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19 Jan 2018, 01:17
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A
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[GMAT math practice question]
If \(xy≠0\), is \(|x|>|y|\)？
1) \(x=-4y\)
2) \(x=y^3\)
If \(xy≠0\), is \(|x|>|y|\)？
1) \(x=-4y\)
2) \(x=y^3\)
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19 Jan 2018, 06:37
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MathRevolution
(1) x = -4y.
Since we are talking about only modulus signs of x and y, we can ignore the negative sign here (just for the sake of comparison between |x| and |y|)
So whatever value of y we take here (positive or negative, integer or decimal, whether 0 < |y| < 1 or |y| =1 or |y| > 1) - the thing is that x will be 4 times of that value. And since we are talking about absolute values only here, |x| will always be > |y|. So Sufficient.
(2) x = y^3.
If y=1, then x=1. Here |x| = |y|.
If y=2, x=8. Here |x| > |y|. So we cant be sure. Not Sufficient.
Hence A answer
21 Jan 2018, 18:38
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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) and 2) together first.
Conditions 1) and 2):
Since \(x = -4y\) and \(x = y^3\), \(|x| = |-4y| = 4|y|\) and \(|x| = |y^3| = |y|^3.\)
It follows that \(4|y| = |y|^3\).
Now,
\(4|y| = |y|^3 ⇔ |y|^3 – 4|y| = 0\)
\(⇔ |y|(|y|^2- 4) = 0\)
\(⇔ |y|(|y|+2)(|y|-2) = 0.\)
Since \(|y| ≠0\) and \(|y| ≠-2\), this implies that \(|y| = 2\).
It follows that |\(x| = 4*2 = 8\).
So, \(|x| > |y|\), and both conditions together are sufficient.
Since this 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)
Since \(|x| = |-4y| = 4|y|\) and \(xy≠0, |x| = 4|y| > |y|.\)
Condition 1) is sufficient.
Condition 2)
If \(x = 8,\) then \(y = 2\), and the answer is ‘yes’.
If \(x = 1,\) then \(y = 1\), and the answer is ‘no’.
Since we have obtained two different answers, condition 2) is not sufficient by CMT 2.
Therefore, A is the answer.
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.
Answer: A
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) and 2) together first.
Conditions 1) and 2):
Since \(x = -4y\) and \(x = y^3\), \(|x| = |-4y| = 4|y|\) and \(|x| = |y^3| = |y|^3.\)
It follows that \(4|y| = |y|^3\).
Now,
\(4|y| = |y|^3 ⇔ |y|^3 – 4|y| = 0\)
\(⇔ |y|(|y|^2- 4) = 0\)
\(⇔ |y|(|y|+2)(|y|-2) = 0.\)
Since \(|y| ≠0\) and \(|y| ≠-2\), this implies that \(|y| = 2\).
It follows that |\(x| = 4*2 = 8\).
So, \(|x| > |y|\), and both conditions together are sufficient.
Since this 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)
Since \(|x| = |-4y| = 4|y|\) and \(xy≠0, |x| = 4|y| > |y|.\)
Condition 1) is sufficient.
Condition 2)
If \(x = 8,\) then \(y = 2\), and the answer is ‘yes’.
If \(x = 1,\) then \(y = 1\), and the answer is ‘no’.
Since we have obtained two different answers, condition 2) is not sufficient by CMT 2.
Therefore, A is the answer.
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.
Answer: A
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