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October 21, 2014

October 21, 2014

Posted by **monica** on Sunday, August 24, 2008 at 10:06am.

- math -
**Reiny**, Sunday, August 24, 2008 at 1:09pmlet the distance around the track be d km

so the time needed to qualify is 2d/100

his time for the first lap is d/75 hours

let his speed for the second lap be r km/h

so the time for his second lap is d/r hours

total time = d/75 + d/r = (dr+75d)/75

then (dr+75d)/75 = 2d/100

divide both sides by d

(r+75)/75 = 2/100

you finish it, let me know what you got

- math -
**monica**, Sunday, August 24, 2008 at 1:26pmi got 3675, but that doesn't make any sense, he has to go 3675 mph?

- math -
**drwls**, Sunday, August 24, 2008 at 5:18pmI believe Reiny meant to write

d/75 + d/r = (dr + 75 d)/(75 r)

=================================

d/75 + d/r = 2d/100

1/r = 2/100 - 1/75 = 6/300 - 4/300 = 1/150

r = 150

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