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11 Jun 2018, 02:42
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B
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Difficulty:
25%
(medium)
Question Stats:
69% (00:49) correct
31%
(00:52)
wrong
based on 45
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If \(x\) and \(y\) are integers greater than \(1\) and \(x>y\), what is the value of \(x\)?
1) \(x+y=10\)
2) \(xy=21\)
1) \(x+y=10\)
2) \(xy=21\)
11 Jun 2018, 02:42
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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.
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):
\(y = 10 - x\) and \(xy= x(10-x) = 21\)
⇔ \(-x^2 + 10x = 21\)
⇔ \(x2 - 10x + 21 = 0\)
⇔ \((x-3)(x-7) = 0\)
⇔ \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y\), we must have \(x = 7\) and \(y = 3\).
Thus \(x = 7\), and conditions 1) and 2) are sufficient, when taken together.
Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).
Condition 1)
There are two possible solutions: \(x = 6\) and \(y = 4\), and \(x = 7\) and \(y = 3\).
Since the solution is not unique, condition 1) is not sufficient.
Condition 2):
Either \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y\), \(x = 7\) and \(y = 3\).
Thus, we have the unique solution, \(x = 7\).
Therefore, condition 2) is sufficient.
Therefore, B is the answer.
Answer: B
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):
\(y = 10 - x\) and \(xy= x(10-x) = 21\)
⇔ \(-x^2 + 10x = 21\)
⇔ \(x2 - 10x + 21 = 0\)
⇔ \((x-3)(x-7) = 0\)
⇔ \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y\), we must have \(x = 7\) and \(y = 3\).
Thus \(x = 7\), and conditions 1) and 2) are sufficient, when taken together.
Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).
Condition 1)
There are two possible solutions: \(x = 6\) and \(y = 4\), and \(x = 7\) and \(y = 3\).
Since the solution is not unique, condition 1) is not sufficient.
Condition 2):
Either \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y\), \(x = 7\) and \(y = 3\).
Thus, we have the unique solution, \(x = 7\).
Therefore, condition 2) is sufficient.
Therefore, B is the answer.
Answer: B
29 Oct 2018, 21:25
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Condition 2 is incorrect, as it results in 2 possible solutions: (a) x=7, y=3, and (b) x=21, y=1. So answer should be C.
