i work really hard for Cat but still i am getting pathetic scores in Mock Cat. I think i have problem in attempting the question paper. so what should be my correct approach to attempt cat?

You still have around 4 and a half months left for CAT and its good that you have started attempting mocks. From mocks, you would have understood your weaknesses. You would know that in which topic you can solve questions quickly and which ones take time. So my suggestion would be to attempt those questions first which you are more confident about. Let me give you can example, arithmetic is my strength so I always start the paper and find arithmetic questions and finish those in first 20-25 minutes. It not only helps you in terms of score but it also gives you the confidence. Same goes for other sections also, start the paper with your strengths. Meanwhile work on your weak topics so that you don't miss out on easy to moderate questions from those topics.Hope it helps.

If the letters of the word "METROPOLITAN" are arranged together,what is the probability that neither 2 T's nor 2 O's are adjacent to each other?

Total words which can be formed using the letters of the word METROPOLITAN = 12!/(2!)(2!)When both T are together then the number of words would be, 11!/2!When both O are together, the number of words would be 11!/2!However, in the above cases there will be a repetition. The cases where both T's as well as both O's are together have been counted twice. So we have to remove such cases. i.e. 9!So, cases without repetition will be 2*11!/2 -9!Hence the required probability would be 1 - (11!-9!)/12!/4 = 1 - 4*11!/12! + 9!*4/12! = 1 - 1/3 + 1/3302/3 + 1/330= 221/330

find the number of ways in which 5 men can watch 4 movies such that each movie is watched by atleast 1 person?

Each movie can be either watched or not watched. There are 4 movies and 5 friends. So each movie has 5 choices that it can be watched by 1 person, 2 persons,3 ,4 or 5.So each movie has these 5 choices. So required number of ways would be 5^4.

how do i solve para jumbles

Solving para jumbles has become a little tricky once they started coming as TITA. Now it is more time consuming than it was earlier.Practice is the only thing which can help you identify links between sentences. So practice a lot of questions and read a lot. One advice would be to attempt para jumbles at the end. It will ensure that you don't miss out on other easy questions.

i gave cat 2015 but did not get a good score. i want to give cat16. how should i start now?

Since you took CAT 15, you would know that CAT doesn't ask very difficult questions. So start with the basics and clear your concepts. Once you are done with this, you can start taking mocks. Mocks will help you understand your strengths and weaknesses so that you can work on your weaknesses and be better prepared when the actual exam comes.We have a quite a few free mocks, you can try them. If you want more, you can also buy our test series to have access to further mocks.Thanks.

find the number of ways in which 3 couples can sit in such a way that no husbands sits next to his wife?1)64 2)48 3)32 4)24

The three wives can be seated in 3! ways. From this, you can rule out two option choices as the final answer has to be a multiple of 3! i.e. 6.There will be 4 places which will be created by seating the 3 women. __ a__b__c__dNow four people can occupy the given seats in 8 ways(when mentioned restrictions are followed).So required number of ways are 6*4 = 24

how to work on RC passages for greater understanding?

The best way would be to practice it without keeping any time constraints. Read the passage again and again till you don't understand it completely. After that try answering the questions. Your focus should be on accuracy. Once you start getting accuracy greater than 80%, you can start to keep track of time and should challenge yourself to reduce that timing every week while maintaining the accuracy. This will surely help as last year I used to get around 12-14 RC;s correct out of 24. This year till now, in 5 mocks that I have taken lowest have been 17 correct.

QUESTION 3..DI TEST 8..In the financial category, the number of spam emails received in September 2002 as compared to March 2003...the solution says that it can't be determined ..but in the second question of the same test which is of similar type to q3..u have used the logic that the no of spam emails is greater so the answer is larger value..But in q3 the same condition is given so why it can't be determined as smaller value...

In the question, the figures are in percentage. Moreover it is also mentioned that sep 2002 had more emails than march 2003.Financial sector in sept 2002 had 25% spam mails and in march 2003 it had 45 % of the total spam mails. So it will depend on how many spam mails were recieved in which month, hence cannot be determined. In 2nd question Dec 2002 and June 2003 are compared and it is given in the question that dec 2002 had more total e-mails than June 2003. And share of health sector in dec 2002 was 19% which was anyway greater than the share of health sector in June 2003. So larger percentage of a larger quantity will always be large.Hope you get the point.Thanks

question no. 3..DI TEST 8...In the financial category, the number of spam emails received in September 2002 as compared to March 2003...

In the question, the figures are in percentage. Moreover it is also mentioned that sep 2002 had more emails than march 2003.Financial sector in sept 2002 had 25% spam mails and in march 2003 it had 45 % of the total spam mails. So it will depend on how many spam mails were recieved in which month, hence cannot be determined. In 2nd question Dec 2002 and June 2003 are compared and it is given in the question that dec 2002 had more total e-mails than June 2003. And share of health sector in dec 2002 was 19% which was anyway greater than the share of health sector in June 2003. So larger percentage of a larger quantity will always be large.Hope you get the point.Thanks

sum of 3/2 +5/4=9/8.....10 terms......???anyone plzz

3/2=1+1/25/4=1+1/49/8=1+1/8So, the series effectively becomes 1+1+1...10 times+1/2+1/4+1/8...10 terms=> $$10+\frac{\frac{1}{2}(1-(\frac{1}{2})^{10})}{1-1/2}$$=> $$10+(1-(\frac{1}{2})^{10})$$