# Morgan Foster

Riddle Count: 268
Riddle: Not far off shore a ship stands with a rope ladder hanging over her side. The rope has 10 rungs. The distance between each rung is 12 inches. The lowest rung touches the water. The ocean is calm. Because of the incoming tide, the surface of the water rises 4 inches per hour. How soon will the water cover the third rung from the top rung of the rope ladder?
Answer: When a problem deals with a physical phenonmenon, the phenonmenon should be considered as well as the numbers given. As the water rises, so does the rope ladder. The water will never cover the rung.
Riddle: Two cyclists began a training run, one starting from Moscow and the other starting from Simferopol. When the riders were 180 miles apart, a fly took an interest. Starting on one cyclists shoulder, the fly flew ahead to meet the other cyclist. After reaching him the fly then turned around and yet back. The restless fly continued to shuttleback and fourth until the pair met; then settled on the nose of one rider. The flys speed was 30 mph. Each cyclist speed was 15 mph. How many miles did the fly travel?
Answer: The cyclists took 6 hours to meet. The fly traveled 6*30=180 miles.
Riddle: I can come in a can, I can come as a punch, I can come as a win, You can eat me for lunch. What am I?
Riddle: Using only brackets, parentheses, and these signs +,-, X, /. How can you express 100 with 5 1's and express 100 three ways with five 5's?
Answer: 111-11=100 (5 x 5 x 5)-(5 x 5)=100; (5+5+5+5)x 5=100;(5 x 5)(5-(5/5)=100.
Riddle: (A) Use two digits to make the smallest possible positive integer. (B) Five 3s can express 37: 37=33+3+3/3 Find another way to do it. (C) Use six identical digits to make 100. (Several solutions are possible.) (D) Use five 4s to make 55. (E) Use four 9s to make 20.
Answer: (A) 1 X 1; 1/1;2/2;ect....;1-0;2-1;and many others. (B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3 (C) 99 + 99/99; 55+55- 5- 5; (666-66)/6 (D) 44 + 44/4=55. (E) 9 + 99/9=20.
Riddle: The title of the problem tells you how to approach these four questions. (A). A bus leaves Moscow for Tula at noon. An hour later a cyclist leaves Tula for mosco, moving slower than the bus. When the bus and cyclist meet, which one of the two will be farther from Moscow? (B). Which is worth more: a pound of \$10 gold peices or half a pound of \$20 gold pieces? (C). At six o'clock the wall clock struck 6 times. Checking with my watch, I noticed the time between the first and last strokes was 30 seconds. How long will the clock take to strike 12 at midnight? (D). Three swallows fly outward from a point. When will they all be on the same plane in space? Now check the answers. Did you fall into any of the traps which lurk in these simple problems?
Answer: (A). Neither (B). A pound of metal is always more than half a pound of the same metal. (C). Six strokes took 30 seconds, therefore 12 strokes will take 60 seconds. But when the clock struck six, there were only 5 intervals between strokes, and each interval was 30/5=6 seconds. Between the first and twelfth strokes there will be 11 intervals of 6 seconds each, therefore 12 strokes will take 66 seconds. (D). There is always a plane that contains any 3 points
Riddle: As I traveled up and down our great glorious country, I found myself in a place where the tempature goes up sharply in the day and down at night. This had an effect on my watch, I noticed it was 1/2 a minute faster at nightfall, but at dawn it had lost 1/3 minute, making it only 1/6 minute fast. One morning- May 1- my watch showed the right time. By what date was it 5 minutes faster?
Answer: In 24 hours the watch gained 1/2-1/3= 1/6 minute. It would seem it would be 5 minutes fast in 5 X 6= 30 days; that is,the morning of May 31. But already on the morning of May 28 the was 27/6= 4 1/2 minutes fast. At the close of the day the watch gained 1/2 minute more, so it was 5 minutes fast on may 28.
Riddle: An alarm clock runs 4 minutes slow every hour. It was set right 3 1/2 hours ago. Now another clock which is correct shows noon. In how many minutes, to the nearest minutes, to the nearest minute, will the alarm clock show noon?
Answer: In 3 1/2 hours the alarm clock has become 14 minutes slow. At noon the alarm clock will fall behind approximently an additional minute. Its hands will show noon in 15 minutes.
Riddle: A watchmaker was telephoned urgently to make a house call to replace the broken hands on a clock. He was sik so he sent his apprentice. The apprentice was thorough. When he finished inspecting the clock it was dark. Assuming his work was done, he attached the new hands and set the clock by his pocket watch. It was sic o'clock, so he set the big hand at the 12 and the little hand at the 6. The apprectice returned, but soon the telephone rang. He picked up to his angry client: "You didn't do the job right. The clock shows the wrong time." Surprised he hurried back. He found the clock showing not much past eight. He handed is watch to the client and showed her that her clock was not even one second late. The client had to agree. Early the nect morning, the client telephoned to say the clock has apparently gone berserk, hands were moving around the clock at will. The apprentice again rushed over, the clock showed a little past seven. After checking his watch he yelled: "You are making fun of me! Your clock shows the right time!" Have you figured out whats going on?
Answer: As the problem says the apprentice mixed up the hands so that the minute hand was short and the hour hand was long. The first time the apprentice returned to the client was about 2 hours and 10 minutes after he had set the clock at six.The long had moved olny from twelve to a little past two. The little made two whole circles and an additional 10 minutes. Thus the clock showed the correct time. The next day around 7:o5 a.m.he came a second time,13 hours and 15 minutes after he had set the clock for six. The long had, acting as the hour hand,covered 13 hours to reach 1. The short hand made 13 full circles and 5 minutes, reaching 7, So the clock showed the correct time again.
Riddle: Can you divide the watch face with 2 straight lines so that the sums of the sums of the numbers in each part are equal?
Answer: The sum of the numbers on the watch face is 78. If the two lines cross, there must be 4 equal parts, but 78 is not divisible by 4. Then the lines do not cross, giving three parts with the sum of 26 each. Once you see the pairs on the face that add to 13(12+1,11+2, and so far) the answer will be easy to find.