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Difficulty:
55%
(hard)
Question Stats:
67% (01:46) correct
33%
(01:31)
wrong
based on 1408
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If x > y > 0, which of the following must be negative?
I. \(y - x^3\)
II. \(x^2y - x^3\)
III. \(x^2y - xy^2\)
A. I only
B. II only
C. III only
D. I and II
E. I and III
GMAT-Club-Forum-g9xzo2iv.png [ 45.15 KiB | Viewed 8446 times ]
I. \(y - x^3\)
II. \(x^2y - x^3\)
III. \(x^2y - xy^2\)
A. I only
B. II only
C. III only
D. I and II
E. I and III
Attachment:
GMAT-Club-Forum-g9xzo2iv.png [ 45.15 KiB | Viewed 8446 times ]
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Updated on: 14 Aug 2023, 09:37
Originally posted by MartyMurray on 14 Aug 2023, 09:05.
Last edited by MartyMurray on 14 Aug 2023, 09:37, edited 5 times in total.
Last edited by MartyMurray on 14 Aug 2023, 09:37, edited 5 times in total.
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If \(x > y > 0\), which of the following must be negative?
I. \(y - x^3\)
If x and y are both fractions, this value could be positive.
For example, \(\frac{1}{3} - (\frac{1}{2})^3 = \frac{1}{3} - \frac{1}{8}\), which is positive.
So, we can eliminate choices (A), (D), and (E).
II. \(x^2y - x^3\)
This value will always be negative because \(x > y > 0\).
So, \((x * x) * x\) will always be greater than \((x * x) * y\).
Thus, \(x^2y - x^3 < 0\)
Keep.
At this point, since we know that (D) is not correct, the only choice we can choose that includes II is (B). Thus, the correct answer must be (B).
III. \(x^2y - xy^2\)
We have already completed the question, but if we want to check this choice, we can proceed as follows.
Since \(x > y > 0, x * x > x * y\).
So, \((x * x) * y > (x * y) * y\).
Thus, \(x^2y > xy^2\) and therefore \(x^2y - xy^2\) is always positive.
A. I only
B. II only
C. III only
D. I and II
E. I and III
I. \(y - x^3\)
If x and y are both fractions, this value could be positive.
For example, \(\frac{1}{3} - (\frac{1}{2})^3 = \frac{1}{3} - \frac{1}{8}\), which is positive.
So, we can eliminate choices (A), (D), and (E).
II. \(x^2y - x^3\)
This value will always be negative because \(x > y > 0\).
So, \((x * x) * x\) will always be greater than \((x * x) * y\).
Thus, \(x^2y - x^3 < 0\)
Keep.
At this point, since we know that (D) is not correct, the only choice we can choose that includes II is (B). Thus, the correct answer must be (B).
III. \(x^2y - xy^2\)
We have already completed the question, but if we want to check this choice, we can proceed as follows.
Since \(x > y > 0, x * x > x * y\).
So, \((x * x) * y > (x * y) * y\).
Thus, \(x^2y > xy^2\) and therefore \(x^2y - xy^2\) is always positive.
A. I only
B. II only
C. III only
D. I and II
E. I and III
14 Aug 2023, 09:09
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Bunuel
On a number, x lies to the right of y, and both lie on the right of zero
--- \(0\) ---------- \(y\) --------------- \(x\) -----------
Note: This is a must-be-true type of question.
Answer choice elimination.
I. \(y - x^3\)
Assume \(x = 3\) and \(y = 2\), \(y - x^3\) will be negative.
However, if \(x = 0.3\) and \(y = 0.2\), \(y - x^3\) will be positive.
Hence, eliminate I.
II. \(x^2y - x^3\)
\(x^2(y - x)\)
\(x^2\) is always positive in this case. \(y - x\) will be always negative, as the value of \(x\) is greater than the value of \(y\). Hence, \(x^2y - x^3\) will always be negative.
Keep.
III. \(x^2y - xy^2\)
\(xy(x - y)\)
\(xy\) is positive, and \((x-y)\) is positive. Hence the product is always positive.
Eliminate III.
Option B
