Geometry is one of the key areas to focus on for CAT 2025, as evident from the previous years’ exam pattern. If you have decided to prepare for CAT, you are at the right place. This article will guide you in solving some of the most common queries such as the important Geometry theorems you should learn first for CAT, the topics most frequently asked from Geometry, the frequency of Geometry questions in the exam, and how mastering this area can impact your overall percentile in CAT.
Tangent from external point, Intersecting circles, relation between chord and tangent, perpendicular to the chord from centre, Sector and Segment
Important Geometry Theorems for CAT - Just 5 You Need
Since Geometry is very vast topic in CAT Quantitative Aptitude, it contains several theorems and concepts to understand. You will find important theorems for CAT Geometry preparation in this article. You should have knowledge of basic concepts like
Relation between angles between two parallel lines
In this section, we are going to discuss 5 important theorems that covers almost 80% of CAT questions from geometry.
Why only 5 theorems cover 80% questions
The five theorems that we are going to discuss in this article covers almost 80% of the CAT geometry questions since these theorems are most versatile.
For example: Pythagoras theorem is important for the questions related to triangle, trigonometry, quadrilateral, and circles.
Theorem 1 - Pythagoras and Its Variations
The Pythagorean theorem states that in a right-angled triangle; the square of the longest side is equal to the sum of the squares of the other two sides.
The longest side in a right-angled triangle is known as Hypotenuse.
Also, if in a triangle, $a^2 + b^2 > c^2$, where '$c$' is the longest side of the triangle, the triangle is obtuse angled triangle.
if in a triangle, $a^2 + b^2 < c^2$, where '$c$' is the longest side of the triangle, the triangle is acute angled triangle.
Direct applications in right triangles
Pythagoras theorem is directly used in a right-angled triangle:
To find the third side if two sides are given.
To check whether the given triangle is right angled triangle or not.
If one of the acute angles is given and we need to relate all the sides of an equilateral triangle.
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For Example:
In a triangle ABC, right angles at A, angle B is 30o, determine the ratio of all the sides. Solution:
For angle B, Base is AB and Perpendicular is AC.
So, $\sin 30^\circ = \frac {AC}{AB}$
⇒ $\frac 12 = \frac {AC}{AB}$
Let $AB = 2, AC = 1$, then $BC^2 = AC^2 + AB^2$
So, $AC = \sqrt {2^2 - 1^1} = \sqrt 3$
Common CAT question types using Pythagoras and PYQ
In CAT, the questions on Pythagoras theorem, are asked on
The application of Pythagoras theorem in right angled triangle to find the unknown side
The application of Pythagoras theorem in quadrilateral and circles
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Q.1) In a right-angled triangle ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is:
A) 2
B) 3
C) 4
D) 8
Solution:-
Area of ∆ABC = $\frac{1}{2} \times 5 \times 12$ = 30 sq cm
So, Area of ∆ABP = $\frac {30}{1.5}$ = 20 sq cm
And, Area of ∆ABQ = $\frac {30 +20}{2}$ = 25 sq cm (Since the areas are in AP)
Now, Area of ∆ABP = $\frac{1}{2} \times 5 \times BP$ = 20 sq cm
⇒ $BP = 8$ cm
Now, Area of ∆ABQ = $\frac{1}{2} \times 5 \times BQ$ = 25 sq cm
⇒ $BQ = 10$ cm
So, PQ = $BQ-BP = 10 -8=2$ cm
Hence, the correct answer is option (1).
Q.2) In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP: PQ: QC is
A) 1:1:2
B) 1:2:4
C) 2:4:1
D) 2:4:3
Solution:-
The areas of the figures $ABP$, $APQ$, and $AQCD$ are in geometric progression.
Area of $ABP = m$.
Area of $APQ = mr$.
Area of $AQCD = mr^2$.
Also, Area of $AQCD = 4 \times$ (Area of $ABP$),
$\Rightarrow mr^2 = 4m$,
$\Rightarrow r = 2$.
