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Is x<1/y?

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Is x<1/y? [#permalink]

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Is x<1/y?
1) y>0
2) xy<1

*An answer will be posted in 2 days.
[Reveal] Spoiler: OA

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Re: Is x<1/y? [#permalink]

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New post 24 Aug 2016, 19:38
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There is no information about x or y so nothing can be done to reduce the initial statement...at first.

Statement 1: y>0, explains nothing about x. NSUF
Statement 2: xy<1, x and y could be different signs or x and y could be fractions or a combination of the two. NSUF

Statement 1 & 2: if y>0 and xy<1 then divide the equation by y and you have the statement. SUF
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Re: Is x<1/y? [#permalink]

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New post 25 Aug 2016, 15:36
S1)\(y>0\)

Pick \(y=1\) and \(x=2\);
=>\(2<\frac{1}{1}\)
=>No

Pick \(y=2\) and \(x=-2\)
=>\(-2<\frac{1}{2}\)
=>Yes

So S1 is insufficient

BCE

S2)\(xy<1\)

Pick \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\)
=>\(\frac{1}{2}<2\)
=> Yes

Pick \(x=\frac{1}{2}\) and \(y=\frac{-1}{2}\)
=>\(\frac{1}{2}<-2\)
=>No

So S2 is insufficient (Note: This statement could be a trap if inequality in question stem is just cross multiplied without checking possible values that \(y\) could take.)

CE

S1 and S2)

Pick \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\);This meets both the statements
=>\(\frac{1}{2}<2\)
=>Yes

Pick \(x=-2\) and \(y=1\); This meets both the statements
=> \(-2<\frac{1}{1}\)
=>Yes

Here \(y>0\) is satisfied from statement 1; both positive and negative values (integer and fraction as well) for \(x\) are tried and that should cover all possible values for \(x\) and the output is an YES in both the cases. Hence Statements 1 and 2 taken together is sufficient. Answer is C.

(S1 says that \(y>0\); So if \(y\), which is greater than 0, is divided both sides of statement 2, then it will not change the sign of the inequality and we get \(x<\frac{1}{y}\) which is exactly what the question stem is asking. Hence C)
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Re: Is x<1/y? [#permalink]

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New post 26 Aug 2016, 06:08
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MathRevolution wrote:
Is x < 1/y?

1) y > 0
2) xy < 1



Target question: Is x < 1/y?

Statement 1: y > 0
No information about x, so statement 1 is NOT SUFFICIENT

Statement 2: xy < 1
IMPORTANT: some students will want to divide both sides by y to get x < 1/y, HOWEVER doing so is incorrect, because we don't know whether y is POSITIVE or NEGATIVE.
If y is POSITIVE, then dividing both sides by y gives us x < 1/y
If y is NEGATIVE, then dividing both sides by y gives us x > 1/y [when we divide both sides by a negative value, we must REVERSE the direction of the inequality]
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that y is POSITIVE
Statement 2 tells us that xy < 1
Since we know that y is POSITIVE, we can safely divide both sides by y to get the inequality x < 1/y.

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer =
[Reveal] Spoiler:
C


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Re: Is x<1/y? [#permalink]

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New post 27 Aug 2016, 05:21
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If we modify the original condition and the question, in case of inequality, square is very important. If we multiply y^2 to both sides, we get xy^2<y?, xy^2-y<0?, y(xy-1)<0?. There are 2 variables (x and y) and there is a high chance that C is the correct answer. Using 1) & 2), the answer is always yes and the condition is sufficient. Thus, the correct answer is C.
 
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Is x<1/y? [#permalink]

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New post 12 Dec 2017, 12:28
\(x < 1/y\) ?
Re-arranging,
=> \(x - (1/y) < 0\) ?
=> \((xy - 1)/y < 0\) ?
For this term to be negative, (\(xy < 1\) AND \(y > 0\)) OR (\(xy > 1\) AND \(y < 0\)) ?
Question becomes => (\(xy < 1\) AND \(y > 0\)) OR (\(xy > 1\) AND \(y < 0\)) ?

Statement 1: \(y > 0\), but we don't know if \(xy < 1\) => InSufficient
Statement 2: \(xy < 1\), but we don't know if \(y > 0\) => InSufficient

Statement 1 + 2: \(y > 0\) AND \(xy < 1\) => Answers our question

Answer (C)
Is x<1/y?   [#permalink] 12 Dec 2017, 12:28
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