Updated On : August 7, 2023

According to the experts, learning formulas and short tricks would help you crack the quantitative section of the IPMAT exam easily.

In this post, we have provided the topic-wise important number system for IPMAT Maths Formulas to help enhance your preparation levels.

So, why late! Scroll through the post to learn important formulae and perform calculations using these formulae in a matter of seconds.

As we all know, the quantitative ability section plays a vital role in the IPMAT exam, and this is because it holds 60% weightage in the exam compared to other areas.

Each one of your preparation strategies may vary. However, focusing on formulae will help enhance your preparation for the number system for IPMAT Maths section.

From the post below, let us go through the formulas based on number system for IPMAT, square roots, cube roots, percentage, profit, and loss, etc.

- 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 = n
^{2}

- Sum of first n even numbers = n (n + 1)
- (a + b) * (a – b) = (a
^{2}– b^{2}) - a + b)*2 = (a2 + b2 + 2ab)
- (a – b)*2 = (a
^{2}+ b^{2}– 2ab) - (a + b + c)*2 = a
^{2}+ b^{2}+ c^{2}+ 2(ab + bc + ca) - (a
^{3}+ b^{3}) = (a + b) * (a^{2}– ab + b^{2}) - (a
^{3}– b^{3}) = (a – b) * (a^{2}+ ab + b^{2}) - (a
^{3}+ b^{3}+ c^{3}– 3abc) = (a + b + c) * (a^{2}+ b^{2}+ c^{2}– ab – bc – ac) - When a + b + c = 0, then a
^{3}+ b^{3}+ c^{3}= 3abc

**Note:**

In the above formulae, "n" or "a" or "b" or "c" are constants or variables with value as an integer > 0.

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 divides 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**

The following are the three methods that are used to find the highest common factor of the given numbers:

- Prime factors method
- Divison method
- HCF of the large number

Let us understand all these methods with examples from the post below.

**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.

**Read More: Expert Recommended Resources for IPMAT Quant**

**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 are found).
- Now, perform the division method to the remaining numbers and find the HCF.

**Methods of finding LCM of two or more numbers**

You can use the prime factors method to find the least common multiple of the given numbers.

Let us understand how to find the LCM using the prime factors method with the following example.

**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 in the given numbers.
- The product obtained will be the LCM.

* Read More*:

**Example:**

LCM of 8,12,15 and 21.

Now, 8 can be written as = 2*2*2 = 23

12 = 2*2*3 = 22*3

15 = 3*5

21 = 3*7

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

LCM = 23*3*5*7

= 840.

You can follow the **Toppers Tips to Crack IPMAT **with a good score. The following are some of the important formulas from the simplifaction topics.

**BODMAS Rule**

- Through this rule, you can understand the correct sequence in which the operations are to be executed.
- This rule depicts the correct sequence in which the operations are executed and can evaluate the sequence.

**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.

**Read More: Download Free Previous Year Quant Question Papers**

**Duplex Combination Method for Squaring**

In this method, we either calculate the square by multiplying the same digit twice or 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 better 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**

Let us understand the cube method with the help of the following example:

**Find out the result of (16) ^{3}**

Here, we will write the 16 power 3 in the following way:

1 6 (6*6) (6*6*6)

6*6 = 36

6*6*6 = 216

Therefore, the final value = 1 6 36 216

The following are some important formulas for the problems based on ages for the upcoming **IPMAT Entrance Exam**.

- 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 the 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)**

- Suppose the present age of the father is y times the age of his son. After t2 years, the father’s age becomes z times the age of his son. Then the present age of son will be?

**Son’s Age = (z-1)t1 / (y-z)**

- t
_{1}years earlier, the age of the father was x times the age of his son. After t_{2}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 _{2 } + (x-1)t_{1}] / (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 ]**

**Read More:** **Know Previous Year Qualifying Marks for IPMAT Exam**

- 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 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 y km/hr, then the 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 y km/hr speed, then the average speed during the up and down journey is 2xy / (x+y) km/hr.
- If a person travels three equal distances at a speed of x km/hr, y km/hr, and z km/hr, respectively, the average speed during the whole journey is 3xyz / (xy+yz+zx) km/hr.

**Percentage = [Value / Total Value * 100]**- If two values are 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 1
^{st}year, increases by y% during the 2^{nd}year, and 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 three years will be –
**P*100+x)(100-y)(100+z) / 100*100*100**

Selling Price (SP) – Cost Price (CP)**Profit =**Cost Price (CP) – Selling Price (SP)**Loss =**(Loss or Gain / CP) * 100 %**Gain or Loss % =**- 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).

- 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 fewer days.

**Important Pipe & Cisterns Formulas for IPMAT**

- If a pipe can fill a tank in x hours, 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 can empty the full tank in y hours, then the net part is filled in 1 hour, when both the pipes are opened = (1/x) – (1/y).
- Time is taken to fill the tank when 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, the net part is 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, the leak takes time to empty the tank = xy / (y-x) hours.

- Speed = Distance / Time
- If the body's speed is changed in the ratio a:b, then the ratio of the time taken changes in the ratio 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) * (Second’s arrival time – First's starting time)] / (Sum of time taken by both)

Solving problems related to trains require time management skills and problem-solving skills. Therefore, attempting ** IPMAT Mock Tests** will help improve your time management skills and speed in the final exam.

- 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 the train & platform.
- 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)

The **Questions asked in the IPMAT Exam** from the boats and streams topic will be based on the following 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 by the man’s rate in still water, y km/hr is the current rate. 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 z km/hr, then the man's speed 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)

- 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)

- 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)

- 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 +(r
^{2}/100)]% of capital - Simple Interest for 3 years = 3*r = 3r% of capital
- Compound Interest for 3 years = [3r +(3r
^{2}/100) + (r^{3}/100^{2})]% of capital.

- 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 4a
^{2}– b^{2} - 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) * r
^{2} - 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 (l
^{2}+ b^{2}+ h^{2}) - Volume of a cube = a
^{3} - Whole Surface Area of cube = (6*a
^{2}) - Diagonal of Cube = Root of (3) * a
- Volume of Cylinder = (22/7) * r
^{2}* h - Curved Surface area of Cylinder = 2*(22/7)*r*h
- Total Surface Area of Cylinder = [2*(22/7)*r*h] + {2*(22/7)*r
^{2}) - Volume of Sphere = (4/3) * (22/7) * r
^{3} - Surface Area of Sphere = 4 * (22/7) * r
^{2} - Volume of hemisphere = (2/3) * (22/7) * r
^{3} - Curved Surface area of hemisphere = 2 * (22/7) * r
^{2} - Whole Surface Area of hemisphere = 3 * (22/7) * r
^{2}.

Mastering the important formulas in the IPMAT Maths section is essential for aspiring students in Indore and Rohtak. These formulas serve as powerful tools that enable students to solve complex problems efficiently and accurately. By understanding and memorizing these formulas, students can gain a competitive edge and improve their overall performance in the IPMAT exam. All the best!

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