One of the hardest GMAT questions

This is from a practice GMAT test. #math #maths #mathematics #shorts

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[gf]

Your questions are getting easier and easier

[psibarpsi]

One of those confidence-builders, huh?

[ch.s]

As someone who got 675(96th) on GMAT focus edition and 90(100th) on quantitative reasoning
This question is very easy to answer. Just simply replace those algebra with -1, 1, 2

[coralreef]

I thought I was a genius for getting this... then I realized it was a CONFIDENCE BUILDER

[protondium_]

Commenting before seeing the end:
The answer must be E because
1. A negative number (<0) multiplied by a positive number is always negative, and two positive numbers multiplied are always positive. Therefore, yz(x) < 0, which is rearranged to xyz < 0
2. x + z = y is a true statement because, since they are evenly spaced, z must be exactly twice the distance from 0 compared to the distance between x and 0 (and in the opposite direction as x is negative). Let u represent the number of units between each number (with negative values of u representing distance to the left). This turns the equation into -u + 2u = u based on distance, and by combining like terms on the left side one finds that u = u, or y = y, meaning that x + z does in fact = y.

Since both of these options are true and there are no options that disambiguate between I and II but not III, and I, II, III all being true, the answer must be E.

[Thirdbase]

Assuming that the number line reads left (negative) to right (positive).

[maxhagenauer]

I simply just said the space between them all were 1 and so I replaced x, y, and z with -1, 1, and 2 like an actual number line. Then just put those in for the variables to see if they work.

[nicholas]

So I get that standard number line notation always has negative numbers to the left of the zero, but this problem does not say to limit the problem to real numbers only. If this is a line of imaginary numbers then the answer would be "B. II only" so that should be the correct answer to this problem. Am I missing something?

[lordbeany]

It does not state which direction would make a value greater or less than zero. One could easily argue x>0 making the first statement false since (+)*(-)*(-) is greater than 0.

[worthysandwich]

option E is ok, but what if we invert the number line and e.g. x = 1, y = -1 and z = -2 then only statements 2 and 3 are correct

[username-urdq]

1 isn't necessarily true since the direction of positivity wasn't stated

[WirableCrown]

FFS I had a panic attack when MATHs showed up in my shorts feed lol

[kulvirsingh]

I thought y-x would = 0 as y is the distance from 0 to Y and the distance from 0 to x so they should cancel right?Could you please explain why this is not the case

[mirkhare]

I was thinking to myself earlier. "If only my math teacher explained this thoroughly rather than expecting me to just get it."

[tamimplayz]

I and III are definitely correct but confused about II. Alright gonna watch the video now

[ferociousfeind]

you can't know that the number line has positive numbers to the right- that's convention, but that's not guaranteed.

[DiarmuidHenry]

Can we necessarily assume that y, z>0? Aren't you making an assumption as to which direction is positive?

[RichardRentrop]

Nowhere does it say that y>0.
They can evenly be spaced with distance 0.
If you allow this only II is true.

The question is not clear

[rafaeldearaujo]

If I choose the negatives to be the right side and the positive the left, the first statement isn’t true anymore…

[SamuelFreitas-vivant]

I couldn't believe that i got the answer in my head without paper.