Santos went on a bike ride of 90 miles. He realized that if he had gone 24 mph faster, he would have arrived 12 hours sooner. How fast did he actually ride?

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Explanation:

distance = rate*time

90 = r*t

where r is the slower speed and t is the time it takes when going that slower speed.

If r is bumped up 24 mph faster, to r+24, then Santos takes t-12 hours to get there. The second equation is

90 = (r+24)(t-12)

We can solve the first equation for r to get r = 90/t

Then plug this into the second equation and do a bit of algebra

90 = (r+24)(t-12)

90 = (90/t+24)*(t-12)

90 = 90 - 1080/t + 24t - 288

90t = 90t - 1080 + 24t^2 - 288t

0 = 90t - 1080 + 24t^2 - 288t-90t

0 = 24t^2 - 288t - 1080

24t^2 - 288t - 1080 = 0

If you apply the quadratic formula, then you should get the two solutions t = -3 and t = 15. Due to time constraints, I'll skip these steps.

We'll ignore the negative t value. It makes no sense to have a negative time.

So we focus on t = 15 as the only solution.

If t = 15, then,

r = 90/t

r = 90/15

r = 6

Santos rode his bike at 6 mph at first. Going this speed means he takes 15 hours.

If he rode 24 mph faster, at 6+24 = 30 mph, then he would ride for 15-12 = 3 hours instead. Note that 90/30 = 3.

answer: what in the world is this

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