$A$ is the average of $10$ given numbers. $B$ is the average after replacing $2$ numbers by $3$ other different numbers. The average of the removed numbers is $48$ and the average of the newly added numbers is $56$. If $A+B=438$, then find the value of $A-B$.

Let the initial sum of the $10$ numbers be $S_{10}$.Since $A$ is the average of these $10$ numbers: $$S_{10}=10A$$
Now, $2$ numbers are removed. The average of these $2$ removed numbers is $48$, so their sum is:
$$\text{Sum of removed numbers}=2\times 48=96$$
Then, $3$ new numbers are added. The average of these $3$ newly added numbers is $56$, so their sum is:
$$\text{Sum of added numbers}=3\times 56=168$$
The total number of items now becomes $10-2+3=11$ numbers.Let the new sum of these $11$ numbers be $S_{11}$.
$$S_{11}=S_{10}-96+168=S_{10}+72$$
Since $B$ is the new average of these $11$ numbers:
$$S_{11}=11B$$
Equating the expressions for $S_{11}$:
$$11B=10A+72$$
$$10A-11B=-72\text{— (Equation 1)}$$
We are also given:
$$A+B=438\text{— (Equation 2)}$$
From Equation 2, we can write $A=438-B$. Substitute this into Equation 1:
$$10(438-B)-11B=-72$$
$$4380-10B-11B=-72$$
$$4380-21B=-72$$
$$21B=4380+72$$
$$21B=4452$$
$$B=\frac{4452}{21}=212$$
Now find $A$ using Equation 2:
$$A=438-212=226$$
We need to find the value of $A-B$:
$$A-B=226-212=14$$