GATE 2023 IN GA Q.4: solution and answer

Official answer: CSource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. "Must be TRUE" means the statement has to hold for every possible set of scores, so one counterexample is enough to rule an option out.

2. Option (A), test with a right-skewed set: scores $1,2,9$ give median $=2$ and mean $=\frac{1+2+9}{3}=4$. The median is smaller than the mean, so (A) is not always true.

3. Option (B), test with a left-skewed set: scores $1,8,9$ give median $=8$ and mean $=\frac{1+8+9}{3}=6$. The mean is smaller than the median, so (B) is not always true.

4. Option (D), same left-skewed set: mean $=6$, and $2$ of the $3$ candidates (scores $8$ and $9$) are above it. $\frac{2}{3}>\frac{1}{2}$, so more than half the candidates score above the mean and (D) is not always true. The mean has no fixed position in the ordered list.

5. Option (C), definition of the median: arrange the $n$ scores in increasing order $x_1\le x_2\le \cdots \le x_n$. The median is the middle value ($n$ odd) or the average of the two middle values ($n$ even).

6. For $n$ odd, every score up to and including the middle one is $\le$ the median, so only the $\frac{n-1}{2}$ scores after it can be larger. For $n$ even, the lower half $x_1,\dots ,x_{n/2}$ are all $\le x_{n/2}\le$ median, so only the $\frac{n}{2}$ scores in the upper half can be larger. Either way the count is $\le \frac{n}{2}$, and scores tied at the median only lower it.

7. So (C) holds for every distribution of scores. Only (C) must be true.