Riddle #973

interviewlogicclever

You have 3 jars that are all mislabeled. One jar contains Apples, another contains Oranges and the third jar contains a mixture of both Apples and Oranges. You are allowed to pick as many fruits as you want from each jar to fix the labels on the jars. What is the minimum number of fruits that you have to pick and from which jars to correctly label them?
Let's take a scenario. Suppose you pick from jar labelled as Apples and Oranges and you got Apple from it. That means that jar should be Apples as it is incorrectly labelled. So it has to be Apples jar. Now the jar labelled Oranges has to be Mixed as it cannot be the Oranges jar as they are wrongly labelled and the jar labelled Apples has to be Oranges. Similar scenario applies if it's a Oranges taken out from the jar labelled as Apples and Oranges. So you need to pick just one fruit from the jar labelled as Apples and Oranges to correctly label the jars.
76.54 %
51 votes

Similar riddles

See also best riddles or new riddles.

cleanlogicsimpleclever

Two fathers and two sons went fishing one day. They were there the whole day and only caught 3 fish. One father said, that is enough for all of us, we will have one each. How can this be possible?
There was the father, his son, and his son's son. This equals 2 fathers and 2 sons for a total of 3!
81.39 %
92 votes
logicsimplecleanclever

You have two jugs, one that holds exactly 3 gallons, and one that holds exactly 5 gallons. Using just these two jugs and a fire hose, how can you measure out exactly 4 gallons of water?
Fill the 5-gallon jug to the top, and then pour it into the 3-gallon jug until the 3-gallon jug is full. You now have 2 gallons remaining in the 5-gallon jug. Pour out the 3-gallon jug, and then pour the 2 gallons from the 5-gallon jug into the 3-gallon jug. Finally, fill the 5-gallon jug to the top and pour it into the 3-gallon jug until it's full. Since there was only space left for 1 more gallon in the 3-gallon jug, you now have exactly 4 gallons in the 5-gallon jug.
80.52 %
49 votes
logiccleversimple

In olden days you are a clever thief charged with treason against the king and sentenced to death. But the king decides to be a little lenient and lets you choose your own way to die. What way should you choose? Remember, you're clever!
I would have chosen to die of "old age". Did you?
80.49 %
160 votes
logicsimpleclever

Pirate Pete had been captured by a Spanish general and sentenced to death by his 50-man firing squad. Pete cringed, as he knew their reputation for being the worst firing squad in the Spanish military. They were such bad shots that they would often all miss their targets and simply maim their victims, leaving them to bleed to death, as the general's tradition was to only allow one shot per man to save on ammunition. The thought of a slow painful death made Pete beg for mercy. "Very well, I have some compassion. You may choose where the men stand when they shoot you and I will add 50 extra men to the squad to ensure someone will at least hit you. Perhaps if they stand closer they will kill you quicker, if you're lucky," snickered the general. "Oh, and just so you don't get any funny ideas, they can't stand more than 20 ft away, they must be facing you, and you must remain tied to the post in the middle of the yard. And to show I'm not totally heartless, if you aren't dead by sundown I'll release you so you can die peacefully outside the compound. I must go now but will return tomorrow and see to it that you are buried in a nice spot, though with 100 men, I doubt there will be much left of you to bury." After giving his instructions the general left. Upon his return the next day, he found that Pete had been set free alive and well. "How could this be?" demanded the general. "It was where Pete had us stand," explained the captain of the squad. Where did Pete tell them to stand?
Pete told them to form a circle around him. All the squad was facing in at Pete, ready to shoot, when they realized that everyone who missed would likely end up shooting another squad member. So no one dared to fire, knowing the risk. Thus at sundown he was released.
80.42 %
81 votes
logiccleanclevermath

At a dinner party, many of the guests exchange greetings by shaking hands with each other while they wait for the host to finish cooking. After all this handshaking, the host, who didn't take part in or see any of the handshaking, gets everybody's attention and says: "I know for a fact that at least two people at this party shook the same number of other people's hands." How could the host know this? Note that nobody shakes his or her own hand.
Assume there are N people at the party. Note that the least number of people that someone could shake hands with is 0, and the most someone could shake hands with is N-1 (which would mean that they shook hands with every other person). Now, if everyone at the party really were to have shaken hands with a different number of people, then that means somone must have shaken hands with 0 people, someone must have shaken hands with 1 person, and so on, all the way up to someone who must have shaken hands with N-1 people. This is the only possible scenario, since there are N people at the party and N different numbers of possible people to shake hands with (all the numbers between 0 and N-1 inclusive). But this situation isn't possible, because there can't be both a person who shook hands with 0 people (call him Person 0) and a person who shook hands with N-1 people (call him Person N-1). This is because Person 0 shook hands with nobody (and thus didn't shake hands with Person N-1), but Person N-1 shook hands with everybody (and thus did shake hands with Person 0). This is clearly a contradiction, and thus two of the people at the party must have shaken hands with the same number of people. Pretend there were only 2 guests at the party. Then try 3, and 4, and so on. This should help you think about the problem. Search: Pigeonhole principle
80.14 %
48 votes
logicmathclever

You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck. Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this. How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started. This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
80.05 %
67 votes
logicmathstorycleanclever

In the land of Brainopia, there are three races of people: Mikkos, who tell the truth all the time, Kikkos, who always tell lies, and Zikkos, who tell alternate false and true statements, in which the order is not known (i.e. true, false, true or false, true, false). When interviewing three Brainopians, a foreigner received the following statements: Person 1: I am a Mikko. Person 2: I am a Kikko. Person 3: a. They are both lying. b. I am a Zikko. Can you help the very confused foreigner determine who is who, assuming each person represents a different race?
Person 1 is a Miko. Person 2 is a Ziko. Person 3 is a Kikko.
79.99 %
73 votes