Puzzle on Logical Reasoning - A blindfolded man is asked

Question:  A blindfolded man is asked to sit in the front of a carrom board. The holes of the board are shut with lids in random order, i.e. any number of all the four holes can be shut or open.

Now the man is supposed to touch any two holes at a time and can do the following.

    Open the closed hole.
    Close the open hole.
    Let the hole be as it is.

After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of what are the holes which are open or closed.

How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes?

Note that whenever all the holes are either open or close, there will be an alarm so that the blindfolded man will know that he has won.

Answer:
The blindfolded man requires 5 turns.

Solution:
1.   Open two adjacent holes.

2.  Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close.

3.  Check two diagonal holes.
      -> If one is close, open it and all the holes are open.
      -> If both are close, open any one hole. Now, two holes are open and two are close. The diagonal holes are in the opposite status i.e. in both the diagonals, one hole is open and one is close.

4.  Check any two adjacent holes.
      -> If both are open, close both of them. Now, all holes are close.
      -> If both are close, open both of them. Now, all holes are open.
      -> If one is open and one is close, invert them i.e. close the open hole and open the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close.

5.  Check any two diagonal holes.
     -> If both are open, close both of them. Now, all holes are close.
     -> If both are close, open both of them. Now, all holes are open.

Puzzle on Logical Reasoning - At the Party there were 9 men and children......

Question: At the Party:

• There were 9 men and children.
• There were 2 more women than children.
• The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.
•  Also, of the three groups - men, women and children - at the party:
• There were 4 of one group.
• There were 6 of one group.
• There were 8 of one group.

Exactly one of the above 6 statements is false.

Can you tell which one is false? Also, how many men, women and children are there at the party?

Answer:    Statement (4) is false. There are 3 men, 8 women and 6 children.

Solution: 

Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false. So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true. From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women. If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible. Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.

Puzzle on Calender - Yesterday in a party, I asked Mr. Shah his birthday.

Question: Yesterday in a party, I asked Mr. Shah his birthday. With a mischievous glint in his eyes he replied. "The day before yesterday I was 83 years old and next year I will be 86."
Can you figure out what is the Date of Birth of Mr. Shah? Assume that the current year is 2000.

Answer:    Mr. Shah's date of birth is 31 December, 1915

Solution: 

Today is 1 January, 2000. The day before yesterday was 30 December, 1999 and Mr. Shah was 83 on that day. Today i.e. 1 January, 2000 - he is 84. On 31 December 2000, he will be 85 and next year i.e. 31 December, 2001 - he will be 86. Hence, the date of birth is 31 December, 1915. Many people do think of Leap year and date of birth as 29th February as 2000 is the Leap year and there is difference of 3 years in Mr. Shah's age. But that is not the answer.

Puzzle on Liner Equation - There are four groups of Mangoes, Apples and Bananas

Question:  There are four groups of Mangoes, Apples and Bananas as follows:
Group I : 1 Mango, 1 Apples and 1 Banana
Group II : 1 Mango, 5 Apples and 7 Bananas
Group III : 1 Mango, 7 Apples and 10 Bananas
Group IV : 9 Mango, 23 Apples and 30 Bananas

Group II costs Rs 300 and Group III costs Rs 390.

Can you tell how much does Group I and Group IV cost?

Answer:    Group I costs Rs 120 and Group IV costs Rs 1710

Solution: 

Assume that the values of one mango, one apple and one banana are M, A and B respectively. From Group II : M + 5A + 7B = 300 From Group III : M + 7A + 10B = 390 Subtracting above to equations : 2A + 3B = 90 For Group I : = M + A + B = (M + 5A + 7B) - (4A + 6B) = (M + 5A + 7B) - 2(2A + 3B) = 300 - 2(90) = 300 - 180 = 120 Similarly, for Group IV : = 9M + 23A + 30B = 9(M + 5A + 7B) - (22A + 33B) = 9(M + 5A + 7B) - 11(2A + 3B) = 9(300) - 11(90) = 2700 - 990 = 1710 Thus, Group I costs Rs 120 and Group IV costs Rs 1710.