Showing posts with label Puzzles. Show all posts
Showing posts with label Puzzles. Show all posts

Thursday, August 4, 2011

Which way is the bus below traveling?

To the left or to the right?



Can't make up your mind?

Look carefully at the picture again.

Still don't know?

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Primary school children were shown this picture and asked the same question.

90% of them gave this answer:

"The bus is travelling to the right.."

When asked, "Why do you think the bus is travelling to the right?"

They answered:

"Because you can't see the door to get on the bus."

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How do you feel now???

Wednesday, July 20, 2011

4 criminals are caught and are to be punished. The Judge allows them to be freed if they can solve a puzzle. If they do not, they will be hung. They agreed.



The 4 criminals are lined up on some steps (shown in picture). They are all facing in the same direction. A wall separates the fourth man from the other three.

So to summarize:

Man 1 can see men 2 and 3.
Man 2 can see man 3.
Man 3 can see none of the others.
Man 4 can see none of the others.

The criminals are wearing hats. They are told that there are two white hats and two black hats. The men initially don't know what color hat they are wearing. They are told to shout out the color of the hat that they are wearing as soon as they know for certain what color it is.

They are not allowed to turn round or move.
They are not allowed to talk to each other.
They are not allowed to take their hats off.

Who is the first person to shout out and why?

Answer:

If both Man 2, Man 3 have same hat color. Man 1 shouts opposite color to that of Man 2 and Man 3.

If Man 2, Man 3 have different hat color. Man 1 is quiet. So Man 2, shouts opposite color to that of Man 3!

Sunday, July 17, 2011

Find the hidden tiger in the below picture!

Monday, July 11, 2011

Car Puzzle

This is an interesting car puzzle. All you have to do is take the car out of this maze!

Wednesday, November 3, 2010

Try to guess the meaning of each picture, first one is "seven seas".



View comments section to find the solution. Do that only after you give a crack at this!

Monday, July 5, 2010

Atlast I am posting something in my blog. I am planning to post my travelogue on Malaysia-Singapore trip and Experience in Infosys after my training is over in Infosys (those are two REALLY BIG posts). Even this post is based on something that I learnt in Infosys training. It is no rocket science that we learn overnight. It was self-realization.

I was sitting in the Personal Effectiveness class(a class of 37) and it was an afternoon. I was feeling really drowsy and would have almost slept if not for this puzzle.



We were asked to connect each color/letter to its corresponding pair using straight lines and the lines are not supposed to intersect each other at any point. The lines should not go above the top A and below the bottom B. Mohan was sitting next to me and both of us were visualizing the solution. We thought it was not solvable(atleast for us).

Pravin, she was a woman with a man's name, was the facilitator. She asked, "How many of you in this hall think it's not solvable?". I knew there must be a solution but then I couldn't visualize it. So my hand went up in a jiffy. Mohan followed me and then few more hands were raised. But, since I was the one who raised the hand first, She called me on to the stage. She asked me to just join the pairs using straight lines without worrying about the intersection part. I drew. She pointed out the problem part and asked me to think.



In a flash I arrived at the solution.


Pravin told me, "I didn't give the solution, you solved it on your own. All that you need is belief that you can solve the problem..." and some blah... blah...

Personally I learnt few lessons from this puzzle.

Lessons learnt:
1. Sometimes you need to bend and adjust to reach the position you want to.
2. Where there is a will there is a way.
3. Better bend than break.

PS: *I would try to increase the frequency of posts.
*Conditions Apply :)

Sunday, December 28, 2008



There are 4 women who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each woman walks at a different speed. A pair must walk together at the rate of the slower woman's pace.
Woman 1: 1 minute to cross
Woman 2: 2 minutes to cross
Woman 3: 5 minutes to cross
Woman 4: 10 minutes to cross

For example if Woman 1 and Woman 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Woman 4 then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. What is the order required to get all women across in 17 minutes? Now, what's the other way?

Sunday, October 5, 2008



Following is a dozen of puzzles that I found very interesting. Try to solve them.

