What Was the Color of Gabbar Singh's Shirt ?

In Rangeelia, a neighbour situated west of our country, the native people can be divided into three types: Lalpilas, Pilharas and Haralals. 

Lalpilas always get confused between red and yellow (i.e. they see yellow as red and vice versa) but can see any other color properly. Pilharas always get confused between yellow and green (i.e. they see yellow as green and vice versa) but can see any other color properly. Haralals always get confused between green and red (i.e. they see green as red and vice versa) but can see any other color properly.

Three people Amar, Akbar and Anthony, who belong to Rangeelia, made the following statements about Gabbar Singh, the famous dacoit of Rangeelia, when he was last seen by them :-

Amar : Gabbar Singh was wearing a green shirt.

Akbar : Gabbar Singh was not wearing a yellow shirt.

Anthony : Gabbar Singh was wearing a red shirt.

If none of Amar, Akbar or Anthony is a Haralal, then what was the color of Gabbar Singh's shirt ?

THIS is the color of Gabbar Singh's shirt! 

Ultimate Test of 3 Logic Masters

Try this. The Grand Master takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the Grand Master's pocket and the two on her own forehead. He asks them in turn if they know the colors of their own stamps:

A: "No."

B: "No."

C: "No."

A: "No."

B: "Yes."

What color stamps does B have? 

'THIS' could be his color combination! 

Earlier logicians had been part of "Spot On The Forehead!" and  "Spot on the Forehead" Sequel Contest

The Wisest Logic Master!

What was the challenge in front of him?

Let's denote red by R and green by G. Then, each can have combination of RR, RG or GG.

So, there are total 27 combinations are possible.

1.  RR RR GG
2.  RR GG RR
3.  GG RR RR

4.  GG GG RR
5.  RR GG GG
6.  GG RR GG

7.  RR RG GG
8.  GG RG RR
9.  RG RR GG
10.RG GG RR
11. RR GG RG
12. GG RR RG

13. RR RG RG
14. GG RG RG
15. RG RR RG
16. RG GG RG
17. RG RG RR
18. RG RG GG

19. RR RR RG
20. GG GG RG
21. RG RR RR
22. RG GG GG
23. RR RG RR
24. GG RG GG

25. RR RR RR
26 .GG GG GG

27. RG RG RG.

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1. Now, obviously (19) to (26) are invalid combinations as those have more than 4 red or green stamps.

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2. In first round, everybody said 'NO' thereby eliminating (1) to (6) combinations. That's because, for example, if C had seen all red (or all green) then he would have known color of own stamps as GG (or RR). Similarly, A and B must not have seen all red or all green.

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3. For (9 - RG RR GG), A would have responded correctly at second round as NO of B had eliminated GG and NO of C had eliminated RR for A in first round. Similarly, (10) is eliminated.

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4. For (11 -  RR GG RG ), C would would have responded correctly immediately after NO of A had eliminated GG and NO of B had eliminated RR for him in first round. With similar logic, (12) also get eliminated.

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5. Remember, B has guessed color of own stamps only in second round of questioning.

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6. For B in (13), his logic would be I can't have RR (total R>4) but GG [ No.(11) - RR GG RG] and RG can be possible. But (11) is eliminated by C's response in first round. That leaves, (13) in contention.

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7. Similarly, if it was (14 - GG RG RG) combination, then B's thought would be - I can't have GG (total G>4) but can have RR as in (12) - GG RR RG which is already eliminated by C's NO response at the end of first round. Hence, I must have RG. That means (14) also remains in contention.

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8. On similar note, (17), (18) remains in contention after A's NO at the start of second round.

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9. If it was (15), then A would have been responded with RG when C's NO in first round eliminates RR (as proved in 2 above) and GG (as proved in 4 above) both. Similarly, (16) is also eliminated after C's NO in first round as above.

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11. For (27), B's logic would be -

"If I had RR then A must had seen RR-RG and had logic - 

"Can't have RR (total R>4); if had GG then C would have answered with RG after I and B said NO in first round itself. Hence, I must tell RG in second round."

Similarly, A's response at the start of second round eliminates GG for me.

Hence, I must have RG combination."

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10. So only possible combinations left where only B can deduce color of own stamps are -

7.  RR RG GG
8.  GG RG RR

13. RR RG RG
14. GG RG RG

17. RG RG RR
18. RG RG GG

27. RG RG RG.

If observed carefully all above, we can conclude that B must have RG color combination of stamps after observing A's and C's stamps as above to correctly answer in the second round.


