How to improve Vocab for CAT - Tips to improve English & RC

Below are some of the ways to improve your verbal skills:

#1 - Reading and Practice

  • Reading good newspapers, preferably: The Hindu, The Economic Times, regularly. (Read it loudly)
  • Referring to a good dictionary, preferably, The Oxford Advanced Learner Dictionary regularly.
  • Watching good news channels likes BBC, NDTV, DD News, CNBC etc regularly
  • Reading good news magazines like Business Line, The Economist, Business World , Business Week etc regularly
  • Reading good novels & recommended management books regularly.
Refer below pages for list of recommended books, novels, newspapers and magazine for CAT aspirants:


#2 - How to make reading interesting?
Furthermore, the other thing which you may try is make your preparation a bit more interesting.
Just take example of reading The Hindu:
-you can try to read a big article exactly for 10 minutes (with the help of stopwatch)
-Remember or mark the line till which you have read in those 10 minutes.
-Close the newspaper & write down all you have understood in those 10 minutes
-Re-read the complete article and mark the words which are new to you  
-Re-read the complete article and see whether your summary matches with the actual content of the article
-Find out the meaning of those new words and maintain a record of it -Keep a count of number of words you have read in 10 minutes.
-Maintain a daily record book and try to improve it gradually

When you will observe that your 
reading abilities are improving, it will boost your confidence and your communication skills.


#3 - How to Overcome speaking Hesitation?
In addition we also recommend CAT aspirants to focus on spoken English. If you are currently doing your graduation it becomes lot easier for you to work on it. 
CAT aspirant who are doing their graduation are advised to form a group within their friend circle. The purpose of this group should be to facilitate and organize mock Extempore/ GDs/ Interviews/ Debates. The aim of the group should be to provide a platform to its group member where they could improve their spoken English and overcome their hesitation of public speaking. Below guidelines would be helpful:
1. Identify serious and mature group members 5-10
2. Identify a place where you will organize the group activities. 
It could be common room of your hostel/college or room of any of the students.
3. Fix a time (1 Hours) at which group will meet everyday
4. Each day one person could be appointment as moderator/facilitator who will moderate the activities such as giving topic for group discussion, extempore or conduct mock interviews.

Participation in such a group will not only help you in your MBA preparation but also for job interviews & will surely help in your overall personality development.


#4 - Have good collection of Software's
Below are some of the good Vocab Builder software's shared for you:

If you like these software's, do not forget to write your thanks below :)

Quantitative Aptitude Tips & Tricks for CAT



For exams like CAT where time management is key to success, its very important to manage your time in efficient way. Learning and mastering handy tips, tricks and shortcuts for quantitative aptitude would surely save lot of time during those crucial hours.

Quantitative Aptitude Tricks for CAT - Page 3


34. -> For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is
0.5*d1*d2, where d1, d2 are the length of the diagonals.
-> For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where s=(a + b + c + d)/2
Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.
-> Area of a Rhombus = Product of Diagonals/2


35. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]


36. Area of a triangle
-> 1/2*base*altitude
-> 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B
-> root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2
-> a*b*c/(4*R) where R is the circumradius of the triangle
-> r*s ,where r is the inradius of the triangle


37. In any triangle
-> a=b*cos C + c*cos B
-> b=c*cos A + a*cos C
-> c=a*cos B + b*cos A
-> a/sin A=b/sin B=c/sin C=2R, where R is the circumradius
-> cos C = (a^2 + b^2 - c^2)/2ab
-> sin 2A = 2 sin A * cos A
-> cos 2A = cos^2 (A) - sin^2 (A)


38. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1


39. Appollonius Theorem
In a triangle ABC, if AD is the median to side BC, then
AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2)


40. -> In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
-> In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.


41. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.


42. Let W be any point inside a rectangle ABCD, then,
WD2 + WB2 = WC2 + WA2


43. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))


44. -> Distance between a point (x1, y1) and a line represented by the equation ax + by + c=0 is given by ax1+by1+c/Sq(a2+b2)
-> Distance between 2 points (x1, y1) and (x2, y2) is given by Sq((x1-x2)2+ (y1-y2)2)



45. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.

