Puzzle on Logical Reasoning - There are six boxes containing

Question: There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in colour. Some boxes contain only red balls and others contain only blue.

One sales man sold one box out of them and then he says, "I have the same number of red balls left out as that of blue."

Which box is the one he solds out?

Answer:

He sold the box which contain 29 balls.

Solution: 

Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89

Total number of balls are odd. Also, same number of red balls and blue balls are left out after selling one box. So it is obvious that the box with odd number of balls in it is sold out i.e. 5, 7 or 29.

Now using trial and error method,
(89-29) /2 = 60/2 = 30 and
14 + 16 = 5 + 7 + 18 = 30

So box with 29 balls is sold out.

SET 2015 ( LAW) Exam - Complete Analysis

The Symbiosis Entrance Test (SET) 2015 for admission to SLS Pune, Noida & Hyderabad was conducted on May 2, 2015, all over India. SET has five sections of Logical Reasoning, Legal Reasoning, Analytical Reasoning, Reading Comprehension and G. K. It has 150 questions, to be attempted in 150 minutes. Each question carries one mark each. There was no negative marking for incorrect attempts.


SET 2015 was on expected lines with few repeat questions and typographical errors. Overall difficulty level for paper was ‘moderate’.

Read Complete Analysis at - http://www.endeavorcareers.com/gurus-speak/549-set-law-2015-analysis

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Puzzle on Logical Reasoning - One side of the bottom layer

Question:  One side of the bottom layer of a triangular pyramid has 12 balls. How many are there in the whole pyramid?

Note that the pyramid is equilateral and solid.

Answer:

There are total 364 balls.

Solution: 

As there are 12 balls along one side, it means that there are 12 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55, 66 and 78 balls in the remaining layers.

Hence, the total number of balls are
= 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78
= 364 balls.

SET ( BBA, BSc, BA, BMS, BCA) Exam Complete Analysis

Total Questions: 150
Total Marks: 150
Total Duration: 120 minutes
Negative Marking: No Negative Marking

Cut-offs: It is very difficult to predict exact cut-off for each college as the number of applicants is yet not disclosed. But different colleges considering SET General scores may consider the score in the range of 70 to 100.

Wishing all the students a great luck for results and for next stage of admission!

Read Complete Analysis at - http://www.endeavorcareers.com/gurus-speak/550-set-2015-analysis-for-bba-bsc-ba-bms-bca-exam

IMPORTANT:

Symbiosis group colleges that consider the SET General scores are:

•    SICSR [BCA / BBA (IT)],
•    SCMS (BBA) - Pune & Noida campus,
•    SIMC (BMS), SSE (B.Sc Economics),
•    SSLA (BA / BSc)

Students wishing to apply to any of the above mentioned colleges must apply to the respective college before the results are declared. Visit: http://set-test.org/payment.html for more details.

Looking for BBA or LAW Exam Preparation such as for SET, AILET or for CLAT Coaching visit www.endeavorcareers.com

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.

Important Do's and Dont's Few Days Before CLAT Exam

From the infancy we are taught that HARD WORK IS THE KEY TO SUCCESS, but it’s the time to break the conventional notion and change it to SMART WORK IS THE KEY TO SUCCESS. Especially the students who are targeting to clear their respective competitive examination with one month preparation, it is their survival mantra.

This article mainly deals with how to crack CLAT in less than one month. CLAT is not tough as medical or engineering exams are. But do not take this exam lightly if you are preparing for it during a crash course.


Read Complete Analysis at - http://www.endeavorcareers.com/gurus-speak/545-last-few-days-before-clat-exam-important-dos-and-dont




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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%.

Preparing for AILET Exam in Less Than 30 Days

AILET or the All India Law Entrance Test can be prepared by laying down a basic plan for 30 days. The plan can be made on a week basis. Three weeks of studying and reserving the remaining days for practicing Mock Tests.

In order to approach the content of the AILET examination systematically, it would be really quick to start with the parts and their marking distribution.

English Comprehension	35 Questions
GK & Current Affairs	35 Questions
Mathematical Ability	10 Questions
Legal Aptitude	35 Questions
Logical Reasoning	35 Questions

Key differentiator is that there is no negative marking for AILET. That means ANSWER ALL THE QUESTIONS

Read Complete Preparation Plan - http://www.endeavorcareers.com/gurus-speak/543-preparation-plan-for-ailet-in-last-30-days

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Puzzle on Probability - What are the chances that at least

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%.

Preparing for CLAT in Last 45 Days

CLAT or the Common Law Admission Test can be prepared by laying down a basic plan for 45 days. The plan can be made on a week basis. Five weeks of studying and reserving the remaining ten days for practicing Mock Tests.

1)    First Week
In order to approach the content of the CLAT examination systematically, it would be really quick to start with the parts and their marking distribution.  

English Comprehension	40 Questions
GK & Current Affairs	50 Questions
Mathematical Ability	20 Questions
Legal Aptitude	50 Questions
Logical Reasoning	40 Questions

Read Complete Analysis at - http://endeavorcareers.com/gurus-speak/542-how-to-prepare-for-clat-in-45-days

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Puzzle on Time, Speed and Distance - Two people enter a race in which you run to a point and back

Question:  Two people enter a race in which you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back.

Who came in first?


Answer:
Person A came in first.

Solution:
Let's assume that the distance between start and the point is D miles.

Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D

Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D

Thus, Person A is the Winner.