Who is faster???

Printable View

June 15th 2006, 09:32 AM
fizsle

Who is faster???

Paul & john leave city A for City B at the same time. After reaching City B, both return to city A immediately. Paul travels from A to B @ 56kmph and returns @ 64kmph. John travels @ 60kmph both ways. Who gets back to A first?

(i) Paul
(ii) John
(iii) Both Arrive at the same time
(iv) Cannot be determined

Now my prof says JOHN arrives first as the average speed of Paul(the harmonic mean) will be less than the arithmetic mean(60kmph)....

My question is if the distance is 60 kms between the cities it will take them both the same time.... Am i right or am i incorrect in assuming the distance as 60kmph?
June 15th 2006, 10:49 AM
ThePerfectHacker

Quote:

Originally Posted by fizsle

Paul & john leave city A for City B at the same time. After reaching City B, both return to city A immediately. Paul travels from A to B @ 56kmph and returns @ 64kmph. John travels @ 60kmph both ways. Who gets back to A first?

(i) Paul
(ii) John
(iii) Both Arrive at the same time
(iv) Cannot be determined

Now my prof says JOHN arrives first as the average speed of Paul(the harmonic mean) will be less than the arithmetic mean(60kmph)....

My question is if the distance is 60 kms between the cities it will take them both the same time.... Am i right or am i incorrect in assuming the distance as 60kmph?

Allow the distance to be $s$
Then the time for 56 and 64 to come back is,
$\frac{s}{56}+\frac{s}{64}\approx .03348s$
The time for 60 and 60 to come back is,
$\frac{s}{60}+\frac{s}{60}\approx .03333s$
Since $s>0$ we see that the second method is faster.
June 15th 2006, 04:29 PM
Soroban

Hello, fizsle!

Quote:

Paul & John leave city A for City B at the same time.
After reaching City B, both return to city A immediately.
Paul travels from A to B @ 56 kph and returns @ 64 kph.
John travels @ 60 kph both ways.
Who gets back to A first?

. (i) Paul . . (ii) John . . (iii) Both arrive at the same time . . (iv) Cannot be determined

Now my prof says JOHN arrives first as the average speed of Paul (the harmonic mean)
will be less than the arithmetic mean (60kmph) ...
Brilliant . . . but it'll take him an extra hour to explain all that.

Here's a baby-talk approach that should make it clear . . .

Suppose it is 448 miles from A to B.

Paul drove from A to B: 448 km at 56 kph.
. . His time: $\frac{448}{56} = 8$ hours.
Then he drove from B to A: 488 km at 64 kph.
. . His time: $\frac{448}{64} = 7$ hours.
Hence, Paul drove for a total of: $8 + 7 \,= \,15 \text{ hours.}$

John drove a total of $2 \times 448\,=\,896$ km at 60 kph.
. . His time: $\frac{896}{60}\,=\,14.9333...\text{ hours} \,=\,14\text{ hours, }56\text{ minutes.}$

Regardless of the distance from A to B, John will return sooner.