Solution

The question could become complex to evaluate theoretically. So the best way is to pick numbers

The stem can be tricky: it does not say that the numbers \(MUST\) be integers. As such, you need to take in account both options:\(X= -2\) and \(Y=-1\) OR \(X=\frac{-3}{2}\) and \(Y=\frac{-1}{2}\)

Testing all the answer choices we will end up that E will be \(-3<4\) and it is true;\(\frac{3}{4}<\frac{9}{4}\) and this is also true

The correct answer is \(E\)

We can see that both x and y are negative, and y is greater than x. So A is not true. B is not true, either, since we don’t know how much y is greater than x. If we divide both sides by x in the inequality in C (and switch the inequality sign since x is negative), we have:

y^2 > 1

However, since we don’t know the value of y, we can’t determine whether its square is in fact greater than 1. So C might not be true. If we divide both sides by y in the inequality in D (and switch the inequality sign since y is negative), we have:

x > y

This is not true since we know y > x. So the correct answer must be E. However, let’s verify it by dividing both sides of the inequality by x:

y > x

This is true since we know y is indeed greater than x.

Answer: E
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Given

\(x < y < 0\)

Here we understand that x is the most negative between x and y. So, whenever we square it , we get the maximum value which is greater than y.

Thus the best answer is E.

Please correct the OA here. It is "E"

is there an explanation

Dear Sir,

I think above your question there are 3 explanations of the same question. Please refer to them .

Regards

The question asks, "What MUST be true?"

So if we can find a case in which an answer choice is NOT true, then we can ELIMINATE that answer choice.

GIVEN: x < y < 0

A) y + 1 < x
If x < y < 0, then it's possible that x = -3 and y = -1
When plug these values into the above inequality, we get: -1 + 1 < -3
Simplify to get: 0 < -3, which is NOT true
ELIMINATE A

B) y - 1 < x
If x < y < 0, then it's possible that x = -3 and y = -1
When plug these values into the above inequality, we get: -1 - 1 < -3
Simplify to get: -2 < -3, which is NOT true
ELIMINATE B

C) xy < x
If x < y < 0, then it's possible that x = -3 and y = -1
When plug these values into the above inequality, we get: (-3)(-1) < -3
Simplify to get: 3 < -3, which is NOT true
ELIMINATE C


D) xy < y²
If x < y < 0, then it's possible that x = -3 and y = -1
When plug these values into the above inequality, we get: (-3)(-1) < (-1)²
Simplify to get: 3 < 1, which is NOT true
ELIMINATE D

By the process of elimination, the correct answer must be E

Cheers,
Brent
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[quote="Carcass"]If which \(x < y < 0\), of the following inequalities must be true?

A \(y + 1 < x\)

B \(y - 1 < x\)

C \(xy^2 < x\)

D \(xy < y^2\)

E \(xy < x^2\)

How about this approach after separating x < y from the given equation:

multiply both the sides with x to get x^2 > xy. -- sign changes because x is negative.

Whenever an inequality gets multiplied on both sides by a negative number, the inequality sign flips. Since x and y are both given as negative numbers, multiply by x once and y once to the given inequality x < y to find an outcome matching one of the answers. In this case, multiplying both sides with x as follows:

\(x < y\)
\(x*x > y * x\)
\(xy < x^2\)

Answer is E.