Also Read




Quantitative Aptitude Tricks for CAT - Page 2


Below are useful tricks & shortcuts for Quantitative Aptitude for CAT and other MBA exams(continued):

21. The relationship between base 10 and base ‘e’ in log is given by log10N = 0.434 logeN


22. WINE and WATER Mixture formula
Let Q - volume of a vessel, q - qty of a mixture of water and wine be removed each time from a mixture, n - number of times this operation is done and A - final qty of wine in the mixture, then,
A/Q = (1-q / Q)^n


23. Pascal’s Triangle for computing Compound Interest (CI)
The traditional formula for computing CI is
CI = P*(1+R/100)^N – P
Using Pascal’s Triangle,
Number of Years (N) 
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
… 1 .... .... ... ... ..1
Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1:
Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331
The coefficients - 1,3,3,1 are lifted from the Pascal's triangle above.

Step 2:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)
If N =2, we would have had,
Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210
CI = 2 * 100 + 1* 10 = Rs.210


24. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:
Final Difference% = X - Y - XY/100
Eg. The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?
Applying the formula,
Final difference% = 40 – 25 - (40*25/100) = 5 %.
So if 5 % = 1,000
Then, 100 % = 20,000.
Hence, C.P = 20,000
& S.P = 20,000+ 1000= 21,000


25. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.


26. -> Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)^2
-> The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003


27. -> If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
-> If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
-> If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
-> If A can finish a work in X time and B in Y time and A, B & C together in S time then
C can finish that work alone in (XYS)/ (XY-SX-SY)
B+C can finish in (SX)/(X-S); and
A+C can finish in (SY)/(Y-S)


28. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5


29.-> When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0
-> When an unbiased coin is tossed even no. (2n) of times, then, P (no. of heads=no. of tails) = 1-(2nCn/22n)


30. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!
Eg:1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?
Here n=10, m=3 (i.e. A, B, C)
Hence, P (A>B>C) = 1/3! = 1/6
Eg:2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?
P (M>S) = 1/2!
= 1/2


31. CALENDAR
-> Calendar repeats after every 400 years.
-> Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
-> Century has 5 odd days and leap century has 6 odd days.
-> In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
-> January 1, 1901 was a Tuesday.


32. -> For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)
-> For any regular polygon, the sum of interior angles =(n-2)*180 degrees
So measure of one angle is (n-2)/n *180
-> If any parallelogram can be inscribed in a circle, it must be a rectangle.
-> If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).


33. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)


Also Read:

Quantitative Aptitude Tricks for CAT - Page 1

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


Also Read:

SNAP 2012 Exam Notification - Symbiosis 2012 Test

Source: snaptest.org
SNAP 2012 exam notification is out. Below are Important Dates for SNAP 2012:
Event
Start Date
Last Date
SNAP 2012 Online Registration Dates
From 2nd September 2012 (Sunday)
Till 20th November 2012 (Tuesday)
Payment of Application Fees
From 2nd September 2012 (Sunday)
23rd November 2012 (Friday)
SNAP 2012 Admit Card Download Period
From 3rd December 2012 (Monday)
Till 16th December 2012 (Sunday)
SNAP 2012 Exam Date
SNAP 2012 Test Date
16th December 2012  (Sunday)
2.00PM - 4.00PM

SNAP 2012 Test Result
SNAP 2012 Exam Result
10th January 2013 (Thursday)
-

Check below the SNAP 2012 notification published in newspapers(click on image to enlarge):
SNAP 2012 Exam Notification - Test Date
SNAP Test 2012 Notification
Note:
Bring the following documents while reporting to the Test Centre:

1.Admit Card.

2.Photo identity: any one of the following:
  • Passport
  • Driving license
  • College/Institute identity card
  • Credit card with photograph
  • Voter ID Card
3. Blue/Black ballpoint pen.

Enter your Email to Subscribe