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CAT 2015 Quantitative Aptitude Preparation Tips

Last Updated - March 31, 2017

The first section as per the new pattern released for CAT Exam  will be of Quantitative Aptitude (QA). Earlier this section also included questions on Data Interpretation (DI) but now it will be separate section For QA there will be 34 questions with a time limit of 60 minutes to solve it.

The new Exam Pattern for CAT will cover following three sections:-

SectionNumber of QuestionsDuration
Quantitative Ability3460
Data interpretation and logical reasoning3260
Verbal and Reading comprehension3460
Total duration180 minutes

Check MCQ Questions for CAT 2016

This has led to decrease in the weightage of Quantitative ability section from 50 to 34 although all the questions will carry equal marks whether it is an objective type or descriptive one. So, the quant section will have following pattern

Name of the SectionNumber of Questions AskedType of QuestionsTime allotted for this section
Quantitative Aptitude (QA)34MCQs and Descriptive60

One major point to keep in mind is that you cannot start from any section you like, the periodicity of the sections is to be maintained   and  you  will not be allowed to swap between the sections till the assigned time of 60 minutes for the section is over.  Once the time of 60 minutes is over, you will be directed  to the next section automatically .So you have to begin your exam from QA section which  will cover  areas like Arithmetic ,Algebra, Geometry, Trigonometry among other topics to be solved in 60 minutes and will fetch a maximum raw score of 102 marks. On the top of it, most of the Quant questions are the single answer questions as against the DI questions which commonly appear in set of 4-5 questions and you could crack at least 1 or 2 correct out of them.There will be Non-MCQ type of questions in Quant section of CAT 2015 with no answer options.


It is important to score well in QA section because a candidate scoring high percentile of 99+ in Verbal Ability section and scoring 75+ in Quant section, despite scoring a very high overall percentile of 95-96 cannot hope  to be shortlisted by IIM A,B,C,L or even by the newer IIMs as the sectional percentile is below the prescribed cut off level.

Tips for attempting QA section in CAT 2015

To score well in QA section you must prepare rigorously , clear your basics thoroughly & take mock tests as much as possible , Below mentioned are some tips that you can follow while attempting the QA section

  • Mensuration , Geometry and Coordinate Geometry  :- These three parts will form around 25-30% of the QA section giving maximum weightage to geometry ,3 to 4 questions on mensuration and few questions on coordinate geometry .These three parts are basically considered as visualization part .

For preparing for geometry part you must be familier with the basic theorems of involving triangles, circles and parallel lines. You may have encounter a question like find the value of certain angles or length of certain sides many a times while going through a CAT’s previous question paper . Therefore, make sure that you cover topics such as congruency and similarity of triangles.

For preparing for coordinate geometry focus mainly on 2 topics straight lines and circles. Don’t give  much emphasis on topics like  conic sections and other advanced topics. Some sample questions on coordinate geometry which will provide you a vague idea about coordinate geometry questions

Que:Determine the equation of a line, the intercepts of which are twice those of the line 2x – 3y – 12 = 0. 
1.  3x – 2y = 24      2.   2x – 3y = 12 
3.  2x – 3y = 24      4.  None of these 

Answer : 3
Solution: If x = 0, y = – 4 
Thus, the y intercept is – 4 
Put y = 0, in that case x = 6 
Hence the x intercept = 12 and -8 
Thus, the intercepts of the required line are 12 and – 8. So the equation of the line is 
(x/12) + (y/-8) = 1-8x + 12y = -96 
2x – 3y = 24  , So 3rd option is correct

Few short cuts to keep in mind while solving coordinate geometry questions

If ABCD is a parallelogram, then D= A-B+C

  • Area of Rhombus formed by ax (+ or -) by (+or -) c = 0 is calculated by 2c2/ab The equation of line parallel to X axis is form = k
  • If a line slope passes through (x1, y1) and (x2, y2) is (y2-y1)(x2-x1) only when x1is not equal to x2
  • The equation of a line running parallel to x- axis is of the form y=k
  • The equation of a line with m slope and intercept y is y= mx+c
  •  The equation of line for non- zero intercept and b on x and y axis is (x/a) + (y/b)= 1
  • The intersection point of two lines, that don’t run parallel can be detected by solving them simultaneously
  • ax+ by + c= 0
  • y= mx +b where m= slope and b =y – intercept
  • (y- y1 = m (x- x1), where m= slope and the lines passes via (x1, y1)
  • Three points x,y and z are collinear if (slope (xy) = Slope (yz). This is the right collinearity test

