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If x and y are positive integers such that
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18 Mar 2021, 10:13
18 Mar 2021, 10:13
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If x and y are positive integers such that \(x > y > 1\) and \(z = \frac{x}{y}\), which of the following must be true?
I. \(z > \frac{(x − 1)}{(y − 1)}\)
II. \(z < \frac{(x − 1)}{(y − 1)}\)
III. \(z > \frac{(x + 1)}{(y + 1)}\)
A. I only
B. I and II
C. II and III
D. II only
E. I and III
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
I. \(z > \frac{(x − 1)}{(y − 1)}\)
II. \(z < \frac{(x − 1)}{(y − 1)}\)
III. \(z > \frac{(x + 1)}{(y + 1)}\)
A. I only
B. I and II
C. II and III
D. II only
E. I and III
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: If x and y are positive integers such that
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18 Mar 2021, 10:37
18 Mar 2021, 10:37
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Carcass wrote:
If x and y are positive integers such that \(x > y > 1\) and \(z = \frac{x}{y}\), which of the following must be true?
I. \(z > \frac{(x − 1)}{(y − 1)}\)
II. \(z < \frac{(x − 1)}{(y − 1)}\)
III. \(z > \frac{(x + 1)}{(y + 1)}\)
A. I only
B. I and II
C. II and III
D. II only
E. I and III
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
I. \(z > \frac{(x − 1)}{(y − 1)}\)
II. \(z < \frac{(x − 1)}{(y − 1)}\)
III. \(z > \frac{(x + 1)}{(y + 1)}\)
A. I only
B. I and II
C. II and III
D. II only
E. I and III
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Notice that statements I and II say the exact OPPOSITE thing.
Also, notice that either I or II appear in ALL 5 answer choices. So, one of them must be true.
Given this, let's see what happens when we assign some values to x and y.
Say x = 4 and y = 2
So, z = x/y = 4/2 = 2
Now test (x − 1)/(y − 1) by plugging in x = 4 and y = 2
We get: (4 - 1)/(2 - 1) = 3/1 = 3
Since z = 2, we can see that z < (x − 1)/(y − 1)
So, statement II is true, and statement I is false.
This means the correct answer is C or D
Now test statement III: z > (x + 1)/(y + 1)
Since z = x/y, we can rewrite III as x/y > (x + 1)/(y + 1)
Let's find out if this statement is true.
Since y is POSITIVE, we can safely multiply both sides by y to get: x > (y)(x + 1)/(y + 1)
y+1 is also POSITIVE, we can safely multiply both sides by y+1 to get: x(y + 1) > (y)(x + 1)
Expand both sides: xy + x > xy + y
Subtract xy from both sides to get: x > y
Is this true? YES! It's given in the question
So statement III is true
Answer: C
Re: If x and y are positive integers such that
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09 Mar 2024, 12:56
09 Mar 2024, 12:56
Hey Brent & Carcass can we also solve this way... x>y>1 subtract 1 from all sides to get x-1>y-1>0. from this we can see that z < x-1/y-1. Also if we divide all sides by -1, it gives us x+1<y+1<0 thus z< x+1/y+1? does this work?
