IPMAT Question Papers: Download Previous Year Papers Free PDF with Answer Keys

The Indian Institute of Management (IIM) organizes the IPMAT (Integrated Program in Management Aptitude Test), which is administered in an electronic format. Reviewing past IPMAT question papers and sample papers helps students grasp the test format, question kinds, subjects covered, and difficulty level. IPMAT 2024 Exam was taken on May 23, 2024. Candidates may download previous question papers to practice and improve their speed and accuracy. This article also tells you where to get IPMAT question papers with answers in PDF format to help you prepare.

New: IPMAT 2025 Rohtak Memory Based Questions and Exam Analysis

IPMAT 2025: Mock Test Series | Preparation Tips | Admission Guide | PYQ's

Admission Alert: IPM/BBA (Hons.) at TAPMI, Bengaluru. Accepts IPMAT Scores

This Story also Contains

1. IPMAT Question Papers
2. Download IPMAT Previous Year Question Papers PDF
3. How to Download IPMAT Question and Sample Papers
4. IPMAT Question Paper: Exam Pattern
5. How to Use IPMAT Question Papers
6. IPMAT Question Paper: Sample QA Questions with Solutions
7. IPMAT Syllabus:
8. IPMAT Previous Year papers
9. Strategies for Solving Previous Papers
10. Benefits of Solving IPMAT Previous Papers

IPMAT Question Papers

To crack IPMAT 2025 with excellent scores, past year papers and practice exams can really assist well. The previous year's IPMAT question papers help candidates assess their aptitude and test-taking skills and provide them with valuable insights on how to approach the IPMAT exam. By solving the IPMAT last year paper and using IPMAT Mock Test PDFs candidates can tremendously improve their time management skills, speed, and accuracy.

Download IPMAT Previous Year Question Papers PDF

Download the free IPMAT exam official question papers and practice with them to ramp up your IPMAT preparation.

How to Download IPMAT Question and Sample Papers

Candidates can follow the simple steps below to download previous IPMAT previous year papers pdf:

• Click on the link to the IPMAT question papers given above
• Select the PDF files.
• The PDF files will open in a new window.
• Download the IPMAT exam papers and save them.

IPMAT Question Paper: Exam Pattern

Candidates who want to appear for the IPMAT Exam must be familiar with the structure of the question paper. Below is an overview of the IPMAT question paper based on the previous year questions.

• The overall difficulty level of the IPMAT exam questions paper would be moderate.
• There will be three sections - two sections have MCQs questions while QA (SA) will have Non-MCQs questions
• The duration of the exam is 120 minutes, with each section having 40 minutes.
• There will be logic-based and a few arithmetic questions. The higher mathematics component is high.
• The exam will also have many language and grammar-based questions.

How to Use IPMAT Question Papers

Using IPMAT previous year papers with answers as a part of your preparation strategy can be highly beneficial. By keeping the following points in mind, you can maximize the benefits:

1. Understand IPMAT Exam Pattern

Analyze the distribution of questions across IPMAT sections (Quantitative Ability, Verbal Ability, and Logical Reasoning) and their respective weightage. This will help you understand which areas/topics are more frequently tested and prioritize your preparation accordingly.

2. Identify Important Topics for IPMAT

By going through IPMAT question paper pdfs, you can identify recurring topics or types of questions that are commonly asked. Focus more on these areas as they are likely to be important. In these papers, you'll find IPMAT Geometry questions, IPMAT Log questions, IPMAT percentage questions, IPMAT algebra questions, and IPMAT trigonometry questions. In the verbal ability questions, you can find IPMAT grammar syllabus questions.

3. Time Management Practice

Solve the IPMAT previous year papers within the stipulated time limit to practice time management. This will help you improve your speed and accuracy, crucial for competitive exams like IPMAT.

4. Make Notes

Good notes can make a good understanding of concepts. While solving the IPMAT previous year's question papers, you must make notes of easy and difficult questions so you can make your preparation ahead accordingly and keep revising difficult concepts. The notes can be used as a quick reference guide.

