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30 Oct 2019, 21:56
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:38) correct
34%
(01:37)
wrong
based on 108
sessions
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If the average (arithmetic mean) of x and y is positive, is x positive?
(1) xy < 0
(2) x = y + 4
(1) xy < 0
(2) x = y + 4
Updated on: 17 May 2021, 08:30
Originally posted by BrentGMATPrepNow on 31 Oct 2019, 14:14.
Last edited by BrentGMATPrepNow on 17 May 2021, 08:30, edited 1 time in total.
Last edited by BrentGMATPrepNow on 17 May 2021, 08:30, edited 1 time in total.
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Bunuel
Given: The average (arithmetic mean) of x and y is positive
Target question: Is x positive?
Statement 1: xy < 0
This statement doesn't feel sufficient, so I'll TEST some values
There are several values of x and y that satisfy statement 1 (AND the given information). Here are two:
Case a: x = 3 and y = -1. In this case, the answer to the target question is YES, x is positive
Case b: x = -1 and y = 3. In this case, the answer to the target question is NO, x not is positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x = y + 4
Rewrite as: y = x - 4
Since the average of x and y is positive, we can write: \(\frac{x+y}{2}>0\)
Replace y with x - 4 to get: \(\frac{x+(x-4)}{2}>0\)
Simplify numerator to get: \(\frac{2x-4}{2}>0\)
Simplify to get: \(x-2>0\)
Add 2 to both sides to get: \(x>2\)
If x is greater than 2, then x must be positive
The answer to the target question is YES, x is positive
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
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01 Nov 2019, 07:38
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Bunuel
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.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The original condition means (x + y)/2 > 0 or x+y>0.
Since we have 2 variable (x and y) and 1 equations, D is most likely to be the answer. So, we should consider each condition on its own first. When a question is an inequality, inequalities should be counted as equations.
Condition 1)
If x = 1 and y = -1, then x is positive and the answer is 'yes'.
If x = -1 and y = 1, then x is negative and the answer is 'no'.
Since condition 1) does not yield a unique solution, it is not sufficient
Condition 2)
Since x + y > 0 and y = x - 4, then we have 2x - 4 > 0 or x > 2.
Thus x is positive and the answer is 'yes'.
Since condition 2) yields a unique solution, it is sufficient
Therefore, B is the answer.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
