Algebra is one of the most consistent and high-scoring topics in the CAT exam, and as CAT 2026 approaches, mastering its core equation patterns can give you a powerful edge. This article will cover the four most important types of algebraic equations that consistently appear in CAT exams, along with clear explanations, exam-focused solving methods, common mistakes to avoid, and smart shortcuts to improve speed and accuracy. By understanding these evergreen equation formats, you can strengthen your problem-solving speed, improve accuracy, and boost your chances of securing a top percentile in CAT 2026. It will also include yearwise and slotwise distribution of algebra questions, practice-level examples and CAT-oriented strategies to help you confidently tackle Algebra questions in CAT 2026.
This Story also Contains
Algebra in CAT Quantitative Aptitude
Overview of Algebra in the CAT Exam
Type 1 – Linear Equations in CAT Algebra
Type 2 – Quadratic Equations in CAT Algebra
Type 3 – Inequalities and Modulus Equations
Type 4 – Functions & Polynomials in CAT Algebra
Advanced CAT Algebra Problem Types
Algebra Important Topics for CAT 2026 & Beyond
CAT 2026 Algebra Preparation Tips
CAT 2026 Preparation Resources by Careers360
Common Mistakes in CAT Algebra Preparation
How to Practice Repeated Algebra Questions in CAT for High Percentile
CAT 2026 Algebra Decoded: 4 Equation Types That Appear Year After Year
Algebra in CAT Quantitative Aptitude
Algebra is one of the most important and high-scoring topics in CAT Quantitative Aptitude, playing a crucial role in boosting your overall CAT percentile. It covers a wide range of concepts such as equations, functions, inequalities, modulus, logarithms, and algebraic expressions, which form the backbone of advanced problem-solving. A strong command over CAT Algebra not only improves your accuracy in direct algebra questions but also strengthens your ability to solve complex problems from other sections of Quant with confidence.
Mastering Algebra for CAT 2026 means building a solid mathematical foundation that helps you tackle both easy and tricky questions efficiently, making it a must-focus area in your preparation strategy.
Why Algebra Dominates CAT Quant Every Year?
Algebra consistently holds the highest weightage in CAT Quantitative Aptitude, making it the most dominant topic in the exam. Its importance lies in its predictable presence and its ability to test deep conceptual understanding.
Here’s why Algebra remains a CAT favorite:
Algebra questions appear every year in significant numbers, making it a high-return topic
It forms the base for solving problems in Arithmetic, Logarithms, Surds and Indices, and Modern Mathematics
It tests conceptual clarity rather than lengthy calculations
Algebra-based questions are ideal for creating moderate to high difficulty levels in CAT
Strong Algebra skills improve overall problem-solving speed and accuracy
Amity University-Noida MBA Admissions 2026
Ranked among top 10 B-Schools in India by multiple publications | Top Recruiters-Google, MicKinsey, Amazon, BCG & many more.
Because of its consistency and versatility, Algebra acts as a powerful scoring weapon in CAT 2026. Focusing on Algebra preparation can dramatically increase your chances of achieving a top percentile in the Quantitative Aptitude section.
Must-know Algebra concepts for CAT candidates
We have analysed the last 5 years of CAT question papers to give you some insight on the concepts that have been asked previously, as listed below:
Topic
Important Concept to understand
Complex numbers
Iota, conjugate of complex number
Linear Equations
Forming and solving linear equations, nature of solutions, graphs
Algebra in the CAT exam forms a key part of the Quantitative Ability section, testing concepts like equations, inequalities, sequences, and functions. Mastery of these topics helps in solving problems quickly and accurately, boosting overall CAT 2026 performance.
Weightage of Algebra in the CAT Quant section
Here is a year-wise and slot-wise distribution of the number of questions asked from Algebra in CAT in the previous years, so that candidates can check the weightage and plan their preparation accordingly:
Year
2020
2021
2022
2023
2024
2025
Slot/
Topic
Slot 1
Slot 2
Slot 3
Slot 1
Slot 2
Slot 3
Slot 1
Slot 2
Slot 3
Slot 1
Slot 2
Slot 3
Slot 1
Slot 2
Slot 3
Slot 1
Slot 2
Slot 3
Equations
4
4
2
1
3
1
1
3
3
5
2
2
2
4
4
3
3
3
Functions
1
1
1
1
1
1
2
1
1
0
0
1
1
1
1
1
1
1
Inequalities & Modulus
1
0
0
1
0
1
0
0
0
0
1
0
1
1
0
1
1
1
Sequence & Series
0
3
1
1
2
2
1
1
0
2
2
3
1
1
1
1
1
1
Logs / Surds / Indices
2
1
3
1
1
1
0
1
0
1
1
0
1
2
1
1
1
0
Others
0
0
0
0
0
0
1
0
1
0
1
1
0
0
0
0
0
0
Total Algebra Qs
8
9
7
5
7
6
5
6
5
8
7
7
7
9
7
7
7
6
CAT 2025 College Predictor
Use CAT 2025 College Predictor to check your chances for IIM and top MBA calls based on CAT percentile, profile, work experience and cut-off trends.
