success tips for CAT

Common Admission Test (CAT) is one amongst the largest MBA entrance exams conducted by Indian Institute of Management every year. Through this examination, prospective management students are shortlisted by six IIMs - IIM Ahmedabad, IIM Bangalore, IIM Calcutta, IIM Lucknow, IIM Indore, and IIM Kozhikode. Besides these IIMs, the score card of this examination is also accepted by more than 102 other institutes. Cracking CAT examination is one of the biggest dreams of MBA aspirants and to make it true one has to do a lot of efforts. Jagranjosh.com has jot down some tips on how to prepare for CAT exam:

• Be focussed on Exam Pattern:  As the test will consist of two sections - (a) Quantitative Ability & Data Interpretation (b) Verbal Ability & Logical Reasoning, the candidates must focus on these two sections only. Aspirants must not go beyond these sections while preparing for CAT examination.
• Set a Time Table: The first step to start the preparation must be setting a time table as it will help the candidates to devote proper time in the daily routine jobs along with the study. Not only setting a time table will help but the aspirants must follow the same to get some fine results.
• Systematic and meticulous study: Candidates must ensure that they are very systematic in their approach as it will help them to wind up their topics according to their time table. The best way to furnish this is to choose a topic first and decide the required number of days for concluding that topic. Try to resolve all your confusions regarding various concepts, formulae and reasoning on that topic within the stipulated time period decided by you.
• Know your strength and weaknesses: Candidate must thoroughly study the syllabus of the examination and do smart work rather than hard work. In CAT examination, the candidates are judged on conceptualized knowledge, analytical skills and decision making capabilities. So, try to know your weaknesses and work hard on that and brush up your strengths to excel in examination.
• Practise Mock test papers to develop speed and accuracy: As there will be 30 questions in each section and time allotted to each section will be 1 hour and 10 minutes (that is 70 minutes), so the candidates must practise mock test papers to develop speed and accuracy. It will also help them to know their pitfalls and mistakes by the time to overcome that.
• Healthy Diet and Proper Sleep: Just study, study and study will not lead a student to crack the exam but the healthy diet and right amount of sleep along with the study will show the good results. Both are very necessary to keep the mind fresh and to improve concentration power.
• Revise before the big Day: On the Examination day, aspirants should remain calm. Revision of the studied topics is very necessary. One should not start studying new topics on the examination day.

analysis of Previous Years' CAT Questions:

With respect to the number of questions asked, CAT follows a dynamic pattern. The section based distribution of the number of questions in the previous years' CAT is represented below:



Section	CAT 2003	CAT 2004	CAT 2005	CAT 2006	CAT 2007	CAT 2008
Quantitative Ability	50	35	30	25	25	25
Data Interpretation and Logical Reasoning	50	38	30	25	25	25
Verbal Ability	50	50	30	25	25	40
Total Questions	150	123	90	75	75	90



 

Topic Based Distribution of Previous Years' CAT Questions: 

Section	CAT 2003	CAT 2004	CAT 2005	CAT 2006	CAT 2007	CAT 2008
Arithmetic	17	11	12	12	8	12
Algebra	8	5	3	5	4	5
Geometry	14	10	8	6	3	5
Word Problems	1	8	4	2	9	1
Miscellaneous	7	0	3	0	1	2
Data Interpretation	34	32	30	25	21	27
Logical Reasoning	16	6	0	10	4	5
Reading Comprehension	25	21	12	15	12	20
Verbal Logic	12	9	7	10	7	4
Vocabulary	13	13	7	0	3	11
Grammar (Sentence Correction)	0	6	4	0	3	5

Quantitative Aptitude Tricks for CAT

Divisibility by 2

-> A number is divisible by 2 if and only if the last digit is divisible by 2.
Example: Not required :)
Divisibility by 3
-> A number is divisible by 3 if and only if the sum of the digits is divisible by 3.
Example: 
Which of the following two numbers is/are fully divisible by 3: 97533222 or 97533322?
Answer:  97533222
Reason:
Sum of digits for 97533222 =   33 / 3 = 11
Sum of digits for  97533322 =   34 / 3 = 11.33  

