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?

THIS should be the maximum sum! 

Source

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.

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.

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
    

Maze Challenge For a Rat?

A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.

What is the expected amount of time it will take for the rat to finish the maze?

This could be the average time that rat needed!
 

A Rat Finishing off The Maze!

The challenge ahead of rat?

For rat, there are 2 paths viz A (5 minutes) and C (3 minutes) leading to the end while paths B (8 minutes) and D (2 minutes) lead to the start again.

Since, there are 4 paths & each having equal chance of being chosen by rat, there is 1/4 th chance for each path for to be chosen by rat.

Let's assume T be the time needed for rat to finish the maze. 

But if rat selects path B or D then rat need T more time again as these paths lead to the start of the maze again.

Hence,

T = (1/4) x A + (1/4) x B + (1/4) x C + (1/4) x D

T =  (1/4) x 5 + (1/4) x (8 + T) + (1/4) x 3 + (1/4) x (2 + T)

T = (5/4) + (2) + (T/4) + (3/4) + (1/2) + (T/4)

T =  (9/2) + (T/2)

T/2 = 9/2

T = 9

That is rat needs 9 minutes to finish the maze. 

 

A World Class Swimmer's Puzzle

A world-class swimmer can swim at twice the speed of the prevailing tide.

She swims out to a buoy and back again, taking four minutes to make the round-trip.

How long would it take her to make the identical swim in still water?

Click here for the answer!

Solution of A World Class Swimmer's Puzzle

What was the puzzle?

 Let C be the speed of water current then the speed of a world class swimmer will be 2C.

When she swims out to a buoy located at a distance say D and back again, the time needed is - 

D/(C+2C) + D/(2C-C) = 4 minutes

D/3C + D/C = 4

Multiplying both the sides by 3, 

D/C + 3D/C = 12

4D/C = 12

D/C = 3  

Now when she swims out to a buoy and back to shore again in still water, she needs only time -

D/2C + D/2C 

Since D/C = 3, then D/2C = 3/2

Hence, 
 
D/2C + D/2C  = 3/2 + 3/2 = 3.

That is, she needs only 3 minutes to swim out to a buoy and back to shore again in still water.

 
 

Puzzle : The Case of Missing Servant

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. 

For example, servant 14 comes in and says "Servant 14, reporting in." 

One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one.

Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. 

Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. 

The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him.

How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?

Mathematical Trick to know the missing servant! 

Solution : The Missing Servant in the Case

What was the case?

When the first servant comes in, the king should write his number on the small piece of paper. For every next servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper while erasing old one.

Addition of numbers from 1 to 100 = 5050.

Hence, 

Missing Servant Number = 5050 -  Addition of ranks of 99 Servants.

So, depending on how far the addition of 99 servants' rank goes to near 5050, the king can easily deduce the rank of missing servant.

For example, if the addition that king has after 99 servants report in is 5000 then the servant having rank = 5050 - 5000 = 50 must be missing. 

 

Puzzle : Ants Walk on a Stick

Twenty-five ants are placed randomly on a meter stick. Each faces east or west. At a signal they all start to march at 1 centimeter per second. Whenever two ants collide they reverse directions. How long must we wait to be sure that all the ants have left the stick?

This sounds immensely complicated, but with a simple insight the answer is immediately clear. What is it?

You need to wait for.....seconds only!
 

Analysing Ants Walk on a Stick

Read the question associated with the walk.

For a moment, let's assume that there are only 2 ants 20 cm away from either end of the stick. Now, after 30 seconds they both will collide with each other & will reverse the direction.

At 50th seconds they will be at the end of the sticks falling off the stick.

So after 80 seconds they will fall off the stick. Now, imagine if ants avoid collision & pass through (or above) each other. Still, both ants would need 80 seconds to leave the stick.

In short, 2 ants' collision & reversal in direction is equivalent to their passing through each other. The other ant continues the journey on the behalf of the first ant & vice versa.

And in case, if they were 100 cm apart, they would need 100 seconds to get off the stick. Again, after collision at halfway mark here, the other ant travels the rest of distance that other ant was supposed to travel.

On the similar note, we can say that even if there are 25 ants on the stick then each ant will cover some distance on the behalf of some other ant. And we need to wait for maximum 100 seconds if 1 of 25 ants is at the edge of the stick. 

All 25 ants together completes the journey of each others in 100 seconds. The ant which is at the edge of stick might complete journey of some other ant which might be only 10 seconds long. But the 100 seconds journey of that ant will be shared by rest of ants. 

7 Additions and 1 Subtraction

Fill the boxes below with only 7 addition (+) and 1 subtraction (-) operators. 

The smartest way to find the right place 

Correct Placement of Subtraction Operator

This was the question!

For a moment. let's put '+' operator in all given boxes.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 33.

Let 'x' be the number ahead of which '-' should be there where x can be any number from 1 to 9. 

In above addition the x is already (unknown value) added so first we need to subtract it twice from LHS; first to compensate the addition operation & second for the mandatory subtraction operator that is to be placed at right place.

So for example, if it's ahead of 4 as right place then

1 + 2 + 3 + 4 - 4 - 4 + 5 + 6 + 7 + 8 + 9 = 33.

