- Problem:
*Two bicyclists start twenty miles apart and head toward each other, each going at a steady rate of $10$ mph. At the same time a fly that travels at a steady $15$ mph starts from the front wheel of the southbound bicycle and flies to the front wheel of the northbound one, then turns around and flies to the front wheel of the southbound one again, and continues in this manner till he is crushed between the two front wheels. What total distance did the fly cover?*## Possible Solution

The relative velocity between two bicycles is $20$ mph. Initially, they were $20$ miles apart. So, it will take one hour for them to hit each other. Therefore the fly must have flied for one hour. Since the fly was flying at $15$ mph, it must have traveled total $15$ miles. - Problem: There are total $12$ coins in a box. All of them are of equal size and shape. But one of them is a counterfeit coin. The weight of the counterfeit coin is different from the weight of the original coins. You are provided with a marker pen and a balance. How can you figure out the counterfeit coin by weighing at most three times?
## Possible Solution

Coming Soon . . . . - Problem: A deck of cards contains $42$ cards which are faced down and $10$ cards which are faced up. The deck is shuffled. You are blindfolded (i.e., you are not allowed to look at the cards). Divide the deck into two stacks such that each stack contain exactly the same number of up-faced cards. (Pay attention to the wording of this problem).
## Possible Solution

Take ten cards from the top of the deck (well, from anywhere is fine). Make a stack using those ten cards in reversed-faced position. i.e., if a card were up-faced, it will now be faced-down and vice versa. Now, you have two stacks (a stack of $42$ cards and a stack of $10$ cards) such that the number of up-faced cards in those stacks are the same ('same' could be anywhere from $0$ to $10$). Why? Let's say your new set of $10$ cards originally contained $3$ up-faced cards. Then the stack of remaining $42$ cards must have $7$ up-faced cards. Now, after reversing the direction of the new stack of $10$ cards, the new stack will also contain $7$ up-faced cards. - Problem: Your mom does not like you smoking cigarettes. One day, after coming back early from work, she caught you red handed while you were smoking cigarettes in the apartment. She became so angry that you were thrown out of the apartment. She told you that she would open the door only if you knock at the door exactly after $45$ mins. She gave you two ropes and your cigarette lighter. The ropes have the property that if you light one end of a rope, it will take one hour to burn the whole rope. How would you measure $45$ mins using those two ropes? Note that the ropes may not burn at a uniform rate. (This problem was given to me by Subhadip Dey)
## Possible Solution

Note that if you light both the ends of any rope, it will burn for $30$ mins. Now, you light both the ends of one rope, and one end of another rope. When the first rope burns out (i.e., exactly after $30$ mins), light up the other end of the second rope. Since the second rope was already burned for $30$ mins, it will now burn for $15$ mins only. - Problem: There are ten boxes with ten coins in each box. One of the boxes contains all counterfeit coins. Original coins weigh $5$ gms each, and the counterfeit coins weigh $3$ gms each. You are provided with a digital balance instrument (i.e., you can measure the exact weight of any object). Find out the box of counterfeit coins by using the balance instrument only once.
## Possible Solution

Arrange the boxes in a row. Now, take one coin from the first box, two coins from the second box . . . . ten coins i.e., all coins from the last box. So, now you have $1+2+\cdots+10=55$ coins in your hand. Weigh the coins using the balance instrument. If all the coins were genuine, then the total weight would be $55\times 5=275$ gms. However, not all the coins are genuine, there are some coins which are counterfeit. Therefore those $55$ coins will weigh less than $275$ gms. If there is only one counterfeit coin in that set of $55$ coins, the total weight will be $2$ gms less i.e., $273$ gms. Since one coin was taken from the first box, the first box must be the box of counterfeit coins. Similarly, if you see the total weight is $8$ gms less i.e., $267$ gms, then there must be four coins which are counterfeit i.e., the fourth box contains the counterfeit coins etc. - Problem: The following shape is divided into four equal congruent shaped parts. Divide it into six congruent equal shaped parts. (This problem was given to me by Subhadip Dey)
## Possible Solution

Don't let those four divisions block your thinking. In fact, the shape can be divided into any number of congruent shapes. Here is the answer. - The five cards magic: There is a deck of $52$ cards in front of you. Choose $5$ cards randomly (or as you like) from the deck and give those to my assistant. At this point, only you and my assistant know all the cards. My assistant gives me four cards (let's say those are 10 of hearts, 9 of club, 2 of spade, and Jack of diamond). I have not seen the fifth card which is in you hand. But my friend, that is the Queen of hearts. How did I know it?
## Possible Solution

Coming soon . . .