MATHEMATICS

 Q1. The next term of the A.P. √6, √24, √54 is:

We can write:

√6, √24, √54 = √(3×2), √(3×8), √(3×18) = √3×√2, √3×√8, √3×√18

This becomes √3 × (√2, √8, √18)

Let’s check differences:

√8 − √2 ≠ √18 − √8, so not in A.P.

But squaring all:

6, 24, 54 → Differences: 18 each → So next = 54 + 18 = 72,

So next term = √72 = Option (b)

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 Q2. If x + 1, 3x and 4x + 2 are in A.P., find x.

Since in A.P.:

2nd – 1st = 3rd – 2nd

⇒ 3x − (x + 1) = (4x + 2) − 3x

⇒ 2x − 1 = x + 2

⇒ x = 3

Answer: x = 3

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 Q3. First term = 5, Common difference = 9

10th term = a + (10−1)d = 5 + 9×9 = 5 + 81 = 86

Answer: (a)

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 Q4. Which term of A.P. 1/2, 1, 3/2, ... is 49?

This is an A.P. with a = 1/2, d = 1/2

Use formula: an = a + (n−1)d

49 = 1/2 + (n−1)(1/2)

⇒ 49 − 1/2 = (n−1)/2

⇒ (97/2) = (n−1)/2

⇒ n−1 = 97 → n = 98

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 Q5. Find a and b so that 7, a, b, 23 are in A.P.

Given 4 terms in A.P., so common difference (d) is same.

Let a = 7 + d, b = 7 + 2d, 23 = 7 + 3d

⇒ 23 = 7 + 3d ⇒ d = (23 − 7)/3 = 16/3

So,

a = 7 + d = 7 + 16/3 = 37/3

b = 7 + 2d = 7 + 32/3 = 53/3

Answer: a = 37/3, b = 53/3

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 Q6. Show that a², ab, b² are in A.P.

Check if 2nd term = average of 1st and 3rd:

2ab = a² + b²?

No. So instead:

Check if 2(ab) = a² + b²

⇒ 2ab = a² + b²

Only true when a = b

Hence, not always in A.P. unless a = b

Alternatively, you may show:

ab − a² = b² − ab

⇒ ab − a² = ab − b²

⇒ a² = b² → a = b or a = −b

So they are in A.P. if a = b or a = −b

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 Q7. How many two-digit numbers are divisible by 3?

Two-digit numbers range: 10 to 99

First 2-digit divisible by 3 = 12

Last = 99

Use formula: an = a + (n−1)d

99 = 12 + (n−1)×3

87 = (n−1)×3

n−1 = 29 ⇒ n = 30

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 Q8. Which term of 3, 15, 27, ... is 120 more than its 21st term?

a = 3, d = 12

21st term = a + 20d = 3 + 240 = 243

So required term = 243 + 120 = 363

Let an = 363 = a + (n−1)d = 3 + (n−1)×12

363 − 3 = 12(n−1) ⇒ 360 = 12(n−1) ⇒ n−1 = 30

⇒ n = 31

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 Q9. In an A.P., a₇ = 4, d = 2. Find a₁.

a₇ = a + 6d ⇒ 4 = a + 12 ⇒ a = 4 − 12 = −8

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 Q10. Common difference of A.P. where a₁ = −4, a₇ = 2

a₇ = a + 6d ⇒ 2 = −4 + 6d ⇒ 6d = 6 ⇒ d = 1

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📘 Chapter: Real Numbers – Class 10 CBSE

🔹 Key Concepts:

1. Real Numbers:

Includes rational numbers and irrational numbers.

Every point on the number line is a real number.

2. Euclid's Division Lemma:

For any two positive integers a and b, there exist unique integers q and r such that:

👉 a = bq + r, where 0 ≤ r < b

Used to find the HCF (Highest Common Factor) of two numbers.

3. Fundamental Theorem of Arithmetic:

Every composite number can be written uniquely as the product of prime numbers, apart from the order.

4. Irrational Numbers:

Numbers that cannot be expressed as a fraction.

Examples: √2, √3, π

5. Rationalization:

Process of removing irrational numbers from the denominator.

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🧠 Important Theorems (to be remembered):

1. √2, √3, √5, etc. are irrational.

2. If p is a prime number and p divides a², then p divides a.

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✍️ Important Questions with Solutions:

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Q1. Use Euclid's Division Lemma to find HCF of 96 and 36.

