Training For The Olympics 100 Meters Freestyle

First, notice that they swim at a combined rate of $5\text{ m/s}$ . Therefore, it takes them $90\div 5=18\text{ s}$ to cover the length of the pool. After their first crossing, however, they must cover twice the length of the pool to cross each other again. Therefore, after the first cross at $18\text{ s}$ , every subsequent crossing takes $36 \text{ s}$ . They swim for a total of $60\times 12=720$ seconds; therefore, the number of crossing they do in total is $\left\lfloor\dfrac{720+18}{36}\right\rfloor = \boxed{20}$ .

This problem was based on a MathCounts Nationals problem #29. I forgot which year.

If we assume the one of the them is stationary, and the other swimmer starts with an effective speed of 5 m/s. To cover 90 m it takes him 18 seconds. Now since the first person is stationary, the second person has to cover the length of swimming pool twice to get to him. Once, when he goes to one end and the other when he returns from the other end towards the swimmer we have assumed stationary all at the speed of 5m/s. Now, since the distance covered is twice, thus the time taken will also be twice the initial meeting time. So a meeting takes place every 36 seconds after the 1st. Out of 720 seconds i.e. 12 minutes 18 seconds have already gone by and now 708 seconds are left. 36 which is the time for every subsequent meeting goes into 708 a maximum of 19 times.

Hence 19+1=20 meetings

For every 3 minutes they meet 5 times. Hence in 12 minutes they meet 5x3/4=20 times

I got annoyed with my stupidity so I cheated

int f(int x){
    return (((3*x)%90)+((2*x)%90));
}

int g(int x){
    return (90-((3*x)%90))+((2*x)%90);
}
int main(){
    int counter=0;
    int a;
    int b;
    for (int i=1; i<=720; ++i){
        a=(((3*i)/90)%2);
        b=((((2*i)/90)+1)%2);
        if ((f(i)==90) && (a!=b)){
            ++counter;
        }
        else if ((g(i)==90) && (a==b)){
            ++counter;
        }
    }
    std::cout<<counter;
} 

*Result : 20

The first time the swimmers meet, the sum of the distances they have covered is 90 metres. When they next meet, the sum of the distances they have covered is 270 metres. At the next crossing, they have covered 180 metres more; etcetera. In 12 minutes, the swimmers will cover 3x720 + 2x720 = 3600 metres. 3600/180 = 20, therefore there are 20 crossings.

NOTHING IS NEEDED.....JUST DRAW 2 WAVEFORMS...TOTAL DURATION OF BOTH WAVEFORM 12 MIN...WAVEFORMS ARE TRIANGULAR...ONE HAS TIME PERIOD 60 SEC STARTING FROM ORIGIN OTHER HAS TIME PERIOD 45 SEC STARTING FROM MAX POINT..NOW CHECK NO. OFINTERSECTIONS..... :) :)

Let john and peter be J and P J's speed-3m/s That means that he covers 90 m in every 30 seconds by unitary method. It means J takes 1 min for 2 rounds. Then, J takes 24 rounds in 12 mins by unitary method. Then it comes to P. P's speed-2m/s P takes 45 secs to complete 90m(1 round). That means it takes 1.5 mins to complete 2 rounds by unitary method. Then, till the time 12 mins complete, P completes 16 rounds by unitary method. Then, add the rounds taken by both Jack and Peter. The total will be 40 rounds. Then divide it by 2. As there are two people(Jack and Peter). It will come 20 which is the answer. They will pass each other 20 times!! Hope it helps you!! :)

We can start by calling John's distance x and Peter's y. We can represent these distances in parametrics as follows.

y = 2t x = 3t

We can then infer from the information given that x + y = 90n where n is an odd positive integer since when their paths cross the distance traveled will equal the length of the pool. In this case n must be odd since at the starting point n = 0 and the intersection happens for n = 1. At this point the remaining total distance is 90 to the 180 mark, but John is 1.5 times faster and has a shorter distance before reaching the end of the pool and Peter has the longest distance and is slower so John will reach his end before Peter reaches his and when x + y = 180 they wouldn't have crossed paths. They will have to travel a combined distance of 90 to do so.

We can substitute 3t and 2t into x + y = 90n to get 5t = 90n. For n = 1, t = 18 so they first cross at t = 18. For n = 3 we have 5t = 270 so t = 54. For n = 5, t = 90. The common difference for each successive t is 36. They swim for 720 seconds so 5(720) = 90n. Here, n = 40, but since that is even, the last time they will cross, is at n = 39 and t = 702. Between 1 and 39 inclusive, there are 20 odd numbers so they cross 20 times in 12 minutes.

20 times