Fair Distribution Of Water

In Sahara desert , 3 men found a big 24L Jar is full of water. Since there is shortage of water so they decided to distribute the water among themselves such that they all have equal amounts of it. But they only have a 13L, a 5L and an 11 liter Jar.

How do they do it? 

Here is how to do it!

Source 

Equal Distribution Of Water

What was the challenge?

1. First pour 24L into 13L and 11L jar.There will be no water in 24L jar.

2. Now pour 13L jar into 5L jar till 5L is filled. So 8L of water will be left in 13L jar.

3. This 8L of water from 13L jar is emptied in empty 24L jar.This will leave 13L jar empty.

4. Now pour 11L water from 11L jar into 13L of jar. There is still space for 2L of water in 13L jar.

5. Pour 5L jar into 13L of jar which had space for 2L only. So 3L of water will be left in 5L of jar.

6. This 3L of water is emptied in 11L jar.

7. A 13L jar full of water is again poured into 5L jar leaving behind 8L of water in it.

8. A 5L jar full of water is finally poured in 11L jar already having 3L of water. 

9. This way, 24L, 11L and 13L of jar would have 8L of water each.

Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 

     SEND
  + MORE
  ----------
 = MONEY

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.

To conclude,

  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2 

Burn The Strings

You have two strings whose only known property is that when you light one end of either string it takes exactly one hour to burn. The rate at which the strings will burn is constant and strings are identical.

How do you measure 45 minutes?

That's how to measure 45 minutes!

Timer Using Burning Strings

What was the challenge? 

1. Burn both the ends of 1st string and 1 end of other string.

2. After 30 minutes, 1st string will be burnt out and half of the other string will be burnt out.So far we have counted 30 minutes.

3. Now burn the other end of 2nd end. It will take 15 minute to burnt out totally. This way, we have counted 30 + 15 = 45 minutes using 2 strings.

"Your Surname Tells My Shirt's Color!"

Mr. Yellow, Mr. Black and Mr. Brown, three best friends since kindergarten meet in a function after 5 years. The three of them are wearing either a Yellow, Black or Brown shirt.
After giving each other a friendly hug, Mr. Black says, "Hey, did you notice that we are wearing a different colored shirt than our names!"
The man wearing a Brown shirt said, "Wow I certainly did not notice that but you are right Mr. Black."
Based on this conversation, can you find our who was wearing which colored shirt? 

 
Know color of each person's shirt here! 
  

Surnames And Shuffling of Shirts

Read this conversation first!

One thing is sure that Mr.Black is not wearing the BLACK shirt. He is also not wearing BROWN as well as we found the other person wearing the BROWN colored shirt is agreed with the statement made by Mr.BLACK in a conversation. So only color left is the YELLOW & Mr. Black must be wearing that.

Now, the colors left are BROWN and BLACK and persons left are Mr. Yellow and Mr.Brown. Since, each of them wearing different color than his surname, Mr. Brown must be wearing the BLACK shirt. And hence, Mr. Yellow is wearing the BROWN shirt! 

To conclude, Mr.Black is wearing YELLOW shirt, Mr. Brown is wearing BLACK shirt & Mr. Yellow must be wearing BROWN shirt.

 

Test of an Examiner

Five students - Adam, Cabe, Justin, Michael and Vince appeared for a competitive exam. There were total five questions asked from them from which were two multiple choice questions (a, b or c) and three were true/false questions. Their answers are given as follows:

Name I II III IV V

Cabe c b True True False

Adam c c True True True

Justin a c False True True

Michael b a True True False

Vince b c True False True

Also, no two students got the same number of correct answers. Can you tell the correct answer? Also, what are their individual score?

Know correct answers and scores here!

Source 

Responding To Test of an Examiner

What was the test?

There are 2 possibilities of scores & that are either 0,1,2,3,4 or 1,2,3,4,5. First of all, let's arrange students' responses in order like below.