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The Green-Eyed Logic Puzzle

In the green-eyed logic puzzle, there is an island of 100 perfectly logical prisoners who have green eyes—but they don't know that. They have been trapped on the island since birth, have never seen a mirror, and have never discussed their eye color.

On the island, green-eyed people are allowed to leave, but only if they go alone, at night, to a guard booth, where the guard will examine eye color and either let the person go (green eyes) or throw them in the volcano (non-green eyes). The people don't know their own eye color; they can never discuss or learn their own eye color; they can only leave at night; and they are given only a single hint when someone from the outside visits the island. That's a tough life!

One day, a visitor comes to the island. The visitor tells the prisoners: "At least one of you has green eyes." 

On the 100th morning after, all the prisoners are gone, all having asked to leave on the night before. 

How did they figure it out?


Here is the solution! 

The Green-Eyed Puzzle Solution

Here is that Puzzle! 

Nobody is going to dare to go the guard unless he is absolutely sure that he is green eyed; otherwise it would be suicidal move.

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For a moment, let's assume there are only 2 prisoners named A & B trapped on island.

On first day, A is watching green eyed B & B can see green-eyed A. But both are not sure what is color of their own eyes. Instead, A(or B) waited B(or A) to escape from island since he is green-eyed. Rather both are sure that other too doesn't know anything about color of own eyes.

On next morning both see each other still on island. Here is what A thinks.

If I was non green-eyed then B would have realized that the person pointed by visitor in his statement ('at least one of you have green eyes') is himself. Hence, B would have realized that he is green-eyed & could have escaped easily. Since B didn't try to escape that means I too must have green eyes.

So A can conclude that he too have green eyes. Exactly same way, B concludes that he too has green eyes. Hence, on next day both can escape from the island.
 
Note that, if the night of the day on which visitor made statement is counted then next day would have 1st morning & 2nd night since visitor's visit. Now since A & B left on 2nd night, we can't see anybody on 2nd morning next day.

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Now let's assume there are 3 prisoners named A,B and C trapped on island.

Let's think from A's point of view as an example.What he thinks.

Let me assume I don't have green eyes.Now each B and C could see 1 green-eyed & other non green-eyed person. But still they don't know color of own eyes.So on that night nobody tries to escape.

On first morning I see both B and C still present there.

Now B can think that if he has no green eyes then C could have concluded that the person pointed by visitor's statement ('at least one of you have green eyes') is C himself (as both A & B are non green-eyed. This way, C would have realized that he is green-eyed.

In a very similar way, B would have realized that he too is green-eyed.

Now both of them could have escaped on that night as they are sure that they are green-eyed.

But on the second morning, I see again both of them are still there. So now I can conclude that I too have green eyes.

If A can conclude then why can't B and C?  So after seeing each other on 3rd day, each of 3 can conclude color of eyes is green. Now on 3rd night they all can escape safely.

This is called as inductive logic.

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If observed carefully, 2 prisoners need 2 nights and 3 prisoners need 3 nights to logically deduce that the each of them is green-eyed.

Hence, 100 prisoners would require 100 nights to absolutely make sure that each of them is green eyed.

That's why on the 100th morning day, there is no prisoner present on island. 

Viral Maths Problem Confusing Students & Parents

There are 73 red, blue and green marbles in a jar. There are twice as many red marbles as blue marbles. There are 19 more marbles than green marbles. 

How many green marbles are there?

Check if you are correct! 

Solution of Viral Maths Problem Confusing Students & Parents

What was the problem?

Using Algebra : 

Let r,b and g be the numbers of red, blue and green marbles in the jar.

There are 73 red, blue and green marbles in a jar.

So r + b + g = 73    .....(1)

There are twice as many red marbles as blue marbles.

r = 2b                  ......(2)

There are 19 more blue marbles than green marbles. 

b = g + 19            

g = b - 19            ......(3)

Putting (2) and (3) in (1), gives

2b + b + b - 19 = 73

4b = 92

b = 23

Putting b = 23 in (3) gives,

g = 23 - 19 = 4

Putting  b = 23 in (2), gives 

r = 2x23 = 46

So there are 46 red,23 blue and 4 green marbles in the jar.

Without Algebra :

In the case, we need to try trail and error method.

