Aaron travels from town x to town y and then back from town y to town

Nice solution Salsanousi.
I've elaborated below so that others can see "1 equation 1 variable we can find the answer" (as I certainly didn't when I started).

Average speed = \(\frac{Total Distance}{Total Time}\)
D1T1V1 and D2T2V2 6 variables
D1T136mph and D2T248mph 4 unknown variables (speed taken from the original condition)

1)0.76D, T1, 36mph and 0.24D, T2, 48mph
T=\(\frac{D}{V}\)

T1=\(\frac{0.76D}{36}\)

T2=\(\frac{0.24D}{48}\)

Now we have the respective Ts. Want to find the respective Ds
D1=36·\(\frac{0.76D}{36}\)=36·0.76D

D2=48·\(\frac{0.24D}{48}\)=48·0.24D

Average speed = \(\frac{Total Distance}{Total Time}\)

Average speed = \(\frac{36·0.76D+48·0.24D}{\frac{0.76D}{36}+\frac{0.24D}{48}}\)

Multiply through by 100 to get rid of the decimals

=\(\frac{3600·76D+4800·24D}{\frac{76D}{3600}+\frac{24D}{4800}}\)

Simplify the denominator

=\(\frac{\frac{D(3600·76+4800·24)}{1}}{\frac{47D}{1800}}\)

=\(\frac{1800D(3600·76+4800·24)}{47D}\)

Cancel D from the numerator and denominator

=\(\frac{1800(3600·76+4800·24)}{47}\)

Shows that average speed is a number, we just don't need to work out what that number is.
Sufficient
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2) D1T136mph and D2100miles48mph 3 unknown variables
We can't solve for 3 unknown variables with one equation


Anyone that reads this answer, kudos please! It took me over half an hour with all of the formatting!