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3 bulbs and 3 switches

October 26, 2012 by asinghal 4 Comments

There are 3 bulbs(B1, B2 and B3) in a room on 1st floor, they can be operated using 3 switches(S1, S2, S3) present on the ground floor. You don’t know which Switch is for which bulb. How can you figure out, which switch is for which bulb if you are allowed to visit 1st floor only once.

Think before you give up, its really possible.

See Solution : 3 bulbs and 3 switchesHide Solution
Assuming all switches are turned off initially, You can do the following:-
1.) Turn on switch S1 for 5 minutes.
2.) Turn Off switch S1.
3.) Turn on switch S2.
4.) Go Upstairs.
5.) Glowing bulb can be assigned to Switch S2.
6.) Touch both other bulbs, hotter one should be assigned to switch S1 and the other one to switch S3.

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if following relationship holds then what is value of 9

September 2, 2012 by Ankur 3 Comments

if following relationship holds then what is value of 9
4=61
5=52
6=63
7=94
8=46
9=?

Tried enough already?

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Confusing one!

May 25, 2008 by Ankur 1 Comment

Q1. Which is the first question where c) is the correct answer

a) Q3
b) Q4
c) Q1
d) Q2

Q2. Which is the first question where a) is the correct answer

a) Q4
b) Q2
c) Q3
d) Q1

Q3. Which is the first question where d) is the correct answer

a) Q1
b) Q2
c) Q4
d) Q3

Q4. Which is the first question where b) is the correct answer

a) Q2
b) Q4
c) Q3
d) Q1

See Solution : Confusing one!Hide Solution

Answers:
d
c
a
b

🙂

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River crossing – the harder one

December 18, 2007 by Ankur Leave a Comment

A dysfunctional family has to cross the river. On one side of the river are a mom and 2 daughters, dad and 2 sons, the maid and the dog. There is a boat only big enough to hold 2 people (counting the dog as 1 person). Only the adults are capable of operating the boat. Everyone has to get to the other side, without anything bad happening.
Now the difficulties are : if the dog is left with anyone and the maid isn’t there to control him, he’ll bite. The dad can’t be left with any of the daughters when the mom isn’t there. Likewise, the mom can’t be trusted alone with either of the sons when the dad isn’t there.
Remember! only an adult can operate the boat, and the boat can’t drive itself.See Solution : River crossing - the harder oneHide Solution

Say
West shore is {W} and East shore is {E}
Lets give the nick names to each family member and dog 😉
Mother – {m}, Father – {F}, Daughters – {d1, d2}, Sons – {s1, s2}
House maid – {h}, Dog – {d}

Initially,
W = {m, d1, d2, f, s1, s2, h, d}
E = {…}

let’s move everyone, over…

housemaid and dog go east, and the housemaid comes back:

W = {m, d1, d2, f, s1, s2, h}
E = {d}

housemaid and s1 go east, h and d come back:

W = {m, d1, d2, f, s2, h, d}

E = {s1}

father and s2 go east, father comes back:

W = {m, d1, d2, f, h, d}

E = {s1, s2}

mother and father go east, mother comes back:

W = {m, d1, d2, h, d}

E = {f, s1, s2}

h and d go east, father comes back:

W = {m, d1, d2, f}

E = {s1, s2, h, d}

father and mother go east, mother comes back:

W = {m, d1, d2}

E = {f, s1, s2, h, d}

mother and d1 go east, housemaid and d come back:

W = {d2, h, d}

E = {m, d1, f, s1, s2}

h and d2 go east, h comes back

W = {h, d}

E = {m, d1, d2, f, s1, s2}

h and d go east

W = {}

E = {m, d1, d2, f, s1, s2, h, d}

And we are done.

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Defective coins

December 7, 2007 by Ankur 1 Comment

We have 10 bags of 1 Rupee coins. One bag contains all the defective coins, the weight of each coin in that bag is 1 gram lesser than the weight of a normal 1 Rupee coin. You have a spring balance, which tells the exact weight. After how many minimum no. of weighs you can separate the bag with defective coins.

Other variant of the same problem:
There are 10 machines in a factory. Each produces coins weighing 10 grams each. One day the factory owner cones to know that one of the machine is not functioning properly and produces coins of weight 9 grams. You have to find out the faulted machine. You ONLY have a weighing machine and you can use it only ONCE.

See Solution : Defective coinsHide Solution
The minimum no. of weighs required is one.