IPMAT Question Paper: Sample QA Questions with Solutions

IPMAT Exam 2019

Q1. The sum of the interior angles of a convex n-sided polygon is less than 2019. The maximum possible value of n is ______

Ans: Sum of the interior angles of polygon = (n-2). 180

=> (n-2). 180 < 2019

=> 180n – 360 < 2019

=> 180n < 2379

=> n < 2379/180

=> n < 13.21 {No. of sides must be an integer}

so maximum possible value of n = 13.

Q2. Three friends divided some apples in the ratio 3:5:7 among themselves. After consuming 16 apples they found that the remaining number of apples with them was equal to largest number of apples received by one of them at the beginning. Total number of apples these friends initially had was _____

Ans: Let three friends are A, B, C, A: B: C = 3:5:7,

So, A = 3x, B = 5x, C = 7x, Total apples = 15x

After consuming 16 apples remaining apples = 15x -16;

As given in the question, remaining apples = largest number of apples received by one of them at the beginning.

So, 15x-16 = 7x

8x = 16; x = 2

So, total number of apples = 15x = 15 x 2 = 30.

Q3. A shopkeeper reduces the price of a pen by 25% as a result of which the sales quantity increased by 20%. If the revenue made by the shopkeeper decreases by x% then x is

Ans: A = B X C

If percentage change in B and C is x% & y% then effective percentage change in A will be equal to (x + y + x.y/100) %. [Successive percentage change]

Revenue = Price of pen x Sales quantity

Percentage change x%+25%+20%

As per above explained formula percentage change in revenue = (25+20+25 x 20/100) %

X % = 50%.

Q4. The average of five distinct integers is 110 and the smallest number among them is 100. The maximum possible value of the largest integer is _____

Ans: Let five distinct integers are A,B,C,D & E, where A is smallest and E is largest.

As given, A=100

Average = 110

• (A + B + C + D + E)/5 = 110
• A + B + C + D + E = 550 ………..Equation (1)

For maximum value of E, other values must be minimum;

As all integers are distinct; If A = 100, B = 101, C = 102, D = 103

On putting value in equation (1), E = 550 – 100 -101- 102 – 103 = 144.

Q5. The number of whole metallic tiles that can be produced by melting and recasting a circular metallic plate, if each of the tiles has a shape of a right-angled isosceles triangle and the circular plate has a radius equal in length to the longest side of the tile (Assume that the tiles and plate are of uniform thickness, and there is no loss of material in the melting and recasting process) is ______

Ans:

Assume that the tiles and plate are of uniform thickness, and there is no loss of material in the melting and recasting process.

Let radius of circle = r

Length of longest side of right-angled isosceles triangle = r {Hypotenuse}

So, area of the triangle = (Hypo.)2 / 4 = r2/4

No. of triangles(N) X Area of triangle = Area of metallic circular plate

• N.r2/4 = πr2
• N = 4.π = 4 x 3.14 = 12.56

So, number of whole metallic tiles that can be produced by melting and recasting a circular metallic plate = 12.

Q6. If the compound interest earned on a certain sum for 2 years is twice the amount of simple interest for 2 years, then the rate of interest per annum is percent
a) 200%
b) 2%
c) 4%
d) 400%

Correct option: a)

Ans: let P= Principal, R = Rate of interest; Time = 2 years (given)

Compound Interest (C.I.) = 2 x Simple Interest (S.I.)

• P [1+R/100]2 – P = 2 x P.R.T/100
• P[{1+R/100}2 - 1] = 2 x P. R. 2/100
• 1 + 2R/100 + R2/1002 – 1 = 4. R/100
• R/100 [2 + R/100] = 4. R/100
• R = 200

Q7. The maximum value of the natural number n for which 21n divides 50! is
a) 6
b) 7
c) 8
d) 9

Correct option: c)

Ans:

To factorize 50!, IGP (Index of greatest power) is found out for prime numbers. But in this case divisor is 21n, (3 x 7)n so only IGP of 3 & 7 is required.