For mensuration- For preparing for this part , learn all the basic formulae on areas, surface areas and volumes of triangles, circles, cylinders, cones, cuboids and spheres from NCERT school text  book of class 8th,9th  & 10th . Mensuration problems are calculation intensive, and require lots of practice so try to attempt them at the last hour of section.


Important formulae to solve mensuration based problems


Important Formulae for Geometry based questions

  • Cube


Total Surface Area6a2
Length of face diagonal b√2a
Length of body diagonal c√3a


CAT-preparation 2

Total Surface Area2(lb+bh+hl)
Length of four equal body diagonals AF√l2+b2+h2
Length of face diagonal AC√l2+h2
Length of face diagonal BF√b2+h2
Length of face diagonal DF√l2+b2
Radius of sphere circumscribed in cuboidDiagonal /2 , √l2+b2+h2/2

CAT-preparation 3


Surface area of the sphere4πr2
Volume of the sphere4πr3

Spherical Shell

CAT-preparation 4

Volume of the shell4/3πr3



Total number of vertices4
Curved Surface Area3√3a2/4
Total Surface Area√3a2
  • Algebra : Most of the questions in CAT QA section comes from this topic, approximately around .From this area lay emphasize on topics like Permutations andCombinations, Probability (e.g. like die and card problems and perhaps Bayes’ theorem), Functions, Progressions (AP, GP,HP and AGP), Logarithms, Equations (Quadratic and Linear/Simultaneous) .

Permutations and Combinations important formulae

Formula 1:-If n distinct items are arranged in a row, then the number of ways they can be rearranged such that none of them occupies its original position is,

n! * ((1/0!) - (1/1!) + (1/2!) - (1/3!) + ... ((-1)n/n!))

Note: De-arrangement of 1 object is not possible.

Dearr(2) = 1; Dearr(3) = 2; Dearr(4) =12 4 + 1 = 9; Dearr(5) = 60 20 + 5 1 = 44

For eg- A person has eight balls and eight bags corresponding to those balls. In how many ways can he put those balls in the bags such that exactly 5 of them get into the correct bag correctly?

Ans:- five balls that gets into the correct bags can be done in 8c5 ways

So the rest three balls must be put into wrong bags which can be done in Dearr(3)= 2

Hence the total ways =2* 8C5=2*56=112 ways

Formula 2:- Partitioning

To arrange n identical items in r distinctive groups and there is no restrictions= n+r-1Cr-1

To arrange

Formula 3:- Number of ways of arranging n items, out of which p are alike, q are alike and r are  alike given that p + q + r = n


Number of ways of dividing n distinct items, in groups of size p, q and r given that p + q + r = n is

n! / (p! * q! * r!)
  • Number Theory-Questions form this topic is very easy to solve and least time consuming if you know how to solve them. Number theory covers 3-4 questions on an average in every CAT question paper.

For preparing this topic first of all you should be well aware of how to frame numbers into algebraic form ,( for e.g. a three digit number having digits xyz can be represented as 100x + 10y + z). You should also learn about divisibility tests and the ‘modulo’ notation and its applications (for programmers, 10%5==0 is also referred to as 10 modulo 5 is 0, that is, the remainder when 10 is divided by 5, is zero).

Different divisibility tests and modulo notations that will help you to solve questions based on number

Basic formulae to solve algebraic equations

  • (a+b)2=a2+b2+2ab
  • (a−b)2=a2+b2−2ab
  • (a+b)2−(a−b)2=4ab
  • (a+b)2+(a−b)2=2(a2+b2)
  • (a2−b2)=(a+b)(a−b)
  • (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
  • (a3+b3)=(a+b)(a2−ab+b2)
  • (a3−b3)=(a−b)(a2+ab+b2)
  • (a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca)
  • If a+b+c=0, then a3+b3+c3=3abc