So, the total area = Area of rectangle = $9 \times 6 = 54$.
So, $m + 2m + 4m = 54$,
$\Rightarrow m = \frac{54}{7}$.
Area of $ABP = m = \frac{54}{7}$,
$\Rightarrow \frac{1}{2} \times AB \times BP = \frac{54}{7}$.
$\Rightarrow BP = \frac{12}{7}$ (since $AB = 9$).
Area of $ABQ = 2m + m = \frac{162}{7}$,
$\Rightarrow \frac{1}{2} \times AB \times BQ = \frac{162}{7}$.
Triangles having exact similar shape but different sizes are similar triangles.
Two triangles are said to be similar if either their corresponding angles are equal or corresponding sides are in equal proportion.
Key properties to remember
Similarity Criteria
Description
AA
If two corresponding angles are equal then the triangles are similar
SSS
If the ratio of corresponding sides is equal then the triangles are similar.
SAS
If the ratio of two corresponding sides is equal and the angle between these two sides are equal then the triangles are similar.
Also,
If triangle ABC is similar to triangle PQR, then
$(\frac {AB}{PQ})^2 = (\frac {BC}{QR})^2 = (\frac {AC}{PR})^2 =\frac {ar(ABC)}{ar(PQR)} $
and $(\frac {AB}{PQ}) = (\frac {BC}{QR}) = (\frac {AC}{PR}) =\frac {ar(ABC)}{ar(PQR)} $
Application in height-distance
Application of similar triangles in the questions of height and distance can be understood with the help of following example:
A boy of height 120 cm is walking away from the base of a tower at a speed at 0.8 m/sec. If the tower is 2.8 m above the ground, find the length of the shadow of boy after 5 seconds.
From the information given in the question
The larger triangle and smaller triangle are similar as one angle of each is of 90 degrees and the other is common.
So, $\frac {280}{120} = \frac {4+x}{x}$
So, $x = 3$ metres
The Basic Proportionality Theorem (Thales’ theorem), states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
If PQ || BC, then $\frac {AP}{PB} = \frac {AQ}{QC}$
Some Important results derive from BPT:
1. $\frac {AP}{AB} = \frac {AQ}{AC}$
2. If P and Q are mid points of AB and AC respectively, then PQ = ½ BC.
How CAT frames tricky questions using BPT
In CAT, direct questions from BPT are rarely asked. They integrate BPT with the other topics like
Complex diagram of quadrilateral in which BPT will be used to relate the sides of triangles formed inside the quadrilateral to find the unknowns.
In cyclic quadrilaterals in which diagonals are intersecting.
In circles to relate chords, radius, and tangents.
In coordinate geometry where equation of two or more intersecting lines are given in a plane and we have to determine the coordinates of intersecting points or lengths of sides forming triangles to find the area etc.
Theorem 4 - Circle Geometry (Tangent & Secant Properties)
Tangent is a line that touches a circle at one point only while a secant intersects a circle at two points.
Tangent-radius relation (most tested)
Radius is perpendicular to the tangent at point of contact. Questions based on tangent and radius relations are frequently asked. An important relation is given below:
$OP$ is perpendicular to $PQ$.
So, $OQ^2 = OP^2 + PQ^2$
Radius - chord relation
Perpendicular to the chord from the centre of a circle bisects the chord. An important relation is given below:
Power of a point theorem in CAT questions
Point theorem is applied to find the length of a line segment, chord, and tangent in CAT questions.
A point outside the circle:
A tangent drawn from point B outside the circle at point A, and a secant from B intersects the circle at C and D. Then $AB^2 = BC \times BD$.
Two secants drawn from a point outside the circle:
Two secants from point C intersect the circle at A, B and D, E respectively. Then CA × CB = CD × CE.
Two intersecting chords inside the circle:
Two chords AB and CD intersecting at a point P inside a circle, then PA × PB = PC × PD.
Tangent - Secant theorem:
A tangent drawn from point P outside the circle at point C, and a secant from P intersects the circle at B and A. Then $PC^2 = PA \times PB$.
