UPSC CDS 2 2021 Math Paper

Given, $P$ is the smallest integer. $Q$ is the largest integer. Now, Arithmetic mean (AM), $\text{AM} = \frac{P+Q}{2}$ (i) $-4,-3,-2,-1,0$ [from previous solution] $P=-4, Q=0$ $\text{AM} = \frac{-4+0}{2} = -2$ (ii) $-3,-2,-1,0,1$ $P=-3, Q=1$ $\text{A M} = \frac{-3+1}{2} = \frac{-2}{2} = -1$ (iii) $-2,-1,0,1,2$ $P=-2, Q=2$ $\text{A M} = \frac{-2+2}{2} = \frac{0}{2} = 0$ (iv) $-1,0,1,2,3$ $P=-1, Q=3$ $\text{A M} = \frac{-1+3}{2}$ $= \frac{2}{2} = 1$ (v) $0,1,2,3,4$ $P=0, Q=4$ $\text{A M} = \frac{0+4}{2} = 2$ Arithmetic mean of $P$ and $Q$ have five different values. As no particular condition is given. Therefore, we cannot mark any of the mean with this insufficient information. Hence, option (d) is correct.