by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
Owner:
eleven pairs in which the first time is a certain number of minutes after nought, or mid-day Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the ten pairs after 1 o'clock. Then there is another "jump," and you will be able to get by addition all the nine pairs after 2 o'clock. And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." In this way we get under the successive hours, 11 +10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 =66 pairs of times, which result agrees with the formula in the first paragraph of this article.
Some time ago the principal of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. At length some of his readers convinced him that the answer is, "At 5 5 A|43 min. past XII;" and this he finally gave as correct, together with the reason for it that at that time the time indicated is the same vihichever hand you may assume as hour hand!
62.?THE CLUB CLOCK.? solution
The positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. 51 1143 /-|427 sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. 52 496 /-|427 sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. 22 106 /-|427 sec. after eleven; but the
question applied to the hands, and the second hand would not be between the others at that time, but outside them.
63.?THE STOP-WATCH.? solution
The time indicated on the watch was 5 5 /^ min. past 9, when the second hand would be at 27 3 /^ sec. The next time the hands would be similar distances apart would be 54 6 /^ min. past 2, when the second hand would be at 32 8 /^ sec. But you need only hold the watch (or our previous illustration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.
64.?THE THREE CLOCKS.? solution
As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1,1898. What day of the month will that be?
I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a p 9 155 leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by four without a remainder is bissextile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered bissextile. Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. We therefore find that 720 days from noon of April 1,1898, brings us to noon of March 22,1900.
65.?THE RAILWAY STATION CLOCK.? solution
The time must have been 43 7 /^ min. past two o'clock.
66.?THE VILLAGE SIMPLETON.? solution
The day of the week on which the conversation took place was Sunday. For when the day after to-morrow (Tuesday) is "yesterday," "to-day" will be Wednesday; and when the day before yesterday (Friday) was "tomorrow," "to-day" was Thursday. There are two days between Thursday and Sunday, and between Sunday and Wednesday.
67.?AVERAGE SPEED.? solution
The average speed is twelve miles an hour, not twelve and a half, as most people will hastily declare. Take any distance you like, say sixty miles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and the average speed is clearly twelve miles an hour.
68.?THE TWO TRAINS.? solution
One train was running just twice as fast as the other.
69.?THE THREE VILLAGES.? solution