Theorem 5 - Area Theorems (Heron’s Formula & Median Theorem)
There are several formulas to find the area of a triangle but Herons formula can be used for any triangle.
If the sides of a triangle are a, b, and c, then the area of a triangle is given by
$\text {Area of a triangle} = \sqrt {s(s-a)(s-b)(s-c)}$
where s is semi perimeter of the triangle given by
$s=\frac {a+b+c}{2}$
Median is a line joining a vertex of triangle to the mid-point of the opposite side. The intersection point of all medians is known as Centroid. Centroid divides a median in the ratio 2 : 1 from vertex to the side.
Fast calculation of triangle areas and some important results:
There are some concepts and formulas used to determine the area of a triangle effectively.
1. For scalene triangle (three sides are known):
Use Herons formula (Discussed above)
2. For an equilateral triangle (Side is known)
Area = $\frac {\sqrt 3}{4} a^2$, where $a$ is the side of the triangle.
3. Two sides and angle between them is known
Area = $\frac{1}{2}$ × product of two sides × sine of the included angle
4. If inradius ($r$) and semi perimeter (s) is known
Area = rs
5. If circumradius ($R$) and three sides are known
Area = $\frac{abc}{4R}$, where R = circumradius
6. Universal formula to determine the area of a triangle
Area = $\frac{1}{2}$ × base × height
Application in quadrilaterals and polygons
Herons’ formula can be used to find the area of a quadrilateral and polygon. To find the area of a quadrilateral and polygon, divide it in to triangles and apply the Herons formula to each triangle.
Example: Dimensions of quadrilateral are as shown in the figure. Find the area.
Area of ΔABD = $\frac 12 \times 6 \times 8$ = 24 sq units
Find area of triangle BCD using Heron’s formula
s = (6 + 10 + 10)/2 = 13
Area of ΔBCD = $\sqrt {13(13-6)(13-10)(13-10)}$
$= 3 \sqrt {91}$ sq units
So, Area of ABCD = 24 + 3 $\sqrt {91}$ sq units
Similarly, we can determine the area of a polygon.
Solved Past 5 Years’ Questions with Solutions of Geometry for CAT
Q.1) From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is:
A) $\mathrm{\frac{\sqrt{3} s^2}{2}}$
B) $\mathrm{{\frac{2 s^2}{\sqrt{3}}} }$
C) $\mathrm{\frac{s^2}{2 \sqrt{4}} }$
D) $\mathrm{\frac{s^2}{2 \sqrt{3}} }$
Solution:-
Based on the question: A D, C E and B F are the three altitudes of the triangle. It has been stated that
$\mathrm{ \{\mathrm{GD}+\mathrm{GE}+\mathrm{GF}=\mathrm{s}\} }$
Now since the triangle is equilateral, let the length of each side be "a".
So the area of the triangle will be
$\mathrm{ \frac{1}{2} \times G D \times a+\frac{1}{2} \times G E \times a+\frac{1}{2} \times G F \times a=\frac{\sqrt{3}}{4} a^2 }$
Now, $\mathrm{ G D+G E+G F=\frac{\sqrt{3} a}{2} \text { or } s=\frac{\sqrt{3} a}{2} \text { or } a=\frac{2 s}{\sqrt{3}} }$
Given the area of the equilateral triangle $\mathrm{ =\frac{\sqrt{3}}{4} a^2 }$
substituting the value of 'a' from above, we get the area { in terms 's'} $\mathrm{ =\frac{s^2}{\sqrt{3}}}$
Hence, the correct answer is option (4).
Q.2) Let C1 and C2 be concentric circles such that the diameter of C1 is 2 cm longer than that of C2. If a chord of C1 has length 6 cm and is tangent to C2, then the diameter, in cm, of C1 is:
A) 10
B) 8
C) 12
D) 16
Solution:-
Now we know that the perpendicular from the centre to a chord bisects the chord. Hence at the point of intersection of tangent, the chord will be divided into two parts of 3 cm each. As you can clearly see in the diagram, a right-angled triangle is formed there.