As we have seen that every year, on average, 3 to 4 questions are asked from equations. Equations can be linear, quadratic, or miscellaneous. Mastering these equations can be the game-changer in CAT-2026 as it comprises of healthy weightage in the quantitative aptitude section.
Apart from this, solving equations mainly focuses on finding the value or the range of values of the unknown variable. These questions have a lower possibility of being wrong. Also, practising these questions will help in developing critical and analytical thinking.
Type 1 – Linear Equations in CAT Algebra
Linear equations can be in one dimension, two dimensions or three dimensions. Linear equations are a fundamental topic in CAT Algebra, involving equations of the first degree with one or more variables. These questions test your ability to solve for unknowns efficiently and form the basis for more complex problem-solving.
where a and b are non-zero coefficients and c is a constant.
Solving Linear Equations:
If there are two sets of linear equations, such as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, then these represent a pair of linear equations in two variables.
The solutions of these equations can be determined by different methods:
1. Graphical Method
2. Elimination Method
3. Substitution method.
Tricks to handle multiple variables quickly
To quickly solve questions having two or more variables, you should apply the method of
1. Elimination
2. Substitution, to reduce the larger equations in to simpler ones or you should apply trial and error methods to solve through options.
3. For two linear equations, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$
Condition
Nature of Solution
Graphical Interpretation
$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$
Unique solution
The two lines intersect at exactly one point. The point of intersection is the solution of the equations.
Example 1: For some real numbers $a$ and $b$, the system of equations $x +y = 4$ and $(a+5)x + (b^2 -15)y = 8b$ has infinitely many solutions for $x$ and $y$. Then, the maximum possible value of $ab$ is: [CAT 2023, slot 3]
Solution:
For the given equations: $x +y = 4$ and $(a+5)x + (b^2 -15)y = 8b$ Condition for infinite many solutions $\frac {1}{a+5} = \frac {1}{ b^2 -15} =\frac {4}{8b}$ Solving $\frac {1}{ b^2 -15} =\frac {4}{8b}$ $⇒ 8b=4(b^2-15)$ On solving this quadratic equation, we get $b = -3, 5$. Solving $\frac {1}{ a+5} =\frac {4}{8b}$ ⇒ 8b = 4(a+5) For b = -3, a = -11 and hence, $ab = (-3) \times (-11) = 33$ For $b = 5, a = 5$ and hence, $ab = (5) \times (5) = 25$ So, the maximum value of $ab$ is 33.
Example 2: The number of distinct integer solutions $(x, y)$ of the equation $|x+y|+|x-y|=2$, is [CAT 2024, slot 3]
Solution:
We’ll analyze all integer pairs $(x, y)$ such that the equation holds.
Let’s denote: $A = |x + y|$; $B = |x - y|$
We are told: $A + B = 2$ Since both $A$ and $B$ are non-negative and integers, possible values of $(A, B)$ are: $(0, 2)$, $(1, 1)$, and $(2, 0)$
Let’s find all integer solutions $(x, y)$ for each case.
Case 1: $|x + y| = 0$ and $|x - y| = 2$
Then $x + y = 0$ and $x - y = \pm 2$
Solving:
$x + y = 0$, $x - y = 2$ Solving: $x = 1$, $y = -1$
$x + y = 0$, $x - y = -2$ Solving: $x = -1$, $y = 1$
So, 2 solutions.
Case 2: $|x + y| = 1$ and $|x - y| = 1$
Then $x + y = \pm 1$, $x - y = \pm 1$
Possible combinations:
$x + y = 1$, $x - y = 1$ Solving: $x = 1$, $y = 0$
$x + y = 1$, $x - y = -1$ Solving: $x = 0$, $y = 1$
$x + y = -1$, $x - y = 1$ Solving: $x = 0$, $y = -1$
$x + y = -1$, $x - y = -1$ Solving: $x = -1$, $y = 0$
So, 4 solutions.
Case 3: $|x + y| = 2$, $|x - y| = 0$
Then $x - y = 0$, $x + y = \pm 2$
Solving:
$x - y = 0$, $x + y = 2$ $x = y$, so $2x = 2 \Rightarrow x = y = 1$
$x - y = 0$, $x + y = -2$ $x = y$, so $2x = -2 \Rightarrow x = y = -1$
Quadratic equations in CAT Algebra involve second-degree equations with one or more variables. Questions on this topic assess your skills in factorization, applying the quadratic formula, and analyzing roots for problem-solving.
An equation of the form $ax^2+bx+c=0$ where a, b, and c are all real and a is not equal to 0, is a quadratic equation.
Key quadratic identities for CAT
There are a few quadratic identities which are frequently used: 1. Difference of Squares: a² - b² = (a + b)(a - b).
