Fred & Frank Race: Who Finishes First?

[BrownianMan]

Fred and Frank
: Fred and Frank are two fitness fanatics on a run
from A to B. Fred runs half the way and walks the other half.
Frank runs for half the time and walks for the other half. They

both run and walk at the same speeds. Who finishes first?

So I solved this. Frank finishes first.

Francis joins them and teaches them to jog. Fred now runs onethird
of the way, jogs one-third of the way and walks the rest,
while Frank jogs for on-third of the time, runs for one-third and
walks the rest. Who finishes first?

For this, I also got that Frank finishes first. Is this correct?

“Has Francis helped them to finish sooner or later than previously?”

For this I got:

Frank finishes sooner with joging included if the rate of jogging is greater than (rate of walking + rate of running)/2 and later if rate of jogging is less than (rate of walking + rate of running)/2. Fred finishes faster with jogging included regardless of the rate of jogging.

Is this right?

 

[tiny-tim]

Hi BrownianMan!

If you want us to tell you whether you're right, you really need to show your equations, so we can check them.

 

[BrownianMan]

I had the first part checked already and it was correct.

For the second part:

r1 = rate running
r2 = rate walking
r3 = rate jogging
t1 = time running
t2 = time walking
t3 = time jogging

Since r1t1 + r2t2 + r3t3 = 1, r1t1=r2t2=r3t3 for Fred, t1=t2=t3 for Frank, I found that the total time for Fred is (1/r1 + 1/r2 + 1/r3)/3, and total time for Frank is 3/(r1 + r2 + r3). Then I showed that [3/(r1 + r2 + r3)] - [(1/r1 + 1/r2 + 1/r3)/3] <= 0. So this shows that Frank wins.

Then for the final part:

Frank finishes sooner with jogging included than he did previous if and only if

3/(r1 + r2 + r3) - 2/(r1 + r2) <= 0; so whether or not Frank finishes faster with jogging included than he did previously depends on the rate of jogging (r3).

He finishes sooner with jogging if r3 => (r1 + r2)/2
He finishes later with jogging if r3 < (r1 + r2)/2

Fred finishes sooner with jogging if and only if (1/r1 + 1/r2 + 1/r3)/3 - (1/r1 + 1/r2)/2 <= 0. So since (1/r1 + 1/r2 + 1/r3)/3 - (1/r1 + 1/r2)/2 will always be negative, Fred finishes sooner with jogging included in the race.

 

[tiny-tim]

(just got up :zzz: …)

Looks fine!

 

[francisjogger]

There is a mistake in your statement:

"Fred finishes sooner with jogging if and only if (1/r1 + 1/r2 + 1/r3)/3 - (1/r1 + 1/r2)/2 <= 0. So since (1/r1 + 1/r2 + 1/r3)/3 - (1/r1 + 1/r2)/2 will always be negative, Fred finishes sooner with jogging included in the race."

Just like Frank, Fred sometimes going to be slower, and sometimes going to be faster.
Check your calculations.