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20 May 2019, 04:56
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E
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FRESH GMAT CLUB TESTS QUESTION
If x, y, and z are positive numbers such that 3x < 2y < 4z , which of the following statements could be true?
I. y = z
II. y > z
III. x > z
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
M37-15
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19 Nov 2021, 04:04
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Official Solution:
If \(x\), \(y\), and \(z\) are positive numbers such that \(3x < 2y < 4z\), which of the following statements could be true?
I. \(y = z\)
II. \(y > z\)
III. \(x > z\)
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
Notice that the question asks: "which of the following statements COULD be true?" NOT "which of the following statements MUST be true?"
I. \(y = z\). From the stem we know that \(2y < 4z\). So, \(y < 2z\). It's certainly possible that \(y = z\), for example, \(y = z=1\).
II. \(y > z\) From the stem we know that \(2y < 4z\). So, \(y < 2z\). It's certainly possible that \(y > z\), for example, \(y = 3\) and \(y=2\).
III. \(x > z\). From the stem we know that \(3x < 4z\). And again, it's possible that \(x > z\), for example, \(x=10\) and \(y=9\).
Answer: E
If \(x\), \(y\), and \(z\) are positive numbers such that \(3x < 2y < 4z\), which of the following statements could be true?
I. \(y = z\)
II. \(y > z\)
III. \(x > z\)
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
Notice that the question asks: "which of the following statements COULD be true?" NOT "which of the following statements MUST be true?"
I. \(y = z\). From the stem we know that \(2y < 4z\). So, \(y < 2z\). It's certainly possible that \(y = z\), for example, \(y = z=1\).
II. \(y > z\) From the stem we know that \(2y < 4z\). So, \(y < 2z\). It's certainly possible that \(y > z\), for example, \(y = 3\) and \(y=2\).
III. \(x > z\). From the stem we know that \(3x < 4z\). And again, it's possible that \(x > z\), for example, \(x=10\) and \(y=9\).
Answer: E
General Discussion
20 May 2019, 05:35
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Bunuel
Let us take each option..
I. y=z....
Let us compare y and z in the inequality 3x<2y<4z, so 2y<4z or y<2z..
So yes...let y=z=2 and x=1...3*1<2*2<4*2...3<4<8...
II. y>z....
Let us compare y and z in the inequality 3x<2y<4z, so 2y<4z or y<2z..
So yes, y can be between z and 2z..let y=3, z=2 and x=1...3*1<2*3<4*2...3<6<8...
III. x>z....
Let us compare x and z in the inequality 3x<2y<4z, so 3x<4z or x<4z/3..
So yes, x can be between z and 4z/3...let z=30 and x=31, and y=...3*31<2*50<4*30...91<100<120.
All three possible
E