Important Divisibility Tests

  • Divisibility by 2- A number is divisible by 2, if its unit's digit is any of 0,2,4,6,8.
  • Divisibility by 3 - A number is divisible by 3, if the sum of its digits is divisible by 3.
  • Divisibility by 4- A number is divisible by 4, if the number formed by the last two digits is divisible by 4
  • Divisibility by 5- A number is divisible by 5, if its unit's digit is either 0 or 5.
  • Divisibility by 6- A number is divisible by 6, if it is divisible by both 2 and 3
  • Divisibility by 8- A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8.
  • Divisibility by 9- A number is divisible by 9, if the sum of its digits is divisible by 9.
  • Divisibility by 10- A number is divisible by 10, if it ends with 0.
  • Divisibility by 11- A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
  • Divisibility by 12- A number is divisible by 12, if it is divisible by both 4 and3
  • Divisibility by 14- A number is divisible by 14, if it is divisible by 2 as well as 7.
  • Divisibility by 15- A number is divisible by 15, if it is divisible by both 3 and 5
  • Divisibility by 16- A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.
  • Divisibility by 24- A given number is divisible by 24, if it is divisible by both 3 and 8.
  • Divisibility by 40 -A given number is divisible by 40, if it is divisible by both 5 and 8
  • Divisibility by 80- A given number is divisible by 80, if it is divisible by both 5 and 16.

Important divisibility rule forCo-primes- If a number is divisible by pas well as q , where p and q are co-primes, then the given number is divisible by pq. If p and q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q.


36 is divisible by both 4 and 6, but it is not divisible by (4∗6 )=24, since 4 and 6 are not co - primes.

Important Results for solving AP based questions

An A.P. with first term a and common difference d is given by a, (a+d),(a+2d),(a+3d),.....

The nth term of this A.P. is given by


The   sum of n terms of this  A.P.

Sn= (n2)[2a+(n−1)d]=(n2)(first term + last term).

Some Important Results:

  • (1+2+3+....+n)= n(n+1)2
  • (l2+22+32+...+n2)= n(n+1)(2n+1)6
  • (13+23+33+...+n3)= n2(n+1)2

Geometric Progressions Results to kept in mind

A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression

Important Results

A G.P. with first term a and common ratio r is :a,ar,ar2,.....

In this G.P. nth term, Tn =arn−1

sum  of n terms, 

Sn=a(1−rn)(1−r) when r<1

Remainder and Quotient problem’s solving tricks

Let us try it using an example , when we divide 7 by 3 we get a remainder 1 which means 7=3∗2+1

Presenting it notationally  let the two numbers be p and q , when we divide p by q we get p=kq+r  where k is quotient and r is remainder

Dividing both sides of p=kq+r by k we get a result


Let us solve a question based in above result:-

The remainder is 57 when a number is divided by 10,00. What is the remainder when the same number is divided by 1,00?

(A) 5 (B) 7 (C) 43 (D) 57 (E) 570


Since it is given that the remainder is 57 when the number is divided by 10,00, the number can be expressed as 10,00n+57, where n is an integer.

Rewriting 10,000 as 1,00∗10 we get


Now, since n is an integer, 10n is also an integer. Let 10n=q , we get


Hence, the remainder is still 57 (by the p=kq+r form) when the number is divided by 1,000.

So,the answer is (D).

Method II (Alternative form):

Since the remainder is 57 when the number is divided by 10,00, the number can be expressed as 10,00n+57. Dividing this number by 1,000 yields

10,00n+57100 =10,00n100+57100 =10n+57100

Hence, the remainder is 57 (by the alternative form pk=q+rk ), and the answer is (D).

Some more sample questions

This question was in CAT 2005 question paper for 2 marks
Que :-What is the righmost non-zero digit of the number 302720?

(1) 1
(2) 3
(3) 7
(4) 9

Correct Answer: Choice (1).

Explanatory Answer
The given number is a multiple of 10, so the last 2720 digits will be 0s.
The right most non-zero digit of the number will be the unit digit of the number 32720.
The unit digit of the powers of "3" follow a cyclicity of "4"

i.e., The units digit of 31 is 3
The units digit of 32 is 9
The units digit of 33 is 7
The units digit of 34 is 1. Then the cyclicity sets in.

