Let X = \ 11, 12, 13, . , 40, 41\ and Y =\ 61, 62, 63, . . . , 90,...

We have,

$X = \left\{11, 12, 13, \ldots . , 40, 41\right\}$

$⇒ n\left(X\right) = 31$

$Y = \left\{61, 62, 63, . . . , 90, 91\right\}$

$⇒ n\left(Y\right) = 31$

So,

$\overset{¯}{x} = \dfrac{{∑}_{i = 11}^{41}i}{31} = \dfrac{1}{31}\left[\dfrac{41 \times 42}{2} - \dfrac{10 \times 11}{2}\right] = 26$

$\overset{¯}{y} = \dfrac{{∑}_{i = 61}^{91}i}{31} = \dfrac{1}{31}\left[\dfrac{91 \times 92}{2} - \dfrac{60 \times 61}{2}\right] = 73$

Combined mean,

$\mu = \dfrac{31 \times 26 + 31 \times 76}{31 + 31}$

$⇒ \mu = \dfrac{26 + 76}{2} = 51$

Also,

${\sigma}^{2} = \dfrac{1}{62} \times \left(\mathop{∑}_{i = 1}^{31}{\left({x}_{i} - \mu\right)}^{2} + \mathop{∑}_{i = 1}^{31}{\left({y}_{i} - \mu\right)}^{2}\right)$

Since, ${x}_{i} \in X$ are in A.P. with $31$ elements common difference $1$ , same is ${y}_{i} \in y$, when written in increasing order.

$∴ \mathop{∑}_{i = 1}^{31}{\left({x}_{i} - \mu\right)}^{2} = \mathop{∑}_{i = 1}^{31}{\left({y}_{i} - \mu\right)}^{2} = {10}^{2} + {11}^{2} + \ldots . + {40}^{2}$

$= {10}^{2} + {11}^{2} + \ldots . + {40}^{2}$

$= \dfrac{40 \times 41 \times 81}{6} - \dfrac{9 \times 10 \times 19}{6} = 21855$

Hence, ${\sigma}^{2} = \dfrac{21855}{62} + \dfrac{21855}{62} = 705$

$∴ \left|\overset{¯}{x} + \overset{¯}{y} - {\sigma}^{2}\right| = | 26 + 76 - 705 | = 603$