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Showing posts with the label money

Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination

What is the maximum sum of money you could accumulate before the drawing ends?


Draw The Maximum Sum


THIS should be the maximum sum! 

The Maximum That You Can Get!


What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.



The Maximum That You Can Get!

Puzzle: A Visit to the Casino!

There is a casino and it has 4 gates, let say A, B, C and D

Now the condition is that every time you enter casino you have to pay $5 and every time you leave the casino, you again have to pay $5. Also, whenever you enter the casino whatever amount you have with you will get double.

Now you enter the casino through gate A and come out through gate B, again you go inside casino from gate C and come out of gate D, at the end of this process you should be left with no money

So calculate how much money you should carry with you when you enter the Casino?

A Visit to the Casino!


THIS should be the amount that you need to carry... 

Solution: Amount Needed for a Casino Visit


What is the puzzle?

Let's suppose you have amount 'x' initially in your wallet.

1. On paying $5 for entry at gate A amount left is x - 5 .

2. After entry into the casino it double and becomes 2 ( x - 5 ) = 2x - 10.

3. For exit at gate B you pay $5 and the amount left with you is 2x - 10 - 5 = 2x - 15.

4. Again, for the entry at gate C, you pay $5 more. So the amount with you will be  - 
    2x - 15 - 5 = 2x - 20.

5. The amount is doubled to become 2(2x - 20) = 4x - 40 after the entry into casino.

6. Now, you have to pay $5 one more time to have exit via gate D. Hence, the amount 
left will be 4x - 40 - 5 = 4x - 45.

7. As per given condition, the amount that you should have on exit must be 0.

Hence, 4x - 45 = 0 i.e. 4x = 45. Therefore, x = 11.25. 

So you should carry $11.25 before you enter into the casino to have $0 after exit out of the casino.

Amount Needed for a Casino Visit


Let's verify this one with the amount at each of stages above.

1. Amount = 11.25 - 5 = 6.25

2. Amount = 2 x 6.25 = 12.50

3. Amount = 12.50 - 5 = 7.50

4. Amount = 7.50 - 5 = 2.50

5. Amount = 2 x 2.50 = 5 

6. Amount = 5 - 5 = 0

7. Amount = 0
    

The Little Johnny's Dilemma

Little Johnny is walking home. He has $300 he has to bring home to his mom. While he is walking a man stops him and gives him a chance to double his money. 

The man says -

 "I'll give you $600 if you can roll 1 die and get a 4 or above, you can roll 2 dice and get a 5 or 6 on at least one of them, or you can roll 3 dice and get a 6 on at least on die. If you don't I get your $300."
 
What does Johnny do to have the best chance of getting home with the money?


The Little Johnny's Dilemma


THIS is the BEST he can do! 

The Loss Making Deal !


What was the deal?


The little Johnny should not accept the deal.

The problem is actually finding the probability of winning in each case dice throw.

---------------------------------------------

CASE 1 :  

Condition - Roll a die and get 4 or above.

The probability in the case is 3/6 = 1/2 (getting 4,5,6 from 6 possible outcomes). 

That is only 50% chances of winning the deal.

---------------------------------------------

CASE 2 :

Condition - Roll 2 dice and a 5 or 6 on at least one of them.

Let's find the probability that none get a 5 or 6. 

Probability that single die doesn't get a 5 or 6 is 4/6 = 2/3 by getting 1,2,3,4 in possible 6 outcomes.

Since, these 2 results are independent, the probability that neither dice gets a 5 or 6 is 2/3 x 2/3 = 4/9.

So, the probability that at least one of them gets a 5 or 6 is 1 - 4/9 = 5/9.

That is, about 56% chances of winning the deal.

By other approach, there are 36 possible combinations in 2 dice throw.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) 

There are total 20 combinations (last 2 rows and 2 columns) where at least one die gets a 5 or 6.

Hence, the probability in the case is 20/36 = 5/9 i.e. about 56% of winning chances.

----------------------------------------

CASE 3 : 

Condition - Roll three dice and get a 6 on at least on die.  

Again finding probability that none gets a 6.

Probability that single die doesn't get a 6 is 5/6 by getting 1,2,3,4,5 out of 6 possible outcomes.

Hence, the probability that three dices doesn't get a 6 = 5/6 x 5/6 x 5/6 = 125/216.

