If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true

algebra-precalculusinequality

This is more or less a question about how many cases I should be trying in general on the quantitative section of the GRE.

I got this question wrong and it is because I did not try enough cases, so I was wondering if anyone has any tips on how to see that a given condition may be true without trying 5 different cases.

Question:

If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true? Indicate all such statements.

a. $x=y$

b. $y=z$

c. $y>z$

d. $x>z$

My work:

For a)

$x=2, y=2, z=3$

$3x=6, 2y=4, 4z=12$

$6<4<12$ is false.

a) cannot be true.

For b)

$x=1, y=4, z=4$

$3x=3, 2y=8, 4z=16$

$3<8<16$ is true.

b) can be true.

For c)

$x=1, y=4, z=3$

$3x=3, 2y=8, 4z=12$

$3<8<12$ is true.

c) can be true.

Now d) was the reason for missing this question. I plugged in one case as I did with a) yet there is in fact a threshold such that this works.

For example:

$x=7, y=11, z=6$

$3x=21, 2y=22, 4z=24$

$21<22<24$ is true.

d) can be true.

However, during the practice test (this was the last question I was crunched for time and had to the second to last because this one looked easier) I only plugged in one case and it didn't work so I quit.

Tips?

Best Answer

Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.

Part (a): $x < \frac{2}{3}y$, so $x$ is never as large as $y$, so $x \neq y$ always.

For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.

For part (d), $3x<4z$ is the given relation (that is, $x < \frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < \frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.

In a bit more detail on (d): $z < x < \frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.

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