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Friday 14 September 2012

A Tricky IQ Question

Q. There was a sheriff in a town that caught three outlaws. He said he was going to give them all a chance to go free. All they had to do is figure out what color hat they were wearing. The sheriff had 5 hats, 3 black and 2 white. Each outlaw can see the color of the other outlaw’s hats, but cannot see his own. The first outlaw guessed and was wrong so he was put in jail. The second outlaw also guessed and was also put in jail. Finally the third blind outlaw guessed and he guessed correctly. How did he know?  

Solution :

• Let us look at it this way. Here are our possibilities:

• 1) BBB
• 2) BBW
• 3) BWB
• 4) WBB
• 5) WWB
• 6) BWW
• 7) WBW

• Now we can eliminate # 6 because in this case the first outlaw would be sure to know that he had on a black hat. # 7 can be eliminated for the same reason for the second outlaw’s guess.

• In # 2, the first outlaw has to see at least 1 black hat (if he saw two while hats he wouldn't have guessed wrong). From this we know that outlaw 2 or outlaw 3 has a black hat (possibly both). Now outlaw 2 has the same dilemma, but he knows that one or both of outlaw 2 and 3 has a black hat. He can see that outlaw 3 has a white hat so in that case he would guess black and be correct, but he didn't (since we know he guessed wrong). Given this, we can remove option 2 from consideration.

• Options 1,3,4,5 all have outlaw 3 wearing a black hat. Thus, assuming that convicts 1 and 2 are as logical as possible, the only options left all have outlaw 3 wearing a black hat.

Friday 7 September 2012

Classic Weighing IQ Questions : Simple, Medium and Hard


Weighing puzzles vary in hardness from simple to extremely mathematical involving either expertise in some fields or extreme ingenuity. Either you know it or you figure out it using extreme intelligence. So here is the chance for some people to burn your grey cells.
There are three puzzles with varying range of hardness. Sometimes you may feel the hardest one is very easy for you have already come across the theory, but I still made it the hardest for the people who will be solving it with out knowing the theory behind it, giving them an option to figure out a small part in evolution of computational history.

There is a shopkeeper who wants to weigh things who has a common balance. He must be in a position to weigh things of all possible integral weighing units from 1 to a given maximum sum. The question will be either about how many weights you will need or how will you weigh.

 Problem 1 : Simple - One Side Only

In this version of the problem shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1,3 and 4 only. Now the question is how many minimum weights and name the weights you will need to measure all weights from 1 to 1000.

 Problem 2 : Medium - Both Sides

This is same as the first problem with the condition of placing weights on only side of the common balance being removed. You can place weights on both side and you need to measure all weights between 1 and 1000.For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

 Problem 3: Hard - Incremental

This is an altogether different one in the same scenario. You have to make 125 packets of sugar with first one weighing 1 kg, second 2 kg, third 3 kg etc ...and 125th one weighing 125kg.You can only use one pan of the common balance for measurement for weighing sugar, the other pan had to be used for weights i.e. weights should be used for each weighing. It has come into notice that moving weights into and out of the pan of the balance takes time and this time depends on the number on the number of weights that are moved. For example - If we need to measure 4 kg using weights 1 and 3 only, it will take twice as much time needed to measure 1 kg. Lets say we want to make sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 kg, with 1 unit of time, we place 3 kg along with 1 kg and measure 4kg with again 1 unit of time, and finally we
move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4kg using weights 1 and 3 only.
Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.