Want to try now ?

solution :

Let the weight of a normal 1 Rupee coin is W.
Now number the bags from 1 to 10 and take out the no. of coins equal to bag’s number from each bag, for example from 1st bag, take one coin, from 5th bag take 5 coins and so on.
Let P is the no. of the bag which contains the defective coins.
Now put all the coins on spring balance and weigh them. Read the spring balance reading, say it is S grams.
Then S = (W grams)*(1+2+3….+10) – P*(1 gram)
P will come out to be a number between 1 to 10 and thats the required bag.

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Hats and IIT students

December 7, 2007 by Ankur 5 Comments

Nine IITians were sitting in a classroom. Their prof wanted to test them. One day he told all of his 9 students that he has 9 hats, colored either in red or black. He also added that he has at least one hat with red color and the no. of black hats is greater than the no. of red hats. He keeps those hats on their heads and ask them tell me how many red and black hats I have? Obviously students can not talk to each other or no written communication, or looking into each other eyes; no such stupid options.
Prof goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

So what is the answer? and why?

See Solution : Hats and IIT studentsHide Solution

After first interval of 20 minutes :

Lets assume that their is 1 hat of red color and 8 hats of black color. The student with red hat on his head can see all 8 black hats, so he knows that he must be wearing a red hat.
Now we know that after first interval nobody was able to answer the prof that means our assumption is wrong.

So there can not be 1 red and 8 black hats.

After second interval of 10 minutes :

Assume that their are 2 hats of red color and 7 hats of black color. The students with red hat on their head can see all 7 black hats and 1 red hat, so they know that they must be wearing a red hat.
Now we know that after second interval nobody was able to answer the prof that means our assumption is again wrong.

So there can not be 2 red and 7 black hats.

After third interval of final 5 minutes :

Now assume that their is 3 hats of red color and 6 hats of black color. The students with red hat on their head can see all 6 black hats and 2 red hats, so they know that they must be wearing a red hat.
Now we know that this time everybody was able to answer the prof that means our assumption is right.

So there are 3 red hats and 6 black hats.

Now as everybody gave the answer so there can be a doubt that only those 3 students know about it how everybody came to know ?
Then here is what i think, the professor gave them FINAL 5 minutes to answer, so other guys will think that the professor expects the answer after 3rd interval (according to prof it must be solved after 3 intervals), so this is the clue for others.

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River crossing

December 6, 2007 by Ankur 1 Comment

Three cannibals and three anthropologists have to cross a river. The boat they have is only big enough for two people. If at any point in time there are more cannibals on one side of the river than anthropologists, the cannibals will eat them. What plan can the anthropologists use for crossing the river so they don’t get eaten? (remember! the boat can’t cross the river by itself, someone has to be in it to row it across)
Note that if a boat with a cannibal and an anthropologist travels to a shore with one cannibal on it, then no. of cannibals > no. of anthropologists, even if you say the anthropologist immediately takes the boat back.See Solution : River crossingHide Solution

Let

A = Anthropologist
C = Cannibal
B = the boat
W = the west shore (which they are all on)
and E = the east shore (where they want to go)Step 1 : A and C crosses
W [A, A, C, C] E [A, C, B]Step 2 : A returns
W [A, A, A, C, C, B] E [C]Step 3 : Two C crosses
W [A, A, A] E [C, C, C, B]Step 4 : C returns
W [A, A, A, C, B] E [C, C]Step 5 : Two A crosses
W [A, C] E [A, A, C, C, B]Step 6 : A and C returns
W [A, A, C, C, B] E [A, C]Step 7 : Two A crosses
W [C, C] E [A, A, A, C, B]Step 8 : C returns
W [C, C, C, B] E [A, A, A]Step 9 : Two C crosses
W [C] E [A, A, A, C, C, B]Step 10 : C returns
W [C, C, B] E [A, A, A, C]

Step 11 : Two C crosses
W [Empty] E [A, A, A, C, C, C, B]

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Average salary

November 30, 2007 by Ankur Leave a Comment

Three coworkers would like to know their average salary. How can they do it, without disclosing their own salaries to other two?

See Solution : Average salaryHide Solution

Let the three persons are A, B and C. Now first A tells the sum of his salary and a random number to B , lets say (SA + A(random no. of A)). Now B adds the sum of his salary and a random number to the number given by A i. e. B makes the total = (SA +A) + (SB+B) and he passes this number to C. (at each step, they don’t show the no. to the third person). Now C does the same, adds his salary and a random number to the amount told by B. Now C passes this total equals to (SA+SB+SC+A+ B+C) to A. Now A subtracts his random no. (A), passes to B then B subtracts his random no. (B) and passes to C. Finally C subtracts his random no. (C) and tells everybody. They divide it by 3. Now they have the average salary.

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