Persons a, b, and c run 2,000m, 6000m, and 10,000m respectively in a medley. their combined time is 66 minutes. person b's time is two minutes more than five times person a's time. the sum of two times person b's time and four times person a; s time is equal to person c's time. how much time did it take person b to run that part of the race?

Let a, b, and c represent the running times of persons a, b, and c, respectively.

The problem statement describes several relationships:

... a + b + c = 66 . . . . . their combined time is 66 minutes

... b = 2 + 5a . . . . . . . . b's time is 2 minutes more than 5 times a's

... 2b + 4a = c . . . . . . . twice b's time plus 4 times a's time is equal to c's

These can be solved a variety of ways. One way is to use the second equation for b to substitute into the other equations. This gives ...

... a + (2+5a) + c = 66

... 2(2+5a) +4a = c

The first of these simplifies to

... 6a +c = 64

The second of these simplifies to

... 9a -c = -4

Adding these two equations together gives

... 15a = 60

... a = 4 . . . . . . . divide by 15

From above, b = 2 + 5a = 2 + 5·4 = 22

It took b 22 minutes to run that part of the race.