Eggs have quite a unique property. They may be extremely fragile that they may break by mere a drop from your hand. However, they may not break even if they are dropped from the 100th floor. That is exactly what you have in this problem. You have two identical eggs and you have access to a 100 story building.
The question is how many drops you will make before you can find out the highest possible floor of the building from which the egg can be dropped without breaking. Remember you only have two eggs to break.
Let us first assume that the number of drops required are N.
If the egg breaks at maximum number of tries, we will have N - 1 drops till it does not break. Thus we must drop the first egg from the height N. Now if the first drop of the first egg does not break the egg, we can have N - 2 drops for the second egg if the first egg breaks in the second drop.
Let us put that into a valid example for better understanding. Suppose we need 16 drops. Now let us drop the egg from the sixteenth floor, if it breaks, we will try all the floors below sixteen. Suppose it does not breaks, then we have 15 drops left and we will drop it from 16+15+1 = 32nd floor. This is because if it breaks at 32nd floor, we can try all the way down to 17th floor in fourteen tries making the total tries to be 16.
Let us assume that 16 is the correct answer
1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45.....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 we can easily do in the end as we have enough drops to complete the task.
Seeking the above case, we must note that we have achieved it in 14 or 14 drops but we still need to find out the optimal one. We know from above that the optimal one will require 0 linear trials in the last step.
First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0
You have successfully split the beer into two equal parts.
Step1. Lets say y = x
Step2. Multiply through by x xy = x2
Step3. Subtract y2 from each side xy - y2 = x2 - y2
Step4. Factor each side y(x-y) = (x+y)(x-y)
Step5. Divide both sides by (x-y) y = x+y
Step6. Divide both sides by y y/y = x/y + y/y
Step7. And so... 1 = x/y + 1
Step8. Since x=y, x/y = 1 1 = 1 + 1
Step9. And so... 1 = 2
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules
Before going to work, Inspector Montalbano got into the fight with his wife. After coming back from the work he found out that the police was in the home and his wife had just killed a burglar.
The police told that she killed the burglar in self-defense. She told her husband the story that she heard a doorbell and thought that it was me and as soon as she opened the door, the burglar jumped into her and she was so scared that she killed burglar immediately with the knife. Inspector Montalbano asked the police to arrest her wife for murder conspiracy. Why?
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)
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.
This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).
No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.
How will all of them reach the other side of the river?
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.
You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.
Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.
So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.
There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3
(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.
That was the easy part.
What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.
Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.
Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.
For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.
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