If g = 1, then b = 20 and r = 2(20) = 40 giving total 40 + 20 + 1 = 61.

If g = 2, then b = 21 and r = 2(21) = 42 giving total 42 + 21 + 2 = 65.

If g = 3, then b = 22 and r = 2(22) = 44 giving total 44 + 22 + 3 = 69.

Total is increasing at the rate of 4. So finally,

If g = 4, then b = 23 and r = 2(23) = 46 giving total 46 + 23 + 4 = 73.


So there are 46 red,23 blue and 4 green marbles in the jar.

Who Lives Where?

There are 4 big houses in my home town.
They are made from these materials: red marbles, green marbles, white marbles and blue marbles.
* Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles.

* Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house.

* Mr Danny lives in the fourth house, while the first house is not made from red marbles.
Who lives where, and what is their house made from ?

Know about location and home about each!

Owners Of Homes in Home Town

What were the clues? 

To simplify the process, let's number the houses as 1,2,3,4 and separate clues given.

a. Mrs Jennifer's house is somewhere to the left of the green marbles one.
b. The third one along is white marbles.
c. Mrs Sharon owns a red marbles house
d. Mr Cruz does not live at either end.
e. Mr Cruz lives somewhere to the right of the blue marbles house.
f. Mr Danny lives in the fourth house
g. The first house is not made from red marbles.

As per clues (b) and (g), 1 and 3 aren't made of red marbles. Since Danny is living in 4 as per (f), the red house owned by Mrs. Sharon must be positioned at 2.

Since 2 is already occupied, as per (d), Mr. Cruz must be living at number 3 which is of white marbles according to (b).

According to (a), 4 must be green marbles since otherwise Jennifer wouldn't be somewhere at left. As per (f) this 4 th house is owned by Mr.Danny.

The only house left for Jennifer is blue positioned at 1.

To conclude,

Mrs Jennifer - blue marbles at Number 1
Mrs Sharon - red marbles at Number 2
Mr Cruz - white marbles at Number 3
Mr Danny - green marbles at Number 4

Clues From Talk About Boats

At the local model boat club, four friends were talking about their boats. There were a total of eight boats, two in each color, red, green, blue and yellow. 

1. Each friend owned two boats. 

2. No friend had two boats of the same color.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

5. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. 

6. Charles had a yellow boat. 

7. Darren had a blue boat, but didn't have a green one.

Can you work out which friend had which colored boats? 

Know here step by step process of finding owners! 

Source 

Owners Of Boats

What's the task given? 

First let's rewrite all the clues here.

1. Each friend owned two boats. 

2. No friend had two boats of the same color.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

5. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. 

6. Charles had a yellow boat. 

7. Darren had a blue boat, but didn't have a green one.

Let's make 2 set of 4 colors of boats. Below is the table with owners in row & color of boats in columns.  

Below are clues which giving clear idea of owner of particular colored boat.

3. Alan didn't have a yellow boat.

4. Brian didn't have a red boat, but did have a green one.

6. Charles had a yellow boat.

7. Darren had a blue boat, but didn't have a green one.

We will fill this table one by one as per clues given.

Now, consider the first part of this clue.

5. One of the friends had a yellow boat and a blue boat.

Now Alan won't have this combination as he doesn't own yellow boat at all. The Brian already had green boat, so he too can't have this combination as in that case he would own 3 boats.

Let's assume Charles had this combination then other would have boats as below.

The 500 Miles Race

Mike, Jimmy, Nader, Kevin and Larry were the top five finishers in the regional 500-mile race. They drove yellow, orange, green, red and blue cars but not necessarily in that order.

1. Neither Kevin nor Larry drove the green car.

2. Kevin finished faster than Mike and Larry.

3. The blue car finished earlier than Larry’s and Nader’s car.

4. The yellow car finished faster than the green car and the orange car.

5. Mike’s and Larry’s car finished ahead of the orange car.

6. Jimmy’s car finished before the blue and the yellow car.

Who drove what color car and what place did each driver finish?

To know that click here! 

Source 

Stats of The 500 Miles Race

Here is the scenario! 

Let's make a bit simple by putting clues into table. Rows representing owner while columns representing their cars. Now we need to use logical deduction by using all the clues & fill the table one by one. Finally, we get this.

Stats of The 500 Miles Race

 Filled circle represents rank of the particular car or participant. e.g. 1 circle means it's first, 2 means second & so on.