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11 Jun 2018, 06:16
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
72% (00:54) correct
28%
(00:44)
wrong
based on 39
sessions
History
Date
Time
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Not Attempted Yet
Is \(x|y|=xz?\)
1) \(x\), \(y\), and \(z\) are positive
2) \(y^2=z^2\)
1) \(x\), \(y\), and \(z\) are positive
2) \(y^2=z^2\)
11 Jun 2018, 06:16
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Official Solution:
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, and then recheck the question.
Modifying the question:
\(x|y| = xz\)
⇔ \(x(|y|-z) = 0\)
⇔ \(x = 0\) or \(|y| = z\)
⇔ \(x = 0\) or \(y = z\) or \(y = -z\)
Since we have 3 variables (\(x\), \(y\), and \(z\)) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.
Conditions 1) & 2):
Condition 2) tells us that \(y = z\) or \(y = -z\).
Since condition 1) states that \(x, y, z > 0\), we can only have \(y = z\).
Thus, both conditions are sufficient, when taken together.
Therefore, the answer is C.
In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
Answer: C
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, and then recheck the question.
Modifying the question:
\(x|y| = xz\)
⇔ \(x(|y|-z) = 0\)
⇔ \(x = 0\) or \(|y| = z\)
⇔ \(x = 0\) or \(y = z\) or \(y = -z\)
Since we have 3 variables (\(x\), \(y\), and \(z\)) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.
Conditions 1) & 2):
Condition 2) tells us that \(y = z\) or \(y = -z\).
Since condition 1) states that \(x, y, z > 0\), we can only have \(y = z\).
Thus, both conditions are sufficient, when taken together.
Therefore, the answer is C.
In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
Answer: C
ShivaniD
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02 Aug 2018, 08:11
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Condition 2) tells us that y=z or y=−z
In both the above possibilities, we can sufficiently answer that x |y| = xz ; because the modulus on y will ensure that both z and -z gives out a positive figure.
Why is it that we need the first condition?
In both the above possibilities, we can sufficiently answer that x |y| = xz ; because the modulus on y will ensure that both z and -z gives out a positive figure.
Why is it that we need the first condition?
