Arithmetic & Algebraic Problems - WESTMINSTER CLOCK

Question: A man crossed over Westminster Bridge one morning between eight and nine o'clock by the tower clock (often mistakenly called Big Ben, which is the name of the large bell only, but this by the way). On his return between four and five o'clock he noticed that the hands were exactly reversed. What were the exact times that he made the two crossings?

Answer: The times were 8 hours 23 71/143 minutes, and 4 hours 41 137/143 minutes. We are always allowed to assume that these fractional times can be indicated in clock puzzles.

Arithmetic & Algebraic Problems - AT RIGHT ANGLES

Question: Rackbrane asked his young friends at the breakfast table one morning this little question:

"How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?"

Answer: To be at right angles the minute hand must always be exactly fifteen minutes either behind or ahead of the hour hand. Each case would happen eleven times in the twelve hours-i.e., every 1 hour 5 5/11 minutes. Starting from nine o'clock, the eighth addition will give the case 5 hours 43 7/11 minutes. In the other case, starting from three o'clock, the second addition gives 5 hours 10 10/11 minutes.

These are the two cases between five and six, and the latter will, of course, be the sooner.

Arithmetic & Algebraic Problems - EQUAL DISTANCES

Question: A few mornings ago the following clock puzzle was sprung on his pupils by Professor Rackbrane. At what time between three and four o'clock is the minute hand the same distance from VIII as the hour hand is from XII?

Answer: At 23 1/13 minutes past three o'clock.

Arithmetic & Algebraic Problems - MISTAKING THE HANDS

Question: "Between two and three o'clock yesterday," said Colonel Crackham, "I looked at the clock and mistook the minute hand for the hour hand, and consequently the time appeared to be fifty-five minutes earlier than it actually was. What was the correct time?"

Answer: The time must have been 5 5/11 minutes past two o'clock.

Arithmetic & Algebraic Problems - WHEN DID THE DANCING BEGIN?

Question: "The guests at that ball the other night," said Dora at the breakfast table, "thought that the clock had stopped, because the hands appeared in exactly the same position as when the dancing began. But it was found that they had really only changed places. As you know, the dancing commenced between ten and eleven oclock. What was the exact time of the start?"

Answer: The dancing must have begun at 59 83/143 minutes past ten, and the hands were noticed to have changed places at 54 l38/l43 minutes past eleven.

Arithmetic & Algebraic Problems - THE AMBIGUOUS CLOCK

Question: A man had a clock with an hour hand and minute hand of the same length and indistinguishable. If it was set going at noon, what would be the first time that it would be impossible, by reason of the similarity of the hands, to be sure of the correct time?

Readers will remember that with these clock puzzles there is the convention that we may assume it possible to indicate fractions of seconds. On this assumption an exact answer can be given.

Answer: The first time would be 5 5/143 minutes past twelve, which might also (the hands being similar) indicate 60/143 minutes past one o'clock.

Arithmetic & Algebraic Problems - WHAT IS THE TIME?

Question: At what time are the two hands of a clock so situated that, reckoning as minute points past XII, one is exactly the square of the distance of the other?

Answer: The time is 6 3/4 minutes past IX, when the hour hand is 45 9/16 (the square of 6 3/4) minutes past XII. If we allow fractions less than a minute point, there is also the solution, five seconds (one-twelfth of a minute) past XII o'clock.