Answers


Some Quick Problems:

  1.   Difference in Change

    A little experimenting should convince you that the answer is $1. What is the simplest way to prove that this must be so? In each of the two cases you end up with the same total amount of money. Since by paying with a dollar bill you end up with $1 less in paper money compared to the other case, this must be compensated for with a dollar more in change.
  2.   The Most and the Least

    Divide the coins into two pairs and weigh each pair. This takes two weighings. Now weigh together the heavier coins from the first two weighings. The heavier of these two is the heaviest coin of the group. Similarly, weigh the two lighter coins. The lighter of these is the lightest of the group.

    This problem is easily generalized to n coins. Divide the coins into n/2 pairs and weigh each pair. Group the n/2 heavier coins together and group the n/2 lighter coins together. It takes (n/2 -1) weighings to find the heaviest of the heavy coins, which will also be the heaviest coin of the group. Similarly, it takes (n/2 - 1) weighings to find the lightest coin. The total number of weighings is 3*(n/2) - 2.

  3.  Town Evacuation

    The easiest way to envision this problem is to think of the road as a conveyor belt with the cars motionless with respect to the road. The number of cars per hour that reach the end of the conveyor belt depends on the speed at which the belt is moving and by how densely the cars are packed together. Thus the equation for the number of cars per hour that can be evacuated is found by multiplying the speed of cars by the number of cars per mile.

    The rule in the driver's manual is that car separation should be one car length for every ten miles per hour of driving speed. Suppose that this rule were actually followed. At 30 mph there would be one car for every 4 car lengths - 3 lengths of separation plus the car's own length. At 60 mph there would be one car for every 7 car lengths. At 60 mph the speed would double but the density would be 4/7 so the comparative rate of evacuation would be 2 *%2