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14 Jun 2016, 02:00
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:21) correct
58%
(02:06)
wrong
based on 202
sessions
History
Date
Time
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Not Attempted Yet
If x and y are positive numbers, is x > y?
(1) \(\frac{x^3}{y} < 1\)
(2) \(\frac{\sqrt[3]{x}}{y} < 1\)
(1) \(\frac{x^3}{y} < 1\)
(2) \(\frac{\sqrt[3]{x}}{y} < 1\)
26 Jun 2016, 22:55
Kudos
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Bunuel
x and y are positive. So we can multiply the inequality by x or y.
(1) \(\frac{x^3}{y} < 1\)
\(x^3 < y\)
How do cubes behave on the right of 0? If x lies between 0 and 1, the cube will be less than x. If x is greater than 1, x^3 will be more than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.
(2) \(\frac{\sqrt[3]{x}}{y} < 1\)
\(\sqrt[3]{x} < y\)
How do cube roots behave on the right of 0? If x lies between 0 and 1, the cube root will be more than x. If x is greater than 1, it will be less than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.
Using both,
x will lie between x^3 and \(\sqrt[3]{x}\) in either case. Since y is greater than both, y must be greater than x. Sufficient.
Answer (C)
General Discussion
14 Jun 2016, 03:04
Kudos
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Bunuel
Statement 1
\(\frac{x^3}{y} < 1\)
x and y are positive numbers so we can cross multiply.
\(x^3 < y\)
x = 2, y = 9
\(x^3\) = 8. This is less than y. So statement (1) is satisfied.
Since x=2, and y =9, x is not more than y.
x = (1/2), y = (1/7)
\(x^3\) = (1/8). This is less than y. So statement (1) is satisfied.
Since x=(1/2), and y =(1/7), x is more than y.
Statement (1) is not sufficient
Statement 2
\(\frac{\sqrt[3]{x}}{y} < 1\)
\({\sqrt[3]{x}} < y\)
x and y are positive numbers so we can raise each of them to cubic powers
so x < \(y^3\)
x = 7, y = 2
\(y^3\) = 8. x =7. x is less than \(y^3\). So statement (2) is satisfied.
Since x=7, and y =9, x is more than y.
x = 1, y = 3
\(y^3\) = 27. x =1. x is less than \(y^3\). So statement (2) is satisfied.
Since x=1, and y =3, x is not more than y.
Statement (2) is not sufficient
Combining Statement 1 and 2
\(x^3 < y\)
x < \(y^3\)
Since x is positive we can cube it. so \(x^3 < y^6\)
Now \(x^3 < y\) and \(x^3 < y^6\)
This means that we can take \(x^3 < y\)
This is nothing but Statement (1) only. We know it was insufficient.
So even after combining - insufficient
E is the answer.
