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!