So, $\mathrm{(r+1)^2=r^2+9 \text { or } r^2+1+2 r=r^2+9 \text { or } 2 r=8 \text { or } r=4 \mathrm{~cm}}$
Therefore, the radius of the larger circle is 5 cm and the diameter is 10 cm.
Hence, the correct answer is option (1).
Q.3) Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the midpoint of CD, then the length of AT, in cm, is:
A) $\sqrt{13}$
B) $\sqrt{14}$
C) $\sqrt{12}$
D) $\sqrt{15}$
Solution:-
Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangles forming the hexagon.
Hence, its length should be twice the side of the hexagon, in this case, 4 cm.
Now, AD divided the hexagon into two symmetrical halves.
Hence, AD bisects angle D, and hence, $\angle$ADC is 60°.
We can find out the value of AT using the cosine formula:
Q.4) A circle of diameter 8 inches is inscribed in a triangle ABC, where $\angle A B C=90^{\circ}$. If BC = 10 inches, then the area of the triangle in square inches is:
A) 120
B) 100
C) 80
D) 90
Solution:-
Diameter = 8 inches; So, radius = 4 inches
Let p be the perpendicular and h be the hypotenuse of the triangle.
We know that inradius = $\frac{(\text { Perpendicular }+ \text { Base }- \text { Hypotenuse })}{2}$
$
⇒4=\frac{(p+10-h)}{2}
$
$
⇒\mathrm{h}-\mathrm{p}=2$
$ ⇒\mathrm{h}=\mathrm{p}+2
$
Now, $\mathrm{P}^2+100=\mathrm{h}^2$
$
\begin{aligned}
&⇒ p^2+100=(p+2)^2 \\
& ⇒p^2+100=p^2+4 p+4 \\
&⇒ 4 p=96 \\
& \therefore p=24
\end{aligned}
$
Hence, area of the $\triangle$ABC $=\frac{1}{2} \times 10 \times 24=120$
Hence, the correct answer is option (1).
Q.5) All the vertices of a rectangle lie on a circle of radius $r$. If the perimeter of the rectangle is P, then the area of the rectangle is:
A) $\frac{P^2}{16}-r^2$
B) $\frac{P^2}{8}-2 r^2$
C) $\frac{P^2}{2}-2 P r$
D) $\frac{P^2}{4}-2 P r$
Solution:-
Let the length and breadth of the rectangle be $l$ and $b$, respectively.
So, the area of the rectangle = $lb$
$\begin{aligned} &\text{Now, } l^2+b^2=4 r^2 \\ &\text{Also, } \mathrm{P}=2(l+b) \\ &⇒ \frac{P}{2}=l+b\end{aligned}$
Squaring on both sides, we get
$\begin{aligned} & \frac{P^2}{4}=l^2+b^2+2 l b \\ & ⇒\frac{P^2}{4}=4 r^2+2 l b \\ &⇒ \frac{P^2}{8}-2 r^2=l b\end{aligned}$
So, the area of the rectangle is $\frac{P^2}{8}-2 r^2$.
Hence, the correct answer is option (2).
Q.6) Suppose the medians BD and CE of a triangle ABC intersect at a point O. If the area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is:
A) 9
B) 12
C) 15
D) 18
Solution:-
Given: The medians BD and CE of a triangle ABC intersect at a point O and the area of triangle ABC is 108 sq. cm.
(Area of ABD):(Area of BDC) $=1: 1$
Therefore, the area of $A B D=54$
Area of $A D E$ : Area of $E D B=1: 1$
Therefore, the area of $A D E=27$
$\mathrm{O}$ is the centroid and it divides the medians in the ratio of 2:1
So, (Area of BEO):(Area of EOD) $=2: 1$
⇒ Area of EOD $=9$
Hence, the correct answer is option (1).