2. Square of a Binomial: (a + b)² = a² + 2ab + b².
3. Square of a Binomial: (a - b)² = a² - 2ab + b².
Approaches to solving quadratic equations
There are a few approaches to solving the quadratic equation:
1. Quadratic Formula (Shridharacharya Formula)
For an equation of the form
$ax^2+bx+c=0$
$x = \frac{-b \pm \sqrt{D}}{2a}$ where $D= b^2-4ac$
Example:Find the roots of $5x^2+8x+3=0$
Solution:
Here
$a=5, b= 8, c =3$
Discriminant, $D=8^2-4 \times 5 \times 3=4$
So, $x= \frac{-8 \pm \sqrt{4}}{10}$ $x=\frac{-3}{5}, -1$
2. Method of splitting middle term:
Example:Find the roots of $5x^2+8x+3=0$
Solution:
$5x^2+8x+3=0$
⇒ $5x^2+5x+3x+3=0$
⇒ $5x(x+1)+3(x+1)=0$
⇒ $(x+1)(5x+3)=0$
⇒ $x = -1, \frac{-3}{5}$
Common traps and shortcuts in CAT Quant
While solving a quadratic equation, a student can make mistakes in
misinterpreting the questions
making calculation errors
While finding discriminants, students often neglect the negative value of the discriminant, which leads to a wrong answer. You must consider each possibility.
Some important points that may help you to solve questions based on quadratic equations effectively:
(i) If D is a perfect square, then the roots are rational and in case it is not a perfect square then the roots are irrational.
(ii) In the case of imaginary roots (D < 0) and if p + iq is one root of the quadratic equation, then the other must be the conjugate p - iq and vice versa (where p and q are real and $i = \sqrt{-1}$)
(iii) For an equation of the form
$ax^2+bx+c=0$
If a > 0, D < 0, roots are imaginary.
If a > 0, D = 0, roots are real and identical.
If a > 0, D > 0, roots are real and distinct.
If a < 0, D > 0, roots are imaginary.
If a < 0, D > 0, roots are real and distinct.
If a < 0, D = 0, roots are real and equal.
(iv) For an equation of the form
$ax^2+bx+c=0$
Sum of roots = $\frac{-b}{a}$
Product of roots = $\frac{c}{a}$
Sample questions and solutions
Q.1) If r is a constant such that |x² - 4x - 13| = r has exactly three distinct real roots, then the value of r is:
A) 17
B) 21
C) 15
D) 15
Solution:-
Alternatively,$|x^2−4x−13|=r$. This can be represented in two parts: $x^2-4x-13=r$ if r is positive. $x^2-4x-13=-r$ if r is negative. Considering the first case: $x^2−4x−13=r$ The quadratic equation becomes: $x^2−4x−13−r=0$ The discriminant for this function is : $b^2−4ac=16−[4×(−13−r)]=68+4r$ Since r is positive, the discriminant is always greater than 0 this must have two distinct roots. For the second case: $x^2−4x−13+r=0$ the function inside the modulus is negative. The discriminant is $16−(4×(r−13))=68−4r$ In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots. So, $68−4r=0$ ⇒ $r=17$ For $r=17$, we can have exactly 3 roots.
Hence, the correct answer is option (1).
Q.2) Suppose one of the roots of the equation $ax^2−bx+c=0$ is $2+3i$, Where a,b and c are rational numbers and a≠0. If $b=c^3$ then |a| equals
A) 1
B) $\frac{2}{13\sqrt{13}}$
C) 3
D) $\frac{2}{13}$
Solution:-
Given one root $x=2+3i$ and a,b,c∈Q,a≠0
Since coefficients are rational, the other root is $2−3i$
Sum of roots: $(2+3i)+(2−3i)=4=\frac{b}{a}$
⇒ $b=4a$
Product of roots: $(2+3i)(2−3i)=4+9=13$
⇒ $\frac{c}{a} =13$ ⇒ $c =13a$
Given $b=c^3$ ⇒ $4a=2197a^3$ ⇒ $a=\frac{2}{13\sqrt{13}}$
Hence, the correct answer is option 2.
Q.3) If the equations $x^2+mx+9=0,x^2+nx+17=0$ and $x^2+ (m+n)x+35=0$ have a common negative root, then the value of $(2m+3n)$ is
Solution:-
Let the common negative root be x
Since x is a root of all three equations:
From first equation: $x^2+mx+9=0$ From second equation: $x^2+nx+17=0$ From third equation: $x^2+(m+n)x+35=0$
Subtract the second from the first: $(x^2+mx+9)−(x^2+nx+17)=0$ ⇒ $(m−n)x−8=0$ So, $(m−n)x=8$ ...(i)
Now subtract the third from the first: $(x^2+mx+9)−(x^2+(m+n)x+35)=0$ ⇒$−nx−26=0$ So, $−nx=26⇒nx=−26$ ...(ii)
Add (i) and (ii) We get, $(m−n)x+nx=8−26⇒mx=−18$ ...(iii)
Now from (ii): $x=−26/n$ Substitute into (iii): $m⋅(−26/n)=−18$ ⇒$−26m=−18n$ ⇒$13m=9n$
So, $m/n=9/13$
Let $m=9k, n=13k$
Then $2m+3n=2(9k)+3(13k)=18k+39k=57k$
We now use the value of x from earlier: From $nx=−26$, ⇒ $13k⋅x=−26⇒x=−2k$
Now substitute $x=−2k$ into the first equation: $x^2+mx+9=0$ ⇒ $(−2k)^2+9k⋅(−2k)+9=0$ ⇒ $4k^2−18+9=0$ ⇒ $4k^2=9$ ⇒ $k^2=4/9⇒k=2/3$
So, $2m+3n=57k=57⋅(2/3)=38$
Q.4) If $(x+6\sqrt2)^{1/2}−(x−6\sqrt2)^{1/2}=2\sqrt2$, then x equals
Squaring both sides again ⇒ $(x−4)^2=x^2−72$ ⇒ $x^2+16−8x=x^2−72$ ⇒ $x=11$
Hence, the correct answer is 11.