IGP of 3 in 50! = 22

IGP of 7 in 50! = 8

50! = 322 x 78

50! / 21n = 50! / (3 x 7)n;

In this case numerator has max 8 power of 7, so denominator can’t have more than 8 power of 7

So, maximum possible value of n = 8

Q8. Placing which of the following two digits at the right end of 4530 makes the resultant six-digit number divisible by 6,7 and 9?

a) 96

b) 78

c) 42

d) 54

Correct option: a)

Ans:

On putting values as per options:

a. 453096- Sum of digits = 27

b. 453078- Sum of digits = 27

c. 453042- Sum of digits = 18

d. 453054- Sum of digits = 21

On the basis of this we can say that option (d) is eliminated as sum of digits is not divisible by 9 but all other numbers satisfy condition as per divisibility rule of 6 & 9.

Now, for 7 we can divide other three options and check.

Q9. The inequality log2⁡3x-12-x<1 holds true for
a) x13,1
b) x13,2
c) x0,13∪(1,2)
d) x∈(-∞,1)

Correct option: d)

Ans: log2⁡3x-12-x<1

=> 3x-12-x < 21

=> 3x-1 < 2 (2-x) => 3x-1 < 4 – 2x

=> 5x < 5 => x < 1

Q10. The number of terms common to both the arithmetic progressions 2,5,8,11,...., 179 and 3,5,7,9,....., 101 is

a. 17

b. 16

c. 19

d. 15

Correct option: a)

Ans: A.P. 1 => 2,5,8,11……..179 {Common Difference d1=3}

A.P. 2 => 3,5,7,9………..101 {Common Difference d2=2}

Common sequence:

Common difference d = lcm {d1, d2} = lcm {3,2} = 6

From both sequences, first common term 5, then next term will be 11.

So, sequence 5,11,17,……..

nth term must be less than or equal to 101 {which is last & least term out of both sequences}

• an <= 101
• 5 + (n-1). 6 <= 101
• (n-1) <=16
• n <=17 ; so, No. of common terms of both sequences = 17

IPMAT Exam 2020

Q1. The probability that a randomly chosen factor of 1019 is a multiple of 1015 is
a) 125
b) 112
c) 120
d) 116

Correct option: d)

Ans: 1019 = 219 x 519

Total no. of factors = (19+1). (19+1) = 400

To find out no. of factors of 1019 which are multiple of 1015, find the factors of remaining part in no. after excluding 1015.

1019 = 1015 x 104 = 1015 x 24 x 54

No. of factors of 1019 which are multiple of 1015= (4+1). (4+1) = 25

Probability = favorable outcomes/ total no. of outcomes

= No. of factors of 1019 which are multiple of 1015 / Total no. of factors

= 25/400 = 1/16

Q2. The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is

a) 28

b) 29

c) 31

d) 33

Correct option: b)

Ans:

If three sides are consecutive, then possible sides of triangle are (1,2,3), (2,3,4), (3,4,5), (4,5,6)…………..

For acute angle triangle, triangle should follow following two conditions:

1. Sum of any two sides must be greater than the third one.

So, as per this condition (1,2,3) can’t be sides of any triangle.

2. Square of the greatest side must be less than the sum of squares of other two sides.

Ex: If sides of triangle are (a,b,c), where c is the greatest side then for acute angle triangle c2 < a2 + b2.

Hence, (2,3,4) & (3,4,5) doesn’t satisfy the condition of acute angle triangle.

Sides Perimeter (Sum of sides)

(1,2,3) 6

(2,3,4) 9

(3,4,5) 12

.

.

.

(32,33,34) 99

Out of these pairs first doesn’t satisfy as per condition (i) and second & third pair doesn’t satisfy the condition (ii) given and required.

So, no. of possible pairs of triangles is 29.

