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Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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Updated on: 26 Mar 2019, 03:59
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Is \(\frac{x}{y} < x*y\)? (1) \(x^3>y^3\) (2) \(y^5>y^4\)
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Originally posted by Asad on 11 Mar 2019, 08:28.
Last edited by Bunuel on 26 Mar 2019, 03:59, edited 1 time in total.
Edited the OA.




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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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11 Mar 2019, 09:26
Is x/ y < x*y ? Squaring both sides and cross multiplying, we get or, x^2 < x^2 *y^4 or, x^2 (1y^4)<0 This would be true if x is not equal to 0, and y >1 or y <1 So, the question reduces to " Is x is not equal to 0, and y >1 or y <1" ? Statement 1: x^3 > y^3 Here x can be zero, and y any negative number. for which the answer can be NO (x,y) can be (2,3), Answer is YES NOT SUFFICIENT Statement 2: y^5 > y^4 or y^4(y1) >0 here y >1, but x can be zero or non zero. if y >1, x =0 , answer is NO if y >1, x is non zero, ans is YES NOT SUFFICIENT Combing Both statements , we get y >1 and x^3 > y^3 , Hence x is NON zero, So, We have y >1 ans X is NON ZERO. Hence SUFFICIENT. Answer is C AsadAbu wrote: Is \(\frac{x}{y} < x*y\)?
(1) \(x^3>y^3\)
(2) \(y^5>y^4\)
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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11 Mar 2019, 10:05
If i wanted to solve this question with the help of numbers how can I solve it? Can anyone help me?



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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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11 Mar 2019, 12:32
I would go for C.
Statement 1: insufficient Case 1: X=0 & y=1 (thus, both sides are equal in main question. So, NO to main question)
Case 2: x=2 & y=3 (yes to main question)
Statement 2: insufficient Case 1: X=0 & y=2 (thus, both sides are equal in main question. So, NO to main question)
Case 2: x=2 & y=3 (yes to main question)
Combining From statement 2, it is clear that y is positive and greater than 1. So in statement 1, x must be positive.
In such situation, multiplication of same direction two number is greater than division of same numbers.
I.e. x=3 & y=2
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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Updated on: 26 Mar 2019, 03:33
Can people make sure that their answers are correct when posting questions? Makes it a lot harder to study when you are trying to work out whether the answer is actually the answer or not



