MathRevolution wrote:
[GMAT math practice question]
(Statistics) a1, a2, …., an are n integers. What is the average of a1, a2, …., an?
1) The minimum value of possible averages of \(n-1\) integers among a1, a2, …., an is \(60.\)
2) a1, a2, …., an are consecutive integers and a1 = 21.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have many variables and \(0\) equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2) together give us:
We have a
1 = 20 + 1, a
2 = 20 + 2, … , a
n = 20 + n from condition 2).
(a
1 + a
2 + … + a
n-1 ) / (n - 1) = [(20 + 1) + (20 + 2) + … + (20 + (n - 1))] / (n - 1) = [20(n - 1) + n(n - 1) / 2] / (n-1) = 20 + n/2 = 60.
We have n/2 = 40 or n = 80.
Thus, the maximum value of possible averages of a
1, a
2, … , a
n is
(21 + 22 + 23 + … + 100) / 80
= [80·(21 + 100) / 2] / 80, (21 + 100) / 2, or 121/2.
The answer is unique, and conditions 1) and 2) together are sufficient according to Common Mistake Type 2, which states that the number of answers must be only one.
Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Let’s look at condition 1).
If a
1 = 100 and a
2 = 60, the average of a
1 and a
2 is 80.
If a
1 = 80 and a
2 = 60, the average of a
1 and a
2 is 70.
The answer is not unique, and the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.
Let’s look at condition 2).
If a
1 = 21, a
2 = 22 and a
3 = 23, then the average of a
1, a
2 and a
3 is (21 + 22 + 23) / 3 = 66/3 = 22.
If a
1 = 21, a
2 = 22, a
3 = 23, a
4 = 24 and a
5 = 25, then the average of a
1, a
2 and a
3 is (21 + 22 + 23 + 24 + 25) / 5 = 115/5 = 23.
The answer is not unique, and the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.
Both conditions 1) & 2) together are sufficient.
Therefore, C is the correct answer.
Answer: 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.
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