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

D

1T

1V

1 and D

2T

2V

2 6 variablesD

1T

136mph and D

2T

248mph

4 unknown variables (speed taken from the original condition)

1)0.76D, T

1, 36mph and 0.24D, T

2, 48mph

T=\(\frac{D}{V}\)

T

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

T

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

Now we have the respective

Ts. Want to find the respective

Ds

D

1=36·\(\frac{0.76D}{36}\)=36·0.76D

D

2=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

AD2) D

1T

136mph and D

2100miles48mph

3 unknown variablesWe 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!