You are driving a car on one big stormy night. You pass a bus station. There are three people who are waiting for the bus: One old sick lady who is dying, One doctor who saved your life before, and one lady who is someone you have been dreaming to be with. You can only take one passenger, which one will you choose?
Give the car key to the doctor, let the doctor take the old lady to the hospital and stay to wait for the bus with the lady of your dreams!
Hussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range.
To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot.
If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower.
He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Hussey choose?
Hussey should have Amy Hussey pull the trigger again without spinning.
We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.
If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
Four people come to an old bridge in the middle of the night. The bridge is rickety and can only support 2 people at a time. The people have one flashlight, which needs to be held by any group crossing the bridge because of how dark it is.
Each person can cross the bridge at a different rate: one person takes 1 minute, one person takes 2 minutes, one takes 5 minutes, and the one person takes 10 minutes. If two people are crossing the bridge together, it will take both of them the time that it takes the slower person to cross.
Unfortunately, there are only 17 minutes worth of batteries left in the flashlight. How can the four travellers cross the bridge before time runs out?
The two keys here are:
You want the two slowest people to cross together to consolidate their slow crossing times.
You want to make sure the faster people are set up in order to bring the flashlight back quickly after the slow people cross.
So the order is:
1-minute and 2-minute cross (2 minute elapsed)
1-minute comes back (3 minutes elapsed)
5-minute and 10-minute cross (13 minutes elapsed)
2-minute comes back (15 minutes elapsed)
1-minute and 2-minute cross (17 minutes elapsed)
I am first in Earth, second in Heaven, I appear two times in a week.
You can only see me once in a year, although I'm in the middle of sea.
Who am I?
Answer is E
The asnwer is E. As In spelling of Earth E comes first, in Heaven it comes second, in week it comes twice, in a year it comes once and finally in spelling of sea E stands in middle of the spelling.
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