Table 1

What we notice here is that, there are few responses to same question by different student matching.

For the Question III, only Justin given different answers than other.

Case 1 : If we assume Justin's answer is correct then rest of all are wrong in response to Question III. That means either maximum score in test is 4 or Justin himself has scored 1 to 5.

Let's test that apart from Justin who can have score of 4. If any body other scores 4 then he must share at least 3 similar answers with other (excluding Answer III; refer image below). Only Adam has exact 3 matching responses with Justin.

Table 2

If Adam's score is 4 (Answers to I, II, IV, V are correct) then, Justin too would score 4 (Answers to II, III, IV,V are correct) since Adam & Justin have same responses to Questions II, IV,V).
  
If nobody scoring as 4 then Justin can have score of 4 or 5.

Case 1.1 : If his score is 4 then there has to be somebody has to be there scoring 0. Now Vince and Adam has at least 2 responses matching with the Justin. That means they can't score 0 since even 1 answer is wrong as Justin the other must be correct as Justin. Michael or Cabe can have 0 score in the case. If anybody of them has score 0 then answer as a TRUE to the Question IV is incorrect i.e. correct Answer IV is FALSE. So Justin is WRONG in Answer IV only. In short, a, c, FALSE, FALSE, TRUE is correct combination of answers. But thing is here in the case both Michael and Cabe would have score 0! Hence Justin's score can't be 4 too.

Case 1.2 : If Justin's score is 5, then a, c, FALSE, TRUE, TRUE are the right answers. No one would score 4 in that case with 3 as second highest by Adam.

An Experimental Professor

An eccentric professor used a unique way to measure time for a test lasting 15 minutes.
He used just two hourglasses. One measured 7 minutes and the other 11 minutes.
During the whole time he turned the hourglasses only few times. 

How did he measure the 15 minutes?

Couple of methods to count 15 minutes. 

Source 

15 Minutes Countdown

What was the challenge?

There were 2 ways for professor to count 15 minutes using 11 & 7 minutes hourglasses.

Method 1 :

1. He started both 7 & 11 minutes hourglasses but didn't begin test.

2. After 7 minutes hourglass ran out, he started the test. Still 11 one yet to count 4 minutes.

3. After conducting test for 4 minutes, 11 hourglass ran out.

4. Now he turns around 11 hourglass & continues. So 11 + 4 = 15 minutes counted.

Method 2 :

1. He started both the hourglasses & started to conduct the test as well.

2. When 7 minutes ran out he turned around it & kept 11 minutes counting.

3. After 4 minutes, 11 minutes hourglass ran out. Meanwhile 7 minutes hourglass also counted 4 minutes. So far 11 minutes counted.

4. Then he again turned around 7 minutes hourglass which had counted 4 minutes. That's how 7 minutes counted 4 minutes again.

5. In this way, 11 + 4 = 15 minutes counted.

A Check Post At Each Mile

A poor villager grows mango in his land and sells them in the town. The town is 1000 miles away from the village. He has rented a truck for transporting the mangoes to the town. The truck can carry 1000 mangoes at one time and this season, he was able to yield 3000 mangoes.

There is a problem. At each mile till the town, there is a check post at which he must give one mango each while traveling towards the town. However, if he is traveling from the town towards his village, he won’t have to give anything.

Transportation Truck

Tell a way in which the villager can take highest possible number of mangoes to the town.

Know here that highest possible number!

Source 

Smart Saving At Check Posts

How much each check post charging?

Obviously, he can't make 3 trips from town to village straightaway as in that case he wouldn't have anything left (3 x 1000 mangoes paid).

So he need to divide the journey into parts. While breaking journey into parts he has to make sure that after each part he will need less trips to complete the next part.

Now if somehow he pays 1000 mangoes in first part of the journey then for next part he has to make only 2 trips to carry 2000 mangoes.

