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                                     A SIDECAR PROBLEM

         Atkins, Baldwin, and Clarke had to go on a journey of fifty-two miles across country. Atkins had a motorcycle with a sidecar for one passenger. How was he to take one of his companions a certain distance, drop him on the road to walk the remainder of the way, and return to pick up the second friend, who, starting at the same time, was already walking on the road, so that they should all arrive at their destination at exactly the same time?

          The motorcycle could do twenty miles an hour Baldwin could walk five miles an hour, and Clarke could do twenty miles an hour. Of course, each went at his proper speed throughout and there was no waiting.

           I might have complicated the problem by giving more passengers, but I have purposely made it easy, and all the distances are an exact number of miles – without fractions.

                                          THE BATH CHAIR

           A correspondent informs us that a friend’s house at A, where he was invited to lunch at 1 P.M., is a mile from his own house at B. He is an invalid, and at 12 noon started in his Bath chair from B towards C. His friend, who had arranged to join him and help push back, left A at 12:15 P.M., walking at five miles per hour towards C. He joined him, and with his help they went back at four miles per hour, and arrived at A at exactly l P.M. How far did our correspondent go towards C?

                                       


   COIN AND HOLE

We have before us a specimen of every American coin from a penny to a dollar. And we have a small sheet of paper with a circular hole cut in it of exactly the size shown. (It was made by tracing around the rim of a penny.) What is the largest coin I can pass through that hole without tearing the paper?   

                                                 





THE COUNTER CROSS

                  Arrange twenty counters in the form of a cross, in the manner shown in the diagram. Now, in how many different ways can you point out four counters that will form a perfect square if considered alone? Thus the four counters composing each arm of the cross, and also the four in the center, four counters marked A, the four marked B, and so on.

                   How may you remove six counters so that not a single square can be so indicated from those that remain?

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