Identify the two sets

$X=\{11,12,13,\dots,41\}, \qquad Y=\{61,62,63,\dots,91\}$

Each is an arithmetic progression with common difference $1$.

Find the means $\bar x$ and $\bar y$

For a consecutive arithmetic sequence, mean is the average of first and last terms.

$\bar x=\frac{11+41}{2}=26$ $\bar y=\frac{61+91}{2}=76$

Hence, $\bar x+\bar y=26+76=102$

Describe the union $X\cup Y$

The combined observations are: $11,12,\dots,41,61,62,\dots,91$

Number of elements in each set: $41-11+1=31, \qquad 91-61+1=31$

So total number of observations is $N=31+31=62$

Find the mean of all observations in $X\cup Y$

Since both sets have equal number of observations, $\mu=\frac{\bar x+\bar y}{2}=\frac{26+76}{2}=51$

Compute variance of all observations

Variance is $\sigma^2=\frac{1}{62}\sum (x_i-51)^2$

Split into the two groups.

For set $X$

Write elements as $11,12,\dots,41$. Their deviations from $51$ are: $-40,-39,\dots,-10$ So contribution is $\sum_{k=10}^{40} k^2$

For set $Y$

Elements are $61,62,\dots,91$. Their deviations from $51$ are: $10,11,\dots,40$ So contribution is again $\sum_{k=10}^{40} k^2$

Therefore, $\sigma^2=\frac{2\sum_{k=10}^{40} k^2}{62}=\frac{\sum_{k=10}^{40} k^2}{31}$

Evaluate $\sum_{k=10}^{40} k^2$

Use $\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$

Then $\sum_{k=10}^{40} k^2=\sum_{k=1}^{40}k^2-\sum_{k=1}^{9}k^2$

$=\frac{40\cdot 41\cdot 81}{6}-\frac{9\cdot 10\cdot 19}{6}$

$=22140-285=21855$

Hence, $\sigma^2=\frac{21855}{31}=705$

Compute the required value

$\left|\bar x+\bar y-\sigma^2\right|=|102-705|=603$

Compare with stored answer

Derived answer = $603$

Stored correct answer = $603$

They match.