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11 Jun 2018, 07:03
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E
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Difficulty:
15%
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Question Stats:
84% (01:10) correct
16%
(01:28)
wrong
based on 45
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If \(x, y and z\) are positive numbers, is \(xy+z > x+yz\)?
1) \(x>1\)
2) \(y>1\)
1) \(x>1\)
2) \(y>1\)
11 Jun 2018, 07:03
Kudos
Bookmarks
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.
Since we have 3 variables (x and y) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.
Conditions 1) & 2):
\(xy + z > x + yz\)
⇔ \(xy - x - yz + z > 0\)
⇔ \(x(y-1)-z(y-1) > 0\)
⇔ \((x-z)(y-1) > 0\)
⇔ \(x - z > 0\) since \(y > 1\).
If \(x = 2, y = 2\), and \(z = 1\), then \(xy + z = 5, x + yz = 4\) and \(xy + z > x + yz\).
So, the answer is 'yes'.
If \(x = 2, y = 2\), and \(z = 3\), then \(xy + z = 7\), \(x + yz = 8\) and \(xy + z < x + yz\).
So, the answer is 'no'.
Since the question does not have a unique answer, both conditions together are not sufficient.
Therefore, E is the answer. 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: 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 3 variables (x and y) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.
Conditions 1) & 2):
\(xy + z > x + yz\)
⇔ \(xy - x - yz + z > 0\)
⇔ \(x(y-1)-z(y-1) > 0\)
⇔ \((x-z)(y-1) > 0\)
⇔ \(x - z > 0\) since \(y > 1\).
If \(x = 2, y = 2\), and \(z = 1\), then \(xy + z = 5, x + yz = 4\) and \(xy + z > x + yz\).
So, the answer is 'yes'.
If \(x = 2, y = 2\), and \(z = 3\), then \(xy + z = 7\), \(x + yz = 8\) and \(xy + z < x + yz\).
So, the answer is 'no'.
Since the question does not have a unique answer, both conditions together are not sufficient.
Therefore, E is the answer. 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: E
