The Common Admission Test  CAT syllabus consists of 3 sections  Verbal Ability and Reading Comprehension, Data Interpretation & Logical Reasoning, Quantitative Ability.
Quantitative Ability and Data Interpretation & Logical Reasoning in CAT syllabus are focused to judge the person's ability to crunch numbers, concepts and applying the same within the limited time span whereas Verbal Ability and reading comprehension, is to know the person's command over the language. We will discuss the marking scheme of CAT exam, sectionwise topics below.
Also read, 15 Must Read Books For CAT Exam 2019 Preparation


A total of 100 questions are to be solved in 180 minutes. Each correct response carries 3 marks, while each incorrect response carries 1 marks. NonMCQ questions will not be negatively marked.
Arithmetic  Algebra  Number Systems  




Geometry & Mensuration  Modern Mathematics  


Data Interpretation  Logical Reasoning  


Verbal Ability  Reading Comprehension  


The CAT 2019 syllabus can ideally be completed in a span of 48 months, depending on the level of preparation of a candidate.
Cyclicity is a basic but a very important concept when it comes to finding the unit digit of a particular number. No matter what the number is or what power the number has, you can use the concept of cyclity to find the unit digit of that number in seconds.
So, starting with the concept, every no. say "x^n" repeats its unit digit in a cycle of 4 (power).
Number  x^1  x^2  x^3  x^4  
1  1  1  3  4  
2  2  4  8  6  
3  3  9  7  1  
4  4  6  4  6  
5  5  5  5  5  
6  6  6  6  6  
7  7  9  3  1  
8  8  4  2  6  
9  9  1  9  1 
For instance, you have to find the unit digit of (348)^21
The unit digit of (348)^21 will solely depend upon 8 and not on any other digit. Now divide the power 21 in the form of (4n + a). Thereby we can write 21 as 4(5) + 1. So, as forestated the unit digit repeats itself in a cycle of 4, we can say that the unit of (348)^21 = (348)^1.
Therefore, unit digit of (348)^21 = 8^1 => 8.
In a n*n chessboard,
The total no. of squares = 1^2 +2^2+3^2+…..+n^2
= n*(n+1)(2n+1)/6 (summation formula)
and the total no. of rectangles are given by: 1^3 +2^3+3^3+…..+n^3
=[{n^2}*{(n+1)^2}/4
So, for instance in a 8*8 chessbosard,
no. of squares are: 8*(8+1)(2*8+1)/6
i.e. 8*9*17/6
=1,224/6 => 204
and no. of rectangles are given by: [{8^2}*{(8+1)^2}/4
i.e. 64*9^2*/4
=5,184/4 => 1296 (this includes the no. of squares as well)
No. of rectangles, which are not squares are: 1296  204 => 1092 learn more