IPMAT is one of the highly competitive entrance exams conducted after 12th for students who wish to get admission into IIM Indore for the Five-Year Integrated Program in Management (IPM). As the IPMAT Exam is around the corner, it is advised to enhance last-minute preparation. If you are IIM IPM aspirant but worried about the maths section, then here are few short tricks to attempt the Arithmetic section in IPMAT 2021. Go through the complete post to know how to prepare for IPM Maths.

How to Prepare for IPM Maths?

  • As per the IPMAT Exam Pattern, there are only three sections in the exam i.e., Quantitative Aptitude Short Answers, MCQs, and Verbal Ability.
  • Quant has the highest weightage in the exam as it includes 40 short answers and 20 MCQs. Each correct answer is awarded 4 marks while 1 mark is deducted for the wrong answer.
  • So, it is important to keep an eye while answering the questions and do not try to guess answers.
  • One of the best methods for last-minute preparation is to solve IPMAT Question Papers to know the different types of questions and difficulty level of the exam.
  • In this post, candidates can know different strategies to solve any question, marking scheme, and also most expected questions.

IPMAT Quantitative Aptitude Important Topics

Before you begin the preparation, go through the entire IPMAT Syllabus to know which are the topics to be studied. Make sure to study only the important topics instead of all topics which are not very important. As per the experts, opting for the best IPMAT Coaching will help students to enhance their Maths preparation and score good marks in the exam.

Algebra Number System
Set Theory Data Interpretation
Geometry, Mensuration Percentage, Profit and Loss
Time, Speed Distance, Time and Work
partnership Pipes & Cisterns
Polygons, Circles Work-Related Problems
Integration & Differentiation Permutations, Combinations, Probability
Vectors Determinants

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IPM Maths Preparation Strategy

Here are a few experts recommended preparation strategies that must be implemented during preparation. Follow these tips & tricks to enhance your preparation for the upcoming IIM IPM Maths section.

  • First, make a list of topics in which you are strong and weak. Be thorough with the entire IPMAT Syllabus and pattern.
  • This helps in planning your preparation and studying topics accordingly.
  • Quantitative Aptitude is the section that completely involves calculations. Hence, aspirants must solve more and more questions from each topic.
  • This way you can improve your speed and attempt different types of questions.
  • The tricky topics in Maths are Algebra, Mensuration, Integration & differentiation. So, try solving more problems from these topics.
  • It is very important to understand the topic in-depth as it helps in answering any type of question asked.
  • Go through the tricks to solve quant questions and enhance IPM Mathematics Preparation.

IPMAT Maths Section Pattern & Marking Scheme

Here is the composition of maths section in IPMAT. Check out the pattern and marking scheme for IPMAT Exam and be well prepared.

  • For each correct answer, the candidate is awarded 4 marks while for each wrong answer 1 mark is deducted.
Section Total Question Marks Time Duration
Quantitative Ability(MCQ) 40 180 40 minutes
Quantitative Ability(SA) 20 80 20 minutes

IPMAT Maths Preparation Tricks & Short Cuts

Go through the short cuts and quick tips to solve the IPM Maths section and score more marks. 

Number System Quick Maths Formulas

  • 1 + 2 + 3 + 4 + 5 + … + n = n (n + 1)/2
  • (12 + 22 + 32 + …. + n2) = n (n + 1) (2n + 1) / 6
  • (13 + 23 + 33 + …. + n3) = (n (n + 1)/ 2)2
  • Sum of first n odd numbers = n2
  • Sum of first n even numbers = n (n + 1)
  • (a + b) * (a – b) = (a2 – b2)
  • a + b)*2 = (a2 + b2 + 2ab)
  • (a – b)*2 = (a2 + b2 – 2ab)
  • (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 – ac)
  • When a + b + c = 0, then a3 + b3 + c3 = 3abc

HCF and LCM Quick Maths Formulas

Product of two numbers a and b

(a*b) = Their HCF * Their LCM.

