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IIM-Bangalore to Conduct CAT 2016 (News)

The CAT (Common Admission Test), one of the most popular management entrance tests in India, will be conducted by IIM-B in 2016. Considered as the gateway to prestigious IIMs and top B-Schools, every year lakh of students across India take up the test. CAT 2015 was conducted on 29th November, and around the same time the exam is likely to be conducted this year. A preliminary discussion will be conducted in the 1st week of June, 2016. Read more…

Exam patterns of CAT 2016

The Common Admission Test had undergone a number of changes in the last few years. However, there are no new changes in CAT exam patterns this year. Like 2015, CAT will consist of 3 Sections – Quantitative Aptitude (QA), Data Interpretation & Logical Reasoning (DILR) and Verbal and Reading Comprehension (VRC). A total of 100 questions would be asked – QA (34 questions), VRC (34 questions) and DILR (32 questions). Read more…

Tips to crack CAT 2016

Preparing for CAT 2016? Get excellent tips, strategies, and expert advice on how to crack the exam. CAT is one of the most coveted management entrance exams in India and the competition is increasing year after year. To give an edge to your CAT preparation, follow these effective tips. Read more…

CAT Coaching Lessons

C.A Part - 1
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Apologizing for small mistakes Whoops! Sorry! Oh! Sorry. Sorry about that. Oh, my bad. My fault, bro. Apologizing when you make a more serious mistake: serious mistake I’m so sorry. I apologize....

Sai Vidya Education | 10/10/2016

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Krishna Saraf | 21 Apr

IMO, start with basic Class IX and X maths, and basis my experience, that should be enough for starters and as stated above, you can then follow up by taking classes.

Saikat Mitra | 16 hrs ago

Join a private tutor.

#CAT 17 QA Q. If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset.

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Ajay Kumar | 16 Apr

Any number can be expressed of the form: 9K+i; where k= {0,1,2,....} and i={0,1,...,8} Now, we have to select numbers from {1,2,...,3000} in a manner so that final set does not contain any two numbers whose sum is divisible by 9. We have to identify the maximum size of such selection. Now, within 1 to 3000, #elements of the form 9k+1 = 334. #elements of the form 9k+2 = 334. #elements of the form 9k+3 = 334. #elements of the form 9k+4 = 333. #elements of the form 9k+5 = 333. #elements of the form 9k+6 = 333. #elements of the form 9k+7 = 333. #elements of the form 9k+8 = 333. #elements of the form 9k = 333. Now, as per condition, if you select any number from the group 9k+i then you can not select number from the group 9k+j where i+j=9. So, in order to maximize the size of selection, you will go for group: 9K+1, 9k+2,9K+3 and 9k+4. So, total (334+334+334+333=1335). Now, you can select exactly one number from the group 9K. (More than one will result their sum divisible by 9). So, total becomes 1336. Now, you have to exclude 1 (from group 9k+1) and 3000 (from group 9k+3). So finally you are with 1334 numbers!

Shantanav Bhowmik | 18 Apr

The question was meant for aspirants- I had written this in the post/ Admins do not chop off portions from the post when that is relevant/important.hope you understand.

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Krishna Bajoria | 01 Apr

Try to solve it by yourself. The answer should be 7A. If you are unable to solve it, I will be happy to provide you with detailed solution but first give it a try yourself.

K R K Karthikeyan Kattamuri | 02 Apr



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Success Career Education Centre | 10 Feb

Approx. Rs. 35000 for 1 year.

Shantanav Bhowmik | 16 Apr

18k- my fees (-non negotiable ) /course completed in 3 months/ daily classes/-100% coverage + 24*7 doubt clearing. Study material + test series = Rs. 4-12 k (all are same) Total expense = 18+ (4-12)= Rs. 22k to 30k. elsewhere you would be spending from Rs. 25k to 55k depending upon location and institute that too you would have access to only one study material.


The first n natural numbers, 1 to n, have to be arranged in a row from left to right. The n numbers are arranged such that there are an odd number of numbers between any two even numbers as well as between any two odd numbers. If the number of ways in which this can be done is 72, then find the value of n?

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Bhargava Vadlamani | 02/12/2016

Correction:- It is (n-1)/2 ! * (N+1)/2 ! So if n is 7 then answer is 144 If n is 5 then answer is 12.

Vikram Prabhakaran | 02/12/2016

It needs to be visualized first that there can be only 1 odd number between any two even numbers and vice versa because if you place 3 odd numbers between 2 even numbers then some pair of even numbers will not have any odd numbers between them lets assume a set of numbers 1, 2, 3, 4, 5, 6 2, 1 ,3 ,5 ,4 ,6 not satisfying the condition thus we have only 2 of types possibility that satisfies the given condition: 1 2 3 4 5 6 2 1 4 3 6 5 the two types of possibility happened because the number of odd and number of even numbers are same but for a set of odd numbers there will be only one type of possibility i.e. starting with odd numbers. e.g.:- 2 3 4 5 1 not satisfying the condition 1 2 3 4 5 Now coming to the question. Lets suppose, n is even and thus n=2t+2 (where t is an integar with t>0) thus t+1 even numbers and t+1 odd numbers 2*(t+1)! (t+1)! = 72 (t+1)! = 6 t=2 n=6. Lets suppose n is odd and thus n=2t+1 (where t is an integar with t>0) thus t even numbers and t+1 odd numbers (2t+1)! (t)! = 72 for t=1 rhs =6 for t=2 rhs =240 thus the above equation will not be satisfied for any value of t. Hence, n = 6

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