Q.7) $A B C D$ is a rectangle with sides $A B=56 \mathrm{~cm}$ and $B C=45 \mathrm{~cm}$, and $E$ is the midpoint of side $C D$. Then, the length, in cm , of radius of incircle of $\triangle A D E$ is
Solution:-
Given: $AB = 56$ cm, $BC = 45$ cm, so $AD = 45$ cm, $CD = 56$ cm.
$E$ is the midpoint of $CD \Rightarrow CE = 28$ cm.
In triangle $ADE$, use Pythagoras theorem:
$DE^2 = AD^2 + CE^2$
$⇒DE^2 = 45^2 + 28^2 = 2025 + 784 = 2809$
$⇒DE = \sqrt{2809} = 53$ cm
Using inradius formula for triangle with sides $a$, $b$, $c$:
$r = \frac{a + b - c}{2}$, where $c$ is the longest side
Q.8) A quadrilateral ABCD is inscribed in a circle such that AB:CD = 2:1 and BC:AD = 5:4. If AC and BD intersect at the point E, then AE:CE equals:
A) 1:2
B) 5:8
C) 8:5
D) 8:7
Solution:-
In $\triangle AED$ and $\triangle BEC$
$\angle A D E=\angle B C E$ (Angles made by same are on the same side of the triangles are equal)
$\angle A E D=\angle B E C$ (Vertically opposite angles are equal)
$\triangle A D E \sim \triangle B C E$ (By AA similarity criteria)
So, $\frac{A D}{B C}=\frac{D E}{C E}$.............(1)
In $\triangle C E D$ and $\triangle BEA$
$\angle E D C=\angle B A E$ (Angles made by same are on the same side of the triangles are equal)
$\angle C E D=\angle B E A$ (Vertically opposite angles are equal)
$\triangle C D E \sim \triangle B A E$ (By AA similarity criteria)
So, $\frac{C D}{A B}=\frac{D E}{A E}$..................(2)
From (1) and (2), we get, $\left(\frac{A D}{B C}\right)\left(\frac{A B}{C D}\right)=\left(\frac{D E}{C E}\right)\left(\frac{A E}{C E}\right)$
So, $\left(\frac{A E}{C E}\right)=\left(\frac{4}{5}\right)\left(\frac{2}{1}\right)=\frac{8}{5}$
Hence, the correct answer is option (3).
CAT Geometry Shortcuts & Smart Approaches
To solve the questions on Geometry effectively, use diagrams to interpret the question, eliminate inappropriate options, learn formulas, understand the questions properly.
Elimination technique in MCQs
In MCQs, try to eliminate wrong options by using
Ratios
Using angle sum property in the questions where an unknown angle needs to determine.
Using the constraints given in the question.
Approximations and diagrams for speed
In the questions related to geometry, drawing diagrams is one of the key aspects.
To increase speed, draw an approximate and rough diagram that should make you understand the problem.
Label the key points, angles, and measurement in the diagram.
Geometry formulas for CAT exam to memorize
There are hundreds of formulas in Geometry. You need to memorize the key formulas of the concepts that are frequently asked in CAT quantitative aptitude.
There are some questions that can be solved using concepts easily. For such questions, avoid the formulas.
In the next section, we are going to summarize the key formulas.
Geometry Formulas for CAT Exam - The Ultimate Revision List
Concept
Important Formulas
Triangle
Sum of the length of any two sides > third side.
Difference between the lengths of any two sides of a triangle < third side.
Side opposite to the greatest angle will be the largest and the side opposite to the least angle is the smallest.
The sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,
where $R$ is the circumradius.
The cosine rule: $a^2 = b^2 + c^2 - 2bc \cos A$.
This is true for all sides and their respective angles.
Area formulas:
Area = $\frac{1}{2} \times \text{base} \times \text{height}$.
Area = $\sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{a+b+c}{2}$.
Area = $r \times s$, where $r$ is the inradius.
Area = $\frac{1}{2} \times (\text{product of two sides}) \times \sin (\text{included angle})$.
