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A series of online riddles

 Puzzles based on GK, Logic and Technical Knowledge.
 First one to solve them all wins.
 Cash prize will be awarded to the winner
 Puzzle kicks off on Friday 4th August at 9:00 a.m.
Will continue till 19th August.
 Register for free at Technovanza website ( www.technovanza.org )
 Based on the lines of Notpron and Klueless.
http://wpc.puzzles.com/links/index.htm
Amit there are lots of such sits.
"pqrst" is a popular puzzle site which runs competitions regularly.
its available at
http://www.otuzoyun.com/pqrst/
quarterly competation...
this one is like pqrst...

http://diogen.h1.ru/english/ipst.html
Puzzle Site
This one is one of my favourites:

http://perplexus.info/
http://www.geocities.com/hnene/puzzles/puzzles.html
http://www.brainvista.com
Word Puzzle Game....
This is one of the coolest games, (I am quiet addicted to this game! LOL)
There is a riddle and u have to solve it to save a KANGAROO,
u wil only be given 4 chances, if u r wrong, then the KANGAROO will be HANGED (HANGAROO....)
Try and tell how far u have gone (I mean the no. of levels)
Dont forget to bookmark, cuz thats the coolest game....

Here is the link, I'll try to load another link too if possible....
http://puzzl.50webs.com/hangaroo.htm
hey guys just give this test .link is
http://web.tickle.com/tests/braintwister/authorize/signin.jsp?url=/tests/braintwister/index.jsp

its a good one
Wackiest picture puzzle
This is the wackiest picture puzzle I have ever seen,
there are 12 members in the pic, and after a swap there are 13!!!
From where the 1 came????? After another swap again its 12....
Find out if u can....

http://tutor2.brinkster.net/puzzle/missing.htm

THE BEST IQ Test U CAN GET......
try this IQ test..!
its gr88

http://www.highiqsociety.org/iq_tests/
http://www.brainbashers.com/

puzzle-loop.com/" target="_blank">http://www.puzzle-loop.com/

http://www.saidwhat.co.uk/games/kakuro/standardkakuro3.php

puzzle-bridges.com/" target="_blank">http://www.puzzle-bridges.com/

puzzle-sudoku.net/" target="_blank">http://www.puzzle-sudoku.net/

http://neutralriddle.50webs.com/

Download soduku software from below link

http://www.ppclink.com/index.php?ppclink=product&do=view&id=46

Maths puzzles

http://www.mathpuzzle.com/

C quiz

http://c-faq.com/questions.html

http://www.cprogramming.com/

http://analysingc.50webs.com/

check this game out

http://www.planetdan.net/pics/misc/georgie.htm

http://www.toilette-humor.com/maze-child.html
PUZZLE
http://discuss.fogcreek.com/techInterview/default.asp
Puzzle
Puzzle...

here's the link

http://www.distortedriddle.com/

http://asptutors.brinkster.net/puzzles/couples.htm

Math puzzles
http://www.mathpuzzle.com/
http://www.maa.org/news/mathgames.html

Mar 7

prashanth

Three Switches


Problem Statement:

You are standing in a hallway next to three light switches, all of which are off. Each switch operates a different incandescet light-bulb in the room at the end of the hall. You cannot see the lights from where the switches are. Determine which light corresponds to each switch. You may go into the room with the lights only once.

Solution:

1) Turn on any one switches. Let them be 1st switch.

2) After some time turn off the 1st switch and turn on 2nd switch.

Now go to the room

- If the bulb glowing then right switch whhich is current on.
- If the bulb not currently glowing but itt is hot(because it was on for some time before turning it off), then the right switch is first one.
- Else right switch is third one.

2 eggs problem



Prolem Statement:

* You are given 2 eggs.
* You have access to a 100-storey building.
* Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
* You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
* Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process



If you are one of the people who likes to solve a puzzle before seeing the answer you must quit the blog now and come back later for checking the answer.


Now that this is a Google interview question I am taking the normal "Interview-Style" of solving a problem. Simply saying thinking aloud through the solution from worst to the best correcting the flows optimizing the solution or taking the 5-minute hard thinking acting pause to a problem, which you know already and just want to make your interviewer think that you are a challenge lover.


Solution

Drop the first egg from 50.If it breaks you can try the same approach for a 50-storey building (1 to 49) and try it from 25th floor. If it did not break try at 75th floor. And use linear search with the remaining portion of storey we need to test. For example if the first egg breaks at 50 we need to try all possibilities from 1 to 49.

Now this looks a feasible solution. In computer student's jargon do a binary search with first egg and linear search with the second one. Best case is log (100) and worst is 50.


