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Director
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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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36% (01:59) correct 64% (01:59) wrong based on 47 sessions
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Is \(\frac{x}{y} < x*y\)? (1) \(x^3>y^3\) (2) \(y^5>y^4\)
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“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ―Henry Wadsworth Longfellow
Originally posted by AsadAbu 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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