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Updated on: 13 Feb 2022, 01:18
Originally posted by MathRevolution on 11 Apr 2019, 01:44.
Last edited by MathRevolution on 13 Feb 2022, 01:18, edited 1 time in total.
Last edited by MathRevolution on 13 Feb 2022, 01:18, edited 1 time in total.
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MathRevolution
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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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Suppose \(x, y\) and \(z\) are different integers with \(x < y < z\).
For their average \(\frac{( x + y + z )}{3}\) to be equal to their median \(y\),
we must have \(z – y = y – x\), and so their range is \(z – x = z – y + y – x = 2(y-x).\)
This implies that \(z – x\) is an even integer.
Condition 1)
Since condition 1) gives an odd value for the range, the answer is ‘no’. Thus, condition 1) is sufficient by CMT (Common Mistake Type) 1.
Condition 2)
If \(x = 10, y = 11\) and \(z = 12\), then the average and the median are the same, and the answer is ‘yes’
If \(x = 10, y = 11\) and \(z = 15\), then the average \(12\) is different from the median \(11\), and the answer is ‘no’.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.
Therefore, A is the answer.
Answer: A
11 Apr 2019, 01:46
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[GMAT math practice question]
(number properties) What is the greatest common divisor of positive integers \(m\) and \(n\)?
1) \(m\) and \(n\) are consecutive
2) \(m^2 – n^2 = m + n\)
(number properties) What is the greatest common divisor of positive integers \(m\) and \(n\)?
1) \(m\) and \(n\) are consecutive
2) \(m^2 – n^2 = m + n\)
Updated on: 05 Sep 2021, 08:48
Originally posted by MathRevolution on 12 Apr 2019, 01:36.
Last edited by MathRevolution on 05 Sep 2021, 08:48, edited 1 time in total.
Last edited by MathRevolution on 05 Sep 2021, 08:48, edited 1 time in total.
Kudos
Bookmarks
MathRevolution
=>
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. We then recheck the question.
The statement “\(m^2-n^2\) is divisible by \(4\)” means that \((m+n)(m-n)\) is divisible by \(4.\) This is equivalent to the requirement that \(m\) and \(n\) are either both even integers or both odd integers.
Since condition 2) tells us that both \(m\) and \(n\) are odd integers, condition 2) is sufficient.
Condition 1)
The square of an odd integer \((2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2 + a) + 1\) has remainder \(1\) when it is divided by \(4\).
The square of an even integer \((2b)^2 = 4b^2\) has remainder \(0\) when it is divided by \(4\).
Thus, if “\(m^2+n^2\) has remainder \(2\) when it is divided by \(4\)”, both \(m\) and \(n\) must be odd integers.
Condition 1) is sufficient.
Therefore, D is the answer.
Answer: D
FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.