Solution: Using Euclid’s algorithm:

96 = 36 × 2 + 24

36 = 24 × 1 + 12

24 = 12 × 2 + 0

👉 HCF = 12

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Q2. Find the LCM and HCF of 6 and 20 using the prime factorization method.

Solution:

6 = 2 × 3

20 = 2 × 2 × 5

👉 HCF = 2

👉 LCM = 2 × 2 × 3 × 5 = 60

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Q3. Prove that √5 is irrational.

Solution: Assume √5 is rational:

So, √5 = p/q (in lowest form, q ≠ 0)

⇒ 5 = p²/q²

⇒ p² = 5q²

So, 5 divides p² ⇒ 5 divides p ⇒ let p = 5k

Then p² = 25k²

So, 5q² = 25k² ⇒ q² = 5k² ⇒ 5 divides q

So, both p and q are divisible by 5 ⇒ contradiction.

So, √5 is irrational.

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Q4. Find the HCF of 184 and 230 using Euclid’s Algorithm.

Solution:

230 = 184 × 1 + 46

184 = 46 × 4 + 0

👉 HCF = 46

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Q5. Find the decimal expansion of 13/125 and state whether it is terminating or non-terminating.

Solution:

13/125 = 0.104

Since the denominator has only powers of 2 and 5, the decimal is terminating.

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Q6. Express 1.23̅ (i.e., 1.23333...) in the form p/q.

Solution:

Let x = 1.23333...

Multiply by 10:

10x = 12.333...

Now subtract:

10x - x = 12.333... - 1.233...

⇒ 9x = 11.1

⇒ x = 111/90

⇒ x = 37/30

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✅ Tips to Remember:

HCF × LCM = Product of two numbers

Terminating decimal ⇨ denominator in form 2ᵐ × 5ⁿ

Non-terminating repeating ⇨ rational

Non-terminating non-repeating ⇨ irrational

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Chapter: Polynomials – Class 10 CBSE

🔹 Key Concepts:

1. Polynomial:

An algebraic expression made up of variables and coefficients using only addition, subtraction, multiplication, and non-negative integer exponents.

👉 General form:

P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where aₙ ≠ 0

Degree of the polynomial = highest power of the variable

2. Types of Polynomials:

Type Degree Example

Constant 0 3, -5, 7

Linear 1 x + 2

Quadratic 2 x² - 5x + 6

Cubic 3 x³ + x² - 1

3. Zeroes of a Polynomial:

The value of x for which P(x) = 0.

A polynomial of degree n has at most n zeroes.

4. Relationship Between Zeroes and Coefficients:

For a quadratic polynomial:

P(x) = ax² + bx + c

Let α and β be the zeroes.

Then:

✅ Sum of zeroes (α + β) = –b/a

✅ Product of zeroes (α × β) = c/a

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🔹 Factor Theorem:

If P(a) = 0, then (x – a) is a factor of P(x).

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🔹 Division Algorithm:

If P(x) and G(x) are polynomials such that G(x) ≠ 0,

then there exist unique polynomials Q(x) and R(x) such that:

👉 P(x) = G(x) × Q(x) + R(x)

Where R(x) = 0 or degree of R(x) < degree of G(x)

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🧠 Important Questions with Solutions:

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Q1. Find the zeroes of the quadratic polynomial: x² – 5x + 6

Solution:

Factorizing:

x² – 5x + 6 = (x – 2)(x – 3)

👉 Zeroes: x = 2, x = 3 ✅

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Q2. Find a quadratic polynomial whose zeroes are 4 and –3.

Solution:

Let the zeroes be α = 4, β = –3

We use:

P(x) = x² – (α + β)x + (α × β)

= x² – (4 – 3)x + (4 × –3)

= x² – x – 12 ✅

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Q3. If one zero of the polynomial x² + 4x + k is 2, find k.

Solution:

Let one zero = 2

Use quadratic identity:

Sum of zeroes = –b/a = –4/1 = –4

So, other zero = –4 – 2 = –6

Product = 2 × (–6) = –12

Now, product of zeroes = c/a = k/1

So, k = –12 ✅

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Q4. Divide: 2x³ + 3x² – 2x – 5 by x – 1

Solution:

Using long division:

Quotient = 2x² + 5x + 3

Remainder = –2

✅ So, 2x³ + 3x² – 2x – 5 = (x – 1)(2x² + 5x + 3) – 2

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Q5. Verify the relationship between zeroes and coefficients for: x² – 7x + 10

Solution:

Factor: x² – 7x + 10 = (x – 5)(x – 2)

Zeroes: α = 5, β = 2

Sum = 5 + 2 = 7 = –(–7)/1 = 7 ✅

Product = 5 × 2 = 10 = 10/1 ✅

Verified!

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📝 Tips to Remember:

For ax² + bx + c,

Sum = –b/a, Product = c/a

Always factor carefully when finding zeroes

Chapter: Pair of Linear Equations in Two Variables – Class 10 CBSE

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🔹 Key Concepts:

1. Linear Equation in Two Variables:

An equation of the form:

👉 ax + by + c = 0,

where a, b, c are real numbers and a ≠ 0, b ≠ 0.

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🔹 Graphical Representation:

Each linear equation represents a straight line on a graph.

The solution of a pair of linear equations is the point of intersection of the two lines.

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🔹 Types of Solutions:

Nature of Lines Condition Number of Solutions Example

Intersecting lines a₁/a₂ ≠ b₁/b₂ Unique solution x + y = 2, x – y = 0

Parallel lines a₁/a₂ = b₁/b₂ ≠ c₁/c₂ No solution x + y = 1, 2x + 2y = 3

Coincident lines a₁/a₂ = b₁/b₂ = c₁/c₂ Infinite solutions x + y = 2, 2x + 2y = 4

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🔹 Algebraic Methods to Solve Pair of Equations:

1. Substitution Method

Express one variable in terms of the other and substitute into the second equation.

2. Elimination Method

Multiply equations (if needed) to eliminate one variable and solve for the other.

3. Cross-Multiplication Method

For equations:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Then:

\frac{x}{b₁c₂ - b₂c₁} = \frac{y}{c₁a₂ - c₂a₁} = \frac{1}{a₁b₂ - a₂b₁}

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🧠 Important Questions with Solutions:

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Q1. Solve using substitution method:

x + y = 5

x – y = 1

Solution:

From equation 1: x = 5 – y

Substitute in equation 2:

(5 – y) – y = 1

⇒ 5 – 2y = 1

⇒ 2y = 4 ⇒ y = 2

Then x = 5 – 2 = 3 ✅

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Q2. Solve using elimination method:

2x + 3y = 12

4x – 3y = 6

Solution:

Add both equations:

(2x + 3y) + (4x – 3y) = 12 + 6

⇒ 6x = 18 ⇒ x = 3

Now put x = 3 in 1st equation:

2(3) + 3y = 12 ⇒ 6 + 3y = 12 ⇒ y = 2 ✅

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Q3. Solve using cross-multiplication method:

3x – 2y = 5

2x + y = 4

Solution:

Write in standard form:

a₁ = 3, b₁ = –2, c₁ = –5

a₂ = 2, b₂ = 1, c₂ = –4

Use cross-multiplication:

x = \frac{(-2)(-4) - (1)(-5)}{(3)(1) - (2)(-2)} = \frac{8 + 5}{3 + 4} = \frac{13}{7}

y = \frac{(-5)(2) - (-4)(3)}{(3)(1) - (2)(-2)} = \frac{-10 + 12}{7} = \frac{2}{7} 

✅ So, x = 13/7, y = 2/7

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Q4. Check the nature of the following pair:

x + 2y = 3

2x + 4y = 6

Solution:

a₁/a₂ = 1/2

b₁/b₂ = 2/4 = 1/2

c₁/c₂ = 3/6 = 1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂ ⇒ infinitely many solutions ✅

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Q5. A word problem:

The sum of two numbers is 60. One number is twice the other. Find the numbers.

Solution:

Let one number be x, other = 2x

x + 2x = 60 ⇒ 3x = 60 ⇒ x = 20

Other = 2 × 20 = 40 ✅

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✅ Tips to Remember:

Always simplify the equations before solving.

Draw rough graphs if the question asks for graphical representation.

Memorize the cross-multiplication formula properly.

Word problems = convert statements into equations carefully.

Chapter: Quadratic Equations – Class 10 CBSE

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🔹 Key Concepts:

1. Quadratic Equation:

A quadratic equation is any equation of the form:

👉 ax² + bx + c = 0, where a ≠ 0 and a, b, c are real numbers.

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🔹 Roots (or Zeroes) of the Quadratic Equation:

The values of x which satisfy the equation ax² + bx + c = 0 are called its roots or solutions.

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🔹 Methods to Solve Quadratic Equations:

1. Factorization Method

Factor the quadratic expression and equate each factor to zero.

2. Completing the Square Method

Rewrite in square form and solve.

3. Quadratic Formula Method

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

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🔹 Discriminant (D):

Discriminant D = b² – 4ac

It determines the nature of roots:

D Value Nature of Roots

D > 0 Real and distinct

D = 0 Real and equal (repeated roots)

D < 0 Not real (complex roots)

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🔹 Condition for Roots:

If  and  are the roots of ax² + bx + c = 0, then:

✅ Sum of roots = α + β = –b/a

✅ Product of roots = α × β = c/a

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🧠 Important Questions with Solutions:

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Q1. Solve: x² – 5x + 6 = 0 using factorization.

Solution:

x² – 5x + 6 = (x – 2)(x – 3) = 0

⇒ x = 2, x = 3 ✅

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Q2. Solve: 2x² – 7x + 3 = 0 using quadratic formula.

Solution:

Here, a = 2, b = –7, c = 3

x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4}

x = \frac{7 \pm 5}{4} \Rightarrow x = \frac{12}{4} = 3 \quad \text{or} \quad \frac{2}{4} = \frac{1}{2} ✅ 

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Q3. Find the nature of the roots of: 3x² – 4x + 2 = 0

Solution:

a = 3, b = –4, c = 2

D = b² – 4ac = 16 – 24 = –8

Since D < 0 ⇒ Roots are not real (imaginary) ✅

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Q4. Solve: x² – 4x + 4 = 0

Solution:

x² – 4x + 4 = (x – 2)² = 0

⇒ x = 2 (repeated root) ✅

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Q5. A word problem:

The product of two consecutive positive integers is 132. Find the integers.

Solution:

Let the smaller number be x. Then the next = x + 1

So, x(x + 1) = 132

⇒ x² + x – 132 = 0

Now factor:

(x + 12)(x – 11) = 0 ⇒ x = –12 or x = 11

Take positive: x = 11 ⇒ next = 12 ✅

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Q6. Find a quadratic equation whose roots are 3 and –2.

Solution:

Let roots be α = 3 and β = –2

Then required equation =

x^2 - (\alpha + \beta)x + (\alpha \cdot \beta) = x^2 - (3 - 2)x + (-6) = x^2 - x - 6 ✅

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✅ Tips to Remember:

Always write the equation in standard form before solving.

For word problems, define variables carefully.

Memorize the quadratic formula and discriminant rule.

Chapter Notes: Arithmetic Progressions (AP)

🔹 1. Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant.

Example: 2, 4, 6, 8, 10... (Here, common difference d = 2)

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🔹 2. General Form of an AP

An AP can be written as:

a, a + d, a + 2d, a + 3d, ..., a + (n - 1)d

Where:

a = First term

d = Common difference

n = Number of terms

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🔹 3. nth Term of an AP (Tn or an)

Formula:

a_n = a + (n - 1)d

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🔹 4. Sum of First n Terms of an AP (Sn)

Formula:

S_n = \frac{n}{2} [2a + (n - 1)d]

S_n = \frac{n}{2} (a + l)

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🔹 5. Important Concepts

If d = 0, all terms are equal.

If d > 0, AP is increasing.

If d < 0, AP is decreasing.

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📘 Extra Questions with Solutions

✳️ Q1. Find the 10th term of the AP: 3, 7, 11, 15,...

Solution:

a = 3, d = 4

a_{10} = a + (10 - 1)d = 3 + 9 × 4 = 3 + 36 = 39

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✳️ Q2. Which term of the AP: 5, 9, 13, 17,... is 89?

Solution:

a = 5, d = 4, let 89 be the nth term.

a_n = a + (n - 1)d = 5 + (n - 1) × 4 = 89  

\Rightarrow (n - 1) × 4 = 84  

\Rightarrow n - 1 = 21  

\Rightarrow n = 22

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✳️ Q3. Find the sum of the first 20 terms of the AP: 2, 5, 8,...

Solution:

a = 2, d = 3, n = 20

S_n = \frac{n}{2} [2a + (n - 1)d]  

= \frac{20}{2}[2 × 2 + 19 × 3]  

= 10 [4 + 57] = 10 × 61 = 610

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✳️ Q4. The 7th term of an AP is 17 and the 10th term is 26. Find the first term and common difference.

Solution:

Let a = first term, d = common difference

a + 6d = 17 \quad \text{(1)}  

a + 9d = 26 \quad \text{(2)}  

\text{Subtracting (1) from (2):}  

(a + 9d) - (a + 6d) = 26 - 17  

3d = 9 ⇒ d = 3  

From (1): a + 6 × 3 = 17 ⇒ a = 17 - 18 = -1

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✳️ Q5. If the sum of first n terms of an AP is , find its nth term.

Solution:

a_n = S_n - S_{n-1}  

S_n = 5n^2 + 3n  

S_{n-1} = 5(n - 1)^2 + 3(n - 1) = 5(n^2 - 2n + 1) + 3n - 3 = 5n^2 - 10n + 5 + 3n - 3 = 5n^2 - 7n + 2  

a_n = S_n - S_{n-1} = (5n^2 + 3n) - (5n^2 - 7n + 2) = 10n - 2

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✳️ Q6. Is 125 a term of the AP 5, 11, 17, ...?

Solution:

a = 5, d = 6

Let 125 be the nth term:

a_n = a + (n - 1)d = 5 + (n - 1) × 6 = 125  

⇒ (n - 1) × 6 = 120 ⇒ n - 1 = 20 ⇒ n = 21  

125 is the 21st term.  

**Answer: Yes, it is the 21st term**

Coordinate Geometry – Class 10 Notes

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🔹 1. Coordinate Plane (Cartesian Plane)

A plane divided into 4 quadrants by two perpendicular lines:

x-axis (horizontal)

y-axis (vertical)

The point of intersection is the origin (0, 0).

Every point is represented as an ordered pair (x, y).

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🔹 2. Distance Formula

Used to find distance between two points:

A(x₁, y₁), B(x₂, y₂)

\text{Distance AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

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🔹 3. Section Formula

If a point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n, then:

P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)

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🔹 4. Midpoint Formula

Special case of section formula when m = n:

\text{Midpoint M} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

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🔹 5. Area of Triangle Formula

If A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are the vertices of a triangle, then:

\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

If area = 0 ⇒ Points are collinear.

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✅ Extra Questions with Solutions

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✳️ Q1. Find the distance between the points A(3, 4) and B(7, 1).

Solution:

AB = \sqrt{(7 - 3)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

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✳️ Q2. Find the coordinates of the point which divides the line joining A(2, 3) and B(4, 7) in the ratio 1:2.

Solution:

Using section formula:

x = \frac{(1 × 4) + (2 × 2)}{1 + 2} = \frac{4 + 4}{3} = \frac{8}{3}  

y = \frac{(1 × 7) + (2 × 3)}{1 + 2} = \frac{7 + 6}{3} = \frac{13}{3}

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✳️ Q3. Find the midpoint of the line joining P(6, -2) and Q(-4, 8).

Solution:

M = \left( \frac{6 + (-4)}{2}, \frac{-2 + 8}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)

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✳️ Q4. Find the area of triangle with vertices A(2, 3), B(4, 6), C(5, -2).

Solution:

Use area formula:

\text{Area} = \frac{1}{2} | 2(6 + 2) + 4(-2 - 3) + 5(3 - 6) |  

= \frac{1}{2} | 2×8 + 4×(-5) + 5×(-3) |  

= \frac{1}{2} | 16 - 20 - 15 | = \frac{1}{2} × |-19| = \frac{19}{2}

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✳️ Q5. If A(1, 2), B(4, 6), and C(x, 10) are collinear, find the value of x.

Solution:

Area = 0 for collinear points

Use area formula:

\frac{1}{2} |1(6 - 10) + 4(10 - 2) + x(2 - 6)| = 0  

⇒ |1(−4) + 4(8) + x(−4)| = 0  

⇒ |−4 + 32 − 4x| = 0  

⇒ 28 − 4x = 0 ⇒ x = 7

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✳️ Q6. Find the length of diagonal of a rectangle with opposite vertices A(1, 2) and B(6, 5).

Solution:

Diagonal AB:

= \sqrt{(6 − 1)^2 + (5 − 2)^2} = \sqrt{25 + 9} = \sqrt{34}

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✳️ Q7. Find the point on the x-axis which is equidistant from (2, −3) and (−4, 5).

Solution:

Let the point on x-axis be P(x, 0)

Distance from P to A(2, −3):

PA = \sqrt{(x - 2)^2 + 9}  

PB = \sqrt{(x + 4)^2 + 25}  

\text{Equating: } (x - 2)^2 + 9 = (x + 4)^2 + 25  

x^2 - 4x + 4 + 9 = x^2 + 8x + 16 + 25  

x^2 - 4x + 13 = x^2 + 8x + 41  

-4x + 13 = 8x + 41 ⇒ -12x = 28 ⇒ x = -\(\frac{7}{3}\)

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