Atrain running between two towns arrives at its destination 10 minutes late when it goes 40 miles per hour and 16 minutes late when it goes 30 miles per hour. the distance between the two towns is

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

This is such a dog of a bear of a problem, I think I ought to check my

== 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

dopamine..

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d eras and periods