If the average of $m$ numbers is $n^2$ and that of $n$ numbers is $m^2$, then the average of $(m + n)$ numbers is

### Concept & Formula This is an algebraic weighted average problem. The core principle remains identical to numerical problems: find the total sum of all items and divide by the total count of items. $$Average = \frac{Sum \ of \ all \ observations}{Total \ number \ of \ observations}$$ ### Step-by-Step Solution * **Given:** * Group 1: Has $m$ numbers, Average = $n^2$. * Group 2: Has $n$ numbers, Average = $m^2$. * **Calculate the Sum of Each Group:** * Sum of the first group = $(\text{Count}) \times (\text{Average}) = m \times n^2 = mn^2$. * Sum of the second group = $(\text{Count}) \times (\text{Average}) = n \times m^2 = nm^2$. * **Calculate Combined Totals:** * Total combined sum = $mn^2 + nm^2$. * Factor out the common term ($mn$) from the sum: $mn(n + m)$. * Total number of elements = $m + n$. * **Calculate the Final Average:** * Overall Average = $\frac{\text{Total Sum}}{\text{Total Count}}$ * Overall Average = $\frac{mn(m + n)}{m + n}$ * Cancel out the common $(m + n)$ term from the numerator and denominator. * Overall Average = $mn$. ### Exam Strategy & Shortcut If you are uncomfortable with pure algebra, use **Number Substitution**. Let $m = 2$ and $n = 3$. * Group 1: 2 numbers, average is $3^2 = 9$. Sum = $2 \times 9 = 18$. * Group 2: 3 numbers, average is $2^2 = 4$. Sum = $3 \times 4 = 12$. * Total Sum = $18 + 12 = 30$. * Total Numbers = $m + n = 2 + 3 = 5$. * Final Average = $\frac{30}{5} = 6$. Now test the options with $m=2, n=3$: (a) $2 - 3 = -1$ (b) $2 \times 3 = 6$ <-- (Matches our result) (c) $2 + 3 = 5$ (d) $\frac{2}{3}$ ### Common Pitfall A common trap is trying to average the algebraic expressions directly: $\frac{n^2 + m^2}{2}$, completely ignoring the weights (the counts $m$ and $n$). Always convert averages into sums before combining them. ### Final Answer **Therefore, the correct answer is $mn$.**