But a*b*c ≠ HCF*LCM

Note:

HCF of two or more numbers is the greatest number which divide all of them without any remainder.

LCM of two or more numbers is the smallest number which is divisible by all the given numbers.

HCF of given fractions = (HCF of Numerator)/(LCM of Denominator)

HCF of given fractions = (LCM of Denominator)/(HCF of Numerator)

If d = HCF of a and b, then there exist unique integer m and n, such that d = am + bn.

Co-primes

Two numbers are said to be co-prime if their H.C.F. is 1.

HCF of a given number always divides its LCM.

Methods of finding HCF of two or more numbers

Method 1: Prime Factors Method

Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF.

Example

If you have to find the HCF of 42 and 70.

Then 42 = 2*3*7

And 72 = 2*5*7

Common factors are 2 and 7 so, HCF = 2*7 =14.

Method 2: Division Method

Divide the greater number by the smaller number, divide the divisor by the remainder, divide the remainder by the next remainder and so on until no remainder is left. The last divisor is required HCF.

Method 3: HCF of Large Numbers

Find the obvious common factor from both the numbers and remove it. Also, remove the prime number (if any found). Now perform the division method to remaining numbers and find the HCF. Check out the example for better understanding.

Methods of finding LCM of two or more numbers

Method 1: Prime Factors Method

Resolve the given numbers into their Prime Factors and then find the product of the highest power of all the factors that occur in the given numbers. The product will be the LCM.

Example

LCM of 8,12,15 and 21.

Now 8 = 2*2*2 = 23

12 = 2*2*3 = 22*3

15 = 3*5

21 = 3*7

So highest power factors occurred are – 23, 3, 5 and 7

LCM = 23*3*5*7 = 840.

Simplification Quick Maths Formulas

‘BODMAS’ Rule

Through this rule, you can understand the correct sequence in which the operations are to be executed and

This rule depicts the correct sequence in which the operations are to be executed and the sequence can be evaluated.

Here are some rules of simplification given below-

B – Bracket

(First of all remove all the brackets strictly in the order (), {} and || and after removing the brackets, you can follow the below sequence)

O – Of

D – Division,

M – Multiplication,

A – Addition and

S – Subtraction

Modulus of a Real Number

Modulus of a real number a is defined as

|a| = a, if a > 0

= -a, if a< 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

Vinculum (or Bar):

When an expression contains Vinculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Vinculum.


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Square roots & Cube Roots Quick Maths Formulas

Duplex Combination Method for Squaring

In this method either we simply calculate the square by multiplying the same digit twice or we have to perform cross multiplication.

Here are the following Duplex rules and formulas, please check below.

a = D = (a*a)

ab = D = 2*(a*b)

abc = D = 2*(a*c)+(b)2

abcd = D = 2*(a*d) + 2*(b*c)

abcde = D = 2*(a*e) + 2*(b*d) + (c)2

abcdef = D = 2*(a*f)2 + 2*(b*e)2 + 2*(c*d)2

Now I’ll make you understand the complete squaring procedure in a better way with the help of examples. Check out the example here –

We have to find out the solution of (207)2 instantly.

Then,

(207)2 = D for 2 / D for 20 / D for 207 / D for 07 / D for 7

(207)2 = 2*2 / 2*(2*0) / 2*(2*7) + 02 / 2*(0*7) / 7*7

(207)2 = 4/0/28/0/49

(207)2 = 4/0/28/0/49

(207)2 = 4 / (0+2) / 8 / (0+4) / 9

(207)2 = 4 / 2 / 8 / 4 / 9

(207)2 = 42849.

Easy Method to calculate Cube

We would like to explain the cube method through example only.

Find out the result of (16)3

Here we’ll write like this –

1  6  (6*6) (6*6*6) = 1 6 36 216

Problems on Ages Quick Maths Formulas

Formulas –

  • If the current age is x, then n times the age is nx.
  • If the current age is x, then age n years later/hence = x + n.
  • If the current age is x, then age n years ago = x – n.
  • The ages in a ratio a : b will be ax and bx.
  • If the current age is x, then 1/n times the age is x/n.

