The average (arithmetic mean) of a list of 5 numbers (arithmetic mean)

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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.

Assume that the five numbers are, \(a, b, c, d\), and \(e\). Then \(\frac{a + b + c + d + e}{5} = 15\), or \(a + b + c + d + e = 75\).

Since we have 5 variables (\(a,b,c,d\), and \(e\)) and 1 equation, E is most likely to be the answer. Thus, we should consider
both conditions together first.

Conditions 1) and 2):
Suppose that a and b are the numbers removed, and that a > 15 and b > 15. Then
\(\frac{( a + b )}{2} > 15\)
⟺\(a + b > 30\)
So,
\(c + d + e = 75 – ( a + b ) < 45\), since \(a + b > 30\). Thus,
\(\frac{c + d + e}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
Therefore, both conditions are sufficient when applied together.

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.

Condition 1)
Assume that \(a\) and \(b\) are the numbers removed, and that \(a > 15\) and \(b > 15\). Then
\(a + b > 30\), and so \(c + d + e = 75 – ( a + b ) < 45\), since \(a + b > 30\).
Therefore,
\(\frac{( c + d + e )}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
This is sufficient.

Condition 2)

Assume that \(a\) and \(b\) are the numbers removed, and that their average is greater than \(15\). That is,
\(\frac{a + b}{2} > 15\). Then \(a + b > 30\), and so
\(c + d + e = 75 – ( a + b ) < 45\).
Therefore,
\(\frac{c + d + e}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
This is sufficient.

Therefore, the answer is D.

Note: By Tip 1) of the VA method, if conditions 1) and 2) provide the same information, we can just select answer D.

Answer: D