A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment.
Can you explain why?
The man is a of short stature. He can't reach the upper elevator buttons, but can push is with his umbrella.
Six jugs are in a row.
The first three are filled with coke, and the last three are empty.
By moving only one glass, can you arrange them so that the full and the empty glasses alternate?
Move and then pour all coke from second glass to fifth glass.
There are five people. One of them shot and killed one of the other five.
We know following clues:
1. Dan ran in the NY City Marathon yesterday with one of the innocent men.
2. Mike consider being a farmer before he moved to the city.
3. Jeff is a top notch computer consultant and wants to install Ben new computer next week.
4. The murderer had his leg amputated last month.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jack brother. They grew up together in Seattle.
Consider yourself to be a famous detective "Sherlock Homles", Can you find the killer?
Jeff Killed Mike
1. Jack is not the murderer because he is the brother of the murderer.
2. Dan can't be the murderer since he ran a marathon, and the murderer recently had his leg amputated, and wouldn't be running a marathon of any magnitude that quickly.
3. Ben is not the murderer if he just met Jack since Jack and the murderer grew up together.
4. This leaves Jeff and Mike.
Since Jeff is still alive (he wants to install a new computer next week, present tense) he must be the murderer. Mike also didn't grow up with Jack. It has been determined that Jack, Dan, and Jeff are all alive. Ben must also be alive since Jeff plans to install Ben computer next week. This means that Jeff killed Mike.
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?
17 mins
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.
Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.
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
If you had an infinite supply of water and a 5 quart and 3 quart pails, how would you measure exactly 4 quarts? and What is the least number of steps you need?
1. Fill 5 quarter pill ( 5p - 5, 3p - 0)
2. Transfer to 3 quart pail (5p - 2, 3p - 3)
3. Empty 3 quarter pill ( 5p - 2, 3p - 0)
4. Transfer 2q from 5 pail to 3 pail (5p - 0, 3p - 2)
5. Fill 5 quarter pail(5p - 5, 3p - 2)
6. Transfer 1q from 5 pail to 3 pail(5p - 4, 3p - 3)
A newspaper is supposed to have 60 pages
but pages 24 and 41 are missing.
Which other pages won't be there?
Pages 19, 20, 23, 37, 38, & 42 will also be missing
A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?
Just 2
Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.
You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.
The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.
You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.
You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.
What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?
10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.
Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.
Here is how you would find one poisoned bottle out of eight total bottles of wine.
Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.
With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.
Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.
Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.
One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.
In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?
The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.
You can place weights on both side of weighing balance 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.
For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.