Story of Farmer's 3 Sons

A farmer’s wife made some chapatis…The farmer had 3 sons…. 

The first son came, gave one chapati to the dog,and made three equal parts of remaining chapatis, ate one part of it and left the other two parts for his brothers…. Other two sons came one after the other and did the same thinking that they came first…

Then at night all three came to the house, one of them gave one chapati to dog and made three equal parts and the three brothers ate one-one part of it… 

If no chapati was broken in pieces then how many minimum number chapatis did the mom made?

Here are MATHEMATICAL steps for solution! 

Mathematics in the Story of Farmer's 3 Sons

What was the story?

Let X be the number of chapatis that farmer's wife made.

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First son gave 1 chapati to dog. Chapatis left = (X - 1)

Made 3 equal parts of remaining. Chapatis in each part = (X - 1)/3

He Ate one part of it. Chapatis left = 2(X - 1)/3.

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Second son gave 1 chapati to dog. Chapatis left = 2(X - 1)/3 - 1

Made 3 equal parts of remaining. Chapatis in each part = [2(X - 1)/3 - 1]/3

He Ate one part of it. Chapatis left = 2[2(X - 1)/3 - 1]/3 

                                                  = [4(X-1) – 6]/9

                                                  = (4X - 10)/9

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Third son gave 1 chapati to dog. Chapatis left = (4x - 10)/9 - 1

Made 3 equal parts of remaining. Chapatis in each part = [(4X - 10)/9 - 1]/3

He Ate one part of it. Chapatis left = 2[(4X - 10)/9 - 1]/3

                                                  = (8X – 20 – 18)/27 
       
                                                  = (8X – 38)/27 

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At that night, one of them gave 1 chapati to dog.

Chapatis left = (8X – 38)/27 - 1

After that they made 3 equal parts of remaining parts and each son ate one part.

If we assume that each son ate Y chapatis then,

Y = [(8X – 38)/27 - 1]/3

Y = (8X - 38 – 27)/81

Y =  (8X – 65)/81

81Y= 8X - 65

X = (81Y + 65)/8 

X = 81Y/8 + 65/8

X = 10Y + Y/8 + 8 + 1/8

X = 10y + 8 + (y+1)/8

For X to be integer, Y has to be 7,15,23.....

So possible values of X are - 79, 160, 241.

Since, question asks minimum number, the farmer's wife must have made 79 chapatis in total.

Flipping The Unusual Coins

You have three coins. One always comes up heads, one always comes up tails, and one is just a regular coin (has equal change of heads or tails). If you pick one of the coins randomly and flip it twice and get heads twice, what is the chance of flipping heads again?

Chances of flipping head again are - .......% Click to know!

Chance of Flipping Head Again

What was the problem?

For a coin to always show head on flip we assume both it's sides are heads and the coin which is showing tail always we assume both of it's sides are tails.

There is no way that you have selected tail only coin since there are 2 heads in first 2 flips.

So it could be either head only coin say D coin or regular fair coin say F.

Let H1 and H2 be the sides of head coin and H, T are side of fair coin.

If it's head only coin D, then possible scenarios on 2 flips are -

DH1 DH1
DH1 DH2
DH2 DH1
DH2 DH2

And if it's fair coin F then possible scenarios on 2 flips are -

FH FH
FH FT
FT FH
FT FT

There are total five combinations (all 4 of head coin + first one of fair coin) where there are 2 consecutive heads on 2 flips.

So, the chances that you have picked a head coin is (4/5) and that you picked fair coin is (1/5).

For head coin, the probability of getting head again is 1 and that for fair coin is (1/2).

Since you holding either head coin or fair coin,

Probability (Head on third flip) = 
Probability (You picked Head coin) x Probability (Head on head coin) + Probability (You picked fair coin) x Probability (Head on fair coin) 


Probability (Head on third flip) = (4/5) x 1 + (1/5) x (1/2)

Probability (Head on third flip) = 9/10.

Hence, the chance of flipping head again on third flip is 90%.

Divide The Cake Into Equal Parts!

I have just baked a rectangular cake when my wife comes home and barbarically cuts out a piece for herself. The piece she cuts is rectangular, but it’s not in any convenient proportion to the rest of the cake, and its sides aren’t even parallel to the cake’s sides. 

 
I want to divide the remaining cake into two equal-sized halves with a single straight cut. How can I do it?



This is how it can be cut! 

Cutting The Cake Into Equal Parts!

What was the problem? 

Generally, a line drawn through the center of rectangle divides it into 2 equal parts.
Hence, a line drawn through the centers of both rectangles would divide each of them into 2 equal parts as shown below. (To get the center of each rectangle, all we need to do is draw diagonals of both).

Different Kind of Dice Game!

Timothy and Urban are playing a game with two six-sided dices. The dice are unusual: Rather than bearing a number, each face is painted either red  or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colors on the second die?

The second die must have 'these' colors!

Die Needed For Different Kind of Dice Game!

What was the different in dice game?

Throwing two six-sided dice produces 36 possible outcomes. Since both Timothy and Urban have equal chances of winning, there are 18 outcomes where top faces are of same color.

Let's assume there are 'x' red faces and (6-x) blue faces on the other dice.

Remember, the first die has 5 red faces and 1 blue face. Then there are 5x ways by which top faces are red & 1(6-x) ways by which they both are blue. But as deduced above, there are 18 such outcomes in total where faces of dice matches color.

Therefore,

5x + 1(6-x) = 18

4x = 12

x = 3

Hence, other die have 3 red colored & 3 blue colored faces.

The Tuesday Birthday Problem

I ask people at random if they have two children and also if one is a boy born on a Tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.

Tip: Don't conclude too early. 

Click here to know the correct answer! 

Finding The Correct Probability

How tricky it was?

If you think that the probability is 1/2 after reading that the couple has equal chance of having child of either sex then you are in wrong direction.

Take a look at the table below.

There are 27 possible combinations when boy is born on Tuesday. Out of which there are only 13 possible combinations where either boy (first or second) is born on Tuesday. 

Hence the probability that the person having at least 1 boy off his 2 boys born on Tuesday is 13/27.

The Coconut Problem

Ten people land on a deserted island. There they find lots of coconuts and a monkeys. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning.

That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him.

Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's slightly bloodied coconut. The monkey conks the second man on the head and kills him.

One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys?

Here is that smallest number! 

Source 

Number Of Coconuts In The Pile

What was the problem? 

Absolutely no need to overthink on the extra details given there. Just for a moment, we assume the number of coconuts in the community pile is divisible by 10,9,8,7,6,5,4,3,2,1.

Such a number in mathematics is called as LCM. And LCM in this case is 2520. Since each time 1 coconut was falling short of equal distribution there must be 2519 coconut in the pile initially. Let's verify the fact for all 10 distributions tried by 10 people.Each time monkey kills 1 person & number of persons among which coconuts to be distributed decreases by 1 each time.