Then the probability that at least one of them get a 6 = 1 - 125/216 = 91/216 

That is only 42% chances of winning the deal.

--------------------------------------------

On the other hand Johnny can be 100% sure that he will take $300 to home if he doesn't accept the man's deal. 

The Loss Making Deal !

Sequel : Story of Distribution of Loot

The five pirates mentioned previously are joined by a sixth, then plunder a ship with only one gold coin.

After venting some of their frustration by killing all on board the ship, they now need to divvy up the one coin. They are so angry, they now value in priority order:


1. Their lives
2. Getting money
3. Seeing other pirates die.



Sequel : Story of Distribution of Loot


How can the captain save his skin?


This is how he should save himself! 

 The Prequel of the story!

Captain's Life Saving Proposal in Sequel


First read the story of sequel!

Let's name all the pirates as Pirate 6,5,4,3,2,1 as per their seniority. Now, the captain should respond with the logic below to save his skin.

Let's consider the cases where there are different number of pirates left on the ship after getting rid of seniors one by one.


----------------------------------------------------------------------------------------------


CASE 1 : 2 Pirates

The captain i.e. Pirate 2 can keep coin with him & obviously vote for himself (1/2 = 50% vote) to approve the proposal.


------------------------------------------------------------------------------------------------

CASE 2: 3 Pirates.

The captain i.e. Pirate 3 offers coin to Pirate 1 to get his support (2/3 = 66%) on proposal. Since, Pirate 1 knows what is going to happen if Pirate 3 dies as crew reduced to 2 Pirates as in CASE 1.


------------------------------------------------------------------------------------------------


CASE 3: 4 Pirates.

The captain i.e. Pirate 4 offers coin to Pirate 2 thereby getting his support (2/4 =50% votes) to get approval on proposal. Again, here Pirate 2 is smart enough to agree on this proposal as he know what will happen if Pirate 4 is eliminated leaving behind 3 pirates on sheep as in CASE 3.


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CASE 4: 5 Pirates.
Now the captain i.e. Pirate 5 always will be in danger as he can give only coin to only 1 of remaining 4. So he can 'earn' only 1 vote in support of his proposal i.e. only 2/5 = 40% votes. Hence, there is no way his proposal get approval & he should be ready to die.


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CASE 5: 6 Pirates.

Here, Pirate 5 has to agree whatever captain i.e. Pirate 6 offers as if Pirate 6 dies the case reduces to CASE 4 where Pirate 5 can't save himself. However, if both Pirate 6 & Pirate 5 die, then CASE 3 comes into reality where Pirate 2 gets coin. So Pirate 2 would never agree on proposal offered by Pirate 6.

However, if Pirate 6 offers coin to Pirate 4 then he would take it happily as he knows that what will happen of both Pirate 6 & 5 are killed as CASE 3 comes into picture where he has to give coin to Pirate 2 to save himself.

If Pirate 6 offers coin to Pirate 3 as well & earn his support as Pirate 3 also knows what is going to be the case of both 6 & hence 5 get eliminated. The case will be reduced to 4 Pirates as in CASE 3 where Pirate 4 will offer coin to Pirate 2.

The Pirate 6 even choose Pirate 1 to offer coin. Again, Pirate 1 smart to think that what will be the case if 6 & hence 5 are killed. It will be the scenario as in CASE 3 where Pirate 4 offers coin to Pirate 2.

In short, the Pirate 6 will always get support from Pirate 5 always in any case & any one from Pirate 4/3/1 if he offers coin to any one of these three. 

That's how he will get support of 50% (3/6) group to get approval for his proposal thereby saving his own life on approval of proposal offered.


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Challenge of Father to Son

A man told his son that he would give him $1000 if he could accomplish the following task. 

The father gave his son ten envelopes and a thousand dollars, all in one dollar bills. He told his son, "Place the money in the envelopes in such a manner that no matter what number of dollars I ask for, you can give me one or more of the envelopes, containing the exact amount I asked for without having to open any of the envelopes. If you can do this, you will keep the $1000."

When the father asked for a sum of money, the son was able to give him envelopes containing the exact amount of money asked for. 

How did the son distribute the money among the ten envelopes?

Challenge of Father to Son


THIS is how son accepts the challenge!

Son's Response to the Father's Challenge


What was the challenge?

For a moment, let's suppose father had given $30 to son and provided 5 envelopes and put the same challenge.
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Here, use of the binary number system helps in matter.

