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http://docs.google.com/Doc?id=dc83hzcs_272kfvps5

can you please give some input on the above problem ?

2007-11-02 05:37:35 · 2 answers · asked by calculus 1 in Science & Mathematics Mathematics

2 answers

Lets look at A and C
Ratio of speeds is 1:3 (ie in time t. C goes 3 times as far as A.

When C has done 1 lap, A will only have done 1/3 of a lap.
In the next half lap for C, A will have done an addional 1/6 of a lap (remember, C will ALWAYS go 3 times as far as A).

So at this point, A has gone 1/2 a lap, and C 3/2 laps and they're at the same point on the track..

Likewise when A has done 1 "lap" around the circle, C will have done 3 "laps" and they will meet at the starting point.

So A and C are at the same point in the track at point L and point L/2 (where L is the start / finish line).

A and E have a ratio of 1:5, so when E has done 1 lap, A has done 1/5 of a lap. When E does an additional 1/4 of a lap, A will have done 1/20 of a lap and again, they are at the same point.

So A and E will meet at L, L/4, 2L/4, 3L/4

Finally for A and G with ration of 1:7 they will meet at L, L/6, 2L/6, 3L/6, 4L/6 and 5L/6

So A meets C E and G at the following points:
L L/6 L/4 L/3 L/2 2L/3 3L/4 5L/6 (8 points)
(remember L/2 = 2L/4 = 3L/6 )

Lets now look at A and B
Speeds are 1:2 but in the opposite direction
When A has gone 1/3 of the way around the track, B will have gone 2/3 in the opposite direction.
So A and B will cross at L, L/3 and 2L/3

Likewise A and D have rations of 1:4 so they meet at L, L/5, 2L/5, 3L/5 and 4L/5 ...
and A and F with ratios of 1:6 will meet at L, L/7, 2L/7, 3L/7, 4L/7, 5L/7 and 6L/7

Here we have A meets with B D and G at
L, L/7, 2L/7, 3L/7, 4L/7, 5L/7, 6L/7, L/3, 2L/3, L/5, 2L/5, 3L/5, 4L/5 (13 points)

So all told A meets runners at L, L/7, 2L/7, 3L/7, 4L/7, 5L/7, 6L/7, L/3, 2L/3, L/5, 2L/5, 3L/5, 4L/5 L/6 L/4 L/2 3L/4 5L/6
or 18 points

2007-11-02 06:07:09 · answer #1 · answered by PeterT 5 · 0 0

C is three times as fast as A. When A has run a half lap, C has run 1.5 laps, overtaking A. When A has run a full lap, C has run 3 laps, again overtaking A.
So, C and A meet at L and L/2, where L is 0, 1, 2, ...

2007-11-02 12:57:54 · answer #2 · answered by DWRead 7 · 0 0

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