1. A man is condemned to death. He is given to say a last statement; if it is truth
he is hanged to death and if it is false he is poisoned. He uttered his last sentence from which he escaped. What is the last statement ?


2. In a 3 way junction there is a house in which there are 2 brothers. Elder one always tell the TRUTH and the younger one alway LIE. Here a man comes from one road and wanted to go for a particular city (say city A) and he don't know which way to take out of the two. He knows that in that house there are 2 brothers who always made TRUE and FALSE statements but he donot know who is who. At one time there is one one brother at home. The traveller asks only one question from one of the brothers who is there at that moment and chooses the correct way for his destination. What is that question and whats the road he chooses ?


3. A,B,C are aligned according to their height. A the shortest. C the tallest. A, B,C are faced left. They cannot see back. C can see A&Bs' head. B can see only A's head and A cannot see anybody's head. After C there is D who holds a gun. ie they stand like A B C D(with the gun)
D has 3 white hats and 2 black hats
ABC are blind fold and hats are put onto their head. After that,
Now C can see A&B's hat colors,
B can see A's hat color

Now respectively C,B,As' chance to predict the color of the hat that they are wearing (one at a time starting with C). If someone made a mistake he will be shot to death(by D). A,B,C are intelligent and try to escape from the death.
They all can hear previous guy's answer and if the previous guy is shot to death the sound of gun fire.
This is what happend.
First C uttered the answer, but shot to death.
Then B uttered the answer, but again shot to death.
Finnally A uttered the answer and it is the correct answer
So he escaped from the death.

What is the answer and explain
how A thought(actually you) that it is the color of
A's hat.



4. You are given two threads which takes 1hr to burn from one end to the other. Using these 2 threads and a matchbox(or lighter) how will you measure 45 minutes You are not allowed to break these threads.


5. A warden of a prison comes back to the prison from a party very drunk. Inhis state, he accidentally turns the key to open all 100 of the jail cells.He realizes what he's done, so he goes back and turns the key on every othercell (thus, now cell 1 is open, cell 2 is closed, cell three is open, etc.).He realizes that he hasn't closed all of them, so he goes back and turns thekey on every third cell (thus, now cell 3 is closed, and cell 6, which heclosed when he turned every other key, is now open). Still drunk, he goesback and turns the key on every 4th, then every 5th, 6th, 7th, etc. until heturns the key on every 100th cell (thus, only on cell #100). After all this,which cells remain open?

6. There are 4 people that need to cross a bridge. Only two can go across at a time and one has to come back with the flashlight each time. They take 1 minute, 2 minutes, 4 minutes and 5 minutes. They only walk as fast as the slowest one. They are able to all cross the bridge in 12 minutes. How?


Very good one

7. Pretend you have two strings which burn at random rates but which will alwaysbe completely burned in exactly one hour. Thus, after 1/2 hour the stringswill NOT necessarily be half burned length-wise, only time wise. How do you use these strings to time exactly 45 minutes without using any othertime-keeping device?
It is inaccurate to assume that each 4th of string A will burn in exactly 15 minutes. Three of the fourths could take one minute, and the other could takea full hour (thus, the entire string burns in one hour, as said).
The string, remember, burns at random rates in random places. If you were tocut it into four pieces and light each SEPARATELY so that one would beginburning when another stopped, the sum of the time would be one hour. That isall you can assume. For example, if you lit all of them together, three ofthe fourths could burn in exactly one minute, and the last fourth could take57 minutes. Thus the entire string burns in one hour. That is the onlyassumption you can make. You must look a little beyond just the string.


8. There are 3 settlers and 3 indians on one side of the creek. How can you get them all over to the other side by using a boat that holds two(they can only get there by boat). You can never have more indians than settlers on one side at a time because the indianswill kill the settler.


9. You ask a person to pick a number. Say they pick the number 20, 20 has six lettres in it, so therefore 20 is 6, then you see that 6 has three lettre's in it, so 6 is 3, 3 has five lettres so, 3 is 5, 5 has four lettre's so 5 is 4, and four is cosmic. this pattern works for any number, at all. so quickly, take the number 13 (thirteen), 13 is 8 (eight), 8 is 5 (five), 5 is four and four is cosmic.