Quicker Methods – To find out son’s age, use this formula

  • If t1 years earlier the father’s age was x times that of his son. At present the father’s age is y times that of his son. Then the present age of son will be?

Son’s Age = t1 (x-1) / (x-y)

  • If present age of the father is y times the age of his son. After t2 years the father’s age become z times the age of his son. Then the present age of son will be?

Son’s Age = (z-1)t1 / (y-z)

  • t1 years earlier, the age of the father was x times the age of his son. After t2 years, the father’s age becomes x times the age of his son. Then the present age of son will be?

Son’s Age = [(z-1)t +  (x-1)t1] / (x-z)

  • Son’s or Daughter’s Age = [Total ages + No. of years ago (Times – 1)] / (Times+1)
  • Son’s or Daughter’s Age = [Total ages – No. of years ago (Times – 1)] / (Times+1)
  • Father: Son
  • Present Age = x : y
  • T years before = a:b
  • Then, Son’s age = y * [ T(a-b) / Difference of cross-product ]
  • And Father’s age = x * [ T(a-b) / Difference of cross-product ]

Average Quick Maths Formulas

  • Average = (Total of data) / (No. of data)
  • Age of New Entrant = New Average + No. of Old Members * Increase
  • Weight of New Person = Weight of Removed Person + No. Of Persons * Increase In Average
  • Number of Passed Candidates = Total Candidates * (Total Average – Failed Average) / (Passed Average – Failed Average)
  • Number of Failed Candidates = Total Candidates * (Passed Average – Total Average) / (Passed Average – Failed Average)
  • Age of New Person = Age of Removed Person – No. of Persons * Decrease in Average Age
  • Average after x innings = Total Score – Increment in Average * y innings
  • If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by – 2xy / (x+y) km/hr.
  • If half of the journey is travelled at a speed of x km/hr and the next half at a speed of y km/hr, then average speed during the whole journey is 2xy / (x+y) km/hr.
  • If a man goes to a certain place at a speed of x km/hr and returns to the original place at a speed of y km/hr, then the average speed during up and down journey is 2xy / (x+y) km/hr.
  • If a person travels 3 equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz / (xy+yz+zx) km/hr.

Percentage Quick Maths Formulas

  • Percentage = [Value / Total Value * 100]
  • If two values are respectively x% and y% more than a third value, then the first is the (100+x) / (100+y) *100% of the second.
  • If A is x% of C and B is y% of C, then A is x/y*100% of B.
  • x% of the quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by the third person. Now is A is left in the fund then there was (A*100*100*100) / (100-x) (100-y) (100-z) in the beginning.
  • x% of the quantity is added. Again y% of the increased quantity is added. Again z% of the increased quantity is added. Now, it becomes A, then the initial amount is given by (A*100*100*100) / (100+x) (100+y) (100+z)
  • If the original population of a town is P and the annual increase is r%, then the population in n years will be – P + P*r/100 = P*(1+r/100)
  • The population of a town is P. It increases by x% during the 1st year, increases by y% during the 2nd year and again increases by z% during the third year. Then, the population after 3 years will be – P*(100+x)(100+y)(100+z) / 100*100*100
  • When the population decreases by y% during the 2nd year, while for the 1st and 3rd years, it follows the same, the population after 3 years will be – P*100+x)(100-y)(100+z) / 100*100*100

Profit and Loss Quick Maths Formulas

  • Profit = Selling Price (SP) – Cost Price (CP)
  • Loss = Cost Price (CP) – Selling Price (SP)
  • Gain or Loss % = (Loss or Gain / CP) * 100 %
  • Gain % = [Error / (True Value – Error)] * 100 %
  • Gain % = [(True Weight – False Weight) / False Weight] * 100 %
  • Total % Profit = [(% Profit + % Less in wt) / (100 – % Less in wt)] * 100 %
  • If CP of x articles is = SP of y articles, then Profit % = [(x –y) / y] * 100
  • Cost Price = (100 * More Charge) / (% Diff in Profit)
  • Selling Price = More Charge * (100+ First Profit%) / (% Diff in Profit).

