Find fastest 3 horses out of 25 horses puzzle

Question

You have 25 horses and you need to pick fastest 3 horses out of those. In each race you can race maximum of 5 horses as there are 5 tracks available. What are minimum number of races needed to find the fastest 3 without using a stopwatch. 

Solution

7 races needed. 

Understanding

Since we don't have a stopwatch only way to find fastest is by racing horses. Lets have 5 races each of 5 horses and let following be the results -

• A > B > C > D > E
• F > G > H > I > J
• K > L > M > N > O
• P > Q > R > S > T
• U > V > W > X > Y

Above are results of each race.  We have 5 races till now. Now lest race between fastest of all previos 5 races i.e A,F,K,P,U. We have 6 races till now .Let's say the result for that is -

• F > K > A > P > U

Since We are interested in top 3 P and U are useless for us. Also since P and U were fastest among their group we can ignore all members of P and U too. So now only horses under consideration are -

• F > G > H > I > J
• K > L > M > N > O
• A > B > C > D > E

Now if we consider A as possible candidate for top 3 then others in group of A are redundant - since we already have F and K faster than A. So we can ignore B,C,D,E

Now horses under consideration are  -

• F > G > H > I > J
• K > L > M > N > O
• A

Now in the group lead by k possible horse candidates are K and L - since F is already faster than K if we consider K and L we already have 3. So we can ignore M,N and O. So remaining horses are -

• F > G > H > I > J
• K > L
• A

 Now lets consider group led by F. We can consider G and H as possible candidates for top 3 since if they are I and J are redundant. So remaining now are -

• F > G > H
• K > L
• A

 Now we already know fastest among all in F since we got that result by running fastest among each group. So only horses we need comparison for 2nd and 3rd position are - G, H, K, L, A

These are 5 horses we can have another race and find the top 2 out of them. Lets say the result was -

• L > H > K > G > A

We have done 7 races now. So fastest onces are L and H. We already the fastest among all - F. So the final answer is -

• F > L > H 

So the answer is 7 races needed.

10 coins puzzle

Question

There are 10 coin placed in front of you 5 of which are heads and 5 of which are tails. You are blind folded so you don't know which ones are which. You need to make two piles of coins so that both have equal heads. You are allowed to flip a coin any number of time. You obviously wont know which is head and which is tails by touching it.

Solution

Make two piles of coins of equal number (5 each) and then flip all the coins on one side.

Understanding

Lets say you split in into two equal piles. 1st pile has 3 heads and 2 tails. Since there were 5 heads and 5 tails other pile will have 3 tails and 2 heads. Now when you flip all in pile 1 then there will be 3 tails and 2 heads same as pile number 2.

Generically if there are n heads and 5-n tails in pile 1 then in pile 2 will have n tails and 5-n heads. When we flip all coins in pile 1 then pile 1 will have n tails and 5-n heads which is same as pile 2.

Burning island puzzle

Question

A man in stranded on an island covered in forest. Wind is blowing from the west. Lightning strikes to the west side of the forest and starts spreading with the wind. The fire will burn the whole forest killing the man in the process. There are cliffs around the island so that man cannot escape. How can man survive the fire?

Solution

Man picks up a logs , lights it up with the fire from the west end. Then he runs towards the east and lights that part of the forest. This will burn up the eastern end of the forest and then man can take shelter in that burnt area while fire from eastern end burns the remaining forest.

Four men in hats puzzle

Question

As shown in picture above there are 4 men looking forward. None of them can see back. There is a opaque wall between man number 3 and man 4 (1,2,3 cannot see pass the wall). Two of the men are wearing a black hat and two of them are wearing a white hat. Each man can see the color of the hat wore by the men in front of him. (1 can see 2,3 and 2 can see 3) but each person does not know the color of the hat he is wearing.

Now one of the man needs to call out the color of his hat else they all die in 10 mins. Which man will callout the color of his hat correctly and why?

Solution

Answer is Man no 2.

Reasoning

 Lets start by eliminating men. Man number 4 is at the other end of opaque wall facing other side. There is no way he can see any men or the color of their hat. So he is eliminated. Now man no 3 also cannot see anyone else - he cannot look back and he cannot see beyond wall. So he is eliminated too. Now man number 1 knows the color of man 2 and 3. Now lets say they (2 and 3) were wearing same color hat then man no 1 would know the color of his hat since there are 2 white and 2 black hat. But he keeps mum which means man 2 and 3 are wearing different hat. S0 man number 2 waits for sometime if he does not hear man 1 calling out that means man 2 and 3 are wearing different color hats. Since man 2 knows the color of hat wore by man 3 he know the color of his hat and calls it out.

Box Ball Logic Problem

Question

There are three bags.

1. The first bag has two blue rocks. 
2. The second bag has two red rocks. 
3. The third bag has a blue and a red rock.

All bags are labeled but all labels are wrong.You are allowed to open one bag, pick one rock at random, see its color and put it back into the bag, without seeing the color of the other rock.

How many such operations are necessary to correctly label the bags ?

Solution

Just one!

You pick a ball from the bag labelled "blue and red" ... if its blue then that's the 2 blue rocks bag and likewise for the red. ( since all bags are labelled wrong the one reading blue and red would not actually contain a blue and red ball. So it label will be whatever color of ball comes from it) Next if we get a blue ball from the blue and red labelled bag we know that the bag labelled red has got to be blue and red since we have already used up the blue label and since all bags are labelled wrongly it cannot be red.

River Crossing Problem

Question 1

Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person:  1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

Answer

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

To reduce the amount of time, we should find a way for 10 and 7 to go together.

So the solution is

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Question 2

A guard is positioned at the one side of bridge say 'A'.
* His task is to shoot all those who try to leave from 'A' to other side say 'B'.
* He also need to welcome the person who come from other side 'B' to his side 'A'.

The guard comes out of his post every 1 hour and looks down the bridge for any people trying to leave.
Monica a brilliant girl is at side 'A' and wish to go to other side 'B'. She also know's it would take her 1:45 hr to cross the river.

She comes with an super idea and able to cross the river.How did Monica cross the river ? 

Answer

Steps
1. Monica go from side A to B for approx 59 minutes
2. Lisa then move back to side A for 2 minutes to fool guard. Now guard will think she is coming to side A.
3. then she again move toward B

Piece of Cake

Question

How would you cut a rectangular cake into two equal pieces when a rectangular piece has already been cut out of it? The cut piece can be of any size and orientation. You are only allowed to make one straight cut.

Possible Solution

1.  The simplest solution that one might think of is to cut the cake horizontally along the height of the cake. That would divide the cake in exact two equal pieces.This solution won't work if there is some frosting on the top which also needs to be divided.
2. Other possible solution would be to make a cut such that it passes through center of both the rectangles. Since the cut halves both the rectangles, the resulting two pieces are guaranteed to have equal area.