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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 02:57
AsadAbu wrote: Is \(\frac{x}{y} < x×y\)?
(1) \(x^3>y^3\)
(2) \(y^5>y^4\) Hi Bunuel and IanStewart, Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\)
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 03:14
AsadAbu wrote: AsadAbu wrote: Is \(\frac{x}{y} < x×y\)?
(1) \(x^3>y^3\)
(2) \(y^5>y^4\) Hi Bunuel and IanStewart, Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Yes. You could reduce by x directly (provided x is not 0). \(\frac{1}{y} < y\) > y^2 > 1 > y < 1 or y > 1. BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). What is the source of the question?
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 03:54
Bunuel wrote: AsadAbu wrote: AsadAbu wrote: Is \(\frac{x}{y} < x×y\)?
(1) \(x^3>y^3\)
(2) \(y^5>y^4\) Hi Bunuel and IanStewart, Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Yes. You could reduce by x directly (provided x is not 0). \(\frac{1}{y} < y\) > y^2 > 1 > y < 1 or y > 1.BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). What is the source of the question? Source: WizakoIn this question, if we can determine that \(y^2>1\) then the data is sufficient. Conversely, if we can determine that \(y^2≤1\) then the data is also sufficient. Isn't it? Statement 2 says: (2) \(y^5>y^4\) Here, y^4 is positive, so we can divide both part by \(y^4\). We get from statement 2: ==> \(y>1\) ==> \(y^2>1\) [multiplying by both side] So, why not B sufficient?
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 03:59
AsadAbu wrote: Bunuel wrote: AsadAbu wrote: Hi Bunuel and IanStewart, Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Yes. You could reduce by x directly (provided x is not 0). \(\frac{1}{y} < y\) > y^2 > 1 > y < 1 or y > 1.BTW, the correct answer is C, not B. (2) give y > 1. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). What is the source of the question? Source: WizakoIn this question, if we can determine that \(y^2>1\) then the data is sufficient. Conversely, if we can determine that \(y^2≤1\) then the data is also sufficient. Isn't it? Statement 2 says: (2) \(y^5>y^4\) Here, y^4 is positive, so we can divide both part by \(y^4\). We get from statement 2: ==> \(y>1\) ==> \(y^2>1\) [multiplying by both side] So, why not B sufficient? Because we have one more variable in the question: x/y[/fraction] < x*y. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO).
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Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 04:08
AsadAbu wrote: Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Bunuel wrote: Because we have one more variable in the question:x/y[/fraction] < x*y. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). But i've deduced the question stem as \(y^4>1\) where there is no existence of x. Thanks__
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 04:11
AsadAbu wrote: AsadAbu wrote: Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Bunuel wrote: Because we have one more variable in the question:x/y[/fraction] < x*y. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). But i've deduced the question stem as \(y^4>1\) where there is no existence of x. Thanks__ Check my post above: You could reduce by x directly ( provided x is not 0). If x = 0, then we cannot reduce and thus only y^5 > y^4 will not be sufficient.
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Re: Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 04:16
Bunuel wrote: AsadAbu wrote: AsadAbu wrote: Can I rearrange the question stem in the following way? \(\frac{x}{y} < x×y\)? ==> \(\frac{x^2}{y^2}<x^2 × y^2?\) ==> \(x^2<x^2 × y^4\)? [dividing by \(x^2\) in both side as \(x^2\) is positive] ==> \(y^4>1?\) Bunuel wrote: Because we have one more variable in the question:x/y[/fraction] < x*y. From (2) we know that y > 1 but know nothing about x. If x ≠ 0, then \(\frac{x}{y} > x×y\) (answer YES) but if x = 0, then \(\frac{x}{y} = x×y\) (answer NO). But i've deduced the question stem as \(y^4>1\) where there is no existence of x. Thanks__ Check my post above: You could reduce by x directly ( provided x is not 0). If x = 0, then we cannot reduce and thus only y^5 > y^4 will not be sufficient. Thank you so much. Better if you edit the OA.
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Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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26 Mar 2019, 11:22
Since division by 0 is not allowed, the prompt should indicate that \(y≠0\): AsadAbu wrote: If \(y≠0\), is \(\frac{x}{y} < x*y\)?
(1) \(x^3>y^3\)
(2) \(y^5>y^4\) Since an absolute value cannot be negative. we can simplify the question stem by multiplying both sides by y: \(\frac{x}{y}y < xyy\) \(x < xy^2\)? If \(x=0\), the answer to the question stem is NO. If \(x≠0\), we can divide both sides by \(x\): \(\frac{x}{x} < \frac{xy^2}{x}\) \(1 < y^2\)? Here, the answer will be YES if \(y>1\) or \(y<1\). Question stem, rephrased: Is it true that \(x≠0\) and that \(y>1\) or \(y<1\)? Statement 1: Case 1: x=3 and y=2 Since \(x≠0\) and \(y>1\), the answer to the question stem is YES. Case 2: x=2 and y=1 Since \(y=1\), the answer to the question stem is NO. INSUFFICIENT. Statement 2: Since the inequality implies that \(y≠0\), we can safely divide both sides by \(y^4\), which must be positive: \(\frac{y^5}{y^4}>\frac{y^4}{y^4}\) \(y > 1\) No information about \(x\). INSUFFICIENT. Statements combined: \(y>1\) implies that \(y^3>1\). Thus: \(x^3 > y^3 > 1\) \(x^3 > 1\) \(x > 1\) Since \(x≠0\) and \(y>1\), the answer to the question stem is YES. SUFFICIENT. .
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Is x/y < xy? (1) x^3 > y^3 (2) y^5 > y^4
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