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If x and y are both negative and xy < y^2

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If x and y are both negative and xy < y^2

by vinni.k » Wed Jan 22, 2014 8:14 am

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If x and y are both negative and xy < y^2, which of the following must be true?

a) x < y < x^2 < y^2
b) x < y < y^2 < x^2
c) y < x < x^2 < y^2
d) x^2 < y^2 < y < x
e) y^2 < x^2 < y < x

OA is C

Why B cannot be the answer. Is this wrong way?
xy - y2 < 0
y(x -y) < 0
y < 0 (negative ,given) So, I am taking it in this way.
x <y (x is negative too as it is given but less than y)

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by GMATGuruNY » Wed Jan 22, 2014 8:46 am
vinni.k wrote:If x and y are both negative and xy < y^2, which of the following must be true?

a) x < y < x^2 < y^2
b) x < y < y^2 < x^2
c) y < x < x^2 < y^2
d) x^2 < y^2 < y < x
e) y^2 < x^2 < y < x

OA is C
Since y<0, dividing each side by y requires that we FLIP the inequality:
xy < y²
xy/y > y²/y
x > y.
Eliminate A and B, which indicate x<y.

The remaining answer choices describe the relationship between x² and y².
Since x<0 and y<0, squaring each side of x>y implies multiplying each side by a negative value.
Thus, when we square each side, we must once again FLIP the inequality:
x > y
x² < y².
Eliminate E, which indicates that x²>y².

Since x<0, x²>0, implying that x<x².
Eliminate D, which indicates that x>x².

The correct answer is C.
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by Brent@GMATPrepNow » Wed Jan 22, 2014 9:26 am
vinni.k wrote: Why B cannot be the answer. Is this wrong way?
xy - y2 < 0
y(x - y) < 0
y < 0 (negative ,given) So, I am taking it in this way.
x <y (x is negative too as it is given but less than y)
Let's begin here:
y(x - y) < 0
If we divide both sides by y we must REVERSE the inequality sign (because we are dividing by a NEGATIVE value.
So, we get: x - y > 0
Add y to both sides to get: x > y

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by Brent@GMATPrepNow » Wed Jan 22, 2014 9:35 am
vinni.k wrote:If x and y are both negative and xy < y², which of the following must be true?

a) x < y < x² < y²
b) x < y < y² < x²
c) y < x < x² < y²
d) x² < y² < y < x
e) y² < x² < y < x
Here's another approach:

We'll begin with the approach that Mitch demonstrated.

We have xy < y²
If we divide both sides by y we must REVERSE the inequality sign (because we are dividing by a NEGATIVE value).
So, we get: x > y
This allows us to ELIMINATE answer choices A and B (since they suggest that x < y)

From here, let's PLUG IN values for x and y such that they are both negative AND x > y
Let's try x = -1 and y = -2
This means x² = 1 and y² = 4
When we arrange the four values in ascending order we get: y < x < x² < y²
Answer: C

Cheers,
Brent
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by vinni.k » Wed Jan 22, 2014 10:01 am
Wow, two amazing answers by two wonderful people. Appreciate you replies.:D

Thanks & Regards
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by [email protected] » Wed Jan 22, 2014 2:52 pm
Hi vinni.k,

Both Mitch and Brent have offered insights into this question. I'm going to focus a bit more on Number Properties and TESTing Values.

We're told that X and Y are both NEGATIVE and XY < Y^2

XY must be positive because (negative)(negative) = positive
Y^2 must be positive because (negative)^2 = positive

So, let's TEST a set of values that fits these facts:

X = -2
Y = -3
(-2)(-3) < (-3)^2 so we're ready to check the answers.

Since each of the answers uses the same four "numbers", we just have to put them in order:
X = -2
Y = -3
X^2 = 4
Y^2 = 9

Final Answer: C

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by Jeff@TargetTestPrep » Mon Feb 12, 2018 4:18 pm
vinni.k wrote:If x and y are both negative and xy < y^2, which of the following must be true?

a) x < y < x^2 < y^2
b) x < y < y^2 < x^2
c) y < x < x^2 < y^2
d) x^2 < y^2 < y < x
e) y^2 < x^2 < y < x
We see that in all the answer choices we have to compare the following four quantities: x, y, x^2 and y^2. We are given that x and y are both negative and xy < y^2.

We see that both xy and y^2 are positive since the product of two negative quantities is positive and the square of a nonzero quantity is always positive. However, we divide both sides of the inequality by y (a negative quantity), we have:

x > y

Since x^2 and y^2 are both positive, we see that y is the smallest of the four quantities. The only answer choice that has y as the smallest quantity is choice C. Thus, it's the correct answer.

(Note: We don't have to analyze, in this case, which is the larger quantity between x^2 and y^2 since y has to be the smallest quantity. However, if we have to, it is always true that if y < x < 0, then y^2 > x^2 > 0. For example, -3 < -2, but (-3)^2 > (-2)^2 since 9 > 4.)

Answer: C

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[email protected]

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