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If x+2y≠0, is x/x+2y<1?

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If x+2y≠0, is x/x+2y<1?  [#permalink]

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New post Updated on: 28 Mar 2018, 01:00
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[GMAT math practice question]

If \(x+2y≠0\), is \(\frac{x}{x+2y}<1?\)

\(1) 10|x|=|y|\)
\(2) x>0\)

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Originally posted by MathRevolution on 25 Mar 2018, 23:55.
Last edited by MathRevolution on 28 Mar 2018, 01:00, edited 2 times in total.
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Re: If x+2y≠0, is x/x+2y<1?  [#permalink]

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New post 26 Mar 2018, 00:28
MathRevolution wrote:
[GMAT math practice question]

If \(x+2y≠0\), is \(\frac{x}{x+2y}<1?\)

\(1) 10|x|=|y|\)
\(2) x>0\)


We are given \(\frac{x}{x+2y}<1?\) .... let's simplify the expression.

\(\frac{x}{x+2y} - 1< 0 ?\)


\(\frac{-2y}{x+2y} < 0 ?\) .. this is easier to work with.

Statement 1) \(1) 10|x|=|y|\)

case 1. x >=0
10 x = y .... substitute in simplified expression and ineq holds.
\(\frac{-2y}{x+2y} = \frac{-20x}{x+20x} = -20/21\)

case 2. x<0
-10x = y ... substitute again and enq holds.
\(\frac{-2y}{x+2y} = \frac{20x}{x-20x} = -20/19\)

Hence suff.

Statement 2) x>0 ... we know nothing about y.. hence insuff.

Hence Option (A) is correct.

Best,
Gladi
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Re: If x+2y≠0, is x/x+2y<1?  [#permalink]

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New post 28 Mar 2018, 00:58
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together:

\(10|x|=|y|\)
\(=> y = ±10x\)

Case 1: \(y = 10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x+20x)} = \frac{x}{21x} = \frac{1}{21}< 1\)

Case 2: \(y = -10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x-20x)} = \frac{x}{(-19)x} = -(\frac{1}{19}) < 1\)

Both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):

\(10|x|=|y|\)
\(=> y = ±10x\)

Case 1: \(y = 10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x+20x)} = \frac{x}{21x} = \frac{1}{21} < 1\)

Case 2: \(y = -10x\)
\(\frac{x}{(x+2y)} = \frac{x}{(x-20x)} = \frac{x}{(-19)x} = -(\frac{1}{19}) < 1\)

Thus, condition 1) only is sufficient.

Condition 2):

Since it gives us no information about y, condition 2) is not sufficient.

Therefore, A is the answer.

Answer: A
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Re: If x+2y≠0, is x/x+2y<1? &nbs [#permalink] 28 Mar 2018, 00:58
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