Mathematical & Logical Puzzles -A2

5.       The Professor On The Escalator

Let n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes the professor to walk down one step.  If he walks down the down-moving escalator in 50 steps, then n - 50 steps have gone out of sight in 50 units of time.  It takes him 125 steps to run up the same escalator taking five steps to every one before.  In this time 125 - n steps have disappeared in 125 ¸ 5 units of time.  Since the escalator can be assumed to run at a constant speed, we have the following linear equation that readily yields n = 100 steps:

                n - 50      125 - n

               =

        50          25

6.       The Encounter On The Bridge

If we call the length of the bridge 3d, the two vehicles will meet when the car is 2d over the bridge and the truck, d.  To reach this point, the car has required t seconds and the truck 2t seconds.  To back up the car would need 2t seconds for the distance 2d; the truck 4t seconds for the distance d.  Now let us consider the possiblities:

Case 1    The car backs up.  It requires 2t seconds, meanwhile the truck proceeds.  It needs 4t seconds to finish the journey across the bridge so the car has to wait 2t seconds until it can start across again.  It requires 1½t seconds to drive over the bridge.  Thus 5½t seconds are required for the operation.

Case 2    The truck backs up.  It requires 4t seconds.  Simultaneously the car proceeds though only with one-eighth its speed because of the truck’s snail pace.  They arrive at the end of the bridge together, then the truck drives over the bridge which takes 6t seconds.  Thus in this case 10t seconds are need to complete the operation.

Therefore, Case 1 is the solution most favourable to both vehicles.

7.       The Early Commuter

The commuter walked for 55 minutes before his wife picked him up.

Since they arrive home 10 minutes earlier than usual, this means that the wife had chopped 10 minutes off her usual travel time to and from the station.  It follows that she must have met her husband five minutes before his usual pick-up time of five o’clock or at 4:55 pm.  He started walking at 4pm and was therefore walking for 55 minutes.

8.       The Two Ferry Boats

When the two ferry boats meet for the first time, the combined distance travelled by the boats is equal to the width of the river.  When they reach the opposite shores, the combined distance travelled is twice the width of the river and when they meet for the second time, the combined distance is three times the width of the river.

Since both boats are moving at a constant speed for the same period of time, it follows that each boat has gone three times as far as when they first met (after they had travelled a combined distance of one river’s width).  Since the white boat, (slow boat), had travelled 720 metres before the first meeting, its total distance at the time of the second meeting must be 3 x 720 or 2160 metres.  This is 400 metres more than the width of the river, so we must subtract 400 from 2160 to obtain 1760 metres as the width of the river.

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