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11 Jun 2018, 06:59
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
53% (01:17) correct
47%
(01:12)
wrong
based on 59
sessions
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If \(x\) and \(y\) are integers, is \(x^2+11x+13y\) an even number?
1) \(x=11\)
2) \(y=13\)
1) \(x=11\)
2) \(y=13\)
11 Jun 2018, 07:00
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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.
Now, \(x^2+11x+13y = x(x+11)+13y\). As \(x(x+11)\) is always an even integer (one of \(x\) and \(x + 11\) must be even), \(x^2+11x+13y\) is even precisely when \(13y\) is even. This is equivalent to \(y\) being even, so the question can be restated as 'Is \(y\) an even number?
Condition 1)
If \(x = 11\) and \(y = 1\), then \(x^2+11x+13y = 255\) is an odd integer.
If \(x = 11\) and \(y = 2\), then \(x^2+11x+13y = 268\) is an even integer.
Since the question does not have a unique answer, condition 1) is not sufficient.
Condition 2)
If \(y = 13\), then \(y\) is odd, and the answer to the question is 'no'.
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
Now, \(x^2+11x+13y = x(x+11)+13y\). As \(x(x+11)\) is always an even integer (one of \(x\) and \(x + 11\) must be even), \(x^2+11x+13y\) is even precisely when \(13y\) is even. This is equivalent to \(y\) being even, so the question can be restated as 'Is \(y\) an even number?
Condition 1)
If \(x = 11\) and \(y = 1\), then \(x^2+11x+13y = 255\) is an odd integer.
If \(x = 11\) and \(y = 2\), then \(x^2+11x+13y = 268\) is an even integer.
Since the question does not have a unique answer, condition 1) is not sufficient.
Condition 2)
If \(y = 13\), then \(y\) is odd, and the answer to the question is 'no'.
Condition 2) is sufficient.
Therefore, B is the answer.
Answer: B
08 Oct 2018, 08:32
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In this question, option D can also be selected as a correct answer. Since by taking values of both statements a definitive answer of the question can be obtained.
