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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?


Knowing Correct Answers And Evaluating Scores

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.

Assessment of students' responses
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.

Assessment of students' responses
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. 


Measure 15 minutes usnig 7 & 11 minutes hourglasses

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.

Counting 15 minutes using 7 & 11 minutes hourglasses

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.

Dealing at the every chech post per mile!
Transportation Truck

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

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.

Taking 3000 Mangoes Across 1000 Miles Smartly!
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.


Taking 3000 Mangoes Across 1000 Miles Smartly!
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. 

Taking 3000 Mangoes Across 1000 Miles Smartly!
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? 


How this could be possible?

 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? 

Use of ! in mathematics

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 ?

How many correct answers?

These should be those numbers! 

  

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.


Analysis of your marks scores in exam

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?  

A Strange Cricket Match

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


Batman 8 will remain NOT OUT!

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

What's Next in the Series?

Can you find the next number in series?
5, 8, 17, 47, 242, ? 

What would be the next number?

The Next Number In The Series


What was the series?

Let's take a look at the series once again.

5, 8, 17, 47, 242, ? 

Here if we observe carefully, we can notice that,

5^2 - 8 = 17
8^2 - 17 = 47
17^2 - 47 = 242



In short, (n the number) = [Square of (n-2) th number] - [(n-1) th number].

Hence,

47^2 - 242 = 1967.  


This should be the next number!
I Found! Did you?

A Mathematical Buy!

In a classic wine shop in Flobecq, Belgium, list of three most popular wines are:

- The cost of 1 French wine bottle: 500$
- The cost of 1 German wine bottle: 100$
- The cost of 20 Dutch wine bottles: 100$


Homer Simpson entered the wine shop and he needs to buy


- All three types of wine shop.
- Needs to buy Dutch wine bottles in multiple of 20.
- Need to buy 100 wine bottles


Simpson has only 10000$. How many wine bottle(s) of each type, Simpson must buy? 


A Mathematical Shopping Challenge!

Simpson must buy.......Read More.... 

Source 
 

A Buy To Be Mathematical...


What was needed to buy?

Let's recollect the data where cost of each kind of wine bottle is listed.

- The cost of 1 French wine bottle: 500$
- The cost of 1 German wine bottle: 100$
- The cost of 20 Dutch wine bottles: 100$ (Cost of 1 bottle = 5$)

 

Now let F be the number of French bottle, G be the number of German bottles and D be the number of Dutch bottle that Simpson should buy.

F + G + D = 100

G = 100 - F - D   .....(1)

Total cost of all bottle must be $10000.

500F + 100G + 5D = 10000

Substituting (1) in above,

500F + 100(100 - F - D) + 5D = 10000

500F + 10000 - 100F - 100D + 5D = 10000

400F - 95D = 0

400F = 95D 
 
80F = 19D

D/F = 80/19

Possible values of D and F are 80 and 19 respectively.

From (1),

G = 100 - 19 - 80 =  1.

Let's verify if all these fits in his budget or not.  

19 French wine bottles would cost 19 x 500 = 9500, 1 German wine would cost = 1 x 100 = 100 and 80 Dutch wine bottles would cost 80 x 5 = 400. Remember we have got number of Dutch bottles in multiple of 20. Hence total cost = 9500 + 100 + 400 = 10000.

Hence with $10000, Simpson should buy 19 French, 1 German and 80 Dutch wine bottles if conditions of buying 100 bottles & Dutch bottles in multiple of 20 are applied.   

A mathematical challenge accepted

 

Just Try To Crack It!

Can you tell the correct key?

Can you find the correct code?

Here is the step-by-step process!

Source 

Cracking of The Code in Steps...


What was the challenge?

Let's number the clues as 1, 2, 3.

Clues numbered for cracking the code

Now following step by step process here onward. 

1. The numbers 3 & 1 are common in first & third combinations. Now both must not be the part of original number as in that case Clue 1 will be invalid. 

2. If numbers 3 & 1 are not correct in third combination then the correct 2 numbers must be among 5,7,9.

3. But it can't be both 7 and 9 as again that would make clue 1 invalid! Hence, the 5 is part of the original key & in correct position as in third combination. So we have got first digit of key as a 5.

4. The 1 correct number in clue 2 is 5 & that's in wrong position. If other is assumed to be 9 & to be in right position then it contradict the clue 3. So the second digit must be 7.

5.The only correct number in clue 1 is 7 & that's in wrong position. That means numbers 1,3,4,9 must not be the part of the code.

6. Since 3,4,9 eliminated in previous steps, the only number that is correct and in right position must be 6 in suggestion made by clue 2. So far we have got 3 digits of the code as 576XX.

7. Last 2 digits can be any combination from 0,2,5,7,6,8. Now addition of all digits is equal to the number formed by last 2 digits. It's impossible that the addition of all digits exceeds 50. Hence, the second last digit must be 2.

8. Now both 57620 or 57628 are perfectly valid where sum of all digits equals to number formed by last 2 digits. 

2 Possible codes discovered!
  
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