The son would distribute 15 dollars into 4 envelops like - 

2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8.


Now, for any amount asked between 1 to 15, son can produce some of these 4 envelops wherever 1 is there in envelop column for that particular amount.

For example, if father asks for $10 (Binary - 1010), son would give envelop 4 and 2 (8+2=10).

After putting $15 dollar in 4 envelops, he puts remaining $15 in 5th envelop so that he can cover rest of amount between 16 to 30.

If father asks amount greater than 15 then he would take envelop of $15 first and depending on how much the amount asked is greater than this $15 he would pick some of those 4 envelops.

For example, if father asks for $24, then he picks envelop 5 having $15 and envelops for amount 24 - 15 = 9 (Binary - 1009) i.e. envelop 4 and 1 (8+1=9)  i.e. total of 15 + 9 = 24.

So, what we observe from this is that the number of envelops needed for such arrangement is equal to the number of binary bits needed to represent the amount itself or nearest power of 2 greater than the amount.

In above case, to represent 30 in binary we need 5 bits or nearest power of 2 greater than 30 is 32 which needs 5 bits for representation.

-----------------------------------------------------------------------------

Now, let's turn to the actual challenge where father has asked son to distribute $1000 rightly in 10 envelopes. 

The reason for selecting 10 as a number of envelops is clear now as 1000 needs 10 bits in binary or nearest power of 2 greater than 1000 is 1024 which needs 10 bits for binary representation.

So, the son puts 256, 128, 64, 32, 16, 8, 4, 2, 1 dollars in 9 envelops (envelop numbered as Envelop 9, Envelop 8.........Envelop 1 in order)
 and 1000 - 511 = 489 dollars in 10th envelop.

First 9 envelops will cover amounts from 1 to 511 and for amounts greater than 511 inclusion of 10th envelop having 489 dollars is mandatory.

Again selection of envelops for the amount 511 to 1000 depends on how much the amount exceeds the $489. The binary representation of that difference and selection of envelop accordingly is all that needed.
 
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Let's make sure this distribution with couple of examples.

If father asks for amount of $109 (binary - 1101101) then son picks 

Envelop 1($1) + Envelop 3 ($4) + Envelop 4 ($8) + Envelop 6 ($32) + 
Envelop 7 ($64) i.e. having amount = 1 + 4 + 8 + 32 + 64 = 109 dollars.

If father asks for $525 then son gives $489 via Envelop 10 and rest of amount 
530 - 489 = 40 (Binary - 101000) in form of Envelop - 6 ($32) and Envelop 4 ($8).   

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Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 
 
     SEND
  + MORE
  ----------
 = MONEY
 
Replace Numbers with letters - Maths Puzzle

Process of decryption is here! 
 

Decrypting The CryptArithmatic Problem


What was the problem? 

Let's take a look at the equation once again.
       S  E N D
  +   M O R E
  -----------------
 = M O N E Y
 
Now letter M must be representing the carry generated & it must be 1. 
 
And if M = 1 then S must be 9 or 8 if carry is generated from hundreds place. 
In any case, O can be either 0 or 1. But can't be 1 as M = 1 hence O = 0.
 
If O = 0 then E + 0 = N i.e. E = N if there is no carry from tens place. 
Hence, N = E + 1.
 
Let's turn towards tens place now. With no carry from units place N + R = 10 + E .
Putting N =  E + 1, we get, R = 9. And with carry from units place, we have,
1 + N + R = 10 + E, gives us R = 8. Hence either R = 9 and S = 8 or R = 8 or S = 9.
 
For a moment, let's assume R = 9 and S = 8 with no carry from units place, then, 
 
   8 E N D
+ 1 0 9  E
=========
1 0 N E Y 
 
For this to be correct, we need carry at thousands place generated from hundreds 
place. That's only possible if E = 9 and carry is generated from tens place
forwarded to hundreds place. Since 9 is already being used for R this combination
is just impossible.
 
Hence, R = 8 and S = 9 with carry 1 from units place.
 
Now it looks like,
 
   9 E N D
+ 1 0 8 E
========
1 0 N E Y 
 
This means D + E >= 10 i.e. D + E = 10 + Y.  And numbers left are 2,3,4,5,6,7.
 