10. Three men, members of a safari, are captured by cannibals in the jungle. The men are given one chance to escape with their lives. The men are lined up and bound to stakes such that one man can see the backs of the other two, the middle man can see the back of the front man, and the front man can't see anybody. The men are shown five hats, three of which are black and two of which are white. Then the men are blindfolded, and one of the five hats is placed on each man's head. The remaining two hats are hidden away. The blindfolds are removed. The men are told that if just one of the men can guess what hat he's wearing, they may all go free. Time passes. Finally, the front man, who can't see anyone, correctly guesses the color of his hat. What color was it, and how did he guess correctly?

11. Ten men stand in a line facing same direction.
Random Red or White hat is placed on their head(one hat per person).
One cannot see his own color(of the hat) but can see everybody else's in front of him.
The first man(Man1) cannot see anybody else's whereas the last man(Man10) can see 9hats.
They know ten men got 10 hats but we do not know how many Reds or Whites were there.
The game organizer starts asking men from the last in line(Man10, Man9,... Man2, Man1)
the color of individual's own hat.
When a man gives correct answer he is awarded a one point.
Individual's answers is heard by everybody else in the line. Their correctness exposed at the end.
What is the strategy the group should take to maximize their group score.
Hint : They can get 9 points.

12. X number of cards are placed on table A. Out of X, Y number of them are face down.
You have to move cards (one at a time) from table A to table B.
After the move both tables should have equal number of cards face down.
You are blind folded and sit in the middle of tableA and tableB.
(means you can touch the cards but you cannot see whether they are face up or down)
You are given the values of X and Y.
How do you accomplish this ?


ANSWERS

1."I should be poisoned" think a bit and analyse. Still if you don't get keep on reading.
Explanation : If "I should be poisoned" is TRUE(by sentense) you have to poison him. But according to the rule of the riddle "He should be hanged". Because of the contradiction no one can hang him or poison him. This solves by CONTRADICTION.

2. "If your brother is here instead of you what would he say for the question "What is the correct way to city A" ? " Then he chooses the other road whatever the brother say.

3.A cannot be black, because if A was black, C or B
would live.
If C sees A is black:
if C sees B is also black
then he knows he is white, and lives.
Cannot happen.
else C sees B is white
B knows this because if he were black
and A were black, C would live.
No matter how C answers, B has to be white, if A
is black, or C would live.

Therefore A cannot be black, and has to be white for
both C and B to die.

4.Burn both ends of one thread and one end of the other. After the first thread is burnt completely burn the other end of the second thread. The total time taken by threads to burn will be 45 minutes.


5. All the cells that are perfect squares (1,4,9,16,25,36,49,64,81,100)remain open because they have an odd number of multiples so the key will be turned an odd number of times on them, making them open at the end. Allothers have an even number of multiples and will stay closed.

6. 1&2 : 2mins, 1back : 1min, 4&5 : 5mins, 2back : 2mins, 1&2 : 2mins = 12mins

7. The first thing you must understand is that the time which it takes the strings to burn is proportional to the number of flames present on the string. If you light both ends of one of the strings, it will take HALF of an hour to burn, by definition. Thus, to time 45 minutes, here's what you'd do. Light string A at both ends while you simultaneously light string Bon one end. When string A is completely burned up, 30 minutes will have passed. At the moment that this happens, you will have 30 minutes left on string B. Right as string A burns out, light the OTHER end of string B, which will now only burn for 15 more minutes because you have two flames on a30-minute string. When it's gone, 45 minutes total have passed.

8. Any number could be ended in four which always has four letters only.

9.This is similar to the priest devil game that I had preciously posted.
Check the first comment of that game or play the game to find the solution.
http://24x7interestingfacts.blogspot.com/2008/09/priests-n-devils.html

10. If black is 1 and white is 0
All possible combinations :
a. 000
b. 001
c. 010
d. 011
e. 100
f. 101
g. 110
h. 111

a. does not happen ; if e fist buy can answer
if c or g, second guy could have answered
In all other cases 3rd guy wears black.

