The GMAT sometimes features quant questions 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:

. What is the value of ?

(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 . Therefore . Because the question has two apostrophe signs, we want to repeat this procedure to get, . 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.

. What the value of p in ?
(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: . Solving for p:

.

Answer (E).

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

. 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 ?

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.

Magoosh Test Prep ExpertMarch 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. 😀

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 ExpertNovember 3, 2016 at 6:29 pm#

Hi Adeniyi,

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.

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.

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.

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.

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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.

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. 😀

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

Hi Adeniyi,

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.

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.

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.

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

Hello and thanks,

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

Thanks, Nelson

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.

The answer to second brutal question is E

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

The answer choice (E) comes to 11.

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

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.

a is the answers

since all the others are greater that 10!

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.

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!