Type 3 – Inequalities and Modulus Equations
Two real numbers or two algebraic expressions related by the symbol > (“Greater Than”) or < (“Less than”) (and by the signs ≥ or ≤) form an inequality.
The inequality consists of two sides, ie. LHS and RHS. LHS and RHS can be algebraic expressions or they can be numbers. The expressions in LHS and RHS have to be considered on the set where LHS and RHS have sense simultaneously. This set is called the set of permissible values of the inequality.
If two or several inequalities contain the same sign (< or >) then they are called inequalities of the same sense. Otherwise, they are called inequalities of the opposite sense.
Now let us consider some basic definitions about inequalities.
For 2 real numbers a and b
The inequality a > b means that the difference a – b is positive.
The inequality a < b means that the difference a – b is negative.
Modulus equations involve an expression with absolute values (e.g., |x|). |always gives a positive value. You require the solutions where the expression inside the modulus is equal to the positive or negative value of the number on the other side of the equation.
For example, |x| = 2 gives x = 2 or – 2. To solve a basic modulus equation like |ax + b| > c, two equations will be formed ax+b>c and ax+b<−c, then solve each for x to find the solutions.
The inequality a≥b means that a>b or a=b, that is, a is not less than b.
The inequality a≤b means that a<b or a=b, that is, a is not greater than b.
Notation of Ranges:
1. Ranges where the ends are excluded:
If the value of x is denoted as (1, 2) it means 1 < x < 2 i.e. x is greater than 1 but smaller than 2.
Similarly, if we denote the range of values of $x$ as (-9, 1) U (8, 23), this means that the value of $x$ can be denoted as -9 < x < 1 and 8 < x < 23.
2. Ranges where the Ends are Included
[2, 5] means $2 \leq x \leq 5$
3. Mixed ranges
(3, 21] means $3 < x \leq 21$
Solving Algebra inequalities in less time
You can use the following results to solve inequalities in less time:
$a^2 +b^2 \geq 2ab$
$|a+b| \leq |a| + |b|$
$|a-b| \geq |a| - |b|$
Arithmetic mean ≥ Geometric mean.
$\frac{a+b}{2} \geq \sqrt {ab}$
$\frac{a}{b}+\frac{b}{a} \geq 2$ if $a>0$ and $b>0$ or if $a<0$ and $b<0$.
$a^2+b^2+c^2 \geq ab+bc+ca$
$(a+b)(b+c)(c+a) \geq 8abc$ if $a \geq 0, b \geq 0$ and $c \geq 0$, the equation being obtained when $a=b=c$.
If $a+b=2$, then $a^4+b^4 \geq 2$
CAT-level practice questions with solutions
Q.1) The number of integers n that satisfy the inequalities |n – 60| < |n – 100| < |n – 20| is:
A) 21
B) 19
C) 18
D) 18
Solution:-
Given: |n – 60| < |n – 100| < |n – 20|
The distance between these two points can be represented by |x – y| or |y – x|.
Let's take the first part of the inequality |n – 60| < |n – 100| This inequality holds good for the values of n below 80. Hence ‘n’ should be less than 80.
Now Let's check for the later part of the inequality. |n – 100| < |n – 20| This inequality holds good for the values of n above 60. Hence ‘n’ should be greater than 60.
∴ Values of ‘n’ range from 61 to 79. So, the total possible integers that satisfy this inequality are 19.
Hence, the correct answer is 19.
Q.2) All the values of x satisfying the inequality 1x+5≤12x−3 are
A) −5<x<32 or 32<x≤8
B) x<−5 or x>32
C) x<−5 or 32<x≤8
D) x<−5 or 32<x≤8
Solution:-
Given: 1x+5≤12x−3
⇒2x−3≤x+5
⇒x≤8
But x≠−5 and x≠32 as at these values fractions are not defined.
So, the set of solutions becomes
x<−5 or 32<x≤8
Hence, the correct answer is option 3.
Q.3) If x and y satisfy the equations |x|+x+y=15 and x+|y|−y=20, then (x−y) equals
A) 15
B) 10
C) 20
D) 20
Solution:-
We are given the equations:
|x|+x+y=15−−−−−−−(1) x+|y|−y=20−−−−−−−−−−(2)
There are 4 cases:
Case 1: Both x and y are positive Then the equations become x+x+y=15⇒2x+y=15 and x+y−y=20⇒x=20 So, y=15−40=−25 which is contradictory. So, this is not possible.