11. If one sees odd number of (say) Reds he calls RED. Then the one in front

12. Pick a card from tableA , flip the side and place on tableB
You do this until you reach Y.

Hope you enjoyed the puzzles!

Monday, September 1, 2008

I found a blog entirely dedicated towards puzzles. I have blogrolled that blog too!
It consists of a crtogram section which is very tough. I was able to solve only one puzzle there and I have posted it here!

The following cryptogram is a quote from a popular movie. It's a very short quote, so it makes the cryptogram a bit more difficult to decypher. It is possible to crack the code as long as you work at it.

81 12 81 415. 50626 39 41 527.

I didn't even work it out. I was able to figure it out.
Being an ardent lover of Starwars how could I forget the quote by Jedi Master Yoda!

Thursday, March 13, 2008



Answers:



Don' worry.....I didn't pass either......

Tuesday, December 11, 2007

Amazing Mind Reader Puzzle

Follow the Five Simple Steps Below to See The Mind Reader Puzzle Work For You!

Step 1: Pick any two-digit number.

Step 2: Add the two digits of your number together. For example, if you picked 35 then add 3 + 5 to get 8.

Step 3: Take your original number, and subtract from it the number you got in step two. For this example, 35 - 8 would give you your ending number!

Step 4: Now, find your ending number below, and concentrate on the symbol next to your number. I'll pick up on your vibrations, and read your mind.

Step 5: Click inside the circle, and I will display your symbol in the circle!



That's the Mind Reader Puzzle! Go ahead and try it again!
Do you want to know how the mind reading works?
Find it yourself :-)

Wednesday, November 28, 2007

You may have seen this quiz before.
If not, test if you have a gift for evaluating people.
This a nice quiz,worth a try....



Hope you liked it...
My score was 7/10....
Post idea from:http://shonzilla.tumblr.com/

Monday, November 26, 2007

While I was surfing through a forum about GRE exams I saw some wonderful puzzles.
From those puzzles,I have posted some here....Solving them involves lateral thinking...




1.There are three houses, and three utilities: gas, electricity and water. Your task is to connect each house to all three utilities. Therefore each house will have three lines and each utility will also have three lines. However, you cannot cross lines. You cannot pass lines through houses or utilities. You cannot share lines. Can you draw the 9 lines required?




2.Three people check into a hotel. They pay £30 to the manager and go to their room. The manager suddenly remembers that the room rate is £25 and gives £5 to the bellboy to return to the people. On the way to the room the bellboy reasons that £5 would be difficult to share among three people so he pockets £2 and gives £1 to each person. Now each person paid £10 and got back £1. So they paid £9 each, totalling £27. The bellboy has £2, totalling £29. Where is the missing £1?


3.Once upon a time, and old lady went to sell her vast quantity of eggs at the local market.
When asked how many she had, she replied:


Son, I can't count past 100 but I know that.
If you divide the number of eggs by 2 there will be one egg left.
If you divide the number of eggs by 3 there will be one egg left.
If you divide the number of eggs by 4 there will be one egg left.
If you divide the number of eggs by 5 there will be one egg left.
If you divide the number of eggs by 6 there will be one egg left.
If you divide the number of eggs by 7 there will be one egg left.
If you divide the number of eggs by 8 there will be one egg left.
If you divide the number of eggs by 9 there will be one egg left.
If you divide the number of eggs by 10 there will be one egg left.
Finally. If you divide the Number of eggs by 11 there will be NO EGGS left!

How many eggs did the old lady have?



Solutions:

Read the solutions only after making a genuine attempt in solving the above puzzles.

1.This puzzle is a classic one which has no solution in 2D. However, if you place the items on a doughnut shape in 3D you can solve it. In the picture below, E is linked to 3 by going over the top and re-entering through the hole in the middle.