Part 1 : Hence, he should first make 3 trips till 333 miles. In this part, he would pay 3 x 333 = 999 mangoes leaving 3000 - 999 = 2001 mangoes in stock.

Part : 1

Part 2 : He should leave 1 mango here & take 2000 mangoes further. For next part, he need to make at least 2 trips for 2000 mangoes. In order to save number of trips in next part some how he need to make mangoes in stock less than 1000. For that he should make 2 trips 500 mile further. So he will pay 2 x 500 = 1000 mangoes but having 2000 - 1000 = 1000 mangoes in stock. Still he has to travel 1000 - 500 - 337 = 167 miles.

Part : 2

Part 3 : For next 167 miles, he need to make only 1 trip of 1000 mangoes where he will pay 167 mangoes leaving 1000 - 167 = 833 mangoes. 

Part : 3

This is how he can save 833 mangoes in entire journey. 

Wrong Looking Correct Mathematical Equation!

The following question it puts forth you:

25 - 55 + (85 + 65) = ?

Then, you are told that even though you might think its wrong, the correct answer is actually 5!

Whats your reaction to it? How can this be true? 

 That's how it's perfectly correct!

That's How Equation is Correct!

Why it was looking wrong? 

If you read the data carefully then you will notice '!' attached to number 5 which is being claimed answer. Actually claimed answer is 5! not 5 Read it again...

"Then, you are told that even though you might think its wrong, the correct answer is actually 5!."

Now use of '!' is not limited to the sentences only. In mathematics it's a 'factorial'.

So 5! = 5 x 4 x 3 x 2 x 1 = 120 and 25 - 55 + (85 + 65) = 120 and hence,

25 - 55 + (85 + 65) = 5! 

Now doesn't it look the correct equation? 

How Accurate You are?

In a competitive exam, each correct answer could win you 10 points and each wrong answer could lose you 5 points. You sat in the exam and answered all the 20 questions, which were given in the exam.

When you checked the result, you had scored 125 marks in the test.
Can you calculate how many answers given by you were correct and wrong ?

These should be those numbers! 

Source 

  

Analysis Of Your Result

What was the test ?

Let C be the number of correct answers and W be the number of wrong answers.

Since there are 20 question in total,

C + W = 20    .....(1)

and the score 125 must be subtraction of marks obtained for correct answer and marks due to wrong answers.

10C - 5W = 125   .....(2)

Multiplying (1) by 5 and then adding it to (2),

5C + 5W + 10C - 5W = 100 + 125

15C = 225

C = 15.

From (1), W = 20 - C = 20 - 5 = 5.

Hence, your 15 answers are correct while 5 answers are wrong.

Who Will Be Not Out?

It is a strange cricket match in which batsman is getting bowled in the very first ball he faced. That means on ten consecutive balls ten players get out.

Assuming no extras in the match, which batsman will be not out at the end of the innings?  

Know that lucky player!

Source 

"He Will Be Not Out!"

What happened in the match?

First let's number all the players from 1 to 11 as Batsman 1, Batsman 2, Batsman 3 & so on with last player as Batsman 11. 

Now let's take a look at what must have happened during 1st over.

1st Ball : Batsman-1 got out
2nd Ball : Batsman-3 got out
3rd Ball : Batsman-4 got out
4th Ball : Batsman-5 got out
5th Ball : Batsman-6 got out
6th Ball : Batsman-7 got out

Batsman 8 comes in 

Batsman 2 is still standing at non-striker end watching fall of wickets. Remember in the match all batsman are bowled out so no change in strike because of run out or before catch is taken.

At the end of first over, the strike is rotated and Batsman 2 comes on strike while Batsman 8 at the non striker end.

Now here is what happens in second over.

1st Ball : Batsman-2 got out
2nd Ball : Batsman-9 got out
3rd Ball : Batsman-10 got out
4th Ball : Batsman-11 got out

So the only batsman left NOT OUT is Batsman 8 standing at the non-striker end.