If E = 2 then D must be 8 or 9 for D + E >= 10.
Since 8 and 9 already taken, this is not possible.
 
If E = 3 then D can be 7 but Y would be 0 in the case. 
Since O = 0 already taken this value of E is also not valid. 
For any other value of D, D + E < 10 .
 
If E = 4 then D = 7 or 6 and Y = 0 or 1.
Both are taken hence this value of E in invalid.
Also,D <= 5 in the case gives  D + E < 10 .
 
If E = 6, N = 7 then, D <= 5. 
With D = 5 or 4, Y = 0 or 1, both are used for O and M already.
And for D = 2 or 3,D + E < 10 .
In short, this value of E is also not valid.   
 
So only value of E left is 5. Hence, N = 6 and D = 7. That gives, Y = 2.
 
Maths Puzzle -  Correct numbers for letters
 
To conclude,
 
  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2 

Generous Devotee

A devotee visits 9 temples when he visits India. All these nine temples have one thing in common - there are 100 steps in every temple. The devotee puts Re.1 coin after climbing up every step. He does the same while climbing down every step. At each temple, the devotee offers half of his money from his pocket to god. In this way, his pocket becomes empty after his visit to 9th temple.

How much donated by the devotee?

Can you calculate the total amount he had initially ? 


Click here to know exact amount! 



Donations By The Devotee


Why to calculate those?

Using algebraic equations in the case can make things complicated unnecessarily. Hence, we would start from backward. Before putting 100 coins on steps while climbing down 9th temple devotee must had 100 coins. That means he had 200 coins when he climbed up the 9th temple half of which i.e. 100 he offered to that temple & 100 put on the 100 steps of 9th temple. Moreover, he must have placed 100 coins while climbing up 9th temple. So before visit to 9th temple he must had, (100 x 2) + 100 = 300 coins.

Same way, finding the amount he had before visit to each temple like below.

Before eight temple: (300+100)*2 + 100 = 900
Before seventh temple: (900+100)*2 + 100 = 2100
Before Sixth temple: (2100+100)*2 + 100 = 4300
Before fifth temple: (4300+100)*2 + 100 = 8900
Before fourth temple: (8900+100)*2 + 100 = 18100
Before third temple: (18100+100)*2 + 100 = 36,500
Before second temple: (36500+100)*2 + 100 = 73300
Before first temple: (73300+100)*2 + 100 = 146900


Calculation of Donations By The Devotee
 
To conclude,  he had Rs. 146900 initially.  


 

Mathematical Coincidence

Messi entered a candy shop and spent half of the money in his pocket. When he came out he found that he had just as many paise as he had rupees when he went in and also half a many rupees as he had paise when he went in. How much money did he have on him when he entered? (1 Rupee = 100 Paise just like 1 Dollar = 100 Cents)



Know the amount that Messy had initially!!

Source 

Money For Mathematical Coincidence


What was the coincidence?

Let X be the rupees & Y be the paise that Messy initially had in his pocket. That means he had 100X + Y paise initially. In shop he paid half of the amount he had i.e. (100X + Y)/2. 

When he came out of the shop he had Y/2 rupees & X paise i.e. 100(Y/2) + X paise.

Equating both,

(100X + Y)/2 = 100(Y/2) + X

Multiply by 2,

100X + Y = 100Y + 2X

98X = 99Y


Hence, X = 99 & Y = 98
 
Putting these values in 100X + Y = 9998. This is the amount in paise. Converting in rupees gives, 9998/100 = 99.98

 
To conclude, Messi had Rs.99.98 initially in his pocket.


 

'The Wisest Son'

Once an old called his three sons. He gave them equal money and ask them to buy something that can fill their living room entirely. He told them that he will give all his money and property to the son who is able to do this task as asked.
The first son buys sticks and tries to fill the room but he falls short of sticks. The second son buys straw but he also falls short of filling the room. The third son buys only two things and he is able to fill the room completely and thus earns all the property and money.

What did he buy?

Test to find the wisest son

Here is what did he buy! 

Source 

The Wisest Son Deserves Rewards


What was the condition? 

The third son was very wise to choose his option (like you did while choosing to read this blog). There are two/three possible things that he might have bought.

1. He bought a candle & match box. Just ignited that candle in room & light filled in entire room.

2. He just bought perfume & sprayed in room. The smell of perfume occupied the entire room. 

 The test passed by the wisest son!
 
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