Answers to Puzzles Updated December 31, 1999

Can you do it in ONE trip?

Solution:

• 2 trips: turn on switches 1 & 2. Go to the lights room. There will be 2 bulbs on [say a & b]. You can now group 1 & 2 with a & b and group 3 & 4 with c & d. Go back to the switch room and turn switch 1 off, and turn switch 3 on. In the lights room, if a is still on, then a=2 and b=1, else a=1 and b=2. If c is on, then c=3 and d=4, else c=4 and d=3.
• 1 trip: turn on switches 1 & 2. Wait 30 minutes then turn off 1 and turn on 3. Hurry to the lights room. the hot bulb that is off is 1, the hot bulb that's on is 2, the not-so-hot bulb that's on is 3, and the cold bulb that's off is 4.

Solution:

• Each person will be labeled by his speed. Have 1 and 2 cross and have 1 come back alone [3 minutes]. Have 5 and 10 cross and have 2 come back alone [12 + 3 minutes = 15 minutes]. Have 1 and 2 cross [2 +15 minutes = 17 minutes].

Solution:

Group balls into 3 groups of 4 thusly: A ={1 2 3 4}, B ={5 6 7 8}, and C ={9 10 11 12}. Weigh group A versus group B.

• If A = B, then the rogue ball is in C. Weight D ={9 10} versus E = {11 3}.
  • If D = E, then the rogue ball is #12 and you can simply weigh #12 versus any of the "good" balls to see if it's light or heavy.
  • If D > E, then either #9 or #10 is a heavy ball, or #11 is light, or...
  • if D < E, then either #9 or #10 is a light ball or #11 is heavy. In either case, weigh #9 versus #10 to determine the rogue ball.
  
  
• If A > B [the solution for A < B is the same, almost, right?], then the rogue ball is either a heavy one in A or a light one in B. Weight F ={1h 2h 5L} versus G = {6L 3h 9}, where 6L indicates that ball #6 is potentially light.
  • If F = G, then the rogue ball is in this subset {4h 7L 8L}. Weigh #7L vs #8L. If #7 = #8, then #4 is heavy, else it's the lighter of #7 or #8 [because we've already determined that those are light]
  • If F < G, then either #5 is light, or #3 is heavy. Weigh #5 or #3 vs any good ball. If #5 is light, then it's #5, else #3 is heavy, and vice versa.
  • If F > G, then either #1 or #2 is heavy, or #6 is light. Weigh #1 vs #2. If #1 = #2, then #6 is light, else the heavier one is the heavy rogue ball.

Solution:

• Label the boxes 1 through 10. Take n balls from box #n. You should have 55 balls in total [1+2+3+...+10] and the 55 balls would weigh 55 oz, but for the light balls. The difference of the weight of your 55 balls and 55 oz will indicate number of light balls in the batch, ie, 54.9 oz would mean 1 light ball and thus box #1 is the light box.

Solution:

The logic behind not being able to is as follows. I cannot give the quiz on Friday because if I have not given the quiz by the end of Thursday, I have to give it on Friday. And it wouldn't be a surprise. Therefore, I cannot give the quiz on Friday. But then if Wednesday passes and I haven't given the quiz yet, then Thursday MUST be the day since it can't be Friday. And so on.

However, this is a topic for discussion, because obviously I can give you a surprise quiz. Even on a Friday. And it'd be a surprise, wouldn't it? [There are no right answer to this logic paradox]

Solution:

I realised that if my face were untainted, my students [seeing only one painted face] would have figured out that they were laughing at each other and stopped laughing. Therefore, my face must also be painted. Else, my students' logic skills were so poor that it made me sad.

Solution:

Another famous puzzle. Some think that staying with door #1 is the same as going to door #3 because each has an original chance of 33.3%. For the original pick, this is true. But they were going to reveal an empty door to you regardless of your original pick, so in repicking, you now have a 50% chance. In other words, your chances were one in three originally [win only if the prize is behind door #1]. If you repick, you win if the prize is behind door #2 or #3.

However, if I, the devil, were running the show, I'd only offer the re-pick if the contestant already initially picked the right door, so you shouldn't re-pick if I am the emcee. A little tip, only to visitors of my page!

Solution:

The most common answer is x=6.5, but the right answer is x=7.0. The experiment can be simplified by using just 13 cards of a suit. Now the ace can be any one of the 13 cards, right? So, the expected number of cards turned over will be the sum [1 + 2 +... +13] = 91 divided by 13 = 7.