The units digits of 35 is 3.
The units digits of 36 is 9 and so on.

In 32720 as 2720 is a multiple of "4", the number will have the same units digit as 34 which is "1".

  • Arithmetic and Miscellaneous: Around 20% of questions in any CAT paper comes from this area. Major topics that you need to cover are Set Theory (especially Venn diagrams) and problems on Time, Speed and Distance,  which are frequently asked.

Miscellaneous problems are those problems which do not fall under any head. They are rarely asked, and even when they do appear in a CAT paper they do not number more than one or two. They are basically used to check the mathematical aptitude of the students, and you cannot learn how to solve them. One way to solve questions based on this topic is to try back-substitution of answer choices, or to avoid these problems altogether.

Although for preparing for QA section you have to clear your basic fundamentals and practice thoroughly  , here are some techniques you can use to solve the questions :-

  • Substitution of numbers for variables in algebraic problems- This will make the question much more simpler however, this method does not work when the options are also in terms of variables.
  • Back-substitution of answers into the problem- Taking one of the options as the final answer  and then solve the problem. If you don’t get any result or the problem reduces to the trivial case, repeat for another choice until you achieve the correct answer .
  • Substituting variables for numbers in the answer choices.-This mainly works for progression problems. Suppose  the nth term of a progression is given in terms of n and some other terms. And you are asked to find the 100th term in the given progression. The answer  are in the form like 2100, 299 – 1 etc. (let). Then, you can start with the first answer choice, assume that the nth term will be 2n, solve the first few terms of the progression and find if this is indeed the case (lets say it’s easy calculating the 3rd term, which you find to be 8 or 23. Hence the 100th term will be 2100). If this doesn’t  work , assume that the nth term is 2n-1 – 1 and repeat, till you find the right answer
  • Solving coordinate geometry algebraically, or vice-versa- A complicated algebra problem involving several equations can be solved very easily if you draw the corresponding figures on an imaginary graph paper. Similarly, coordinate geometry problems can often be solved by writing corresponding algebraic equations.
  • If you can eliminate all options except two, guess- Suppose you have two questions, probability states that you will get one of these wrong and the other right. The expected number of marks a correct answer is is +4 and for a wrong answer is -1, which combines to +3 for both the questions  or +1.5 per question answered.

Last Year QA section weightage

Name of topicWeightage in %
Number Systems, LCM HCF15
Mixtures and Alligations6
Time, distance and speed4
Set theory5
Pipes and Cisterns4
Permutation Combination4
Ratio and Proportion4
Races and Games4
Boats and Streams4
Time and Work4
Boats and Streams4
Problems on Trains4
Calendar and Clocks2
Profit, Loss and Discounts2
Progressions (AP, GP & HP)2
Problems on Age2
Average, Percentage, Inequalities1
Simple and Compound Interest1
Binomial Theorem1
Coordinate Geometry1
Chain Rule2
Linear and Quadratic Equations2

Last year difficulty level of QA section was average , questions were basically formed to test the basic fundamentals of the students ,However you need to prepare really well for this section if you want to get enrolled in A,B,C, level IIM’s because QA is one of the section in which students had outshown in last 2 CATs

Reference Books for preparing QA for CAT 2015

Book NameAuthor/Publisher NameISBNPriceBook Reviews
How to Prepare for Quantitative Aptitude for the CAT Common Admission Test 5th Edition (Paperback)Arun Sharma1259027015650/-
  • Best Book for Quant Preparation
  • Content is self expalanatory , No need of more guidance
  • Content is explained in very understandable manner
The Pearson Guide To Quantitative Aptitude And Data InterpretationNishit Sinha9332528829650/-
  • Book is useful and content is plain an well written
  • Best book for preparation of GRE & CAT
  • The book explains all the concepts in a very transparent manner
Quantitative Aptitude For Competitive Examinations Tips, Techniques, And Short-cut MethodsGuha Abhijit0070706352450/-
  • The book helps in developing the base in QA
  • The book is helpful for beginners to clear their concepts
  • It provides many problem solving tricks and shortcut methods
Quantum CATSarvesh K Verma9350944146550/-
  • It is a must have for every CAT aspirant
  • Level of problems is very good

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