A car is travelling at a uniform speed.
The driver sees a milestone showing a 2-digit number. After travelling for an hour the driver sees another milestone with the same digits in reverse order. After another hour the driver sees another milestone containing the same two digits as in the first one but the two digits separated by a zero(0).

What is the speed of the car?

Let 'ab' represent the number on the first milestone - ie the first digit is 'a' and the second digit is 'b'.
Then 'ba' represents the number on the second milestone.
And 'a0b' represents the number on the third milestone

Since the speed is uniform we know that a0b-ba=ba-ab

ba-ab<90, hence a0b-ba<90 hence a=1
(if a>1 then a0b-ba>100)

Above we had the equation a0b-ba=ba-ab. Since we now know that the numbers are 1b, b1 and 10b, this can be re-written as:

b1 + b1 = 10b + 1b

b1 = (10b + 1b) / 2

The min value of (10b + 1b)/2 is 56 (when b=1)
The max value of (10b + 1b)/2 is 64 (when b=9)

Therefore 56 <= b1 <= 64

So b1 = 61

Thus the three milestones showed 16, 61 and 106, which means that the car was travelling at 45 mph.

Posted by fwaff on 2003-10-03 08:42:30