This puzzle from the Scottish Mathematical Council (SMC) Senior Mathematics Challenge seems at first to have insufficient information to solve.

“Ant and Dec had a race up a hill and back down by the same route. It was 3 miles from the start to the top of the hill. Ant got there first but was so exhausted that he had to rest for 15 minutes. While he was resting, Dec arrived and went straight back down again. Ant eventually passed Dec on the way down just half a mile before the finish.

Both ran at a steady speed uphill and downhill and, for both of them, their downhill speed was one and a half times faster than their uphill speed. Ant had bet Dec that he would beat him by at least a minute.

Did Ant win his bet?”

See the Close Race Puzzle for solutions.

(**Update 1/2/2023**) **Alternative Solution** from Oscar Rojas

Regarding the Close Race Puzzle I have found a very simple way to solve it using just arithmetic:

Since for both runners the speed down is 1.5 the speed up, then going up 3 miles is equivalent to going down 4.5 miles (in terms of time). So the problem can be restated as a 7.5-mile run at a single speed, with a checkpoint at 7 miles, and Ant’s rest can be thought of as an advantage in time.

At checkpoint Ant has a 15 minute lead, so by rule of three at 7.5 miles his lead will be 7.5*15/7=16.07 minutes. In other words, despite the break, he arrived with more than a minute of advantage.

Oscar Rojas, Costa Rica