Q3. The equation of the straight line passing through the point P(-5,1), such that the portion of it between the axes is divided by the point P in to two equal halves, is
a) 10y-8x=80
b) 8y+10x=80
c) 10y+8x=80
d) 8y+10x+80=0

Correct option: a)

Ans:

The point P divides AB into 2 halves.
AP = PB
A is on the x axis, Hence Point A (h,0)
B is on the Y axis, Hence Point B (0,k)

Using midpoint formula.

[(ℎ+0)/2,(?+0)/2] = (-5 , 4)

h/2 = - 5 => h = -10

k/2 = 4 => k = 8

Hence the Y intercept is 8 and X intercept is - 10.
x/(-10) + y/8 = 1
-8x + 10y = 80
The equation is -8x + 10y = 80

Q4. A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is
a) 49
b) 59
c) 415
d) 215

Correct option: d)

Ans: Probability of speaking truth P(T) = 4/5

Probability of appearing 5 on dice P(5) = 1/6

He throws a die and reports that it is a five, the probability that it is actually a five is
= P(T) x P(5) = 4/5 x 1/6 = 2/15

Q5. Consider the following statements:
(i) When 0<x<1, then 11+x<1-x+x2.
(ii) When 0<x<1, then 11+x>1-x+x2.
(iii) When -1<x<0, then 11+x<1-x+x2.
(iv) When -1<x<0, then 11+x>1-x+x2.
Then the correct statements are
a) (i) and (ii)
b) (ii) and (iv)
c) (i) and (iv)
d) (ii) and (iii)

Correct option: c)

Ans:

Let’s compare these 2 inequalities, 11+x<1-x+x2.

• 1-x + x2 – (1/1+x) > 0

• (1 + x3 -1)/(1+x) > 0

• x3/(1+x) > 0

This inequality holds good when x is positive and when both numerator & denominator are negative.
Hence the correct answer is (i) and (iv).

Q6. In a division problem, product of quotient and the remainder is 24 while their sum is 10. If the divisor is 5 then dividend is ______

Ans: Dividend (N) = Quotient(Q) x Divisor(D) + Remainder (R)

Given- Q x R = 24………. Equation (1)

Q + R = 10………Equation (2)

By solving both equations (Either proper method or hit & trial)

Two possible values, (i) Q = 6 & R = 4

(ii) Q = 4 & R = 6 {This is not possible as remainder must be less than divisor}

So, only possible values are Q=6 & R=4,

So, N = Q X D + R

N = 6 x 5 + 4 = 34.

Q7. Ashok purchased pens and pencils in the ratio 2:3 during his first visit and paid Rs. 86 to the shopkeeper. During his second visit, he purchased pens and pencils in the ratio 4:1 and paid Rs. 112. The cost of a pen as well as a pencil in rupees is a positive integer. If Ashok purchased four pens during his second visit, then the amount he paid in rupees for the pens during the second visit is ______

Ans:

First Visit

Ratio of pen: pencil is 2:3 and he spends Rs 86

Let price of pen be 'x' and price of pencil be 'y', as nothing is given related to no. of pen and pencils, so there are multiple possibilities-

2x + 3y = 86 (or)
4x + 6y = 86 (or)
6x + 9y = 86 ………..

Second Visit

Ratio of pen and pencil is 4:1 and he spends Rs 112
It is given that he purchased 4 pens.
Hence, he purchased 4 pens and 1 pencil (as per ratio)
Then,
4x + y = 112

Since, from second visit, 4 pen and a pencil costs Rs 112
so, if more than 4 pens and 1 pencil are bought then the cost must be more than 112.

As per above discussion, only 2x + 3y = 86 is possible in first visit.

From first visit---- 2x + 3y = 86
From second visit---- 4x + y = 112
On solving both equations, We get y = 12, x = 25.

He paid Rs 100 during his second visit for the pens.

