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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8123
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   75% (hard)

Question Stats: 49% (02:24) correct 51% (01:55) wrong based on 53 sessions

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[GMAT math practice question]

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

$$1) 10|x|=|y|$$
$$2) x>0$$

_________________

Originally posted by MathRevolution on 26 Mar 2018, 00:55.
Last edited by MathRevolution on 28 Mar 2018, 02:00, edited 2 times in total.
Senior PS Moderator D
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If x+2y≠0, is x/x+2y<1?  [#permalink]

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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,
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Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8123
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If x+2y≠0, is x/x+2y<1?  [#permalink]

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=>

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.

_________________ Re: If x+2y≠0, is x/x+2y<1?   [#permalink] 28 Mar 2018, 01:58
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