Puzzle on Line and Angle - What are the chances that at least two out of a group of fifty people

Question:  What are the chances that at least two out of a group of fifty people share the same birthday?

Answer:
The probability of at least two out of a group of 50 people share the same birthday is 97%

Solution:
Probability of at least two share the same birthday = 1 - probability of all 50 have different birthdays

Probability of all 50 have different birthday
= 365/365 * 364/365 * 363/365 * ... * 317/365 * 316/365
= (365 * 364 * 363 * 362 * ... * 317 * 316)/36550
= 0.0296264

Probability of at least two share the same birthday
= 1 - 0.0296264
= 0.9703735
= 97% approx.

Thus, the probability of at least two out of a group of 50 people share the same birthday is 97%.

Strategic Tips on How to Crack GD-PI of Top B-Schools

Most of you would have successfully managed to clear through the maze of several MBA entrance exams such as CAT, XAT, IIFT, CMAT, SNAP and other exams and now staring at a seemingly terrifying session of GD/PI ahead. Yes the process is tough, the panelists are extremely learned people who pick out the best for their college. Though it may be intimidating at first, a little practice and a lot of preparation can take you through. 

Here are 10 points that can make sure that you have a smooth sailing ahead.

1) Know Thyself
The importance of this cannot be stressed enough. Know yourself, your strengths, weakness, specialties and shortcomings and have a reason for any particular answer.


2) Tell about Thyself
Prepare a very solid answer to tell me about myself. Cover things like your family background education, work experience, extra circulars, hobbies and also why an MBA.

Read Complete 10 Points at - http://www.endeavorcareers.com/gurus-speak/505-best-tips-to-crack-gd-pi-of-top-b-schools

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.

Puzzle on Logical Reasoning - Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name....

Question:  Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name as follow.

Four of them have a first and middle name of Paresh.

Three of them have a first and middle name of Kamlesh.

Two of them have a first and middle name of Naresh.

One of them have a first and middle name of Elesh.

Pocketwala and Talawala, either both are named Kamlesh or neither is named Kamlesh.

Either Batliwala and Pocketwala both are named Naresh or Talawala and Chunawala both are named Naresh.

Chunawala and Natakwala are not both named Paresh.

Who is named Elesh?

Answer:    Pocketwala is named Elesh.

Solution: 

From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named Paresh. From (6) and (5), if Pocketwala or Talawala both are named Kamlesh, then either of them will have three names i.e. Paresh, Kamlesh and Naresh. Hence, Pocketwala and Talawala both are not named Kamlesh. It means that Batliwala, Chunawala and Natakwala are named Kamlesh. Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala is named Elesh.

Puzzle on Number System - What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?

Question:  What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?

Answer:    After Dividing we Get 5 as remainder

Solution: 

7! onward all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.

The only part to be consider is

= 1! + 2! + 3! + 4! + 5! + 6!

= 1 + 2 + 6 + 24 + 120 + 720

= 873

The remainder left after dividing 873 by 7 is 5

Hence, the remainder is 5.