Well, let's see. I'll call the distance between the two towns ' D '.
Time to cover the distance = (distance/speed) . I'll call the scheduled number of hours ' t ' .
16 minutes = 4/15 of an hour, and 10 minutes = 1/6 of an hour.
So . . .
(t + 1/6) = D/40. Multiply by 4: 4t + 4/6 = D/10
(t + 4/15) = D/30. Multiply by 3: 3t + 12/15 = D/10
Great. The left sides are both equal to D/10, so we can say
4t + 4/6 = 3t + 12/15
Subtract 3t from each side: t + 4/6 = 12/15
Subtract 4/6 from each side: t = 12/15 - 4/6
Simplify the right side: t = 4/5 - 2/3
Make a common denominator: t = 12/15 - 10/15
t = the scheduled time = 2/15 of an hour (8 minutes)
Looking back up above, we found that 3t + 12/15 = D/10
Plug in 2/15 in place of 't' : 3(2/15) + 12/15 = D/10
Simplify the left side: 18/15 = D/10
Multiply by 10: D = 180/15 . . . . D = 12 miles
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This is such a dog of a bear of a problem, I think I ought to check my answer:
== Somewhere up above, I calculated that the scheduled time is supposed to be 8 minutes.
== Time = (distance) / (speed)
== At 40 mph, time = (12 miles) / (40 mi/hr) = 12/40 hr = 18 minutes ... 10 minutes late !
-- At 30 mph, time = (12 mi) / (30 mi/hr) = 12/30 hr = 24 minutes ... 16 minutes late !
YAY !
QED !
Unbelievable
