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.
David and Albert are playing a game. There are digits from 1 to 9. The catch is that each one of them has to cut one digit and add it to his respective sum. The one who is able to obtain a sum of exact 15 will win the game?
You are a friend of David. Do you suggest him to play first or second?
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.
Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.
If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.
Therefore, you should suggest David to play second.
In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.
In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.
If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?
At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.
Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.
In a closed jar, there are three strawberry candies, two mango candies and five pineapple candies. You can't see inside the jar. Now, how many toffees you must take out from the jar to make sure that you have one of each flavor?
2 < 3 < 5
To find out the required number of candies, take one in place of the least number (i.e. take one mango candy) and then add all the greater numbers (i.e. three strawberry and five pineapple candies) to it.
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
There are two glasses in front of you. One of the glasses is full of coke and the other glass is full of lemonade. You take a spoonful of coke and mix it into the glass of lemonade. Now the lemonade glass has a mixture of coke and lemonade. You take a spoonful of that mixture and mix it inside the coke glass.
Now what do you think? - The glass with coke has more quantity of lemonade or the glass with lemonade have more quantity of coke mixed with it?
This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.
Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.
Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.
A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats.
The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home.
All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation?
The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.
Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.
In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.
Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.