Mathematics Bridge Course
Master the foundational concepts of Real Numbers, Polynomials, and Linear Equations to kickstart your Class 10 journey with confidence!
M.RajaRao
MSc. MEd
30-Day Course Schedule
| Day | Date (2026) | Topic to Cover |
|---|---|---|
| 1 | 13.03 - Fri | Syllabus analysis |
| 2 | 16.03 - Mon | Introduction - Real numbers |
| 3 | 17.03 - Tue | Fundamental theorem of arithmetic with examples |
| 4 | 18.03 - Wed | Example questions about Fundamental theorem |
| 5-7 | 21.03 - 24.03 | Exercise 1.1 problems |
| 8 | 25.03 - Wed | Revisiting irrational numbers, Proof of $\sqrt{2}$ |
| 9-10 | 26.03 - 28.03 | Proofs of $\sqrt{3}$, $\sqrt{5}$, $5-\sqrt{3}$, $3\sqrt{2}$ |
| 11-13 | 30.03 - 01.04 | Exercise 1.2 problems |
| 14 | 02.04 - Thu | Slip test on Real Numbers |
| 15 | 04.04 - Sat | Introduction of Polynomials |
| 16-17 | 06.04 - 07.04 | Geometrical meaning of zeroes, Exercise 2.1 |
| 18-20 | 08.04 - 10.04 | Relationship between zeroes & coefficients, Ex 2.2 |
| 21 | 11.04 - Sat | Revision of Polynomials |
| 22 | 13.04 - Mon | Slip test on Polynomials |
| 23 | 15.04 - Wed | Pair of Linear Eq. - Intro & Graphical Method |
| 24-26 | 16.04 - 18.04 | Algebraic methods (Substitution & Elimination), Ex 3.1-3.3 |
| 28 | 21.04 - Tue | GRAND TEST |
| 29-30 | 23.04 - 24.04 | Review on test results & Guidance |
Real Numbers
📚 Number Systems
- Natural Numbers (N): Counting numbers $\{1, 2, 3, ...\}$
- Whole Numbers (W): Natural numbers + zero $\{0, 1, 2, 3, ...\}$
- Integers (Z): Positive & negative numbers including zero $\{..., -2, -1, 0, 1, 2, ...\}$
- Rational Numbers (Q): Form $\dfrac{p}{q}$, where $q \ne 0$. Ex: $\dfrac{2}{7}, -2$
- Irrational Numbers (Q'): Cannot be expressed as $\dfrac{p}{q}$. Ex: $\sqrt{2}, \sqrt{5}, \pi$
⭐ Core Concepts
Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
$HCF(a,b) \times LCM(a,b) = a \times b$
Example Proof: Prove that $\sqrt{2}$ is irrational.
Let us assume $\sqrt{2}$ is a rational number.
Let $\sqrt{2} = \dfrac{a}{b}$ where $a, b$ are co-primes.
$\Rightarrow \sqrt{2}b = a$
$\Rightarrow 2b^2 = a^2$ ...... (1)
Since 2 divides $a^2$, so 2 also divides $a$.
Let $a = 2k$ (where $k \in Z$)
From (1), $2b^2 = (2k)^2 \Rightarrow 2b^2 = 4k^2 \Rightarrow b^2 = 2k^2$
Since 2 divides $b^2$, so 2 also divides $b$.
Practice Problems & Answer Keys
Example 1 & 5: Check $4^n$ / $6^n$ ending with zero ▼
Q: Check whether $4^n$ or $6^n$ can end with the digit zero for any natural number $n$.
Sol ($4^n$): $4^n = (2 \times 2)^n$. If a number has both 2 and 5 in its prime factorization, it ends with zero. 4 has only 2 in its prime factorization. $\therefore 4^n$ cannot end with the digit 0.
Sol ($6^n$): $6^n = (2 \times 3)^n$. 6 has only 2 and 3 in its prime factorization but no 5. $\therefore 6^n$ cannot end with the digit 0.
Example 2 & 3: LCM and HCF using Prime Factorization ▼
Ex 2: Find the LCM and HCF of 6 and 20.
$6 = 2^1 \times 3^1$
$20 = 2^2 \times 5$
$HCF(6, 20) = 2^1 = 2$
$LCM(6, 20) = 2^2 \times 3 \times 5 = 60$
Ex 3: Find HCF of 96 and 404, hence find LCM.
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
$HCF(96, 404) = 2^2 = 4$
$LCM = \dfrac{96 \times 404}{4} = 9696$
Exercise 1.1 (Q1 to Q4): Prime Factors, HCF, & LCM ▼
Q1: Express as product of prime factors:
(i) $140 = 2^2 \times 5 \times 7$
(ii) $156 = 2^2 \times 3 \times 13$
(iii) $3825 = 3^2 \times 5^2 \times 17$
(iv) $5005 = 5 \times 7 \times 11 \times 13$
(v) $7429 = 17 \times 19 \times 23$
Q2: LCM & HCF of 26 and 91:
$26 = 2 \times 13, \quad 91 = 7 \times 13$
$LCM = 182, \quad HCF = 13$
Verification: $LCM \times HCF = 182 \times 13 = 2366$. Product = $26 \times 91 = 2366$. Verified.
Q3: LCM & HCF of groups:
(i) 12, 15, 21 $\rightarrow HCF = 3, LCM = 420$
(ii) 17, 23, 29 $\rightarrow HCF = 1, LCM = 11339$
(iii) 8, 9, 25 $\rightarrow HCF = 1, LCM = 1800$
Q4: Given HCF(306, 657) = 9, Find LCM:
$LCM = \dfrac{306 \times 657}{9} = 22338$
Exercise 1.1 (Q6 & Q7): Composite Numbers & Circular Path ▼
Q6: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite.
$7 \times 11 \times 13 + 13 = 13 \times (77 + 1) = 13 \times 78 = 13 \times 2 \times 3 \times 13$
$7! + 5 = 5 \times (1008 + 1) = 5 \times 1009$
Both are products of primes, hence they are composite numbers.
Q7: Sonia takes 18 mins, Ravi takes 12 mins. When will they meet?
The time is the LCM of 18 and 12.
$18 = 2 \times 3^2, \quad 12 = 2^2 \times 3$
$LCM = 2^2 \times 3^2 = 36$ minutes.
Proofs: Irrationality of $\sqrt{3}, \sqrt{5}, 5-\sqrt{3}, 3\sqrt{2}$ ▼
Prove $\sqrt{3}$ & $\sqrt{5}$ are irrational:
Let $\sqrt{p} = \dfrac{a}{b}$ (a, b co-primes). $\Rightarrow pb^2 = a^2$. Thus $p$ divides $a$. Let $a = pk \Rightarrow pb^2 = p^2k^2 \Rightarrow b^2 = pk^2$. Thus $p$ divides $b$. Contradiction: they share factor $p$. Thus irrational.
Prove $5-\sqrt{3}$ is irrational:
Let $5-\sqrt{3} = \dfrac{a}{b} \Rightarrow \sqrt{3} = \dfrac{5b-a}{b}$. RHS is rational, implying $\sqrt{3}$ is rational. Contradicts the fact that $\sqrt{3}$ is irrational.
Prove $3\sqrt{2}$ is irrational:
Let $3\sqrt{2} = \dfrac{a}{b} \Rightarrow \sqrt{2} = \dfrac{a}{3b}$. RHS is rational, implying $\sqrt{2}$ is rational. Contradicts the fact that $\sqrt{2}$ is irrational.
Prove $3+2\sqrt{5}$ is irrational:
Let $3+2\sqrt{5} = \dfrac{a}{b} \Rightarrow \sqrt{5} = \dfrac{a-3b}{2b}$. RHS is rational. Contradicts fact that $\sqrt{5}$ is irrational.
Polynomials
An algebraic expression in which the exponent of the variable is a whole number. Ex: $2+3x^2$, $5m+8$.
Types according to Degree
| Degree | Polynomial Name | General Form |
|---|---|---|
| Not defined | Zero polynomial | 0 |
| 0 | Constant polynomial | a |
| 1 | Linear | ax + b |
| 2 | Quadratic | ax² + bx + c |
| 3 | Cubic | ax³ + bx² + cx + d |
Quadratic Zeroes ($\alpha, \beta$)
Cubic Zeroes ($\alpha, \beta, \gamma$)
Graphical Meaning of Zeroes
The zeroes of a polynomial $p(x)$ are exactly the x-coordinates of the points where the graph of $y = p(x)$ intersects the x-axis. A quadratic graph makes a shape called a Parabola (Upward if $a > 0$, Downward if $a < 0$).
Practice Problems & Answer Keys
Example 1 & Observations: Finding Zeroes from Graphs ▼
Example 1 Graphs (i to vi): Number of zeroes is the number of times the graph intersects the x-axis.
- (i) Does not intersect x-axis $\rightarrow$ 0 zeroes
- (ii) Intersects at one point $\rightarrow$ 1 zero
- (iii) Intersects at three points $\rightarrow$ 3 zeroes
- (iv) Intersects at two points $\rightarrow$ 2 zeroes
- (v) Intersects at four points $\rightarrow$ 4 zeroes
- (vi) Intersects at three points $\rightarrow$ 3 zeroes
Observation 1: Graph of $y = x^2 - 3x - 4$
Shape: Parabola. Intersects at $(-1, 0)$ and $(4, 0)$. Zeroes are $-1$ and $4$. Sum = 3, Product = -4.
Observation 2: Graph of bent wire $y = x^2 - 8x + 12$
Shape: Parabola. Intersects at $(2, 0)$ and $(6, 0)$. Zeroes are $2$ and $6$. Sum = 8, Product = 12.
Example 2 & Ex 2.2 (Q1): Find Zeroes & Verify Relationship ▼
Example 2: $p(x) = x^2 + 7x + 10$
$x^2 + 5x + 2x + 10 = 0 \Rightarrow x(x+5) + 2(x+5) = 0 \Rightarrow (x+5)(x+2) = 0$
Zeroes: $\alpha = -5, \beta = -2$
Sum: $-5 + (-2) = -7$. Formula: $\dfrac{-b}{a} = \dfrac{-7}{1} = -7$ (Verified)
Product: $(-5)(-2) = 10$. Formula: $\dfrac{c}{a} = \dfrac{10}{1} = 10$ (Verified)
Ex 2.2 Q1: $p(x) = x^2 - 2x - 8$
$x^2 - 4x + 2x - 8 = 0 \Rightarrow x(x-4) + 2(x-4) = 0 \Rightarrow (x-4)(x+2) = 0$
Zeroes: $\alpha = 4, \beta = -2$
Sum: $4 + (-2) = 2$. Formula: $\dfrac{-(-2)}{1} = 2$ (Verified)
Product: $4 \times -2 = -8$. Formula: $\dfrac{-8}{1} = -8$ (Verified)
Ex 2.2 (Q2): Form Quadratic Polynomials ▼
General Formula: $k[x^2 - (\alpha+\beta)x + \alpha\beta]$
(i) Sum = $\dfrac{1}{4}$, Product = -1
$k\left[x^2 - \left(\dfrac{1}{4}\right)x - 1\right]$. If $k=4 \Rightarrow 4x^2 - x - 4$
(vi) Sum = 4, Product = 1
$k[x^2 - 4x + 1]$. If $k=1 \Rightarrow x^2 - 4x + 1$
Linear Equations in Two Variables
General Form: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$
| Ratio Comparison | Graphical Rep. | Algebraic Solution | Consistency |
|---|---|---|---|
| $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$ | Intersecting Lines | Exactly ONE unique solution | Consistent |
| $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$ | Parallel Lines | NO solution | Inconsistent |
| $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ | Coincident Lines | Infinitely MANY solutions | Dependent / Consistent |
Substitution Method
Express one variable in terms of the other from one equation, and substitute this expression into the second equation to solve.
Elimination Method
Multiply the equations by suitable non-zero constants to make the coefficients of one variable equal, then add or subtract to eliminate it.
Practice Problems & Answer Keys
Example 1 & 2: Graphical Method (Consistency) ▼
Ex 1: Check $x + 3y = 6$ and $2x - 3y = 12$
$\dfrac{a_1}{a_2} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{3}{-3} = -1$. Since $\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$, they are consistent (Intersecting lines).
Solution from graph intersection: $(6, 0)$.
Ex 2: Check $x + y = 5$ and $2x + 2y = 10$
$\dfrac{a_1}{a_2} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{1}{2}$, $\dfrac{c_1}{c_2} = \dfrac{-5}{-10} = \dfrac{1}{2}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, they are consistent and dependent (Coincident lines).
Solution: Infinitely many solutions like $(5,0), (0,5)$ etc.
Example 3: Word Problem (Graphical) ▼
Q: 10 students took part. Number of girls is 4 more than boys. Find numbers.
Let boys = $x$, girls = $y$.
Total: $x + y = 10$
Difference: $y = x + 4 \Rightarrow x - y = -4$
Graphing both lines yields an intersection at $(3, 7)$.
Solution: Boys = 3, Girls = 7.
Exercise 3.2: Substitution Method ▼
Q4: Solve $2x - y = 5$ and $3x + 2y = 11$
From (1), $y = 2x - 5$. Substitute into (2):
$3x + 2(2x - 5) = 11 \Rightarrow 3x + 4x - 10 = 11 \Rightarrow 7x = 21 \Rightarrow x = 3$
Substitute $x = 3$ into $y = 2x - 5 \Rightarrow y = 2(3) - 5 = 1$
Solution: $x = 3, y = 1$
Q5: Solve $x + y = 14$ and $x - y = 4$
From (2), $y = x - 4$. Substitute into (1):
$x + (x - 4) = 14 \Rightarrow 2x = 18 \Rightarrow x = 9$
Substitute $x = 9$ into $y = x - 4 \Rightarrow y = 9 - 4 = 5$
Solution: $x = 9, y = 5$
Exercise 3.3: Elimination Method ▼
Q6: Solve $8x + 5y = 9$ and $3x + 2y = 4$
Eq(1) $\times 2 \Rightarrow 16x + 10y = 18$
Eq(2) $\times 5 \Rightarrow 15x + 10y = 20$
Subtracting equations: $(16x - 15x) = 18 - 20 \Rightarrow x = -2$
Substitute $x = -2$ in (1): $8(-2) + 5y = 9 \Rightarrow -16 + 5y = 9 \Rightarrow 5y = 25 \Rightarrow y = 5$
Solution: $x = -2, y = 5$
Q7: Solve $3x + 2y = 11$ and $2x + 3y = 4$
Eq(1) $\times 3 \Rightarrow 9x + 6y = 33$
Eq(2) $\times 2 \Rightarrow 4x + 6y = 8$
Subtracting equations: $(9x - 4x) = 33 - 8 \Rightarrow 5x = 25 \Rightarrow x = 5$
Substitute $x = 5$ in (1): $3(5) + 2y = 11 \Rightarrow 15 + 2y = 11 \Rightarrow 2y = -4 \Rightarrow y = -2$
Solution: $x = 5, y = -2$