2.We have to be careful what we are adding together. Originally, they paid £30, they each received back £1, thus they now have only paid £27. Of this £27, £25 went to the manager for the room and £2 went to the bellboy.(This puzzle was used in a tamil film Rassaiyah for a comedy scene involving vadivel!!!)

3.25,201 eggs.

This puzzle has a few different methods for finding the solution, one of which is:

Find a number X into which all of the numbers from 2 to 10 divide evenly. You can do this by simply using 2*3*4*5*6*7*8*9*10, but you can find a smaller number by finding the prime factors, a subset of which can be used to form any number from 2 to 10. 2*2*2*3*3*5*7 will do. This comes out to be 2520, and is the lowest number into which all the numbers 2-10 divide evenly.

We can add 1 to this number to satisfy the first 9 constraints of the puzzle (the remainder of 2521/2, 2521/3 ... 2521/10 is one), but this does not satisfy the last constraint, divisibility by 11.

Fortunately, we can multiply X (=2520) by any integer and add 1 and we will still satisfy constraints 1-9. So what Y do we multiply X by so that (X*Y) + 1 is divisible by 11. 2520/11 has a remainder of 1. Thus two 2520s divided by eleven would have a remainder of 1+1 = 2, and so forth...so ten 2520s divided by 11 would have a remainder of 10. This number plus one would divide eleven evenly, as well as also satisfy the first 9 constraints - thus 25201 is the answer.

Hope you liked these puzzles...

Puzzles were taken from:http://www.brainbashers.com/

Sunday, October 21, 2007

Right from the day I started this blog.
Which was on 19th of october.
(Just couple of days before.....)
I wanted to post on chess.
What to post on chess?
What to post on chess?
I was thinkin so hard that i forgot what i was thinking?
Lemme post other articles i decided.
And posted 12 other articles.
Nothing came to my mind for an article on chess.
So i didn't post anything....
Suddenly when I woke up this morning I remebered
the 8 queen puzzle.
As far as I remember,this was my first chess puzzle.
The first puzzle i solved......
This is quite famous through out the world.
Ah! i got a post....My brain...knew this puzzle for the past
6 years.....And still i dint remember it for 3 days.......
This is the 13th post of mine.
And this has turned out to be the most lucky post of mine....

THE EIGHT QUEEN PUZZLE

The eight queens puzzle is the problem of putting eight chess queens on an 8×8 chessboard such that none of them is able to capture any other using the standard chess queen's moves. The colour of the queens is meaningless in this puzzle. The queens must be placed in such a way that no two queens would be able to attack each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general n queens puzzle of placing n queens on an n×n chessboard(n ≥ 4).

Constructing a solution

There is a simple algorithm yielding a solution to the n queens puzzle for n = 1 or any n ≥ 4:

1.Divide n by 12. Remember the remainder (it's 8 for the
eight queens puzzle).

2.Write a list of the even numbers from 2 to n in order.

3.If the remainder is 3 or 9, move 2 to the end of the list.

4.Append the odd numbers from 1 to n in order, but, if the remainder is 8, switch pairs (i.e. 3, 1, 7, 5, 11, 9, …).

5.If the remainder is 2, switch the places of 1 and 3, then move 5 to the end of the list.

6.If the remainder is 3 or 9, move 1 and 3 to the end of the list.

7.Place the first-column queen in the row with the first number in the list, place the second-column queen in the row with the second number in the list, etc.

For n = 8 this results in the solution shown above. A few more examples follow.

  • 14 queens (remainder 2): 2, 4, 6, 8, 10, 12, 14, 3, 1, 7, 9, 11, 13, 5.

  • 15 queens (remainder 3): 4, 6, 8, 10, 12, 14, 2, 5, 7, 9, 11, 13, 15, 1, 3.

  • 20 queens (remainder 8): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 1, 7, 5, 11, 9, 15, 13, 19, 17.
Solutions to the eight queens puzzle

The eight queens puzzle has 92 distinct solutions. If solutions that differ only by symmetry operations (rotations and reflections) of the board are counted as one, the puzzle has 12 unique solutions, which are presented below: