Quantitative Aptitude is a vast subject that involves a lot of concepts. You might be worried about how to remember formulas and short tricks for these concepts? To help you ease your preparation for the IPMAT Maths section, this post has listed all the important formulas.

Read through the complete post to know topic-wise important formulas IPMAT Quantitative Aptitude.

IPMAT Quantitative Ability Formulas & Short Tricks

Go through the shortcuts and quick tips to solve the IPM Maths section and score more marks. Utilize these short tricks to enhance your preparation for the IPMAT Maths section.

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)

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.

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 that 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.

We would like to explain the cube method by 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)

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 ]

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 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 speed of y km/hr, then the average speed during the 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 1^{st} year, increases by y% during the 2^{nd} 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

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