On the number line, 0 lies between x and y. Is x>y?

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[GMAT math practice question]

On the number line, 0 lies between x and y. Is x > y?

1) The distance between x and 0 is 2 times the distance between y and 0
2) x+y > 0

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by GMATGuruNY » Mon Apr 30, 2018 3:36 am

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Max@Math Revolution wrote:[GMAT math practice question]

On the number line, 0 lies between x and y. Is x > y?

1) The distance between x and 0 is 2 times the distance between y and 0
2) x+y > 0
Since 0 lies between x and y, x and y have DIFFERENT SIGNS.

Statement 1: The distance between x and 0 is 2 times the distance between y and 0
Thus:
|x| = 2|y|

Case 1: y=1, implying that |x| = 2
Since x have must have different sign from y, x=-2.
In this case, x < y, so the answer to the question stem is NO.
Case 2: y=-1, implying that |x| = 2
Since x have must have different sign from y, x=2.
In this case, x > y, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: x + y > 0
Thus:
x > -y

Case 1: y=1, implying that x > -1
Since x must have a different sign from y, x must be a negative fraction.
In this case, then x < y, so the answer to the question stem is NO.
Case 2: y=-1, implying that x > 1
In this case, x > y, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statements combined:
Of the two cases that satisfy Statement 1, only Case 2 satisfies Statement 2.
Implication:
To satisfy both statements, y must be negative and x must be positive.
Thus, x > y, so the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.
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by Max@Math Revolution » Wed May 02, 2018 1:52 am

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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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question.
We then recheck the question. Since 0 lies between x and y, we must have xy < 0.

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):
Condition 1) tells us that |x| = 2|y|.
Thus x = ±2y. However, since xy < 0, x = -2y.

Condition 2) tells us that x + y = (-2y) + y = -y > 0. So, y < 0 and x > 0.
Thus, x > y.
Both conditions together are sufficient.

Since this question is an absolute value 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): |x| = 2|y|
If x = 2 and y = -1, the answer is "yes".
If x = -2 and y = 1, the answer is "no".

Condition 1) is not sufficient.

Condition 2): x + y > 0
If x = 2 and y = -1, the answer is "yes".
If x = -1 and y = 2, the answer is "no".

Condition 2) is not sufficient.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 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.

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by deloitte247 » Tue May 08, 2018 12:49 pm

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Statement 1: The distance between 'x' and '0' is 2 times the distance between 'y' and '0'.
Therefore, /x/=2 /y/
Case1: y=1, this means that /x/=2
since 'x' must have different signs (because 0 lies in between them) x=-2
'x' is less than 'y', it doesn't answer the target question. x>y

case2: y=-1, meaning /x/=-2
since 'x' must have different sign from 'y'
x=2
x>y
2>-1, it answer the target question.
Therefore, since case 1=NO and case 2= YES, It is not certain. Hence, it is INSUFFICIENT

Statement 2: x+y>0
x>-y
case 1: y=1 means that x>-1
since x must have different sign from y, x must be negative. This makes x<y. Hence, it doesn't answer the target question.

case 2: y=-1 means that x>1
in case 2, x>y answers the target question.
Since case 1=NO
case 2 = YES
it is therefore INSUFFICIENT

Therefore, to satisfy statement 1 and 2, 'y' must be negative and 'x' must be positive which will thus answer the target question x>y.

This certainty makes the statement SUFFICIENT

In conclusion, both statement are sufficient but neither statement alone is sufficient.
option c is the correct answer