Now the optimal solution for the problem is that you figure out that you will eventually end up with a linear search because you have no way of deciding the highest floor with only one egg (If you broke one egg and you have to find the answer among 10 all you can do is start from the lowest to the highest and the worst is the total number of floors). So the whole question grinds up to how to make use of the first egg to reduce the linear testing of the egg.


(For strict computer science students, well this problem can be solved using binary search on the number of drops needed to find the highest floor.)

Now let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

So we could write it as

(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100).

Coins on the table


Problem Statement:

There is a table on which a number of coins are placed. You also know that there are as many coins with Head up as many coins with Tail up. Now you have to divide the coins (number of coins is even) into two equal piles such that number of coins with Heads up and Tails up in either piles be the same. The catch is you are blind folded and you cannot determine the sides (for sure) if you are blinded.

Solution:

Divide the coins in half by quantity (easy to count coins while blindfolded)

Then, flip all the coins in one pile over.

Bridge Crossing

 Problem Statement:

A party of four travelers come to a ricety bridge at night. The bridge can hold the weight of at most two of the travelers at a time, and it cannot be crossed without using a flashlight. The tavelers have one flashlight among them. Each traveler walks at a different speed: The first can cross the bridge in 1 minute, the second in 2 minutes, the third in 5 minutes, and the fourth taes 10 minutes to cross the bridge . If two travelers cross together, they walk at the speed of slower traveler.

What is the least amount of time in which all the travelers can cross from one side of the bridge to the other.

Solution:

To minimize the total time, we need to send those traveler together who takes more time.

- first(1) and second(2) traveler go acrooss     (2 minutes)
- first traveler(1) goes back             (3 minuttes)
- thord(5) and fourth(10) traveler cross      (13 minutes)
- Second Traveler (2) goes back         (15 minnutes)
- First(1)  and second(2) traveler cross      (17 minutes)

Ans: 17 minutes

Camel and Bananas


Prblem Statement:

A camel must travel 1000 miles across a desert to the nearest city. She has 3000 bananas but can only carry 1000 at a time. For every mile she walks, she needs to eat a banana. What is the maximum number of bananas she can transport to the city?


Solution Description:

Let's begin by imagining how the camel might get the bananas across the desert. If she starts off with 1000 bananas and carries them 1000 miles, she won't be able to return for the rest of them because she won't have any bananas to fuel her trip back. She won't have any bananas to give to her friends either, because she will have eaten them all during the journey. This suggests that the camel needs to cache piles of bananas at certain points in the desert so she can actually move some of them instead of using them all for fuel.

Suppose the camel does this:

 1. Pick up 1000 bananas.
 2. Carry them one mile into the desert, eating one on the way.
 3. Keep one of the bananas for the trip back.
 4. Repeat steps (1-3).
 5. Repeat steps (1-2).

At the end of this, there will be 2995 bananas, which are 999 miles from the city. The camel will also be 999 miles from the city.

So in effect, she now has a smaller version of the same problem. Of course, it cost her 5 bananas to move the rest a mile. If she keeps this up, she won't have any bananas left when she gets to the city.

However, at this point, she can pick up 1000 bananas, eat 999 on the way to the city, and have 1 left over when she gets there. It's not nothing, but it's hard not to think that she could do better. And she still won't be able to go back for the 1995 bananas that she left behind.

Can we generalize the strategy? Suppose she does this:

 1. Pick up 1000 bananas.
 2. Carry them N miles into the desert, eating N on the way.
 3. Keep N of the bananas for the trip back.
 4. Repeat steps (1-3).
 5. Repeat steps (1-2).

Again, she ends up with a smaller version of the problem. Now she has (3000-5N) bananas, and is (1000-N) miles from the city.

At this point, we might ask a few questions:

 1. If she simply repeats this strategy, what value of N would allow her to get the most uneaten bananas to the city?

 2. Would she have to repeat the strategy exactly? For example, at some point, she might get down to 2000 bananas. Then it would only cost her 3N bananas (rather than 5N) to move the rest N miles.

 3. Should she always pick up 1000 bananas when she wants to move some of them farther into the desert? Could there be an advantage to making more trips with smaller loads? Could there be an advantage to leaving some of the bananas at the starting point, and ignoring them completely?

Solution:

Camel will need 5 bananas to transfer all the bananas 1 mile away...
5 as for 1st 1000 he'll go n come back then for another 1000 he'll go n come back but for last 1000 he'll only go as nothing will be left behind...

For second mile too it'll take 5 bananas...

