Let the time for which Mary drove at 70 mph be $t$ hours. Then the total distance covered was $55 \times 2 + 70 \times t$ miles. Also, as her average speed over $2+t$ hours was $60$ mph, so the total distance travelled was $60(2+t)$ miles.
Therefore $110 +70t = 120 + 60 t$, that is $10t = 10$, that is $t=1$.
So in total Mary's journey took $3$ hours.
Alternatively, after two hours, Mary has travelled 110 miles. If she had been travelling at 60mph she would have covered 120 miles so she is 10 miles behind schedule. This means that travelling for one more hour at 70mph (10mph faster than the desired average) allows her to catch up that 10 miles. So her total journey time is 3 hours.
This problem is taken from the UKMT Mathematical Challenges.