Tuesday Puzzle

Good puzzle!

I like this kind. No math beyond middle school algebra required, but still quite difficult, IMHO. My initial reaction was "No way! That's two equations in three unknowns", which of course, it is. Took a while to convince myself that I had the answer, and now that I've done that work, the answer is all mine--I'm not going to share! [yet]

Hint

As Alex mentioned, it is two equations in three unknowns. However, look at what is being asked in the question, not the three unknowns.

Kind of a hint

Nothing in the problem requires that the stretch from A to B will contain all 3 types of road.

Re: Kind of a hint

Correct, and that is how I came to my answer, at which point I found out that the question COULD HAVE required that all three types of road were there.

True, but...

It also may contain all. There is another thing going on that helps make a unique solution possible.

someone post answer ;-)

Re: someone post answer ;-)

273 miles distributed as follows (in the A to B direction):

52.5 miles uphill

220.5 miles downhill

273 is right

But it is impossible to determine the portions that are uphill, level, and downhill.

You breakdown of 52.5 uphill and 220.5 downhill works, but is only one of many.

Another is 168 downhill and 105 level.

There are infinitely many possibilities, all adding to 273.

All take the form

x uphill, 0<=x<=52.5,

x+168 downhill, and

105-2x level.

;-) One method to solve.

For those that didn't work it out, but wonder how, here is one way.

Let the total distance traveled downhill, on the level, and uphill, on the outbound journey, be x, y, and z, respectively.

The time taken to travel a distance s at speed v is s/v.

Hence, for the outbound journey

x/72 + y/63 + z/56 = 4

While for the return journey, which we assume to be along the same roads

x/56 + y/63 + z/72 = 14/3

It may at first seem that we have too little information to solve the puzzle. After all, two equations in three unknowns do not have a unique solution. However, we are not asked for the values of x, y, and z, individually; but for the value of x + y + z.

Multiplying both equations by the least common multiple of denominators 56, 63, and 72, we obtain

7x + 8y + 9z = 4 � 7 � 8 � 9

9x + 8y + 7z = (14/3) � 7 � 8 � 9

Now it is clear that we should add the equations, yielding

16(x + y + z) = (26/3) � 7 � 8 � 9

Therefore x + y + z = 273; the distance between the two towns is 273 miles.

Remarks

A unique solution is possible because the speeds are chosen so that a round trip over a sloping section of road takes the same time as that over a flat section of the same length. Had we chosen to write down an equation for the round trip, the answer would have been immediately apparent.

Much more elegant

than my solution! I subtracted the two equations, which gave me x in terms of z. Then I went to meta-logic, figuring that if that is all the information I had, it must not matter what the individual pieces were, as long as none were negative. So I set y to zero and solved. Then set z to zero and solved, confirming my suspicion. Only then did I cotton to the point that the reciprocals of the three speeds were linear.