Divisibility by 4
-> A number is divisible by 4 if and only if the last 2 digits is a number divisible by 4.
Example: 
Which of the following two number is fully divisible by 4: 5555444446 or 5555444448 ?
Answer:  5555444448
Reason:
Last 2 digits for  5555444446  are 46, 46 is not fully divisible by 4 (46/4 = 11.5)

Last 2 digits for  5555444448  are 48, 48 is fully divisible by 4  (48/4 = 12)

Divisibility by 5
-> A number is divisible by 5 if and only if the last digit is divisible by 5.
Example: If a number ends in 5 or 0 then it would be divisible by 5. Such as 777995, 13170

Divisibility by 6
-> A number is divisible by 6 if and only if it is divisible by 2 and 3.
Example: 
Which of the following numbers is/are fully divisible by 3: 97533222, 97533322, 97533225, 97533228?
Answer:  97533222 and 97533228
Reason:

For a number to be fully divisible by 6 it should satisfy following two conditions:

1. It must be a even number (which means out of 4 options we can decide that 97533225 is surely not divisible by 6 as it is not an even number)

2. It must be divisible by 3(sum of all digits of particular even number should be divisible by 3)

Sum of digits for 97533222 = 33 (fully divisible by 3)

Sum of digits for 97533322 = 34 (not fully divisible by 3)

Sum of digits for 97533228 = 39 (fully divisible by 3)

Divisibility by 7

-> To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number, if the resulting number is divisible by 7 then original number is divisible by 7. If you don't know the new number's divisibility, you can apply the rule again.

Example: 
Which of the following numbers is/are fully divisible by 7:  203, 3192, 3197, 38241?

Answer : 203, 3192, 38241

Reason:

For 203, you would double the last digit to get six, and subtract that from 20 to get 14. Hence 203 is divisible by 7.

Applying the same rule to check if 3192 is divisible by 7:

3192 => 319 - 2*2 = 315 => 31 - 2*5 = 21, 21 is divisible by 7  hence 3192 is divisible by 7

Applying the same rule to check if 3197 is divisible by 7:

3197 => 319-2*7 = 305 => 30 -2*5 = 20, 20 is not divisible by 7 hence 3197 is not divisible by 7

Applying the same rule to check if 38241 is divisible by 7:

38241 => 3824 -2*1 = 3822 => 382 - 2*2 = 378 => 37 -2*8 = 21, 21 is divisible by 7  hence  38241  is divisible by 7

Divisibility by 8
-> A number is divisible by 8 if and only if the last 3 digits of a number divisible by 8.
Example: 
Which of the following numbers is/are fully divisible by 8: 97533224, 97533328, 97533222, 97533228?
Answer:  97533224 and 97533328
Reason: 224 and 328 are divisible by 8 whereas 222 and 228 are not divisible  by 8.



Divisibility by 9
-> A number is divisible by 9 if and only if the sum of the digits is divisible by 9.
Example: 
Which of the following numbers is/are fully divisible by 9: 111111111, 111222, 2222244, 66669 699999?
Answer:   111111111, 111222  and 2222244
Reason: Sum of digits for 111111111, 111222  and 2222244 are 9, 9 and 18 respectively all of which are divisible by 9. Whereas sum of digits for 66669 & 699999 are 33 & 51 respectively both of which are not divisible by 9. 


Divisibility by 10
-> A number is divisible by 10 if and only if the number ends in n zeros.Example: If a number ends in 0 then it would be divisible by 5. Such as 111110, 876543200  
Divisibility by 11
-> A number is divisible by 11 if the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11.

Example: 
Which of the following numbers is/are fully divisible by 11:  6050, 3035362, 90002, 900002?
Answer:   6050, 3035362, 90002
Reason:   Sum of digits at even place in 6050 = 11 and sum of digits at odd place is 0. Difference between odd and even = 11. Hence 6050 is divisible by 11.
Similarly sum of digits at even place in 3035362 = 11 and sum of digits at odd place is 11. Difference between odd and even = 0. Hence 3035362 is divisible by 11.
Similarly Difference between odd and even digits for 90002 is 11 whereas for 900002 difference between odd and even digits is 7. Hence 90002 is divisible by 11 but 900002 is not.


