[Math Revolution GMAT math practice question] 8.28
Is xy>0?
1) x+y>0
2) |x|+|y|<1
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Is xy>0?
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Max@Math Revolution
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Brent@GMATPrepNow
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Target question: Is xy>0?Max@Math Revolution wrote: Is xy>0?
1) x + y > 0
2) |x| + |y| < 1
Statement 1: x + y > 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. Here are two:
Case a: x = 1 and y = 1. In this case, xy = (1)(1) = 1. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = 2 and y = -1. In this case, xy = (2)(-1) = -2. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Aside: For more on this idea of testing values when a statement doesn't feel sufficient, read my article: https://www.gmatprepnow.com/articles/dat ... lug-values
Statement 2: |x| + |y| < 1
Let's test values again.
Case a: x = 0.2 and y = 0.2. In this case, xy = (0.2)(0.2) = 0.04. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = -0.8 and y = 0.1. In this case, xy = (-0.8)(0.1) = -0.08. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
There are still several values of x and y that satisfy BOTH statements. Here are two:
Case a: x = 0.2 and y = 0.2. In this case, xy = (0.2)(0.2) = 0.04. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = 0.8 and y = -0.1. In this case, xy = (0.8)(-0.1) = -0.08. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Answer: E
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Brent
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fskilnik@GMATH
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The explanations above are excellent. I would like to add some comments on the matter, just that.
If we BELIEVE x+y > 0 does not imply xy > 0, we must offer a counterexample to this "conjecture".
If we BELIEVE x+y > 0 does not imply the negation of (xy > 0) , i.e., (xy=0 or xy<0 ) , we again must offer a counterexample to this (other) "conjecture".
In this case:
? : xy > 0
Take, for instance, (x,y) = (2,-1) , that satisfies x+y > 0 , to refute the conjecture given by the FOCUS, because xy > 0 is false.
Take, for instance, (x,y) = (1,1), that satisfies x+y > 0, to refute the negation of the conjecture given by the FOCUS (in the sense that we would have a unique answer in the negative if xy>0 would always be false).
In short:
? : xy > 0
Take (x,y) = (2,-1) <NO> , that is, this ordered pair answers the question (=focus) in the negative.
Take (x,y) = (1,1) <YES> , that is, this ordered pair answers the question (=focus) in the affirmative.
Important: each ordered pair MUST respect the statement(s) considered (in this case x+y>0), otherwise the "test" (an expression I respectfully disapprove) does not apply.
That´s what our method call a BIFURCATION ... I took this expression from the ODE (ordinary differential equations) subject.
I hope I have made things clearer.
Regards,
fskilnik.
If we BELIEVE x+y > 0 does not imply xy > 0, we must offer a counterexample to this "conjecture".
If we BELIEVE x+y > 0 does not imply the negation of (xy > 0) , i.e., (xy=0 or xy<0 ) , we again must offer a counterexample to this (other) "conjecture".
In this case:
? : xy > 0
Take, for instance, (x,y) = (2,-1) , that satisfies x+y > 0 , to refute the conjecture given by the FOCUS, because xy > 0 is false.
Take, for instance, (x,y) = (1,1), that satisfies x+y > 0, to refute the negation of the conjecture given by the FOCUS (in the sense that we would have a unique answer in the negative if xy>0 would always be false).
In short:
? : xy > 0
Take (x,y) = (2,-1) <NO> , that is, this ordered pair answers the question (=focus) in the negative.
Take (x,y) = (1,1) <YES> , that is, this ordered pair answers the question (=focus) in the affirmative.
Important: each ordered pair MUST respect the statement(s) considered (in this case x+y>0), otherwise the "test" (an expression I respectfully disapprove) does not apply.
That´s what our method call a BIFURCATION ... I took this expression from the ODE (ordinary differential equations) subject.
I hope I have made things clearer.
Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Max@Math Revolution
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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)
If x = 1/2 and y = 1/4, then xy > 0 and the answer is 'yes'.
If x = 1/2 and y = -(1/4), then xy < 0 and the answer is 'no'.
Since the answer is not unique, both conditions are not sufficient, when taken together.
Therefore, E is the answer.
Answer: E
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)
If x = 1/2 and y = 1/4, then xy > 0 and the answer is 'yes'.
If x = 1/2 and y = -(1/4), then xy < 0 and the answer is 'no'.
Since the answer is not unique, both conditions are not sufficient, when taken together.
Therefore, E is the answer.
Answer: E
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