Case 2: x>0 and y<0 are positive Then the equations become x+x+y=15⇒2x+y=15 and x−y−y=20⇒x−2y=20 So, 4x+2y+x−2y=30+20⇒x=10 and y=−5 So, x−y=10−(−5)=15
Case 3: Both x and y are negative Then the equations become −x+x+y=15⇒y=15 which is contradictory. So, this is not possible.
Case 4: x<0 and y>0 Then the equations become −x+x+y=15⇒y=15 and x+y−y=20⇒x=20 which is contradictory. So, this is not possible.
From Case 2: x−y=15
Hence, the correct answer is option 1.
Q.4) The number of distinct real values of x, satisfying the equation max{x,2}−min{x,2}=|x+2|−|x−2|, is
Solution:-
We are given the equation:max{x,2}−min{x,2}=|x+2|−|x−2|
Left-hand side: max{x,2}−min{x,2}=|x−2| (since difference of max and min of two numbers is just their absolute difference)
So, the equation becomes:
|x−2|=|x+2|−|x−2|
Bring all terms to one side:
2|x−2|=|x+2|
Case 1: x≥2
Then |x−2|=x−2, |x+2|=x+2
The equation becomes:
2(x−2)=x+2⇒2x−4=x+2⇒x=6
This is valid for x≥2
Case 2: −2≤x<2
Then |x−2|=2−x, |x+2|=x+2
⇒2(2−x)=x+2⇒4−2x=x+2⇒3x=2⇒x=23
This is valid in [−2,2)
Case 3: x<−2
Then |x−2|=2−x, |x+2|=−x−2 ⇒2(2−x)=−x−2⇒4−2x=−x−2⇒−x=−6⇒x=6
But x=6 does not lie in x<−2, so discard this solution.
Thus, valid real solutions are:
x=6 and x=23
There are 2 distinct real values.
Hence, the correct answer is 2.
Type 4 – Functions & Polynomials in CAT Algebra
Functions and polynomials in CAT Algebra focus on understanding expressions, relationships, and transformations.
Polynomials:
A polynomial is an expression in x or other variable which may contain one or more terms. Also, the power of the variable should be a whole number.
The degree of a polynomial: It is the highest power of the variable in that polynomial.
It is important to note that the degree of a polynomial can never be negative and it should be a non-negative integer.
For instance, in the polynomial $5x^3−2x^2+3x−7$, the degree is 3, because the highest power of x is 3.
Introduction to Functions:
A function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. The set of inputs is called the domain and the set of possible outputs is called the codomain.
Notation:
A function f from set A to set B is denoted by f: A ➔ B.
Understanding functions and graph-based questions
Graph of some important functions:
Function
Description
Graph
Constant Function
f(x)= k
Domain: R
Range: k
Modulus Function
$f(x) = |x|$
Where:
f(x) = x, if x > 0 and -x if x < 0
Domain: R
Range: Non-negative real numbers
Signum Function
sgn(x) = 1, if x > 0, sgn(x) = 0, if x = 0, and sgn(x) = -1, x < 0
Domain: R
Range: {- 1, 0, 1}
Identity Function
f(x)= x
Domain: R
Range: R
Greatest Integer Function or step function
f(x) = [x]
Domain: R
Range: Integer
Reciprocal Function
f(x) = 1/x
Domain: R – {0}
Range: R – {0}
Function of a Function (Composite Function)
A composite function is the function of another function. If f is a function from A in to B and g is a function from B in to C, then their composite function denoted by (gof) is a function from A in to C defined by
gof (x) = g[f(x)]
Also, composite function fog (read as "f of g") is defined as:
fog (x) = f[g(x)]
Q.1) A function f maps the set of natural numbers to whole numbers, such that $f(xy)=f(x)f(y)+f(x)+f(y)$ for all x,y and $f(p)=1$ for every prime number p. Then, the value of $f(160000)$ is
A) 8191
B) 2047
C) 4095
D) 4096
Solution:-
Given: $f(xy)=f(x)f(y)+f(x)+f(y)$ for all x,y∈N And $f(p)=1$ for every prime p
Since $f(p) =1$, So $f(2)=1$ $f(4)=f(2×2)=f(2)f(2)+f(2)+f(2)=1+1+1=3$ $f(8)=f(2×4)=f(2)f(4)+f(2)+f(4)=3+1+3=7$ $f(16)=f(2×8)=f(2)f(8)+f(2)+f(8)=7+1+7=15$ Similarly, $f(2^8)=f(256)=255$
Q.2) For any non-zero real number $x$, let $f(x)+2f(1/x)=3x$. Then, the sum of all possible values of $x$ for which $f(x)=3$, is
A) -2
B) 2
C) -3
D) 3
Solution:-
We are given: $f(x)+2f(1/x)=3x$−−−−−−−−(1) Also, we are told: $f(x)=3$
Substituting into (1): $3+2f(1/x)=3x⇒f(1/x)=(3x−3)/2$−−−−−−−−(2)
Now use (1) again but switch $x$ with $1/x$: $f(1/x)+2f(x)=3x$ Substitute $f(x)=3$: $f(1/x)+6=3x⇒f(1/x)=3x−6$−−−−−−(3)
Equating (2) and (3): $(3x−3)/2=3x−6$
Multiply both sides by 2: $3x−3=6x−12$
Bring all terms to one side: $3x+9−6x=0$
Multiply through by x: $3x^2+9x−6=0$
Divide by 3: $x^2+3x−2=0$
Sum of all possible values of $x=-3$.