• Example: 12 = M. in a Y. Answer: 12 = Months in a Year. Ready?



• 1 = H. on a U. ==> 1 Horn on a Unicorn
• 1 = W. on a U. ==> 1 Wheel on a Unicycle
• 2 = is C., 3's a C. ==> 2 is Company, 3's a Crowd
• 2 = # it T. to T. ==> 2 = # it Takes to Tango
• 3 = B. M. (S. H. T. R.) ==> 3 Blind Mice (See How They Run)
• 4 = S. and 7 Y. A. ==> 4 Scores and 7 Years Ago
• 4 = Q. in a G. ==> 4 Quarters in a Game
• 4 = H. of the A. ==> 4 Horsemen of the Apocalypse
• 5 = D. in a Z. C. ==> 5 Digits in a Zip Code
• 6 = D. of S. ==> 6 Degrees of Separation
• 7 = # of D. S. ==> 7 Deadly Sins
• 7 = B. for 7 B. ==> 7 Brides for 7 Brothers
• 7 = W. of the W. ==> 7 Wonders of the World
• 8 = S. on a S. S. ==> 8 Sides on a Stop Sign
• 9 = P. in the S. S. ==> 9 Planets in the Solar System
• 11 = P. on a F. T. ==> 11 Players on a Football Team
• 12 = I. in a F. ==> 12 Inches in a Foot
• 12 = S. of the Z. ==> 12 Signs of the Zodiac
• 12 = D. of C. ==> 12 Days of Christmas
• 13 = S. on the A. F. ==> 13 Stripes on the American Flag
• 13 = D. in a B. D. ==> 13 Doughtnuts in a Baker's Dozen
• 14 = D. in a F. ==> 14 Days in a Fortnight
• 18 = H. on a G. C. 18 Holes on a Golf Course
• 24 = H. in a D. ==> 24 Hours in a Day
• 26 = L. of the A. ==> 26 Letters ofthe Alphabet
• 29 = D. in F. in a L. Y. ==> 29 Days in February in a Leap Year
• 32 = D. F. at which W. F. ==> 32 Degrees Fahrenheit at which Water Freezes
• 40 = D. & N. of the G. F. ==> 40 Days & Nights of the Great Flood
• 50 = S. on the A. F. ==> 50 Stars on the American Flag
• 50 = W. to L. Y. L. ==> 50 Ways to Leave Your Lover
• 54 = C. in a D. (with the J.) ==> 54 Cards in a Deck (with the Joker)
• 57 = H. V. ==> 57 Heinz Varieties
• 64 = S. on a C. B. ==> 64 Squares on a Chess Board
• 76 = T. in the B. P. ==> 76 Trombones in the Big Parade
• 80 = D. to G. A. the W. ==> 80 Days to Go Around the World
• 88 = P. K. ==> 88 Piano Keys
• 90 = D. in a R. A. ==> 90 Degrees in a Right Angle
• 92 = the A. N. of U. ==> 92, the Atomic Number of Uranium
• 99 = B. of B. on a W. ==> 99 Bottles of Beer on the Wall
• 101 = D. ==> 101 Dalmations
• 200 = D. For P. G. in M. ==> 200 Dollars for Passing Go in Monopoly
• 1,000 = W. that a P. is W. ==> 1,000 Words that a Picture is Worth
• 1,000 = # of S. L. by the F. of H. of T. ==> # of Ships Launched by the Face of Helen of Troy
• 1,001 = A. N. ==> 1,001 Arabian Nights
• 20,000 = L. U. the S. ==> 20,000 Leagues Under the Sea

• a. A child is born in Boston, Massachusetts to parents who were both born in Boston, Massachusetts. The child is not a United States citizen. How is this possible?
• I am not a big fan of this one. You can always opt to give up your US citizenship, so this question is a non-event. In any case, a child is always human [a baby is not]. And a still-born American child is still American. So, the child needs to be born before 1776.


• b. Before Mount Everest was discovered, what was the highest mountain on Earth?
  • Mt Everest.
  
  
• c. Clara Clatter was born on December 27th, yet her birthday is always in the summer. How is this possible?
  • She lives in Australia.
  
