Logic Problems                                                      The Answers:
1.  A ship is at anchor.  Over the side of the ship is a 
rope ladder 6 feet high, with rungs 1 foot apart,  reaching down to the water level.  The tide is rolling in at the rate of 1 foot per hour.  In how many hours will the tide overflow the ship?
 Never.  The ship floats.
2.  A snail is at the bottom of a 30 foot well.  During the daytime, the snail crawls up 3 feet, but at night, it slides down 2 feet.  In how many days will the snail be able to reach the top of the well and escape?
 28 days.  On the 28th day, the snail reaches the top of the well, and crawls out, and thus, does not slip back down.
3.  Four men at at one end of a bridge that they must all cross in no more than 17 minutes.  The bridge can accommodate at most two men crossing at the same time.  It is night time, and to cross the bridge, they must use the one flashlight that they share.  They must walk the flashlight back and forth; they cannot throw the flashlight.  The men walk at different speeds - One can cross the bridge in one minute, Two in two minutes, Five in 5 minutes, and Ten in 10 minutes.  When two men cross together, they can only go at the speed of the slower man.

For example, if Two and Five cross together, and Two then returns with the flashlight, the round trip time is 5 + 2 = 7 minutes total; Five would then be at the other side, and Two would be back at the origin side since he has to return with the flashlight for the next trip.  Any man can cross any number of times, but the flashlight must be used on all crossings.   Is it possible and how can all four men get across in no more than 17 minutes total?

One possible solution is:

First crossing:  One+Two cross, One returns:  3 minutes
Second crossing: Five+Ten cross, Two returns:  12 mins
Final crossing:  One+Two cross, all on other side: 2 mins.

            Total time:  3+12+2 = 17 minutes total

A slightly different solution is: Two returns on First crossing, and One then returns on Second crossing.  So the total time there would be  4+11+2 = 17 minutes total.

If anyone has any other solution, please [ Contact Us ]

4.  There are three boxes of fruits, with one box containing only apples, the second box only oranges, and the third box containing both apples and oranges.  The boxes are labeled APPLES, ORANGES, and  APPLES/ORANGES.  You know that the label currently on each box has been labeled incorrectly.  Without being allowed to look inside the box, and if you are allowed to pick only one fruit from any box of your own choosing, by knowing the one fruit you pick is apple or orange, are you able to move the labels and correctly label the boxes as to the fruits that the boxes contain?
Pick one fruit from the APPLES/ORANGES box.  Assume it's an apple.  Then the box must contain at least apples, and since you know the label as it is now is incorrect, then the box cannot be apples and oranges; so the box must contain only apples, and thus should be labeled APPLES.  Since the box currently labeled ORANGES is also incorrectly labeled, then the correct labeled has to be APPLES/ORANGES.  This then leaves only the originally labeled APPLES box to be labeled correctly as ORANGES.
5.  What is the next letter in the sequence  OTTFFSS_ ? E is the next letter.  One, Two, Three, Four, Five, Six, Seven

Now, what is the next letter in the sequence..EOEREXN_?

Got the answer?  It's a  " T "

6.  What is unusual about the following sentence?


The sentence reads the same in either direction, forwards or backwards.
7.  You are in jail guarded by two jailers; one of who always tells the truth, and the other always lies.  You do not know which jailer tells the truth, or which jailer is the liar. There are two exit doors; one leads to freedom, and the other is a return to jail.  You do not know which door leads to freedom, and which one is a return to jail, but the jailers know which door is which.

You win a lottery and are allowed one chance to gain your freedom, which is, you may ask one question to either jailer of your choice, and to exit either door of your choice.  It's up to you whether or not you decide to use the answer the jailer gives you to the question that you pose to him.

What question would you ask to which jailer, and how would you select which door to exit?  The jailer must give you some answer.  There is a way to ensure that you exit the freedom door. 

You ask the following question to either jailer, doesn't matter which one:  "Which door would the other jailer say is the freedom door?"

Then, depending on which door this jailer (the one you directed your question to) says is the freedom door, you exit the other door.

Work out the possibilities.

8.  What is the chance that in a group of 50 people, that at least two people (out of the 50) would have the same birthday (month and date of the month)? The answer to this simple question is:  you should bet that yes, in a group of 50 people, there would be at least two people with the same birthday.  The chance of that happening is better than 97%.

How to calculate that probability of 97%.  The simplest way is to do it backwards, i.e. calculate the probability that there would be no match in a group of first 1 person, then 2 persons, then 3, then 4, and so on...going up to 50 people.

For just 1 person, the probability of no match is 
         365/365 days, or 1, since there is no other person to match with.

For a group of 2 persons, the probability of no match is 
         (365/365) x (364/365)
This is because the 2nd person has 364 days with which not to match the birthday of the first person.

You can continue doing this for the 3rd, 4th person, and so on, until you expand the group to 50 people, and your probability is then:
       (365/365) x (364/365)....((365-50+1)/365)

And if you use a calculator or computer to calculate this formula, you would find that the result is about 0.03, or 3%.

Therefore if the chance of a no match is 3%, then the chance that there is a match, two people with the same birthday, is (100 - 3) = 97%

So, the chance is very good that in a group of 50 people, two people would have the same birthday.  Of course, there is no guarantee of winning; there still is the 3% chance of a no match.

In fact, my friend Mr. Bergt computed that in a group of only 23 people, there is a 50% chance that two people would have the same birthday.

9.  Two rooms, three switches, three lights problem.  You may enter/exit each room just once, and you may turn on/off each switch just once, though you don't have to do either if you don't wish.  How can you determine which switch goes with which light? You enter the room with the switches, and turn on any two of the three switches, say switch A and B, leaving switch C turned off.  Wait a minute or so, then turn off either A or B, let's say B.  Then you go into the room with the lights, and the single light which is on belongs with switch A.  Place your hand on the two unlit lights, and the light which is warm to the touch is the light which belongs with switch B, since that light has been on for some time, and just now turned off.  The third light, which has not been turned on at all, and therefore should be cool to the touch, then goes with switch C.
10.  Two ropes of differing thickness and length.  Each burns in exactly one hour, though the burn rate is random and not even.  Given sufficient matches, how can you measure out exactly 45 minutes of time, using only the ropes and matches.  You do not have a clock or watch. You select one of the two ropes, and start to burn the rope from both ends of that rope.  At the same time, you start to burn the second rope from one end only.  When the first rope burns completely, 30 minutes will have elapsed.  At that time, you start to burn the second rope from the end which is not burning.  When the second rope burns completely, 45 minutes will have elapsed from the starting time.

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