Now it'll require 5 bananas per mile for total banana count > 2000
and for banana count < 2000 it'll eat 3 bananas...

and for < 1000 it'll eat 1 banana....

so 1st 1000 bananas will be finished in 1000/5 = 200 Miles
next 1000 in 1000/3 = 334 Miles...

Finaly he'll have 1000 bananas and 466 miles to go...

At B camel will have 534 bananas

25 Horses Race


Problem Statement:

There are 25 horses and in what min no of races can u find top three of them
The constraints are that in one race u can race 5 of the horses and u dont have a timer...

Solution:

divide 25 horses in 5 groups for 5 races
Now take toppers of all the races for another race....  6th Race

Now in this race the 4th and 5th are out because we want only top 3....

2nd and 3rd of the group of third horse in 6th race are also out...as they will have position lower than 3 in 25 horses togethar...

3rd horse of the group of second horse in 6th race is also out for same reason
take tooper of 6th race out as this will be topper of all 25 horses

Now we have 5 horses and we want top 2 of them.... 1 more race

Ans: 7 races

50 black and 50 white ball probability problem


Problem Statement:

There are two empty urns in a room. You have 50 white balls and 50 black balls. After you place the balls in the urns, a random ball will be picked from a random urn. Distribute the balls (all of them) into the urns to maximize the chance of picking a white ball.

Solution:

Place one white in one urn, and all the others in the other urn. This gives a 74/99 probability of picking white.

explanation:

probability of picking white ball from which only one white ball is there = 1
probability of picking white ball from which 49 white and 50 black ball is there = 49/99

probability of picking white ball = (1 + 49/99)/2

13 balls problem


Problem Statement:

The general problem is you will be given n balls and one of them is either heavier or lighter and you are asked to find out the minimum number of weighings using a common balance required to find out the odd one out.

Example 1: 8 Balls,Odd ball being heavier

This is the simplest question among the lot and is often used during phone screenings.The question is to find out the minimum number of weighings required to spot the odd heavier ball among 8 identically looking balls using a common balance.Answer is 2 and the solution is given in the next paragraph.

For simplicity let us name the balls as A1,A2,A3,B1,B2,B3,C1 and C2.We weigh with A's on one side and B's on the other side of the balance.If both sides are equal this means the heavier odd ball is among C1 and C2.We can figure out the heavier ball with one more weighing.If A's are heavier than B's then the heavier odd ball is among A's.Now we balance A1 and A2.If both are same then A3 is the odd ball or else the heavier one of A1 and A2 .Thus, we can find out the heavier ball using only 2 weighings.



Example 2: 8 Balls,Odd ball can be either heavier or lighter than the rest.

This is a little trickier to solve.Let us name the 8 balls as A1,A2,A3,B1,B2,B3,C1 and C2.Now weigh with A1,A2,A3 on one side and B1,B2,B3 on the other side.If both weigh equal then the odd ball is among C1 and C2.Now we know that A's and B's are all normal ones we can weigh C1 with A1 and check whether C1 weighs the same as the normal ball.In this way, we can figure out in one weigh which of C1 and C2 is odd.Now if A's weighed more than B's then we know for sure C's are normal ones.Now lets assume A's heavier than B's and we still don't know whether the odd is among A's or B's.

We know
A1 A2 A3 <>

prashanth

Coins on the table	6	itsme	240	Mon May 28, 2007 5:32 pm sam
	sutable seat	2	rahul	144	Fri May 11, 2007 5:18 pm Guest
	nterview puzzles used at Google	7	guest	299	Fri Apr 13, 2007 6:55 am rekha
	for u guddu	1	sairam	101	Wed Apr 11, 2007 7:47 am guddu
	whats that	1	sairam	66	Wed Apr 11, 2007 7:45 am Guest
	10 gud puzzel	1	guddu	141	Tue Apr 10, 2007 8:56 am guest
	water gallon	1	guddu	68	Tue Apr 10, 2007 8:55 am guest
	choose socks	1	guddu	56	Tue Apr 10, 2007 8:53 am guest
	what is the initial amount held with the merchant	2	sairam	61	Tue Apr 10, 2007 6:40 am guddu
	Camel and Bananas	3	itsme	187	Tue Mar 27, 2007 12:35 pm rakesh
	WEDDING	3	Ajit	118	Tue Mar 27, 2007 8:05 am Aksah
	13 balls problem	1	itsme	169	Tue Mar 27, 2007 6:35 am pramod_jnumca@yahoo.com
	25 Horses Race	2	itsme	239	Tue Mar 27, 2007 6:31 am pramod_jnumca@yahoo.com
	2 egg puzzle

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