  
• d. Captain Frank and some of the boys were exchanging old war stories. Art Bragg offered one about how his grandfather led a battalion against a German division during World War I. Through brilliant maneuvers, he defeated them and captured valuable territory. After the battle he was presented with a sword bearing the inscription: "To Captain Bragg for Bravery, Daring and Leadership. World War I. From the Men of Battalion 8." Captain Frank looked at Art and said, "You really don't expect anyone to believe that ridiculous yarn, do you?" What's wrong with the story?
  • World War I was not called World War I until WWII
  
  
• e. In what year did Christmas and New Year's fall in the same year?
  • Every year.
  
  
• f. A woman from New York married ten different men from that city, yet she did not break any laws. None of these men died and she never divorced. How was this possible?
  • She's a justice of the peace
  
  
• g. Why are 1990 American dollar bills worth more than 1989 American dollar bills?
  • $1990 [1,990 one dollar bills] is $1 more than $1989
  
  
• h. How many times can you subtract the number 5 from 25?
  • Once, after that, you're subtracting from 20, 15...
  
  
• i. A taxi driver was called to take a group of passengers to the train station. The station is normally an hour away, but with traffic being extra heavy, it took a full hour and a half. On the return trip the traffic was still as heavy and yet it took only 90 minutes. Why?
  • Same time each way, 90 minutes = 1.5 hours.
  
  
• j. How could you rearrange the letters in the words "new door" to make one word? Note: There is only one correct answer.
  • "one word"
  
  
• k. Even if they were starving, natives living in the Arctic would probably never eat a penguin's egg. Why not?
  • Never is too strong, because of global trade, but penguins are native to the Antarctic.
  
  
• l. Which is correct to say, "The yolk of the egg are white" or "The yolk of the egg is white"?
  • Neither, yolks are yellow.
  
  
• m. There were an electrician and a plumber waiting in line for admission to the "International Home Show". One of them was the father of the other's son. How could this be possible?
  • One of them was a female.
  
  
• n. After the new Canon Law that took effect on November 27, 1983, would a Roman Catholic man be allowed to marry his widow's sister?
  • For his widow to be a widow, he must be dead, so no.
  
  

Solution:

You need to move 1 mile, drop your cargo, back track, and reload. The key is to always be moving forward with the maximum load of 1,000 bananas, ie, picking up / eating from a dropped load so that what's on the camel's back is 1000 bananas. Doing this, for the first 1000 bananas, when you'll be transporting 3 batches or bunches of bananas, you'll need 5 bananas per mile [forward 3, back 2, etc...]. Then once you are down to 2000 bananas, you'll need 3 bananas per mile [forward 2, back 1, etc...]. You'll wind up then with 533 bananas on the other side of the desert.

Solution:

You will need 2 bearers. So your party of 3 start off with 12 rations altogether. At the end of the first day, you & your bearers each eat 1 ration and one bearer returns with 1 ration, handing off 1 ration a piece to you and the other bearer. So each of the two remaining member still has the maximum 4 rations. At the end of the second day, you two each eat a ration and he gives you one, and returns with 2. You now have 4 rations to last for 4 more days. Oh, you owe 6 days of wages or $3,000, hard currency please.

Solution:

At the first glance, one would expect that the probabilities are the same, but they are not. There are only 4 permutations for a two children household. B-B, B-G, G-B, & G-G. Assuming that the distribution of boys and girls are even, each of these permutations have an equal probability. Now Mr Smith has at least one boy, so that elliminates permutation G-G, so he has a 1/3 chance of having all boys [either B-B, B-G, or G-B]. Mr Jones's eldest child is a boy so he is either B-B, or B-G or a 1/2 chance.

Solution:

The key pick up here is that even though Joe arrives at the station at random times, the trains do not arrive at random times. In fact, the train heading to Brunettesville arrives at 10 minutes past the hour, while the train heading to Blondesville arrives on the hour. Therefore, if Joe gets to the station within the time window between 1.00 and 1.10, he'll see Joan, the brunette. But if he arrives at any other time, or the time window between 1.10 and 2.00, he'll see Jill, the blonde. The chances are 50 [minutes] to 10 [minutes] or 5-1 that he sees Jill.

Solution:

They are impure to the same degree. Think about it: if there is 1 teaspoon of water in the dye glass, and the ending volumes are the same, then there must be 1 teaspoon of dye in the water glass. If you can't see this, do the math with the glasses containing [for example] 10 teaspoons of liquid originally, and you'll see.

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