Area = $\frac{abc}{4R}$, where $R$ is the circumradius.
Area of a quadrilateral = $\frac{1}{2} \times (\text{product of diagonals}) \times (\sin \text{ of the angle between them})$.
A parallelogram circumscribed about a circle is a rhombus.
Median (for a trapezium) = $\frac{1}{2} \times$ sum of the parallel sides (median is the line equidistant from the parallel sides).
Properties related to diagonals
Square
Rhombus
Rectangle
Parallelogram
Isosceles
Trapezium
Diagonals bisect each other
Yes
Yes
Yes
Yes
Yes
Diagonals are Equal
Yes
No
Yes
No
Yes
Diagonals bisect at 90 degrees
Yes
Yes
No
No
No
Properties of a Regular Polygon
Number of diagonals in an $n$-sided regular polygon = $\frac{n(n-3)}{2}$.
Measure of an interior angle of a regular polygon = $\frac{(n-2)180^\circ}{n}$.
Measure of an exterior angle of a regular polygon = $\frac{360^\circ}{n}$.
Circles
Area of a circle = $\pi r^2$, where $r$ is the radius of the circle.
Circumference = $2\pi r$.
Length of arc = $\frac{\theta}{360} \times 2\pi r$, where $\theta$ is the angle made by the arc at the centre.
The perpendicular from the centre of a circle to a chord bisects the chord. The converse is also true.
If two circles intersect in two points then the line through the centres is the perpendicular bisector
of the common chord.
Equal chords of a circle or congruent circles are equidistant from the centre.
The degree measure of an arc of a circle is twice the angle subtended by it at any point on the alternate segment of the circle.
Properties related to Cyclic Quadrilateral
Sum of opposite angles is 180 degrees.
Properties related to Tangents and Secants
If a circle touches all the four sides of a quadrilateral, then the sum of the two opposite sides is equal to the sum of other two
Tangents to a circle form an external point are equal.
Radius is perpendicular to the tangent at point of contact.
Tangent-Secant Theorem
Common Mistakes in CAT Geometry Preparation
Mistakes
How to Rectify
Over-relying on memorisation vs practice
Do not ignore basic concepts, during practice try to solve questions conceptually rather than using formula.
Ignoring Visual diagram practice
Practice by drawing approximate diagrams to solve geometry questions.
Failing to connect theorems with real CAT questions
Include real CAT problems in your practice to understand the application of theorems in real CAT questions.
CAT Geometry Preparation Strategy - Step by Step
We divide our strategy to prepare Geometry for CAT in 3 steps:
Step 1: Work on to build fundamentals:
First, understand the concepts related to angles between two parallel lines and intersecting lines such as corresponding angles, alternate interior angles, vertically opposite angles etc. Also, learn complementary angles and supplementary angles.
Understand theorems related to triangles, quadrilaterals, and Circles.
Step 2: Gain mastery in formulas and drawing diagrams:
Prepare a list of all formulas and practice questions using diagrams. This will help you in saving time and increase accuracy.
Step 3: Taking Mock tests and analyse them:
Include practice tests and CAT Mock tests in your preparation to identify your weak areas to improve. Also make a log to note down the mistakes you are making regularly. Work on the mistakes to rectify. Also, monitor your score after each test and give a direction to your preparation accordingly.
Best books and online resources for Geometry CAT prep
1. NCERT Mathematics Class 9 and 10
2. How to prepare quantitative Aptitude for CAT by Arun Sharma.
3. Quantitative Aptitude for CAT by Nishit K Sinha
CAT 2025 Preparation Resources by Careers360
The candidates can download the various CAT preparation resources designed by Careers360 using the links given below.
eBook Title
Download Links
CAT 2025 Arithmetic Important Concepts and Practice Questions
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The basic eligibility criteria for CAT are:
You must hold a bachelor’s degree in any discipline from a recognised university with at least 50 percent marks (45 percent for SC, ST and PwD categories). Final-year graduation students are also eligible to apply.
