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Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

If \(x=\frac{1}{4}\) then \(a=\sqrt{x} * x - x=\frac{1}{2}*\frac{1}{4}-\frac{1}{4}=-\frac{1}{8}\). Now, \(-\frac{1}{8}\) is not an integer at all (hence not even) and also not positive, so none of the options MUST be true.

I think the answer could be B (II -- a is positive) if we consider sqrt of 2 and 3 as irrational (or sqrt of any other prime number).

or D (I and II -- a is positive and even). Anyway, I feel that the specification of x is not fully given.

The Q is \sqrt{x}*x - x = a

and x is an integer.

If x is -ve, then we have an equation contains sqrt of negative integer- which is not covered in GMAT. So let's rule out -ve integer. If x is -ve, then the answer is "E- none of the above"

If x is positive then: If x is even (E), then SQRT(E)*E is even and E-E is also even. and so a is even and it is positive.

x= 2 then 2*sqrt(2)-2 = not integer, but if we consider sqrt (2) as an irrational number then a has to be positive. x= 16 then 16*sqrt(16)-16 = 16*4-16 = 48 - +ve and even.

If x is odd (O), then SQRT(O)*O is odd and O-O is even. So a is even and again it is positive. x=3, then 3*sqrt(3)-3 = not integer, but a has to be positive again {sqrt (3) is irrational}. x=9, then 9*sqrt(9)-9 = 18 --> +ve and even.

So if one considers sqrt of 2 and 3 are irrational numbers then we can have choice B (a is positive). Otherwise, among the choices the answer is D (I & II - i.e. a is even and positive).

I think the correct answer is "E" because if x=1, a=0 and 0 is not a positive. If x=2, \(\sqrt{2} - 1\) is not a integer or even.

ihero06, you seem to forget about \(x=0\). If \(x=0\), \(a\) is not positive. The question asks "which of the following must be true", meaning for all possible \(x\). You're right that GMAT doesn't cover negative numbers under the radical sign. We have to remember that 0 is neither positive nor negative.

I've updated the OE a bit. Hopefully it's better now.
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As stated by the founder, the key of this question is to show that a is not an integer, if so, a non integer number can not be even nor odd and therefore one is left to prove whether a is positive or negative. Since 0 is and integer and mathematically speaking it has not sign convention i.e. zero has no distance, then given that in this case a = 0, and one can not tell what the sign of a is, then the right answer must be E.

If we prove that \(a\) is not an integer for any \(x\), we'll be able to rule out A, C, and D. \(\sqrt{x} * x - x = a\) can be rewritten as \(x(\sqrt{x}-1)=a\). If \(x=2\) the expression turns into \(2(\sqrt{2}-1)=a\). It's clear that \(a\) is not an integer as \((\sqrt{2}-1)\) is not an integer. Now we need to prove that \(a\) is not necessarily positive. If \(x=0\), the equation turns \(0=a\). 0 is neither positive, nor negative. We're left with E as the answer.

Thanks for the explanation. Got correct answer but did more work than necessary. 1) Didn't rephrase; 2) plugged in 0, 1, 4, and 3. Had I just plugged in 2 and 0 (with the intention of ruling out answer choices), would have gotten to E much faster. Didn't spend a lot of time on it. Nonetheless, my goal should be to get the correct answer as quickly as possible.
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As stated by the founder, the key of this question is to show that a is not an integer, if so, a non integer number can not be even nor odd and therefore one is left to prove whether a is positive or negative. Since 0 is and integer and mathematically speaking it has not sign convention i.e. zero has no distance, then given that in this case a = 0, and one can not tell what the sign of a is, then the right answer must be E.

i always thought zero was positive...since it was 1 less than "1" anyhow...thanks for this note.

1st I factorized: x(\sqrt{x} - 1) = a the stem did not specified whether x is +ve /-ve; odd or even integer. \sqrt{9} - 1 => even; but \sqrt{1/9) - 1 = 1/3 - 1 = -2/3 => the result could be either even, rational, or -ve. So, I, II & III out. Hence, E.
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KUDOS me if you feel my contribution has helped you.

if u notice in the question it is written that x is an integer i.e it can have any value from minus infinity to plus infinity. now if we take x as a negative number (for ex : -5)then the solution turns out be a=-5*sqrt(-5)+(-5)

we allknow that sqrt of -ve number is an imaginary value which can neither be positive ,nor even nor an integer.

Artabandhu(Its my first post feel free to post comments)

Is it also possible to solve this one without factoring? \sqrt{x} * x - x = a and say for every x you put in you'd have to multiply it with zero thus making the outcome none of the one mentioned in the answers? :S

if x = 4 we get sqrt of 4 x 4 - 4 = 2 x 4 - 4 = 8 - 4 = 4. Which is positive, even and an integer. Am i missing something? I saw you guys trying it with 2 and 3 and 0, but when you try it with an integer that has a sqrt root then you dont get this answer... can SOME PLEASE ASSIST?

if x = 4 we get sqrt of 4 x 4 - 4 = 2 x 4 - 4 = 8 - 4 = 4. Which is positive, even and an integer. Am i missing something? I saw you guys trying it with 2 and 3 and 0, but when you try it with an integer that has a sqrt root then you dont get this answer... can SOME PLEASE ASSIST?

The answer has to work in 100% of the situation, so even though 4 works, the equation does not work with others, and therefore, the answer is E. You have to try multiple numbers and if even one of them works, it is out.
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if x = 4 we get sqrt of 4 x 4 - 4 = 2 x 4 - 4 = 8 - 4 = 4. Which is positive, even and an integer. Am i missing something? I saw you guys trying it with 2 and 3 and 0, but when you try it with an integer that has a sqrt root then you dont get this answer... can SOME PLEASE ASSIST?

HERE, you have considered only one case.

answer should fulfill all the case. Consider X= 2 , 8 ,10, all are even but for these values, a is neither odd nor Even. a is not even an integer. same is applicable if you consider X = 2,3,5,6,7,8,10,11,12,13,14,15.... you can generalize it by saying that any value of X, which is not perfect square irrespective of odd or even , will yield a as decimal form (not Integer).

So, the Ans is E.

Hope it ll help. you can post here still you have any doubt.
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kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

GMAT Diagnostic Test Question 3 Field: Arithmetic, Roots Difficulty: 700-750

Rating:

If \(x\) is an integer and \(\sqrt{x} * x - x = a\) , which of the following must be true? I. \(a\) is Even II. \(a\) is Positive III. \(a\) is an Integer

A. I only B. II only C. III only D. I and II E. None of the above

\(x (\sqrt{x} -1) = a\)

I. a is even when X= 9, but a is fraction when X=5 (say, \(5 (\sqrt{5} -1)\). So no II. a is positive when X=9, but a is "0" when X=1. so no III. a is not an integer when X=5. so no