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Source: Official Guide Revised GRE 1st Ed. Part 8; Section 5; #9

5

x and m are positive numbers and m

x and m are positive numbers and m is a multiple of 3. Quantity A : x/x Quantity B : x Quantity A is greater., Quantity B is greater., The two quantities are equal., The relationship cannot be determined from the information given.

5 Explanations

1

Flavio Ariza Malagutti

If m is a multiple of 3, can we assume both in this and all other cases that m could also equal 0?

Aug 30, 2018 • Comment

David Recine

We can't really assume that here, because the problem says that m is positive, and 0 isn't a positive number. Without the "m is positive" constraint, though, we could. That would be a significantly different math problem then, wouldn't it? ;)

Sep 2, 2018 • Reply

1

Isaac Silberstein

The question stem explicitly states, "x and m are positive numbers." How am I to know in this case, then, that x is an integer, as Chris mentions in the video? Why do we assume "positive number" for x means a counting number, not a positive decimal or fraction for x?

Nov 2, 2016 • Comment

Sam Kinsman

Hi Isaac, that's a good question! You're right that we don't know for sure whether x is an integer or not. I think that Chris mean to say that "x and m are positive numbers" in the video!

Note, though, that m has to be an integer. It is a multiple of 3, so it could be 3, 6, 9, 12, etc. All of these are integers. So we know that m is an integer.

For m, it is not entirely clear if it is an integer. However, note that this doesn't make the problem more difficult. As Chris shows in the video, if m = 3, Column A (x^m / x^3) is equal to 1. This is true for any value of x.

So if m = 3, Column A is 1. And Column B would become x^1. Since x can be any value, we don't know if x is equal to or greater than 1.

Again, all of this is true regardless of whether x has to be an integer or not. The fact that x can be any positive number doesn't change things.

I hope this helps! :)

Nov 2, 2016 • Reply

Emma Muhleman

So ETS essentially answered their own question wrong, no? Shouldn't we report this?

Feb 1, 2018 • Reply

David Recine

ETS answered the question correctly. The answer really is "(D) the relationship cannot be determined from the information given."

The only error here is on the part of Magoosh-- Chris shouldn't have said that x is an integer. However, this is a minor error, as the ETS question would still have the same answer even if we knew that x *was* an integer. It doesn't actually matter if x is an integer or not, since the answer depends only on the minimum value of m, which is 3.

This is reflected both in Chris's video explanation and ETS's official text explanation on pages 402-403 of the OG.

Feb 6, 2018 • Reply

1

I solved this slightly differently, is this valid?

X^M -X^3 = M-3 vs M/3 and just tried different numbers to show that when you get to higher multiples of M, a becomes greater.

Jan 1, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

For this question, it's important to keep in mind that we're comparing the whole expression, and not just the power to which x is raised. For that reason, while it is ok to plug in values to solve this problem (which is what Chris does in the explanation video), one key observation is that when m=3, then

x^m/x^3 = x^(3-3) = x^0 = 1

and

x^(m/3) = x^(3/3) = x^1

Since we don't know the value of x, only that it is a positive integer, x could be equal to 1, in which case A and B would be equal. However, if x were equal to another positive integer, such as 2, then A > B. Since A could be equal to or greater than B, the answer is D, the relationship cannot be determined.

Hope this helps :)

Jan 3, 2016 • Reply

2

Newaz Sharif

I think this problem will be at number 8 not at number 9.

Nov 4, 2015 • Comment

Cydney Seigerman, Magoosh Tutor

Thanks for noticing this, Tuhin! You're completely right. I've made a note to our content editors that this question and video correspond to #8 of the OG, so that this can be fixed :) Thanks again!

Nov 11, 2015 • Reply

4

Chris Lele

Oct 11, 2012 • Comment

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