Time and Work Quick Maths Formulas

  • If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days then the formula will be – M1 * D1 * W1 = M2 * D2 * W2
  • If we add Time for both the groups T1 and T2 respectively, then the formula will become – M1 * D1 * T1 * W1 = M2 * D2 * T2 * W2
  • And if we add efficiency for both the groups E1 and E2 respectively, then the formula becomes – M1 * D1 * T1 * E1 * W1 = M2 * D2 * T2 * E2 * W2
  • If A can do a piece of work in x days and B can do it in y days, then A and B working together will do the same work in [(x*y)/(x+y)]
  • If A, B and C can do a work in x, y and z days respectively, then all of them working together can finish the work in [(x*y*z) / (xy + yz + zx)]
  • If A and B together can do a piece of work in x days and A can do it in y days, then B alone can do the work in (x*y) / (x-y)
  • Original Number of Workers = (No. of more workers * No. of days taken by the second group) / No. of less days

Pipe & Cisterns Quick Maths Formulas

  • If a pipe can fill a tank in x hours, then the part filled in 1 hour = 1/x.
  • If a pipe can empty a tank in y hours, then the part of the full tank emptied in 1 hour = 1/y.
  • If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled in 1 hour, when both the pipes are opened = (1/x) – (1/y).
  • Time taken to fill the tank, when both the pipes are opened = xy / (y-x).
  • If a pipe can fill a tank in x hours and another can fill the same tank in y hours, then the net part filled in 1 hour, when both the pipes are opened = (1/x) + (1/y).
  • Time taken to fill the tank = xy / (x+y).
  • If a pipe can fill a tank in x hours and another can fill the same tank in y hours, but a third one empties the full tank in z hours and all of them are opened together, then the net part filled in 1 hour = (1/x) + (1/y) + (1/z).
  • Time taken to fill the tank = xyz / (yz + xz – xy) hours.
  • A pipe can fill a tank in x hours. Due to a leak in the bottom, it is filled in y hours. If the tank is full, then the time is taken by the leak to empty the tank = xy / (y-x) hours.

Time and Distance Quick Maths Formulas

  • Speed = Distance / Time
  • If the speed of a body is changed in the ratio a:b, then the ratio of the time taken changes in the ration b:a.
  • If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then the average speed during the whole journey is 2xy / (x+y) km/hr.
  • Meeting point’s distance from starting point = (S1 * S2 * Difference in time) / (Difference in speed)
  • Distance travelled by A = 2 * Distance of two points (a / a+b)
  • Distance = [(Multiplication of speeds) / (Difference of Speeds)] * (Difference in time to cover the distance)
  • Meeting Time = (First’s starting time) + [(Time taken by first) * (2nd’s arrival time – 1st’s starting time)] / (Sum of time taken by both)

Problem on Train Quick Maths Formulas

  • When x and y trains are moving in the opposite direction, then their relative speed = Speed of x + Speed of y
  • When x and y trains are moving in the same direction, then their relative speed = Speed of x – Speed of y
  • When a train passes a platform, it should travel the length equal to the sum of the lengths of train & platform both.
  •  Distance = (Difference in Distance) * [(Sum of Speed) / (Diff in Speed)]
  • Length of Train = [(Length of Platform) / (Difference in Time)] * (Time taken to cross a stationary pole or man)
  • Speed of faster train = (Average length of two trains) * [(1/Opposite Direction’s Time) + (1/Same Direction’s Time)]
  • Speed of slower train = (Average length of two trains) * [(1/Opposite Direction’s Time) – (1/Same Direction’s Time)]
  •  Length of the train = [(Difference in Speed of two men) * T1 * T2)] / (T2-T1)
  •  Length of the train = [(Difference in Speed) * T1 * T2)] / (T1-T2)
  • Length of the train = [(Time to pass a pole) * (Length of the platform)] / (Diff in time to cross a pole and platform)

