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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
Re: If x, y, and z are positive numbers such that 3x < 2y < 4z, [#permalink]
14 Aug 2018, 15:05
1
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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.
greprepclubot
Re: If x, y, and z are positive numbers such that 3x < 2y < 4z,
[#permalink]
14 Aug 2018, 15:05
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38% (01:08) correct
