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        riddles >> easy >> Racecars
        
(Message started by: THUDandBLUNDER on Jul 4^th, 2005, 11:55am)

Title: Racecars
Post by THUDandBLUNDER on Jul 4^th, 2005, 11:55am

Three cars A,B,C race around a circular track at 290, 130, and 70 kph, respectively.

If one lap is 6.9 km, when all three cars are next abreast of each other
1) how many laps will each car have clocked up?
2) how much time will have elapsed?

Title: Re: Racecars
Post by JiNbOtAk on Jul 4^th, 2005, 9:02pm

Do they all start from the same point ? And at the same time ?

Title: Re: Racecars
Post by THUDandBLUNDER on Jul 4^th, 2005, 9:36pm

Yes. (Sorry, I forgot to mention that.)

Title: Re: Racecars
Post by Ajax on Jul 5^th, 2005, 12:23am

[hide]A generalized solution would be: (for A, B, C the speeds and L the track length)

We’d have to find such a time t for which
decimal(A/L*t)= decimal(B/L*t)= decimal(C/L*t) or
decimal(A/L*t-B/L*t)=decimal(A/L*t-C/L*t)=decimal(B/L*t-C/L*t)=0 or
decimal((A-B)/L*t)+decimal((A-C)/L*t)+ decimal((B-C)/L*t)=0 or
(A-B)/L*t=integer and (A-C)/L*t=integer

In the present case: (290-130)/6.9*t=integer and (290-70)/6.9*t=integer and (130-70)/6.9*t=integer
Or 160/6.9*t=integer and 220/6.9*t=integer and 60/6.9*t=integer
Obvious solution: t=6.9 hours
However, maximum common divider of 160, 220,60: 20
Better solution t=6.9/20= 0.345 hours
[/hide]

Title: Re: Racecars
Post by THUDandBLUNDER on Jul 5^th, 2005, 1:29am

on 07/05/05 at 00:23:06, Ajax wrote:

I'm changing it...

Changing what? ???

Title: Re: Racecars
Post by markr on Jul 5^th, 2005, 1:40am

[hide]1)
A: 14.5
B: 6.5
C: 3.5

2) 0.345 hours[/hide]

Title: Re: Racecars
Post by Ajax on Jul 5^th, 2005, 2:09am

on 07/05/05 at 01:29:14, THUDandBLUNDER wrote:

Changing what? ???

Sorry, I had made a mistake and as I was changing it, I had to do something else, by which time you and markr posted.

Title: Re: Racecars
Post by raprap on Jul 7^th, 2005, 12:18pm

;D
[hide]In 6,9 hrs Car A does 290 laps, Car B does 130 laps, and Car C does 70 laps.

In the same 6.9 hrs
Car A laps Car C 220 times
Car A laps Car B 160 times
Car B laps Car C 80 times

220=11*20
160=8*20
80=4*20

6.9hrs/20=0.345 hrs

so in 0.345 hrs
Car A laps Car C 11 times
Car A laps Car B 8 times
Car B laps Car C 4 times

in 0.345 hrs
Car A does 14.5 laps
Car B does 6.5 laps
Car C does 3.5 laps

so every 0.345 hrs Cars A,B, and C find themselves nose to nose

AND

every 0.690 hrs Cars A, B, and C find themselves nose to nose on the starting line.[/hide]

Rap