Hence, the correct answer is option 3.
Q.3) Let a, b, and c be non-zero real numbers such that $ b^2 <4ac$, and $f(x)=ax^2+bx+c$. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be:
A) the set of all positive integers
B) the set of all integers
C) either the empty set or the set of all integers
D) either the empty set or the set of all integers
Solution:-
It is given that $f(x)=ax^2+bx+c$ and $b^2<4ac$ This means that f(x) has imaginary roots and therefore, no real roots at all. If a function has no real roots, the graph of the function can never touch the x-axis, because touching the x-axis means, for some real value of x, the value of f(x) is 0. This means that the function has roots in some real value of x. When we graph this quadratic function f(x), we get a parabola that should never touch the x-axis. Such a parabola should be completely above the x-axis or completely below the x-axis.
If it is completely above the x-axis: ● It means that the value of f(x) is always positive for any value of x. The set of values of x that satisfies the condition that f(x) < 0 is an empty set. ● If it is completely below the x-axis: It means that the value of f(x) is always negative for any value of x. The set of values of x that satisfies the condition that f(x) < 0 is the set of all real numbers. Since Set S contains all the integers ‘m’ that satisfy the above conditions, So, set S is either an empty set or the set of all integers. Hence, the correct answer is option (3).
Q.4) Let $0≤a≤x≤100$ and $f(x)=|x−a|+|x−100|+|x−a−50|$. Then the maximum value of f(x) becomes 100 when a is equal to:
A) 25
B) 100
C) 50
D) -50
Solution:-
$x≥a$, so, $|x−a|=x−ax<100$, so $|x−100|=100−xf(x)=(x−a)+(100−x)+|x−a−50|=100⇒|x−a−50|=a$
From the graph, we can see that when x = a Then, $|x - a - 50| = a$ $⇒ a = 50$ Similarly when $x = a + 100$ $|x – a - 50|= a$ $⇒ a = 50$ So, the value of a is 50 when f(x) is 100.
Hence, the correct answer is option (3).
High-frequency CAT Algebra problems on polynomials
Questions on Polynomials that are asked in CAT mainly based on following concepts:
Sum of zeros and product of zeros
1. Sum of Zeroes (α and β): For a quadratic polynomial $ ax^2+bx+c$, the sum of its zeroes (roots) is given by $\frac {-b}{a}$.
For a cubic polynomial (zeros α, β and γ) $ ax^3+bx^2+cx+d$, the sum of its zeroes (roots) is given by $\frac {-b}{a}$.
2. Product of Zeroes:
For a quadratic polynomial $ ax^2+bx+c$, the product of its zeroes (roots) is given by $\frac{c}{a}$.
For a cubic polynomial $ ax^3+bx^2+cx+d$, the product of its zeroes (roots) is given by $\frac {-d}{a}$.
Also, for cubic polynomial (zeros α, β and γ)
αβ +βγ + γα = c/a
Questions based on factor theorem:
Factorization (Factor theorem)
If a polynomial P(x) has a factor (x -a), then P(a) = 0 and conversely, if P(a) = 0 then (x -a) is a factor of P(x). This is the factor theorem.
Q.1) $f(x)=(x^2+2x−15)(x^2−7x−18)$ is negative if and only if
A) $−5<x<−2$ or $3<x<9$
B) $−2<x<3$ or $x>9$
C) $x<−5$ or $−2<x<3$
D) $x<−5$ or $−2<x<3$
Solution:-
$f(x)=(x^2+2x−15)(x^2−7x−18)$ $⇒f(x)=(x+5)(x−3)(x−9)(x+2)<0$ We have four inflection points −5,−2,3, and 9 . For $x<−5$, all four terms (x+5),(x−3),(x−9),(x+2) will be negative.
The overall expression will be positive. Similarly, when $x>9$, all four terms will be positive. When x belongs to $(−2,3)$, two terms are negative and two are positive.
The overall expression is positive again. We are left with the range $(−5,−2)$ and $(3,9)$ where the expression will be negative.
Thus, the correct answer is option 1) $−5<x<−2$ or $3<x<9$.
