[GMAT math practice question]
If x^2>y^2, is x>y?
1) x>0
2) y>0
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If x^2>y^2, is x>y?
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Max@Math Revolution
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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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The original condition x^2 > y^2 is equivalent to |x| > |y|.
Since we have 2 variables and 1 equation, D is most likely to be the answer. In inequality questions, inequalities are counted as equations. So, we should consider each of the conditions on their own first.
Condition 1)
x^2 > y^2
=> |x| > |y|
=> x > |y|, since x > 0
⇒ x > |y| ≥ y
⇒ x > y
Thus, x > y.
Condition 1) is sufficient.
Condition 2)
Since y > 0, we have x < -y or x > y.
If x = 2 and y = 1, then x > y, and the answer is "yes".
If x = -2 and y = 1, then x < y and the answer is "no".
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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.
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.
The original condition x^2 > y^2 is equivalent to |x| > |y|.
Since we have 2 variables and 1 equation, D is most likely to be the answer. In inequality questions, inequalities are counted as equations. So, we should consider each of the conditions on their own first.
Condition 1)
x^2 > y^2
=> |x| > |y|
=> x > |y|, since x > 0
⇒ x > |y| ≥ y
⇒ x > y
Thus, x > y.
Condition 1) is sufficient.
Condition 2)
Since y > 0, we have x < -y or x > y.
If x = 2 and y = 1, then x > y, and the answer is "yes".
If x = -2 and y = 1, then x < y and the answer is "no".
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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.
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Jeff@TargetTestPrep
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Statement One Alone:Max@Math Revolution wrote:[GMAT math practice question]
If x^2>y^2, is x>y?
1) x>0
2) y>0
x > 0
We see that x is positive. However, since we don't know anything about y, it can be positive, negative or 0. If y is negative or 0, of course x > y. Therefore, we need to determine whether x > y when y is also positive. We are given that x^2 > y^2, so let's take the square root of both sides:
√(x^2) > √(y^2)
|x| > |y|
Since x > 0 and we also assume that y > 0, then |x| = x and |y| = y. Therefore, we have do have x > y.
Statement one is sufficient.
Statement Two Alone:
y > 0
We see that y is positive. However, since we don't know anything about x, x can be positive, negative or 0. If x is negative or 0, of course x is not greater than y. Therefore, we need to determine whether x > y when x is also positive. In statement one, we've shown that x > y when both x and y are positive. Therefore, we see that x is greater than y (when x is positive) but x is less than y (when x is negative or 0). Since we are not given any information about x, we don't know which case it will be.
Statement two is not sufficient.
Answer: A
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