Boats and Streams Quick Maths Formulas

  •  If the speed of the boat is x and if the speed of the stream is y while upstream then the effective speed of the boat is = x – y
  • And if downstream then the speed of the boat = x + y
  •  If x km/hr be the man’s rate in still water and y km/hr is the rate of the current. Then
  • Man’s rate with current = x + y
  • Man’s rate against current = x – y
  •  A man can row x km/hr in still water. If in a stream which is flowing at y km/hr, it takes him z hrs to row to a place and back, the distance between the two places is = z * (x2 – y2) / 2x
  • A man rows a certain distance downstream in x hours and returns the same distance in y hours. If the stream flows at the rate of z km/hr, then the speed of the man in still water is given by – z* (x + y) / (y – x) km/hr.
  •  Man’s rate against current = Man’s rate with the current – 2 * rate of current
  •  Distance = Total Time * [(Speed in still water)2 – (Speed of current)2] / 2 * (Speed in still water)
  • Speed in Still Water = [(Rate of Stream) * (Sum of upstream and downstream time)] / (Diff of upstream and downstream time)

Allegation Quick Maths Formulas

  • If the gradients are mixed in a ratio, then
  • [(Quantity of cheaper) / (Quantity of dearer)] = [(CP of dearer) – (Mean Price)] / (Mean price) – (CP of cheaper)]
  • Quantity of Sugar Added = [Solution * (Required% value – Present% value)] / (100 – required% value)
  • Required quantity of water to be added = [Solution * (Required Fractional Value – Present Fractional Value)] / 1 – (Required Fractional Value)

Simple Interest Quick Maths Formulas

  • SI = p*t*r/100
  • The annual payment that will discharge a debt of INR A due in t years at the rate of interest r% per annum is = (100 * A) / [(100 * t) + r*t* (t-1)]/2
  • P = (Interest * 100) / [(t1*r1) + (t2*r2) + (t3*r3) + …..]
  • Rate = [100 * (Multiple number of principal – 1)] / Time
  • Sum = (More Interest * 100) / (Time * More Rate)

Compound Interest Quick Maths Formulas

  • When Interest is compounded annually –  Amount = P [1 + (r/100)]t
  • When Interest is compounded half-yearly – Amount = P [1 + (r/200)]2t
  • When Interest is compounded quarterly – Amount = P [1 + (r/400)]4t
  • When rate of Interest is r1%, r2% and r3% then – Amount = P [1 + (r1/100)] * [1 + (r2/100)] * [1 + (r3/100)]
  • Simple Interest for 2 years = 2*r = 2r% of capital
  • Compound Interest for 2 years = [2r +(r2/100)]% of capital
  • Simple Interest for 3 years = 3*r = 3r% of capital
  • Compound Interest for 3 years = [3r +(3r2/100) + (r3/1002)]% of capital.