1 + 2 + 3 + (4 - 4) - 4 + 5 + 6 + 7 + 8 + 9 = 33.

For x, which is already there & can be any from 1 to 9,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9  - x - x = 33. 

45 - 2x = 33

2x = 12

x = 6.

Hence, the subtraction (-) sign should be ahead of 6. 

A Brillian Deception

A witch owns a field containing many gold mines. She hires one man at a time to mine this gold for her. She promises 10% of what a man mines in a day, and he gives her the rest. Because she is blind, she has three magic bags who can talk. They report how much gold they held each day, and this is how she finds out if men are cheating her. 

Upon getting the job, each man agrees that if he isn't honest, then he will be turned into stone. So around the witch's mines, many statues lay! 

Now comes an honest man named Garry. He accepts the job gladly. 

The witch, who didn't trust him said, "If I wrongly accuse you of cheating me, then I'll be turned into stone."  

That night, Garry, having honestly done his first day's job, overheard the bags talking to the witch. He then formulated a plan... 

The next night, he submitted his gold, and kept 1.6 pounds of gold. Later, the witch talked with her bags.

The first bag said it held 16 pounds that day. The second one said it held 5 pounds. The third one said it held 2 pounds. 

Beaming, the witch confronted Garry. "You scoundrel, you think you could fool me. Now you shall turn into stone!" the witch cried. One second later, the witch was hard as a rock, and very grey-looking.

How did Garry brilliantly deceive the witch? 

Here is the Garry's Master Plan!
 

Mathematics Behind The Brilliant Deception!

What was the deception?

As per the honest man, he must have mined 16 Pounds of gold since he kept 1.6 Pounds (10% of 16) gold for himself.

And as per magical bags, since Bag no.1 said it had 16 Pounds, Bag no.2 said 5 Pounds and Bag no.3 said 2 Pounds the total gold mined 16 + 5 + 2 = 23 Pounds.

We know, honest man Garry would never do any fraud and neither of magical bag would lie. 

So, it's clear that some of pounds are counted multiple times by magical bags. There are 23 - 16 = 7 Pounds gold extra found by those bags.

Now, Bag No.3 must have 2 Pounds in real.

And to make count of Bag No.2 as 5, Garry must had put 3 Pound + Bag No.3 itself in Bag No.2. Hence, the Bag No.2 informed witch that it had total 5 Pound of gold.

Finally, to force Bag No.1 to tell it's count as 16, Garry must had put 
11 Pounds + Bag No.2 (5 Pounds = 3 + Bag No.1) = 11 + 5 = 16 Pounds.

In short, Garry put 2 pounds of gold in Bag No.3 and put that Bag No.3 in Bag No.2 where he had already added 3 Pounds of gold. After that, he put this Bag No.2 in Bag No.3 where he had already added 11 Pounds of gold (somewhat like below picture).

 

This way, 2 Pounds of gold in Bag No.3 are counted 2 extra times and 3 Pounds gold of Bag No.2 are counted 1 more extra time. That is 2 + 2 + 3 = 7 Pounds of extra gold found by bags.

A Warden Killing The Boredom

A warden oversees an empty prison with 100 cells, all closed. Bored one day, he walks through the prison and opens every cell. Then he walks through it again and closes the even-numbered cells. On the third trip he stops at every third cell and closes the door if it’s open or opens it if it’s closed. And so on: On the nth trip he stops at every nth cell, closing an open door or opening a closed one. At the end of the 100th trip, which doors are open?

'These' special cells will have doors open!

Similar Puzzle! 

Open Doors in an Empty Prison!

Read the story behind the title!

Let's take few cells into consideration as representatives.

The warden visits cell no. 5 twice - 1st & 5th trip.1 & 5 are only divisors of 5.

He visits cell no. 10, on 1st, 2nd, 5th, 10th trips i.e. 4 times. Here, divisors of 10 are 1,2,5,10.

He stops cell no. 31 only twice i.e. on 1st and 31st trip.

He visits cell no.25 on 1st, 5th, 25th trips that is thrice.

In short, number of divisors that cell number has, decides the number of visits by warden.

For example, above, cell no.5 has 2 divisors hence warden visits it 2 times while 10 has 4 divisors which is why warden visits it 4 times.

But when cell number is perfect square like 16 (1,2,4,8,16) or 25 (1,5,25) he visits respective cells odd number of times. Otherwise for all other integers like 10 (1,2,5,10) or like 18 (1,2,3,6,9,18)  or like 27 (1,3,9,27) he visits even number of time.

For prime numbers, like 1,3,5,....31,...97 he visits only twice as each of them have only 2 divisors. Again, number of visits is even.

Obviously, the doors of cells to which he visits even number of times will remain closed while those cells to which he visits odd number of time will remain open.

So the cells having numbers that are perfect square would have doors open. That means cells 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 would be having their doors open.

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?

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.

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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. 

How much does each container weigh?

There are five containers of oil of different weight. They are weighed in pairs of two with all possibilities. The weights in kgs are 165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . 

How much does each container weigh?

Weight Calculation of each container is here!