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If x, y, and z are positive numbers such that 3x < 2y < 4z, [#permalink]
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Expert's post 00:00

Question Stats: 38% (01:08) correct 62% (01:40) wrong based on 50 sessions

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) $$x > z$$
[Reveal] Spoiler: OA

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Re: If x, y, and z are positive numbers such that 3x < 2y < 4z, [#permalink]
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Expert's post
Explanation

The best strategy is to inspect each option one by one and substitute values

x = y:

If $$x = 1$$, $$y = 1$$ $$3x < 2y$$ or $$(3)(1) < (2)(1)$$ or $$3 < 2$$ This is incorrect

Will it work with a larger number?

If $$x = 100$$, $$y = 100$$ $$3x < 2y$$ or $$(3)(100) < (2)(100)$$ or $$300 < 200$$ also incorrect.

This one cannot ever be true, no matter how large the number we supply for x an y.

y = z:

If $$y = 1$$, $$z = 1$$ $$2y < 4z$$ or $$(2)(1) < (4)(1)$$ or $$2 < 4$$

This statement could be true.

y > z:

If y = 1.1, z = 1 2y < 4z or (2)(1.1) < (4)(1) or $$2.2 < 4$$

This statement could be true.

x > z:

If $$x = 1.1$$, $$z = 1$$ $$3x < 4z$$ or $$(3)(1.1) < (4)(1)$$ or $$3.3 < 4$$

This statement could be true.

Hence option B, C and D are correct.
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Re: If x, y, and z are positive numbers such that 3x < 2y < 4z, [#permalink]
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Expert's post
Carcass wrote:

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

[Reveal] Spoiler: =A
B,C, and D

A) x = y
If this is true, we can take the given inequality (3x < 2y < 4z) and replace x with y to get: 3y < 2y < 4z
Now let's focus on 3y < 2y
Subtract 2y from both sides to get: y < 0
Hmmm, this says that y is less than 0, HOWEVER, the question tells us that y is POSITIVE
So, it cannot be the case that x = y
ELIMINATE A

B) y = z
If this is true, we can take the given inequality (3x < 2y < 4z) and replace y with z to get: 3x < 2z < 4z
Now let's focus on 2z < 4z
This seems to check out.
So, let's see if we can find some values that satisfy this answer choice.
How about x = 1, y = 3 and z = 3
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 3 < 6 < 12 (perfect!)
So, it is POSSIBLE that y = z
KEEP B

C) y > z
This is a little trickier.
So, let's just see if we can find some values that satisfy this answer choice.
How about x = 1, y = 3 and z = 2
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 3 < 6 < 8 (perfect!)
So, it is POSSIBLE that y > z
KEEP C

D) x > z
This is tricky too.
So, let's just see if we can find some values that satisfy this answer choice.
How about x = 10, y = 16 and z = 9
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 30 < 32 < 36 (perfect!)
So, it is POSSIBLE that x > z
KEEP D

[Reveal] Spoiler:
B, C, and D

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Re: If x, y, and z are positive numbers such that 3x < 2y < 4z, [#permalink]
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B, C and D are the answer.
The trick in this question is, the question has asked for Could be true option.
Which means in any range of numbers whether the condition satisfies.
Had it been a Must be true question, then it would be, in every condition whether any number in range satisfies. Re: If x, y, and z are positive numbers such that 3x < 2y < 4z,   [#permalink] 14 Aug 2018, 15:05
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