Data Sufficiency

[GMAT math practice question]

What is the median of m, n and 5?

1) m+n=10
2) m=5

=>

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 2 variables and 0 equations, C is most likely to be 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:
Since m + n = 10 and m = 5, we have m = n = 5.
Thus, the median of m, n and 5 is 5.

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): m + n = 10
If one of m and n is less than 5, the other one must be greater than 5. For example, if m < 5, then n > 5.
If one of m and n is 5, the other must also equal 5. For example, if m = 5, then n = 5.
In both cases, the median is 5.
Condition 1) is sufficient.

Condition 2) : m = 5.
If n > 5, the numbers are 5, 5, n in ascending order.
If n < 5, the numbers are n, 5, 5 in descending order.
If n = 5, the numbers are 5, 5, 5.
In each of the possible cases, the median is 5.
Condition 2) is sufficient.

Therefore, D is the answer.

Answer: D

Note: The VA approach tells us that the answer is most likely to be D, since this is a CMT(Common Mistake Type) 4B question.
Condition 1) is easy to check, but condition 2) is more difficult to work with. If you can't figure out condition 2), you should choose D as the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.