Q.2) The roots $\alpha, \beta$ of the equation $3 x^2+\lambda x-1=0$, satisfy $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=15$. The value of $\left(\alpha^3+\beta^3\right)^2$, is
Q.2) When Rajesh's age was same as the present age of Garima, the ratio of their ages was 3:2. When Garima's age becomes the same as the present age of Rajesh, the ratio of the ages of Rajesh and Garima will become
A) 5:4
B) 4:3
C) 2:1
D) 3:2
Solution:-
Let present ages of Rajesh and Garima be $R$ and $G$ respectively. When Rajesh's age was $G$, the ratio of their ages was $3:2$. So, at that time: $R - G$ years ago, Rajesh's age = $G$, Garima's age = $G - (R - G) = 2G - R$
Now, when Garima's age becomes $R$, the time passed = $R - G = \frac{4G}{3} - G = \frac{G}{3}$ At that time, Rajesh's age = $R + \frac{G}{3} = \frac{4G}{3} + \frac{G}{3} = \frac{5G}{3}$
Required ratio = $\frac{5G/3}{R} = \frac{5G/3}{4G/3} = \frac{5}{4}$
Hence, the correct answer is option 1.
Q.3) A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
Solution:-
Let the initial number of fruits be $x$ and apples be $y$.
The stock of mangoes is given to be 40% of x, which shows that the number of mangoes will be $\frac{2x}{5}$.
The total number of sold fruits is given by including the selling of all the fruits- mango, apple and banana.
Now, the total no. of fruits sold
$=\frac{2x}{10}+\frac{4y}{10}+96=\frac{x}{2}$
$⇒\frac{x}{5}+\frac{2y}{5}+96=\frac{x}{2}$
On simplifying, we get,
$2x+4y+960=5x$
$⇒x=\frac{960+4y}{3}$
For $x$ to be a positive integer, let us check for values of $y$.
$4y+960$ has to be divisible by $3$, which means $4y$ will also be divisible by $3$.
For $\frac{4y}{10}$ to be an integer, $y$ has to be divisible by $5$.
We can say, that the smallest value of $y$ has to be multiple of both $3$ and $5$, i.e. $15$.
Now, put the value of $y=15$.
We get, $x=\frac{960+4 \times 15}{3}$
$⇒x=\frac{1020}{3}$
$⇒x=340$
So, the smallest possible number of fruits in stock at the starting of the day will be $340$.
Hence, the correct answer is $340$.
Q.4) For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is:
A) 34
B) 35
C) 36
D) 38
Solution:-
It is given, y(x + z) = 19 y cannot be 19. If y = 19, then x + z = 1 which is not possible when both x and z are natural numbers. Therefore, y = 1 and x + z = 19 It is given, z(x + y) = 51 z can take values 3 and 17. Case 1: If z = 3, y = 1 and x = 16 xyz = 3 × 1 × 16 = 48 Case 2: If z = 17, y = 1 and x = 2 xyz = 17 × 1 × 2 = 34 So, the minimum value xyz can take is 34. Hence, the correct answer is option (1).
Q.5) The number of integer solutions of the equation $\left(x^2-10\right)^{(x^2-3 x-10)}=1$ is:
A) 4
B) 6
C) 3
D) 8
Solution:-
Case 1: When $x^2-3 x-10=0$ and $x^2-10 \neq 0$
$x^2-3 x-10=0$ $ ⇒(x-5)(x+2)=0$ $\therefore x=5 \text { or }-2 $ Case 2: $x^2-10=1$ $⇒ x^2-11=0 $ $\therefore$ No integer solutions
Case 3: $\mathrm{x}^2-10=-1$ and $\mathrm{x}^2-3 \mathrm{x}-10$ is even. $x^2-9=0$ $⇒(x+3)(x-3)=0$ $\therefore\mathrm{x}=-3$ or 3 for $\mathrm{x}=-3$ and $+3$, $\mathrm{x}^2-3 \mathrm{x}-10$ is even.
$\therefore$ In total 4 values of $x$ satisfy the equations.
Hence, the correct answer is option (1).
Algebra Important Topics for CAT 2026 & Beyond
Algebra consists of many topics, as discussed earlier in this article. Based on CAT's previous papers analysis, we have divided these topics into two categories:
High-priority concepts
Lesser Priority concepts
High-priority concepts based on previous CAT 2026 papers
Topic
Important Concepts
Function
Questions based on fog and gof, value of a function, graphs
Polynomials
Relation between zeros, factor theorem
Linear Equations
Word problems, Nature of solutions, solving linear pair of equations, equations involving modulus, equations involving sequence and series, equations involving logarithm
Quadratic Equations
Nature of roots, relation between roots, equations involving sequences and series, equations involving logarithms
Linear Inequalities
Inequalities involving modulus
Lesser-seen but tricky Algebra concept
There are a few topics which are less seen but very tricky and important for the CAT exam
Greatest integer function
Questions based on the signum function
Questions involving higher-order polynomials
Harmonic Progressions
Questions involving even and odd functions
CAT 2026 Algebra Preparation Tips
In this section, we will focus on a few but important tips to prepare for CAT Algebra.
How to build strong basics in Algebra equations
To build a strong foundation, first learn the basic concepts from the NCERT classes 11 and 12. Start with very basic questions and increase the difficulty level gradually.
