Earnings Of People

Twenty men, women and children earn twenty coins between them. Each man earns 3 coins, each woman 1.5 coins and each child 0.5 coin.

How many men, women and children are there?

Find number of each of them here!

Number Of Earning People

What was the question? 

Let's suppose there are m men, w women and c children.

As per given data,

m + w + c = 20   ......(1)

3m + 1.5w + 0.5c = 20   .....(2)

Multiply equation (1) by 3 and then subtracting from (2), we get,

5m + 2w = 20.

For m and w to be whole numbers, m must be 2 and w must be 5 satisfying above equation.

Hence, from (1),

c = 20 - 2 - 5 = 13.

 
To conclude, there are 2 men, 5 women and 13 children.

A Mathematical Clue From The Merchant

A rich merchant had collected many gold coins. He did not want anybody to know about them. 

One day, his wife asked, “How many gold coins do we have?”

After pausing a moment, he replied, “Well! If I divide the coins into two unequal numbers, then 32 times the difference between the two numbers equals the difference between the squares of the two numbers.”

The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins they have?

Here are mathematical steps to find those!

Using The Mathematical Clue

What was that clue?

Since when divided into 2 unequal numbers difference won't be 0. Let x and y be the 2 unequal numbers.

As per merchant,

32 (x - y) = x^2 - y^2

32 (x - y) = (x - y) (x + y)

Dividing both sides by (x - y) which is non zero as x is not equal to y,

32 = x + y

x + y = 32.

Let's verify with x = 30 and y = 2. So 32 (x - y) = 32 ( 30 - 2) = 896. And x^2 - y^2 = 30^2 - 2^2 = 900 - 4 = 896.

Hence, Merchant had 32 coins in total.

Flip The Triangle

Here you see a triangle formed by ten coins. The triangle points upwards. How can just three coins be moved to make the triangle point downwards?

Flip It By Moving 3

 Here are pointed those 3 coins 

Flipped The Triangle!

What's the challenge? 

Here should be those 3 coins that you need to move so that the triangle point towards down.

And after moving those coin, it will look like,

Round Table Coin Game

You are sitting with one opponent at an empty, round table. Taking turns, you should place one dollar on the table, in such a way that it touches none of the coins that are already on the table. The first player that is not able to place a dollar on the table has lost. By tossing a coin, it has been decided that you may start.

Which strategy will you follow to make sure you are guaranteed to win?

  
Trick to win this game always! 

Never Loose Round Table Coin Game

What was the game? 

There is little trick with which you will always end on winning side in this Round Table Coin Game. Since you have got first chance to place the coin you should place the coin right at the center of the round table. Now for every next coin placed by opponent you need to place coin in such a way that it 'mirrors' opponent's coin.

Imagine line from the center to opponent's coin. Place your coin exactly opposite to that coin at distance equal to distance between center & opponent's coin. Or imagine a circle (with the center fixed at the round table) with opponent's coin lying on it's border.  And place your coin at diagonally opposite point of point where opponent placed coin on that imaginary circle. (Assume imaginary circle though it's not appearing perfectly in the image above)

In this way for every move of your opponent, you will have 'answer' in form of space for placing the coin. This will continue until last place left on the table with your turn of placing the coin in the end. 

This is how to make sure you always on winning side in this 'Round Table Coin Game'!

Equate Number of Heads or Tails

You are blindfolded and 10 coins are placed in front of you on the table. You are allowed to touch the coins but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tail up but not which ones are which.

How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.

This is how it can be done! 

Trick To Equate Number of Heads or Tails

What was the task? 

Without thinking too much we need to make 2 piles of 5 coins each. Now there are 3 possibilities here depending on number of heads in either pile. One of the pile might have either 0 or 1 or 2 heads (other having 5 or 4 or 3 heads).

Case 1 : 

P1 : T T T T T
P2 : H H H H H

Case 2 : 

P1 : H T T T T
P2 : H H H H T

Case 3 :

P1 : H H T T T
P2 : H H H T T

Now just flipping all the coins from single pile will make number of heads (or say tails) in both piles equal. So we can flip coins of either P1 or P2. Let's flip all coins of P2.


Case 1 : 

P1 : T T T T T         Number of heads - 0
P2 : T T T T T         Number of heads - 0

Case 2 : 

P1 : H T T T T         Number of heads - 1
P2 : T T T T H         Number of heads - 1

Case 3 :

P1 : H H T T T         Number of heads - 2
P2 : T T T H H         Number of heads - 2

Challenge of Grouping The Coins

You are given a unlimited number of coins and 10 pouches. Now, you have to divide these coins in the given pouches in a manner that if someone asks you for any number of coins between 1 to 1000, you should be able to give the amount by just giving the pouches. You are not allowed to open pouches for that.

How will you do it? 

Know here the only efficient way to do that! 

Source 

Grouping The Coins in Binary Numbers

What was the challenge? 

Once again here binary number system comes in handy. Similar kind of use of binary system in day to day life is here! Another intelligent use is here!  We are already provided 10 pouches which is exactly equal to the number of bits required to represent any number from 1 to 1000. Let's number the pouch as Pouch 0 to Pouch 9. So we need to group coins in 10 pouches like below.

Pouch 0 : 1
Pouch 1 : 2
Pouch 2 : 4
Pouch 3 : 8
Pouch 4 : 16
Pouch 5 : 32
Pouch 6 : 64
Pouch 7 : 128
Pouch 8 : 256
Pouch 9 : 512


Now if somebody asks us for 30 coins then we should give Pouch 4, Pouch 3, Pouch 2, Pouch 1. (11110) That's the binary representation of 30 if we assume Pouches as a bits. If another asks for 828 (binary - 1100111100) then we should give Pouch 9, Pouch 8, Pouch 5, Pouch 4, Pouch 3,
Pouch 2.