Divisibility by 12
-> A number is divisible by 12 if the number is divisible by both 3 and 4
Example: 
Which of the following two numbers is/are divisible by 12: 22584 and 22756?
Answer: 22584
Reason: 22584 is divisible by 3 because sum of all digits =21 which is divisible by 3 and 22584 is also divisible by 4 because last 2 digits of 22584 i.e. 84 is divisible by 4. Hence 22584 is divisible by 12
22756 is not divisible 3 as sum of digits are 22 which is not divisible by 3.


Divisibility by 13
-> To find out if a number is divisible by 13 take the last digit of number and multiple it by 4 and add it to number formed with remaining digits check if the resultant number is divisible by 13. In case resultant number is big repeat process of multiplying last digit by 4 and adding it to number formed by remaining digits.
Example: 
Which of the following numbers is/are divisible by 13: 11013, 110006, 110026 ?
Answer: 110006
Reason: Applying the rule above of multiplying last digits by 4 and adding it to number formed by remaining digit we get : 1100+4*4= 11024
Continuing the same way with resultant number we get : 1102+4*4= 1118
=> 1118 = 111+8*4 = 143
=> 14+3*4 = 26. 26 is divisible by 13. Hence 110006 is divisible by 13
11013 is not divisible by 13 as per results below:
11013= 1101+3*4 = 1113 => 1113 = 111+3*4 = 123 => 123 = 12+3*4 =24
24 is not divisible 13 hence 11013 is not divisible by 13.

-> If n is even , n(n+1)(n+2) is divisible by 24

50 useful tricks & shortcuts for Quantitative Aptitude for CAT and other MBA exams:

Below are 50 useful tricks & shortcuts for Quantitative Aptitude for CAT and other MBA exams:

1. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)


2. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 <>23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]


3. (m + n)! is divisible by m! * n!


4. When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.


5. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)


6. To Find Square of a 3-Digit Number
Let the number be XYZ
Steps
a.Last digit = Last digit of Sq(Z)
b. Second last digit = 2*Y*Z + any carryover from STEP 1
c. Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2
d. Fourth last digit is 2*X*Y + any carryover from STEP 3
e. Beginning of result will be Sq(X) + any carryover from Step 4
Eg) Let us find the square of 431
Step
a. Last digit = Last digit of Sq(1) = 1
b. Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6
c. Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1
d. Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2
e. Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18
THUS SQ(431) = 185761


7. If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only.
-> The sum of first n natural numbers = n(n+1)/2
-> The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
-> The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
-> The sum of first n even numbers= n (n+1)
-> The sum of first n odd numbers= n2


8. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then
-> the total number of factors is (x+1)(y+1)(z+1) ....
-> the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c)....
-> the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....
-> the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)
-> Total no. of prime numbers between 1 and 50 is 15
-> Total no. of prime numbers between 51 and 100 is 10
-> Total no. of prime numbers between 101 and 200 is 21
-> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6
-> The number of rectangles in n*m board is given by n+1C2 * m+1C2


9. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.


10.The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6
-> The number of rectangles in n*m board is given by n+1C2 * m+1C2


11. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.


12. Certain nos. to be remembered
-> 2^10 = 4^5 = 32^2 = 1024
-> 3^8 = 9^4 = 81^2 = 6561
-> 7 * 11 * 13 = 1001
-> 11 * 13 * 17 = 2431
-> 13 * 17 * 19 = 4199
-> 19 * 21 * 23 = 9177
-> 19 * 23 * 29 = 12673


13. Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.


14. If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.


15. To find out the sum of 3-digit nos. formed with a set of given digits
This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)
Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.
Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)
= 25 * 24 * 11111
=6666600


16. Consider the equation x^n + y^n = z^n
As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.


17. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then, N^(p-1) – 1 is always divisible by p.


18. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.
145 = 1! + 4! + 5!


19.Where a no. is of the form a^n – b^n, then,
The no. is always divisible by a - b
Further, the no. is divisible by a + b when n is even and not divisible by a + b when n is odd

20. Where a no. is of the form a^n + b^n, then,
The no. is usually not divisible by a - b
However, the no. is divisible by a + b when n is odd and not divisible by a + b when n is even