Bunuel wrote:
(n-x)+(n-y)+(n-z)+(n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, z, and k is n.
(2) x, y, z, and k are consecutive integers.
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 5 variables and 0 equation, E is most likely to be the answer.
As E is the most likely answer, we should consider both conditions 1) and 2) together before considering each of them individually. If they are not sufficient when taken together, E is the answer.
Conditions 1) & 2)
\(\frac{(x+y+z+k)}{4} = n\)
\(x + y + z + k = 4n\)
Thus \(( n - x ) + ( n - y ) + ( n - z ) + ( n - k ) = 4n - ( x+ y + z + k ) = 4n - 4n = 0\).
Both conditions together are sufficient.
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)
Actually, the condition 1) only is used in the conditions 1) & 2).
\(\frac{(x+y+z+k)}{4} = n\)
\(x + y + z + k = 4n\)
Thus \(( n - x ) + ( n - y ) + ( n - z ) + ( n - k ) = 4n - ( x+ y + z + k ) = 4n - 4n = 0\).
The condition 1) only is sufficient.
Condition 2)
We can't derive anything \(4n - ( x+ y + z + k )\) from the condition, since we don't know \(n\).
The condition 2) is not sufficient.
Therefore, the answer is A.
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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