$\text{Rate} \times \text{Time} = \text{Distance}$ problem

Question

To drive to the supermarket, Mable drives for $m$ miles, then drives $12$ miles per hour faster for the remaining $\frac{4m}3$ miles. The amount of time Mable spent driving at each of the two speeds was equal. What was Mable’s average speed during her drive to the supermarket, in miles per hour?

I get $$m=rt~\text{and}~\frac{4m}3=(r+12)t.$$ Next, WLOG let $m=60.$ I then find $r=36$ after some manipulation. Since the total distance is $m+\frac{4m}{3}=60+80=140,$ and the average speed is $\frac{240}{140},$ I get the answer $\frac{120}{7}.$ However, this is incorrect. I'm just a beginner at $D=R\cdot T$ problems, so please forgive me, since a lot fo my work is most likely incorrect. Help?

Answer 1

Indicating with $v$ the uknown speed, we have that

$$t=\frac{m}{r}=\frac{\frac 43m}{r+12} \implies \frac13 mr=12m \implies r=36 \, (mph)$$

which corresponds to your result, then the average speed is given by

$$\bar r=\frac{m+\frac 43 m}{2 t}=\frac76 r=42\, (mph)$$

Answer 2

WLOG suppose Mable's total journey time is 2 hours. Then in the 2nd hour she travels 12mi further than in the 1st hour. The distances travelled are in the ratio 3:4 so she travelled a total of 7 x 12mi in 2 hours making the average speed 42 mph.