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Thursday, 11 April 2013

A Mind Boggling IQ Question

Q. Four prisoners escape from a prison. 
The prisoners, Mr. East, Mr. West, Mr. South, Mr. North head towards different directions after escaping.  The following information of their escape was supplied: 

• The escape routes were North Road, South Road, East Road and West Road
• None of the prisoners took the road which was their namesake
• Mr. East did not take the South Road
• Mr.West did not the South Road
• The West Road was not taken by Mr. East

What road did each of the prisoners take to make their escape ?

Answer :

Thursday, 14 February 2013

A Mind Boggling IQ Question

Q. Two friends, a programmer and a mathematician, get together for drinks after work one day at the programmer’s house.  The mathematician asks the programmer how his three children are doing.  The programmer replies that one of his three children just had a birthday.  The mathematician asks, "How old are your children now?"  The programmer replies, "The product of their ages is 36."  The mathematician thinks for a moment and says, "That’s not enough information."  The programmer says, "OK, the sum of their ages equals my house address."  The mathematician steps outside to check the address number,
comes back inside, and says, "That’s still not enough information."  The programmer then says, "Well my oldest child has red hair."  The mathematician immediately responded, "Oh, now I know the ages."  What are the ages of the programmer’s three children?

Solution :

The children are aged 2, 2, and 9 years old.  The mathematician’s logic was that since there are three children whose ages multiply to 36, the possible combinations are:

1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4

Initially, any of these could be the correct ages.  After the programmer says that the sum of the ages is the same as the house address, the mathematician mentally computed the sum of each possible combination:

1, 1, 36 -> 38
1, 2, 18 -> 21
1, 3, 12 -> 16
1, 4, 9  -> 14
1, 6, 6  -> 13
2, 2, 9  -> 13
2, 3, 6  -> 11
3, 3, 4  -> 10

Notice that all the sums are different except for (1, 6, 6) and (2, 2, 9).  If the programmer’s address was anything except for 13, then the mathematician would know the ages, so the three ages must be one of those two combinations.  But after the programmer said that the oldest child has red hair, the mathematician knew that there was a single oldest child which eliminates the (1, 6, 6) combination which has oldest twins.

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.

Monday, 20 August 2012

A Mind Boggling IQ Question

Q. You are given two 60 minute long fuse ropes (i.e. the kind that you would find on the end of a bomb) and a lighter. The fuses do not necessarily burn at a fixed rate. For example, given an 8 foot rope, it may take 5 minutes for the first 4 feet of the fuse to burn, while the last 4 feet could take 55 minutes to burn (a much slower rate) (5+55=60 minutes). Using these two fuses and the lighter, how can you determine 45 minutes?

HINT: You can use the lighter any number of times

 Solution :

Light the first fuse at both ends and the other fuse at one end simultaneously. When the first fuse is burned out, you know that exactly half an hour has passed. You also know that the second fuse still has exactly half an hour to go before it will be burned completely, but we won't wait for that. Now also light the other end of the second fuse. This means that the second fuse will be burned completely after another quarter of an hour, which adds up to exactly 45 minutes since we started lighting the first fuse.

Friday, 6 July 2012

A Mind Boggling IQ Question

Q. There are three wise men in a room: A, B and C. You decide to give them a challenge. Suspecting that the thing they care about most is money, you give them $100 and tell them they are to divide this money observing the following rule: they are to discuss offers  and counter-offers from each other and then take a vote. The majority vote wins. Sounds easy enough... now the question is, assuming each person is motivated to take the largest amount possible, what will the outcome be?

Solution :

• It is unlikely that one wise man would be voted out because counter-offers are allowed.
Consider: A and B decide to leave C out and split halfway. C offers B to leave A out instead, and as incentive offers B $60 as compared to $50 (Note: for C, $40 is better than nothing at all).

• Now A will try to raise that figure, but keep no less than $33 for himself. It would also happen that as B starts to get more and more of the share, A and C would decide to keep B out instead and split amongst each other. This would go on and on until they reach equilibrium with each agreeing to take their rightful share of 33 dollars.

Thursday, 14 June 2012

Some Mind Boggling IQ Questions

Q. You have 5 jars of pills. Each pill weighs 10 grams, except for contaminated pills contained in one jar, where each pill weighs 9 grams. Given a scale, how could you tell which jar had the contaminated pills in just one measurement?

Solution :

• Follow the steps outlined below:
• Step-1: Mark the jars with numbers 1, 2, 3, 4, and 5.
• Step-2: Take 1 pill from jar 1, take 2 pills from jar 2, take 3 pills from jar 3, take 4 pills from jar 4 and take 5 pills from jar 5.
• Step-3: Put all of the jars on the scale at once and take the measurement.
• Step-4: Now, subtract the measurement from 150 (1*10 + 2*10 + 3*10 + 4*10 + 5*10)
• Step-5: The result will give you the jar number which has the contaminated pills.

Q. One train leaves Los Angeles at 15 MPH heading for New York . Another train leaves from New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide, how far will the bird have traveled?  
Solution :
The distance traveled by the bird can be calculated as:

1. Let's say the distance between LA and NY is d miles.
2. The time before which the trains will collide: d / (15+20) hours.
3. The distance traveled by the bird in that time: (d / 35) * 25 = 5*d/7 miles.

Assumptions made: The bird must follow the line of the track, remain at the same altitude, and the speed must be relative to the ground and not air speed.