Nine brilliant most students were summoned by an inspecting professor. He gave them a situation and told them that he is having nine hats. The hats are either red colored or blue colored as told by the professor. He also tells them that he have at least one red colored hat and the number of blue hats are greater than the red hats.
He places one hat on each of their heads. The students are not allowed to talk with each other and no means of communication is feasible. He asks them how many hats are blue and how many are red. He gives them half an hour to deduce and moves out of the room. Nobody is able to answer when he returns back and thus he gives them another fifteen minutes. But when he returns, no one can answer again. Thus he gives them final five minutes. On returning back this time, everyone was available with an accurate answer.
How could the students have deduced the right answer? What is the right answer?
In the first wave, let us make an assumption that there is only 1 red hat and other 8 hats are blue. The student who can see all the other 8 blue hats on others' head will certainly be clear that there are 8 blue hats and 1 red hat since the professor told them that there is at least 1 red hat among the nine. But no one was able to answer after the first half an hour. Thus this can't be possible and our assumption is wrong.
Now let us assume that there are 2 red hats and 7 blue hats. The student who is having a red hat on his head can definitely see 7 blue hats and one red hat. He can deduce that he is wearing a red hat. But nobody was able to answer in the second interval as well. Thus our assumption is again wrong.
In the last wave of final five minutes, let us assume that there are 3 hats and 6 blue hats. The student with a red hat can see 2 other red hats and 6 blue hats and he may deduce that he is wearing a red hat. In this interval, everyone was able to answer which states that our assumption is correct.
There can be two possibilities for the given situation.
One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.
Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.
You are given with two ropes with variable width. However if we start burning both the ropes, they will burn at exactly same time i.e. an hour. The ropes are non-homogeneous in nature. You are asked to measure 45 minutes by using these two ropes.
How can you do it?
Please note that you can’t break the rope in half as it is being clearly stated that the ropes are non-homogeneous in nature.
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.
Thus you have successfully calculated 30+15 = 45 minutes with the help of the two given ropes.
Husband has prepared for a candle light dinner on the honeymoon for his wife. While they were having the dinner, a strong breeze flew through the open window and four candles out of ten were extinguished. After that, the husband closed the window.
We all know that New Year occurs after a week from Christmas and thus it falls on the same day as of Christmas. But this will not happen in 2050. In 2050, Christmas will appear on Sunday while New Year will appear on Saturday.
Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.
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.
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.
Submit your Email Address to get latest post directly to your inbox.