Is (y - x) < 0 ? (1) y > x^(1/2) (2) (y - x^2) < 0

gmatophobia

Any suggestion to think of values in a particular manner so that when combining both the information, it doesn’t take much time to solve the question

Rickooreo - I know that you're not looking for this answer, so my apologies upfront if the reply disappoints and doesn't answer your question.

I avoid picking up numbers as much as possible and try to solve questions based on inequalities (or quant in general) purely based on reasoning (first because GMAT is a test of reasoning, and second because I am very bad in picking up numbers ). That being said, this is how I would solve this question.

Question

y - x < 0

y < x

Inference : : Does "y" lie to the left of "x" on a number line. At this point, I would probably even draw one on my sheet as I did while working on this question.

Statement 1

y > \(\sqrt{x}\)

The first thing that pops out of the page here is \(\sqrt{x}\). Hence, x has to be non negative, and we know that y > \(\sqrt{x}\), i.e. y lies to the right of \(\sqrt{x}\) on the number line.

Let's see which region(s) can x lie and on the basis of the position of x, how can we possibly infer the positions of \(\sqrt{x}\) and y.

Case 1 : 0< x <1

If x lies between 0 and 1, \(\sqrt{x}\) lies to the right of x.

We know that y lies to the right of \(\sqrt{x}\), hence the y is to the right of x. This is shown by the red arrow.

Case 2 : 0< x <1

If x lies beyond 1, \(\sqrt{x}\) lies to the left of x.

We know that y lies to the right of \(\sqrt{x}\). But there can be two positions now that y can have-

y can be between \(\sqrt{x}\) and x - in this case y is to the left of x

OR

y can lie to the right of x

These two positions are shown using blue arrow.

As we are getting two positions of y, we can eliminate A and D.

Statement 2

y - \(x^2\) < 0

y < \(x^2\)

Inference: y lies to the left of \(x^2\)

Just as in statement 1- the first thing that I observe is \(x^2\). \(x^2\) is non - negative, so I will first begin by plotting that. As I have already analyzed for positive values of x in statement 1 (and have determined that knowing x > 0, doesn't help), my focus will be on negative values of x. We will come to positive values of x if need be.

Case 1 : 0< \(x^2\) <1

If \(x^2\) is between 0 and 1, x is between -1 and 0.

We know that y lies to the left of \(x^2\), so y can be at any positions as shown by the red arrow.

So it can lie to the left of x or y can lie to the the right of x.

This is enough to reject the statement 2. However, one can also observe the same behavior when \(x^2\) > 1

So eliminate B.

Combined

When we combine the statements we still have the entire positive region of x. The region between 0 and 1 can be rejected as in that region statement 1 and statement 2 cannot together hold true.

When x > 1 -

• \(\sqrt{x}\) lies to the left of x
• \(x^2 \) lies to the right of x.

On the basis of the information presented in St.1 & in St2, we still can have two position of y as shown by the red arrows. In one position, x is to the right of y and in other position x is the left of y.

Hence the ambiguity remains and the answer is Option E

My final take- This may seem a lot (looking at length of the explanation) and one may doubt that if all the analysis can be done under 2 mins. My answer to that - Yes !

I took 1:18 seconds to solve the question (of course the typing took way too long ). I am sure I would have taken way longer if I had to pick up numbers.

Hope this helps !

can I put it this way?

Statement 1
y > √x
square both side
y^2>x, meaning y can be +/-

Statement 2
y - x^2 < 0
y < x^2, meaning x can be +/-

so 1/2 alone is insufficent, and combining both since x & y can be both +/-, also insufficient?

No, that's not correct. The statement y > √x implies that y is positive because it is greater than the square root of a number, and the square root is always nonnegative.