You are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which.
How do you make two piles of coins each with the same number of heads up?
You can flip the coins any number of times.
Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile.
How this will work? lets take an example.
So initially there are 5 heads, so suppose you divide it in 2 piles.
P1 : H H T T T
P2 : H H H T T
Now when P1 will be flipped
P1 : T T H H H
P1(Heads) = P2(Heads)
P1 : H T T T T
P2 : H H H H T
Now when P1 will be flipped
P1 : H H H H T
P1(Heads) = P2(Heads)
Tarun Asthnaiya go to his office by local train. However nearby train station is quite far from his place and he used to drive his bike to train station daily with an average speed of 60km/hr. One day at halfway point he relized that due to heavy traffic he got late having average speed of just 30km/hr. How fast he must drive for the rest of the way to catch my local train?
The train is just about to leave the station and there is no way Tarun will be able to catch it this time.
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races.
First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.
We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).
This table shows our remaining horses:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).
We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).
Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).
Let's look at our updated table, having removed these horses that can't be in the top 3:
A1 B1 C1
A2 B2 C2
A3 B3 C3
We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:
A1 B1 C1
We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.
Fernando + Alonso + McLaren = 6
Fernando x Alonso = 2
Alonso x McLaren = 6
McLaren x Fernando = ?
3 or 0.75
Rewriting the last 2 equations in terms of Alonso,
Fernando = 2/Alonso
McLaren = 6/Alonso
Replacing above values in equation "Fernando + Alonso + McLaren = 6"
2/Alonso + Alonso + 6/Alonso =6
(2 + Alonso^2 + 6)/Alonso = 6
8 + Alonso^2 = 6Alonso
Alonso^2 - 6Alonso + 8 = 0
(Alonso - 4) (Alonso - 2) = 0
Alonso = 4 or 2
Let's take value of Alonso as 2
Fernando = 2/2 = 1
McLaren = 6/2 = 3
McLaren x Fernando = 3 x 1 = 3
Let's take value of Alonso as 4
Fernando = 2/4 = 0.5
McLaren = 6/4 = 1.5
McLaren x Fernando = 1.5 x 0.5 = 0.75
A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. For example, servant 14 comes in and says "Servant 14, reporting in."
One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one.
Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time.
The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him.
How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?
When the first servant comes in, the king should write down his number. For each other servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper.
Once the final servant has reported in, the number on the paper should equal
(1 + 2 + 3 + ... + 99 + 100) - MissingServantsNumber
Since (1 + 2 + 3 + ... + 99 + 100) = 5050, we can rephrase this to say that the number on the paper should equal
5050 - MissingServantsNumber
So to figure out the missing servant's number, the king simply needs to subtract the number written on his paper from 5050:
MissingServantsNumber = 5050 - NumberWrittenOnThePaper
A monk leaves at sunrise and walks on a path from the front door of his monastery to the top of a nearby mountain. He arrives at the mountain summit exactly at sundown. The next day, he rises again at sunrise and descends down to his monastery, following the same path that he took up the mountain.
Assuming sunrise and sunset occured at the same time on each of the two days, prove that the monk must have been at some spot on the path at the same exact time on both days.
Imagine that instead of the same monk walking down the mountain on the second day, that it was actually a different monk. Let's call the monk who walked up the mountain monk A, and the monk who walked down the mountain monk B. Now pretend that instead of walking down the mountain on the second day, monk B actually walked down the mountain on the first day (the same day monk A walks up the mountain).
Monk A and monk B will walk past each other at some point on their walks. This moment when they cross paths is the time of day at which the actual monk was at the same point on both days. Because in the new scenario monk A and monk B MUST cross paths, this moment must exist.