Means of interchanged data sets

Let the data in the first set be $a_1,a_2,...,a_n$ . Let the data in the second be $b_1,b_2,...,b_k$ .

We have that $\dfrac{a_1+...+a_n}{n}=15\implies a_1+...+a_n=15n$ .

Similarly, $b_1+...+b_k=20k$ .

Assume, without loss of generality, that $a_1$ and $b_1$ are swapped.

This gives $b_1+a_2+...+a_n=16n$ and $a_1+b_2+...+b_k=17k$ .

Then $n=b_1-a_1$ and $3k=b_1-a_1\implies n=3k$ .

Finally, the mean of the combined data is $\dfrac{\left (a_1+...+a_n\right )+\left (b_1+...+b_k\right )}{n+k}=\dfrac{15n+20k}{n+k}=\dfrac{65k}{4k}=\color{#20A900}{\boxed{16\dfrac{1}{4}}}$ .

Let $r$ be the fraction of the data that is in the first set. Then the mean of the combined data set is $15r + 20(1-r),$ and it also equals $16r + 17(1-r).$

Simplifying, we get $20-5r = 17-r,$ or $r = 3/4.$ So the mean of the combined data set is $15(3/4) + 20(1/4) = 16\frac14.$ Note that this method still works if we had interchanged any number of elements from the two sets, as long as the first and second sets maintained their sizes after the changes.

Same method as @Matthew Christopher