Q8. In a four-digit number, the product of thousands digit and unit digit are zero while their difference is 7. Product of the middle digits is 18. The thousands digit is as much more than the unit digit as the hundreds digit is more than the tens digit. The four-digit number is_____

Ans:

In the given 4-digit number, the product of thousands digit and unit digit is zero. If product of two digits is zero that means either both digits are zero or one of them is zero. As thousandth digit cannot be zero, then unit digit is zero.
__ __ __ 0

Given: Thousands digit and digit difference is 7, that means thousands digit is 7
7 __ __ 0

Given: Product of middle digits is 18 and difference is same as between thousands and unit digit. So, product of middle digits is 18 and difference is 7. Only two digits are possible 9 and 2.

As given hundreds digit is more than the tens digit. So hundreds digit is 9 and tens digit is 2

So, no. is 7920.

Q9. Ashok started a business with a certain investment. After few months, Bharat joined him investing half amount of Ashok's initial investment. At the end of the first year, the total profit was divided between them in ratio 3:1. Bharat joined Ashok after

Ans:

Ashok Bharat

Investment 2x (let) x

Amount 12 months y months

Profit Ratio = (Investment x Amount) Ratio

Ashok Bharat

Profit Ratio = 2x. 12 : x.y

3 :1 = 24 : y

On solving this, y = 8 months.

Q10. Out of 13 objects. 4 are indistinguishable and rest are distinct, The number of ways we can choose 4 objects out of 13 objects is ____

Ans:

Distinct objects= 9, Identical(indistinguishable) objects = 4

No. of ways of choosing any no. of items out of identical items =1.

Case 1: All 4 are distinct- No. of ways = 9c4

Case 2: 1 identical & 3 objects are distinct- No. of ways = 9c3 x 1 {1 way of choosing 1 identical}

Case 3: 2 identical & 2 objects are distinct- No. of ways = 9c2 x 1 {1 way of choosing 2 identical}

Case 4: 3 identical & 1 objects are distinct- No. of ways = 9c1 x 1 {1 way of choosing 3 identical}

Case 5: 4 identical & 0 objects are distinct- No. of ways = 9c0 x 1 {1 way of choosing 4 identical}

So, total no. of ways = 9c4 + 9c3 + 9c2 + 9c1 + 9c0 = 126 + 84 + 36 + 9 + 1 = 256.

IPMAT Exam 2021

Q1. The number of positive integers which divide (1890)⋅(130)⋅(170) and are not divisible by 45 is _____

Ans:

N = (1890).(130).(170) = 23.33.53.7.13.17

Total no. of factors of N = (3+1) (3+1) (3+1) (1+1) (1+1) (1+1) = 512

To find out no. of factors of N which are divisible by 45, find no. of factors of the remaining part of no. after excluding 45-

N = 45 x 23.31.52.7.13.17

No. of factors of N divisible by 45 = no. of factors of remaining part (23.31.52.7.13.17)

= (3+1) (1+1) (2+1) (1+1) (1+1) (1+1) = 192

No. of positive integers which divide N (1890)⋅(130)⋅(170) and are not divisible by 45 is = Total no. of factors of N - No. of factors of N divisible by 45 = 512 – 192 = 320

Q2. The sum up to 10 terms of the series 1.3+5.7+9.11+… is

Ans:

= 1*3 + 5*7 + 9*11 …..
=(2 - 1)(2 + 1) + (6 - 1)(6 + 1) + (10-1)(10+1) + …..
= (22 - 12) + (62 - 12) + (102 - 12) + ……
In this series, the common difference is 4.

2,6,10, ……..

The 10th term can be found to be , Sn = a + (n-1).d
= [2 + 9*4]
= 38
So the 10th term in the series can be rewritten as (382 - 12)

= (22 - 12) + (62 - 12) + (102 - 12) + ……+(382 - 12)

= 22 + 62 + 102 + … + 382 -1-1-1…..10 times
= 4 [12 +32 +52 …..192 ] – 10 {Sum of squares of n odd terms= n(2n-1)(2n+1)/3}
= 4 [1330] -10
=5310.

Q3. There are 5 parallel lines on the plane. On the same plane, there are n other lines which are perpendicular to the 5 parallel lines. If the number of distinct rectangles formed by these lines is 360, what is the value of n ?