After learning the basics, apply the concepts in CAT-level questions.
Shortcuts and time-saving tricks for CAT Quant
After building a strong foundation on the topics, the next step is to learn the shortcuts and time-saving techniques. For this, learn some effective calculation methods with the help of Vedic Maths.
Formula sheet for quick revision
Prepare a CAT formula sheet for linear equation, quadratic equation, polynomials, graphs, etc for quick revision.
Best resources & practice material for CAT Algebra
1. Arun Sharma: A Quantitative Approach for CAT (7th Edition) 2. Quantitative Aptitude for CAT by Nishit K Sinha 3. NCERT class 11 and class 12
CAT 2026 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 2026 Arithmetic Important Concepts and Practice Questions
Common mistakes in CAT Algebra preparation include ignoring fundamentals, skipping practice on tricky problems, mismanaging time, and overlooking shortcuts, which can lower accuracy and speed in the exam.
Common mistakes
How to Rectify?
Ignoring the Basics while preparing
Learn fundamentals from NCERT
Solving the modulus inappropriately
Practice more questions and do not avoid any possible case.
Over-dependencies on formulas
Try to solve conceptually
Ignoring domain and restrictions
Solve the questions within specifies domain
Values within square roots should be non-negative
The square root of negative values is not defined. So, you should better take care of it
Poor time management
Prepare a proper timetable
Do not checking the previous year’s pattern
You should check at least the previous 5 years' papers to know the pattern of the question
How to Practice Repeated Algebra Questions in CAT for High Percentile
Algebra is a game-changer in CAT Quantitative Aptitude and one of the most reliable topics for scoring marks. Since CAT follows a pattern-based approach, many algebra question types are repeated every year with slight variations. Practising these recurring Algebra questions strategically can significantly improve your accuracy, speed, and confidence in CAT 2026.
Follow this proven four-step strategy to master repeated Algebra questions in CAT.
Step 1: Analyze CAT Previous Year Papers to Identify Algebra Patterns
Start by analyzing the last 4 to 5 years of CAT question papers.
Identify frequently asked Algebra topics such as linear equations, quadratic equations, inequalities, functions, and modulus
Categorize questions concept-wise
Observe repetition in question structures and difficulty levels
Create a concept-wise list of high-frequency Algebra question types
This step helps you understand what CAT actually tests in Algebra.
Step 2: Strengthen Algebra Basics and Learn Smart Shortcuts
Strong fundamentals are the backbone of CAT Algebra preparation.
Revise all basic formulas and identities
Memorize important concepts and standard results
Learn CAT-specific tricks and time-saving shortcuts
Focus on clarity rather than rote learning
This builds speed and reduces calculation errors in the actual exam.
Step 3: Practice with CAT-Specific Algebra Resources
Use only CAT-oriented study material for best results.
Solve previous year CAT Algebra questions
Attempt sectional tests focused on Algebra
Use high-quality mock tests and topic-wise practice sets
Practice under time constraints to simulate real exam pressure
This ensures your preparation stays exam-focused and result-driven.
Step 4: Evaluate Performance and Improve Weak Areas
Evaluation is the most important step for improvement.
Analyze every wrong answer
Identify conceptual gaps and calculation mistakes
Revise weak Algebra topics regularly
Track accuracy and time spent per question
Frequently Asked Questions (FAQs)
Q: With which books should I start preparing for Algebra?
A:
If you are a beginner,
Start with NCERT classes 11 and 12. After learning the basics, follow Arun Sharma for Quantitative Aptitude for CAT.
Q: How much time should I give to prepare for Algebra?
A:
Algebra is very vast having numerous concepts. It will require a lot of time to practice depending upon your capabilities and previous conceptual knowledge. What I suggest is, prepare a proper time table including all subjects and topics. If you do well in only one section, it will not help you to get into IIMs. So, manage your time and prepare well for all sections.
Q: What kind of questions are asked from Algebra in CAT?
A:
In CAT, questions are generally framed by involving two or more concepts in a single question like:
Sequence and series with log
Inequality with modulus
Equations involving log, surds, and indices etc
Q: Which Algebra topics are most frequently repeated in CAT?
A:
The most repeated Algebra topics in CAT include linear equations, quadratic equations, inequalities, modulus functions, and functions and graphs. These areas form the core of CAT Algebra preparation.
Q: Can I score high in CAT Quant if my Arithmetic is weak but Algebra is strong?
A:
Yes, a strong command over Algebra can compensate for weaker areas in Arithmetic. Many high-percentile scorers rely on Algebra and DILR to balance their overall Quant performance.
Hi, while some of the Agri-business management courses need prior education or experience in the agricultural education background, but for others it is not always mandatory to have an agricultural education background. However with 62 percentile in CAT is it difficult to get into the premium B-Schools in India. For
FMS Udaipur admission for 2026 is not open. The MBA 2026 admission will likely commence in May 2026. You can keep checking the official website of FMS.
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CUSAT CAT cutoff 2025
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Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space
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Question:
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