The most often answer is 0 which is completely wrong.
Let us begin like this. The 12 hours in a clock makes a total of 360 degrees. Thus one hour will equal to 30 degrees.
Now as per the question, the hour hand will directly be on 3 when the minute hand is at 12. After 15 minutes or after 1/4 of an hour i.e. at 3:15, the hour hand will be 1/4 * 30 degrees = 7.5 degrees away from the minute hand.
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
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?
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.
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?
There are three bags.The first bag has two blue rocks. The second bag has two red rocks. The third bag has a blue and a red rock. All bags are labeled but all labels are wrong.You are allowed to open one bag, pick one rock at random, see its color and put it back into the bag, without seeing the color of the other rock.
How many such operations are necessary to correctly label the bags ?
You have been given three jars of 3 liters, 5 liters and 8 liters capacity out of which the 8 liters jar is filled completely with water. Now you have to use these three jars to divide the water into two parts of 4 liters each.
How can you do it making the least amount of transfers?
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.
A couple had to take shelter in a hotel for they could not proceed their journey in the rain. Having nothing to do at all, they started playing cards. Suddenly there was a short circuit and the lights went off. The husband inverted the position of 15 cards in the deck (52 cards normal deck) and shuffled the deck.
Now he asked his wife to divide the deck into two different piles which may not be equal but both of them should have equal number of cards facing up. There was no source of light in the room and the wife was unable to see the cards.
For a certain amount of time, she thought and then divided the cards in two piles. To the husband’s astonishment, both of the piles had equal number of cards facing up.
The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.
Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.
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.