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Is |x+y|>|x+z|?

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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
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Is |x+y|>|x+z|?  [#permalink]

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New post 21 Mar 2018, 02:50
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

53% (01:32) correct 47% (01:27) wrong based on 62 sessions

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[GMAT math practice question]

Is \(|x+y|>|x+z|?\)

\(1) y>z\)
\(2) x>0\)

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Re: Is |x+y|>|x+z|?  [#permalink]

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New post 21 Mar 2018, 03:18
2
MathRevolution wrote:
[GMAT math practice question]

Is \(|x+y|>|x+z|?\)

\(1) y>z\)
\(2) x>0\)


1)
y>z
we know nothing about x - insufficient

2)
we know nothing about y and z - insufficient

combining

1 and 2

if y and z are both positive the statement holds true
if y and z are negatve the statement isn't true

a clear (E)
Math Revolution GMAT Instructor
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Re: Is |x+y|>|x+z|?  [#permalink]

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New post 23 Mar 2018, 00:20
1
=>

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 \(3\) variables (\(x\) and \(y\)) and \(0\) equations, E 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:
If \(x =1, y = 2\), and \(z = 3\), the answer is ‘yes’.
If \(x = 1, y = -2\), and \(z = -3\), the answer is ‘no’.

Since we don’t have a unique solution, both conditions together are not sufficient.

Therefore, the answer is E.

Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is |x+y|>|x+z|?  [#permalink]

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New post 23 Mar 2018, 02:28
The fastest way would be to test the numbers:

(1) y>z
case 1:
x= 1
y= 2
z=-17

substituing we get:
3<16 so inequality doesn't hold true,

case 2:
x= 1
y= 15
z=2
substituting, we get:
16>3 now the inequality holds true

(2) x>0
same example as in (1) works well, so Insufficient

(1)+(2) Again, applying above example shows that it is still Insufficient

Answer E.
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Re: Is |x+y|>|x+z|? &nbs [#permalink] 23 Mar 2018, 02:28
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