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

Question

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. $x>z$

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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?

Answer 1

Simplify the inequality to three separate inequalities:

$$(1)\quad x<\dfrac{2}{3}y$$

$$(2)\quad y<2z$$

$$(3)\quad x<\dfrac{4}{3}z$$

Now we conclude easily

(a) false by (1)

(b) true by (2) since $z=y<2z$ still holds.

(c) true by (2) since $z<1.5z=y<2z$ is possible

(d) true by (3) since $z<1.1z=x<\dfrac{4}{3}z$ is possible.

Answer 2

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 $y=11$.

Answer 3

In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then $$ 3x<2y<2y+y=3y$$ and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.

Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...