Word Problems : Relative Speed of Swimmers

(#48) Two swimmers start at opposite ends of a pool 89 feet long. One
person swims at the rate of 19 feet per minute and the other swims at a
rate of 53 feet per minute. How many times will they meet in 33 minutes?


First swimmer:

Speed: 19 ft/min. Time: t min. Distance: 19t ft.

Second swimmer:

Speed: 53 ft./min. Time: t min Distance: 53t ft.

Since the swimmers start at opposite ends of the pool, when they meet
the first time, the total distance passed would be the length of the
pool, 89 feet. If they meet again, for the second time, the total would
actually be 3 times the length of the pool. The next time the swimmers
meet it would be five times the length of the pool, etc. This pattern
would continue. So, if they meet n times in 33 minutes, we know that
the total distance they have passed is (2n-1)x89. Now, I just have to
set up an inequality, such as this:

≤ 33

By solving for n, I will be able to find the solution to this problem.
The following are my steps in solving this problem:

(2n-1)x89/(19+53)≤ 33

(2n-1)≤ 33x72/89

(Since n is an integer, the above inequality means that 2n-1≤26.7,

so n≤13.85 . We choose n=13. So we know that they can meet a
complete 13 times in 33 minutes.

(#33) Two swimmers start at opposite ends of a 90-foot pool. One swims
30 feet per minute and the other at 20 feet per minute. If they swim for
30 minutes, how many times will they meet each other?

My initial thought was to set the variables of the problem as follows:

First swimmer:

Speed: 30 ft/min. Time: t min. Distance: 30t ft.

Second swimmer:

Speed: 20 ft./min. Time: t min Distance: 20t ft.

I realized this was the pattern I was looking for in order to put their
meetings in general terms, n times. SO, if they meet n times in 30
minutes, we know that the total distance they have passed is (2n-1)x90.
At this point, I was able to set up the inequality:

(2n-1)x90/(20+30)≤ 30

Now, all I had to do was solve for n.

Final solution:

r

2

T

Ь

the swimmers meet the total distance passed will be 3 times the length
of the pool – not 2! The third time the swimmers meet the total
distance will be 5 times the distance of the pool, etc. This pattern
gives us the information that in n times passing, the total distance the
swimmers will have passed will be

(2n-1)x90. Now we can set up an inequality and solve the problem!

≤ 30 Now start to solve for n.

(2n-1)≤ 30x50/90

(Since n is an integer, the above inequality means that 2n-1≤ 16.67,

so n≤ 8.835. We choose n=8. So we know that they can meet a complete
8 times in 30 minutes.