Problem Solving

If X^3 > Y^4, which of the following CANNOT be true?

• A. X<1
  B.X+Y<0
  C.0<XY<1
  D.X(Y^2)<0
  E.X=|X|

sivaelectric wrote:If X^3 > Y^4, which of the following CANNOT be true?

• A. X<1
  B.X+Y<0
  C.0<XY<1
  D.X(Y^2)<0
  E.X=|X|

Hi!

We know that y^4 must be non-negative. So, from the original inequality we know that x^3 must be positive. Accordingly, we can conclude that x is positive.

With that in mind, since we want an answer that CANNOT be true, let's look for a choice that requires x to be non-positive.

A) nope
B) nope
C) nope
D) ding ding ding! In order for x(y^2) to be negative, one of the two terms would have be negative. Since y^2 is definitely non-negative, in order for (D) to be true x must be negative.

However, we've already shown that x must be positive, so (D) CANNOT be true... choose (D)!

* * *

Note that we also could have attacked this question by picking numbers. When a question asks which of the following CANNOT be true, we want to pick numbers to show that an answer COULD be true; as soon as we do so, we can eliminate that choice.

Statistically, (D) and (E) show up more often than they should on problem solving questions with "which of the following" in the stem. So, when we work with the choices we should go bottom up.

E) x=|x|

If x=10 and y=1, then 10^3>1^4 and 10=|10|. (E) could be true: eliminate it.

D) x(y^2)<0

we know that y^2 can't be negative, so x must be negative to make this statement true.

if x = -1 and y = 1, then x^3 = -1 and y^4 = 1. However, -1 is NOT > 1, so we can't pick these numbers.

We should pretty quickly see that there are no numbers which satisfy both (D) and the original inequality, making (D) our CANNOT be true... choose (D)!

Look at case by case

A. This one is trying to trick you into thinking that if X is 0 negative, then X^3 cannot be larger than Y^4, which must be positive. However, the question does not specify that the numbers are integers. If X and Y are equal fractions less than 1, then X^3>X^4
B. X<-Y. If Y is positive, then X is negative and X^3>Y^4 does not hold. If Y is a negative integer, then X^3>Y^4 does not hold. However, if Y is a negative fraction, there are cases where X^3>Y^4 (think of extremes, where X is as close to Y as possible, e.g. Y=1/2 and X=9999/20000).
C. This states that at least one of the numbers must be a fraction less 1. Consider the case where X=Y, and X^3>Y^4.
D. X and Y^2 have different signs. Since Y^2 must be positive, X must be negative. Therefore, X^3 will always be negative, and since Y^4 is also always positive, X^3<Y^4. This is the answer.
E. X is positive. We've already seen in cases where X and Y are positive fractions less than one, it is possible that X^3>Y^4.

Now that I just typed all that up, I've realized the best way to think about it is to find in which case X MUST be negative.

I would go with D.

Awesome explanation Stuart, Better than the Veritas Prep App I should say

I tried to think of X and Y as positive fractions before going through the answers to make things easier, though they would not have to be. Just makes it easier to think of them as a number rather than nameless letters. Now I look at the options:

A. X<1: True if X and Y are positive fractions
B. X+Y<0: True if X>Y
C. 0<XY<1: True if X and Y are positive fractions
D. X(Y^2)<0: X would have to be a negative integer, which would not agree with X^3>Y^4.
E. X=[X]: True if X is a positive fraction

Answer is D.

Great post here guys ... Quick question @ Stuart:

I'm using the second method you referred to in your post, in which we pick numbers to disprove the answer choice. I just want to clairify, answer choice E, what you were getting at was that +10 ccould be used in the phrase [X]=X and still satisfy the criteria outlined by the question?

factor26 wrote:Great post here guys ... Quick question @ Stuart:

I'm using the second method you referred to in your post, in which we pick numbers to disprove the answer choice. I just want to clairify, answer choice E, what you were getting at was that +10 ccould be used in the phrase [X]=X and still satisfy the criteria outlined by the question?

Hi,

that's 100% correct. Since we can pick values that satisfy both the original inequality (which we know is true) and choice (E), we've proven that choice (E) COULD be true. Since it COULD be true, it's the wrong answer to a CANNOT be true question.