Mensuration Quick Maths Formulas

  • Area of Rectangle = Length * Breadth
  • (Diagonal of Rectangle)2 = (Length)2 * (Breadth)2
  • Perimeter of Rectangle = 2 * (Length + Breadth)
  • Area of a Square = (Side)2 = 1/2 * (Diagonal)2
  • Perimeter of Square = 4 * Side
  • Area of 4 walls of a room = 2 * (Length + Breadth) * Height
  • Area of a parallelogram = (Base * Height)
  • Area of a rhombus = 1/2 * (Product of Diagonals)
  • Area of a Equilateral Triangle = Root of (3) / 4 * (Side)2
  • Perimeter of an Equilateral Triangle = 3 * Side
  • Area of an Isosceles Triangle = b/4 * root of 4a2 – b2
  • Area of Triangle = 1/2 * Base * Height
  • Area of Triangle = root of [s(s – a) * (s – b) * (s-c)]
  • Area of Trapezium = 1/2 * (Sum of parallel sides * perpendicular distance between them)
  • Circumference of a circle = 2*(22/7)*r
  • Area of a circle = (22/7) * r2
  • Area of a parallelogram = 2 * root of [s(s – a) * (s – b) * (s-d)]
  • Volume of cuboid = (l*b*h)
  • Whole Surface of cuboid = 2 * (lb + bh + lh) sq. units
  • Diagonal of Cuboid = Root of (l2 + b2 + h2)
  • Volume of a cube = a3
  • Whole Surface Area of cube = (6*a2)
  • Diagonal of Cube = Root of (3) * a
  • Volume of Cylinder = (22/7) * r2 * h
  • Curved Surface area of Cylinder = 2*(22/7)*r*h
  • Total Surface Area of Cylinder = [2*(22/7)*r*h] + {2*(22/7)*r2)
  • Volume of Sphere = (4/3) * (22/7) * r3
  • Surface Area of Sphere = 4 * (22/7) * r2
  • Volume of hemisphere = (2/3) * (22/7) * r3
  • Curved Surface area of hemisphere = 2 * (22/7) * r2
  • Whole Surface Area of hemisphere = 3 * (22/7) * r2.

FAQ's

Into how many sections does the IPMAT Question paper is divided?

The IPMAT Question paper is divided into two main sections-quantitative ability and verbal ability.The Quantitative Ability section will have short question answers and Objective type Multiple Choice Questions and the Verbal ability section will include Multiple Choice Questions only.

How many questions come from the IPMAT Question Paper?

The total number of questions in the IPMAT Question paper is 100. Total of 20 questions come from the Quantitative Section(short questions) of the IPMAT Question paper and 40 questions come in the Quantitative Section(MCQ). And 40 questions come from the Verbal Ability Multiple Choice Questions.

Can I get FREE IPMAT Books PDF?

Yes, you can study from the free IPMAT Books provided in this post. Enhance your preparation from the best books and score more in the exam.

Which is the best IPMAT Book for Verbal Ability?

For cracking verbal ability section, experts suggest Wren & Martin book. It is the highly recommended book by experts and toppers to crack English section easily.

Which is the best IPMAT Study Material for Quantitative Ability?

For cracking quantitative ability section, refer to R.D.Sharma or R.S.Aggarwal book. It includes all the concepts from 12th standard explained in depth. It is also easily understandable without any tutor.

What is the selection process of the IPMAT Exam for Domestic Applicants?

The selection process of the IPMAT Exam for Domestic Applicants include 5 stages and are as follows:

  • Registration
  • IPMAT Aptitude Test
  • Written Ability Test (WAT) & Personal Interview (PI)
  • Merit list generation
  • Final Merit List

What is the exam pattern of the IPMAT exam?

The aptitude exam consists of three sections: Quantitative Ability (Multiple Choice), Quantitative Ability (Short Answer Questions), and Verbal Ability (Multiple Choice Questions). The total number of questions in the exam is 100 and each question carries 4 marks. There will be a negative marking of 1 for each wrong answer. However, there is no negative marking for the Quantitative Ability section.

Can I prepare for IPMAT Exam in two months?

Yes. You can easily prepare for the IPMAT exam in two months with proper preparation methods and the following expert tips. Solving previous year papers will help you answer any type of question easily. Ensure to study all the topics from the highly recommended books by experts.

Is there a sectional time limit in the IPMAT Exam?

Yes. There is a sectional time limit in the IPMAT Exam. Quantitative Section will be given 80 minutes and for the Verbal Ability Section, 40 minutes will be provided. The candidates are required to answer the sections in a pre-specified order and the pre-specified order is the same for all exam appearing candidates.

Is the calculator allowed in the IPMAT Exam?

No. Calculator is not allowed in the IPMAT Exam. However, candidates can use a calculator present on the screen itself while taking a computer-based examination.