Ans:

Let, the number of lines that are perpendicular to given 5 parallel lines is ‘n’.
To make a rectangle we need 2 lines from each group.

Total no. of rectangles = 5C2 x nC2
=> 5C2 xnC2 = 360
=> 10 x n(n−1)/2 = 360
=> n(n-1) = 72
=> n = 9.

Q4. There are two taps, T1 and T2, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T1 is opened keeping T2 closed, the water tank (initially full) becomes empty in half an hour. If both T1 and T2 are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T2 is opened while T1 is closed is

Ans: Time Work (LCM of time) Efficiency

T1 30 min. 2

60 unit

T1+T2 20 min. 3

So, efficiency of T2 = (T1+T2 ) – T1 = 3 -2 = 1

Time taken by T2 to empty the full tank alone = work/efficiency = 60/1 = 60 min.

Q5. A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is

Ans:

Total marks in 1 course = 200

Total courses = 5

Total marks of all courses = 1000

Total marks of all students = 30 x 1000 = 30000

The change in the percentage of average marks scored is (80.02 - 80.00) %
= 0.02%
So the change in the marks is 30 x 1000 x 0.02%
= 3 * 2
= 6 marks.

Q6. Let Sn be sum of the first n terms of an A.P. an. If S5=S9, what is the ratio of a3:a5
a) 9:5
b) 5:9
c) 3:5
d) 5:3

Correct option: a)

Ans:

Let terms of A.P. are, a1, a2, a3, a4, a5, a6, a7, a8, a9

Sum of A.P. = average x no. of terms

=> If no. of terms in A.P. are odd then average = middle-term.

a1, a2, a3, a4, a5 => Average = a3 S5 =a3 x 5

a1, a2, a3, a4, a5, a6, a7, a8, a9 => Average = a5 S9 = a5 x 9

as given, S5 = S9

=> a3 x 5 = a5 x 9

=> a3: a5 = 9 : 5

Q7. It is given that the sequence {xn} satisfies x1 = 0, xn+1 = xn + 1 + 2√(1+xn) for n = 1,2, . . . . . Then x31 is _______

Ans:

xn+1 = xn + 1 + 2√(1+xn)
So x2 = x1 + 1 + 2√(1+x1)
= 0 + 1 + 2√(1+0)
x2 = 3
Similarly, x3 = x2 + 1 + 2√(1+x2)
= 3 + 1 + 2√(1+3)
x3 = 8
We see a pattern here that the nth term in the series is n2 - 1
And so the 31st term is 312 - 1
961 - 1
= 960.

Q8. The unit digit in (743)85-(525)37+(987)96 is
a) 9
b) 3
c) 1
d) 5

Correct option: a)

Ans:

For unit digit, only unit digits are required.

= 385 - 537 + 796

{Cyclicity of 3 is 4, power 85/4 = rem. 1, so power is equivalent to 1}

{Cyclicity of 5 is 1, so all powers of 5 are equivalent to 1}

{Cyclicity of 7 is 4, power 96/4 = rem. 0, so power is equivalent to 4}

= 31 – 51 + 74

= 3 – 5 + 2401 = 3 – 5 +1 = -1 {If it is negative, then add 10 to get unit digit}

= -1 + 10 = 9.

Q9. If the angles A, B, C of a triangle are in arithmetic progression such that sin(2A + B) = 1/2 then sin(B + 2C) is equal to

a.−1/2

b. 1/2

c. −1/√2

d. 3/√2

Correct option: a)

Ans:

The angles A, B and C are in arithmetic progression and so the angles can be written as,
(B-d), B, (B + d), where d is the common difference between the angles.
The sum of the three angles of a triangle is 180 degrees and so,

B-d + B + B + d = 180°
Or
3B = 180°
B = 60°

Now, we also know that sin(2A + B) = 1/2
So, 2A + B = 30° or 150°
But since we already know B = 60o and the angles cannot be negative,
2A + B cannot be 30°
2A + B = 150°
2A + 60o = 150°
Or A = 45°
=> C = 75°

Now that we know all the angles we can find out the value of sin(B + 2C)
B + 2C = 60 + 150 = 210°
sin(210°) = sin(180° + 30°)
= sin(30°) [sin(180° + θ) = -sin θ]
= −1/2.