Being a commerce graduate can actually be an advantage in areas such as accounting, finance, economics and business studies, especially during MBA coursework in finance, marketing and operations.
After qualifying CAT, admission depends on multiple factors including CAT percentile, academic background, work experience (if any), performance in GD, WAT and personal interview rounds. Many IIMs and B-schools follow a diverse academic background policy, which means non-engineering candidates, including commerce graduates, often receive additional weightage.
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So, as a commerce graduate, you are fully eligible to appear for CAT and pursue an MBA, provided you meet the minimum academic requirements and perform well in the selection process.
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With a CAT 2025 raw score of 84 and strong sectionals, your expected percentile may fail in the high 98 range , depending on scaling. Your excellent academics and 2 years of work experience strengthen your profile .IIM Lucknow calls are possible but competitive for EWS engineers , while FMS Delhi looks mainly at CAT score and VARC , so you have a reasonable chance there . Final calls depend on cutoffs and composite score.
For MANAGE Hyderabad MBA , the OBC cutoff in CAT 2024 was generally around 85-90 percentile for shortlist consideration , tough exact figures vary by section and profile . For CAT 2025, if fewer candidates appeared , the cutoff might slightly adjust , but percentile based cutoffs dont change drastically year to year. A minor decrease is possible , but you should still aim for 90+ percentile to be competitive. Factors like VARC, QA-LRDI and overall profile influence the final shortlist.
For a raw score of 56 in CAT 2025 Slot 3, your expected overall percentile is likely to be in the range of the 90th-95th percentile. The exact percentile can vary slightly on the final normalization process and the process and performance of all test-takers. In this
article
you'll find more about the CAT result.
Hello,
With a projected CAT percentile of 87% but not clearing sectional cutoffs, your chances at top IIMs are limited because they require both overall percentile and sectional minimums. However, you still have a good shot at other reputed management institutes and non-IIM B-schools. Consider colleges like NMIMS, SPJIMR, IMT, TAPMI, Great Lakes, and other well-ranked private or state-level B-schools that accept CAT scores and weigh your profile holistically. Your academic record, BSc in Animation with 80%, and 5 years of work experience at Ubisoft India are strong points and may help in institutes that value work experience in their selection process. Also, explore institutes that accept XAT, MAT, or CMAT, where your profile can be competitive.
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A Sales Manager leads a sales team to meet targets, formulates strategies, analyses performance, and monitors market trends. They typically hold a degree in management or related fields, with an MBA offering added value. The role often demands over 40 hours a week. Strong leadership, planning, and analytical skills are essential for success in this career.
A marketing manager is a person who oversees a company or product marketing. He or she can be in charge of multiple programmes or goods or can be in charge of one product. He or she is enthusiastic, organised, and very diligent in meeting financial constraints. He or she works with other team members to produce advertising campaigns and decides if a new product or service is marketable.
A Marketing manager plans and executes marketing initiatives to create demand for goods and services and increase consumer awareness of them. A marketing manager prevents unauthorised statements and informs the public that the business is doing everything to investigate and fix the line of products. Students can pursue an MBA in Marketing Management courses to become marketing managers.
An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.
Digital marketing is growing, diverse, and is covering a wide variety of career paths. Each job function aids in the development of effective digital marketing strategies and techniques. The aims and objectives of the individuals who opt for a career as a digital marketing executive are similar to those of a marketing professional: to build brand awareness, promote company services or products, and increase conversions. Individuals who opt for a career as Digital Marketing Executives, unlike traditional marketing companies, communicate effectively through suitable technology platforms.
Individuals who opt for a career as a business analyst look at how a company operates. He or she conducts research and analyses data to improve his or her knowledge about the company. This is required so that an individual can suggest the company strategies for improving their operations and processes.
In a business analyst job role a lot of analysis is done, things are learned from past mistakes and the successful strategies are enhanced further. A business analyst goes through real-world data in order to provide the most feasible solutions to an organisation. Students can pursue Business Analytics to become Business Analysts.
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