A Hard Maths Question.?

Question

Helen runs faster than Jane but slower than Lisa. They started at the same time from the same place and ran around a circular route. After a while they stopped simultaneously at the same place from which theystarted. It turned out that Lisa overtook Jane ten times during this run. (How many times did it occur that one of the girls overtook another? Assume that each of the girls ran at a constant speed. (The final stop together is not counted as an overtake).

Please show how you got the answer.

Answer 1

It depends on the speed they run, the only thing given is this

Jane < Helen < Lisa  in speed

Without the actual mph or a rate of speed, the problem could be changed dramatically

example: Lisa could be running 10mph, Helen could be running 9mph and Jane could be running 4mph.

Depending on how the speeds change, so does the answer.

Answer 2

Let L, J and H be the number of laps they each ran. Then J < H < L and each must be an integer.

Now, the number of times Helen passed Jane is (H - J - 1). We subtract one because we don't count the final stop. Likewise, Lisa passed Helen (L - H -1) times. Finally, Lisa passed Jane (L - J - 1) times, which we are told is equal to 10.

Sum the three together:

H - J - 1 + L - H - 1 + L - J - 1

2L - 2J - 3

2(L - J - 1) - 1

2(10) - 1

20 - 1

19

The total is 19.