Q10. The highest possible value of the ratio of a four-digit number and the sum of its four digits is:
a) 1000
b) 277.75
c) 900.1
d) 999

Correct option: a)

Ans:

Fraction = Four-digit number / the sum of its four digits

For making the fraction largest we need to have a big number in the numerator and a small number in the denominator.
Let’s try 9999 in the numerator so the fraction becomes 9999/36
We can try to make denominator least, which can be 1.

The fraction becomes 1000/1 = 1000

IPMAT Syllabus:

The IPMAT exam generally assesses a candidate's aptitude in three main areas: Quantitative Ability (QA), Verbal Ability (VA), and Logical Reasoning (LR). Here's a detailed overview of the IPMAT syllabus for each section:

1. Quantitative Ability (QA):

2. 
   

   • Number System, Basic Arithmetic

   • Profit and Loss

   • Interests (SI, CI)

   • Ratio and Proportion

   • Time, Speed and Distance

   • Time and Work

   • Algebra

   • Geometry & Mensuration

   • Mensuration

   • Permutation & Combination

   • Probability

   • Set Theory

   • Functions

   • Trigonometry

   • Determinants

   • Vectors

   • Integration and Differentiation

2. Verbal Ability (VA):

   • Reading Comprehension

   • Grammar

   • Vocabulary

   • Para-jumbles

   • Sentence Completion

   • Error Spotting

3. Logical Reasoning (LR):

   • Data Interpretation

   • Logical Reasoning

   • Seating Arrangement

   • Blood Relation

   • Syllogism

   • Coding-Decoding

   • Direction Sense

   • Clock and Calendars

   • Analogy

   • Series Completion

   • Statement and Assumptions

IPMAT 2025 Section-wise Important Topics

The IPMAT exam paper will have three sections - QA (multiple choice), VA (multiple choice) and QA (short answer). There is negative marking and each question carries 4 marks. For every wrong answer, one mark will be deducted.

IPMAT Previous Year papers

Practicing IPMAT previous year papers is essential for effective exam preparation as it gives candidates a clear understanding of the exam pattern, question types, and difficulty level. By working through these papers, candidates can identify important topics and areas where they need to focus more. It also helps in improving time management skills, as candidates can practice completing the paper within the allotted time, thereby enhancing their speed and accuracy

Strategies for Solving Previous Papers

When addressing previous papers, it is critical to approach them strategically to maximize their usefulness. Start by establishing a timer to imitate the real test situation, which will help you improve your time management abilities. As you go through the paper, concentrate on recognizing the pattern and sorts of questions that arise frequently in exams. After finishing the task, carefully evaluate your answers, paying specific attention to errors. Analyze why you answered a question incorrectly and review the suggestions to improve your knowledge. It's also useful to divide questions into difficulty levels—easy, moderate, and hard—so you can concentrate on improving areas where you struggle. Regular practice with prior papers can help you identify patterns, enhance your speed, and gain confidence for the test.

Benefits of Solving IPMAT Previous Papers

Solving IPMAT past problems has various advantages that might greatly improve your test preparation. First, it gives a comprehensive grasp of the test format, allowing you to get acquainted with the various sorts of questions and the general framework. This familiarity alleviates exam-day worry and increases confidence. Regular practice with previous exams also helps with time management, since it allows you to determine how much time to devote to each part. Identifying repeated question patterns allows you to concentrate on important areas that are regularly evaluated. Reviewing your performance on these exams also helps to identify shortcomings and areas that need further focus, allowing for targeted revision. Overall, answering prior problems is an important step toward developing the skills and self-assurance required to succeed on the IPMAT test.