien March 1, 2014 at 5:11 pm #

For question 1, as Manuel said, neither x or y can be greater than 15.

This leaves x = 0, 1, 4, 9, 16, 25, 36, 49 and y = 0, 1, 2, 3, -1, -2, -3

The possibilities for x+y < 15 is 30. Include values x = 16 and y = -2, -3

10. eddy January 31, 2014 at 3:57 am #

the second Problem:

i got a Problem, maybe you can help:
[[-pi]] ->3,1xxx the two integeres are 3 and 4, the lesser one is 3, so -3

[[-sqrt{37}]] -> the two integers are 6 and 7, the lesser one is 6, so -6

the absolute sum of it is +9

since the got only anwers with [[ ]] we have to solve for the answer which is [[ ]] = 9
that should be the first one.

since all the others are greater that 10!

11. Manuel August 25, 2013 at 3:49 pm #

The answer to the first brutal question is B.

Explanation; For the sum of x and y to be less than 15 neither part of the sum can be greater than 15. This leaves us with values of 1,2,3,4 for y and 4,9,16,25,36 for x.

Now we just have to sum them all up and count the amount of sums that are below 15.

• Chris Lele September 3, 2013 at 5:13 pm #

Hi Manuel,

Almost :). But you forgot to account for ‘0’ and negative integers for ‘y’. Also, the greatest value for ‘y’ is 3, and the greatest possible value of ‘x’ is 49.

Hope that helps!

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