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If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y|

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If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y|  [#permalink]

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New post Updated on: 22 Nov 2017, 19:43
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A
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C
D
E

Difficulty:

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Question Stats:

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[GMAT math practice question]

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

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

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Originally posted by MathRevolution on 21 Nov 2017, 18:21.
Last edited by MathRevolution on 22 Nov 2017, 19:43, edited 1 time in total.
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Re: If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y|  [#permalink]

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New post 21 Nov 2017, 22:41
MathRevolution wrote:
[GMAT math practice question]

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

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


Hi MathRevolution

Not Sure if the \(OA\) is correct as per the question. We need to find if \(y<1/2\)?

From Stmnt 1: X = 10, y = -1 and X = -10, y =1 So, Not Sufficient.
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Re: If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y|  [#permalink]

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New post 22 Nov 2017, 19:49
2
=>

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.

Conditions 1) & 2)
\(|x|=10|y|
⇔ x = ±10y\)

Case 1: If \(x=10y\), then \(\frac{x}{2x+y}= \frac{10y}{20y+y}= \frac{10y}{21y}=\frac{10}{21}<1\),
and the answer is ‘yes’.

Case 2:If \(x= -10y\), then
\(\frac{x}{2x+y} = \frac{-10y}{-20y+y} = \frac{-10y}{-19y} =\frac{10}{19}<1\),
and the answer is ‘yes’.

Both conditions, applied together, are sufficient.

Since this is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4 (A) tells us
that we need to also consider conditions 1) and 2) separately.

Condition 1)
\(|x|=10|y|
⇔ x = ±10y\)

Case 1: If \(x = 10y\), then
\(\frac{x}{2x+y}= \frac{10y}{20y+y}= \frac{10y}{21y}=\frac{10}{21}<1\),
and the answer is ‘yes’.

Case 2: If \(x = -10y\), then
\(\frac{x}{2x+y}= \frac{-10y}{-20y+y}= \frac{-10y}{-19y}=\frac{10}{19}<1\)
and the answer is ‘yes’.

Since the answer is ‘yes’ in both cases, this condition is sufficient.

Condition 2)
If \(x = -2\) and \(y = 1\), then \(\frac{x}{2x+y} = \frac{-2}{-4+1} = \frac{-2}{-3} = \frac{2}{3} <1\), and the answer is ‘yes’.
If \(x=-2\) and \(y=3\), then \(\frac{x}{2x+y} = \frac{-2}{-4+3} = \frac{-2}{-1} =2>1\), and the answer is ‘no’.
This is NOT sufficient.

Note: Since this condition is so trivial, it is unlikely to be sufficient by Tip 4) of the VA method.

The answer is A.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E).

Answer: A
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Re: If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y|  [#permalink]

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New post 09 Dec 2017, 12:14
\(x / (2x + y) < 1\) ?

rearranging , \(1 / ( 2 + y/x) < 1\) ?

Statement 1: \(|x| = 10|y|\)
=> \(|y|/|x| = 1/10\)
so max value of y/x is 1/10 => \(1 / (2+ (1/10))\)=> 1/2.1 is less than 1
and min value of y/x is - 1/10 => \(1 / (2 - (1/10))\)=> 1/1.9 is also less than 1

Sufficient

Statement 2: y> 0, no idea about x => Insufficient

Answer (A)
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Re: If 2x+y≠0, is x/2x+y<1 ? 1) |x|=10|y| &nbs [#permalink] 09 Dec 2017, 12:14
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