Four friends need to cross a dangerous 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?
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)
This is a famous probability puzzle in which you have to choose the correct answer at random from the four options below.
Can you tell us whats the probability of choosing correct answer in this random manner.
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.
In a recreational activity, you are given four different jars of 2 liters, 4 liters, 6 liters and 8 liters respectively with an unlimited water supply. Then you are asked to measure exactly 5 liters of water using them.
A thief was running from the police after the biggest theft the town saw. He took his guard in one of the thirteen caves arranged in a circle. Each day, the thief moves either to the adjacent cave or stay in the same cave. Two cops goes there daily and have enough time to enter any two of the caves out of them.
How will the cop make sure to catch the thief in minimum number of days and what are the minimum number of days?
Three college toppers are summoned by the inspecting faculty. To identify the best from them, the faculty takes them into a room and places one hat on each of their heads. Now all of them can see the hats on other’s heads but can’t see his own. There are two colored hats – green and red.
Now the faculty announces that he had made sure that the competition is extremely fair to all three of them. He also gives them a hint that at least one of them is wearing a red hat. Now the first one who is able to deduce his own hat color will be awarded the most intelligent student of all award. After a few minutes, one of them raises his hand and is able to deduce the color correctly.
There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.
Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.
Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.
Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.
For my anniversary, I decided to surprise my wife. Since she is a voracious reader, I decided to collect a lot of books for her. On the first day of the month, I bought one book, on the second, I bought two and on the third, I bought three. This process went on till the anniversary and on the day, I had 276 books with me to gift her.
Can you calculate, on which day is our anniversary?
There stand nine temples in a row in a holy place. All the nine temples have 100 steps climb. A fellow devotee comes to visit the temples. He drops a Re. 1 coin while climbing each of the 100 steps up. Then he offers half of the money he has in his pocket to the god. After that, he again drops Re. 1 coin while climbing down each of the 100 steps of the temple.
If he repeats the same process at each temple, he is left with no money after climbing down the ninth temple. Can you find out the total money he had with him initially?
Whenever you face such type of questions, it is wise to begin from the last thing. Here in this question the last thing will be the 9th temple. He climbed down 100 steps and thus you know, he had Rs. 100 before beginning climbing down. Thus, he must have offered Rs. 100 to the god in that temple too (he offered half of the total amount). Also, he must have dropped Rs. 100 while climbing the steps of the ninth temple. This means that he had Rs. 300 before he begand climbing the steps of the ninth temple.
Now, we will calculate in the similar manner for each of the temples backwards.
Before the devotee climbed the eight temple: (300+100)*2 + 100 = 900
Before the devotee climbed the seventh temple: (900+100)*2 + 100 = 2100
Before the devotee climbed the Sixth temple: (2100+100)*2 + 100 = 4300
Before the devotee climbed the fifth temple: (4300+100)*2 + 100 = 8900
Before the devotee climbed the fourth temple: (8900+100)*2 + 100 = 18100
Before the devotee climbed the third temple: (18100+100)*2 + 100 = 36,500
Before the devotee climbed the second temple: (36500+100)*2 + 100 = 73300
Before the devotee climbed the first temple: (73300+100)*2 + 100 = 146900
Therefore, the devotee had Rs. 146900 with him initially.
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