CBSE Class 10 Maths Mid-Term 2026: Important Questions, Formulas & Must-Practise Topics

You may have completed the chapters for your CBSE Class 10 Maths Mid-Term Exam, but the real challenge begins when you sit down to revise. Should you solve another set of questions, revise formulas, focus on numericals, or go back to topics where you are losing marks? With multiple concepts and question types to prepare, knowing what to practise and what not to miss becomes important, especially when you have limited time before the exam.

To address this revision challenge, PW has introduced its Class 10 Mid-Term Marathon, covering important concepts, formulas, and question types in a dedicated revision series. The marathon also includes 3-marker, 5-marker, and 7-marker questions with solutions, giving you practice with different question formats and helping you understand how concepts are applied while solving problems.

Important Formulas for Class 10 Maths Mid-Term

Before attempting the practice questions, revise these important formulas from the chapters covered in the marathon. Keeping these formulas fresh will help you solve numerical, proof-based, and application-based questions more confidently. 

Real Numbers

• HCF and LCM: HCF × LCM = Product of the two numbers.

• Prime Factorisation: Use the smallest powers of common prime factors for HCF and the greatest powers of all prime factors for LCM.

Polynomials

• Sum of Zeroes: For ax² + bx + c, sum of zeroes = −b/a.

• Product of Zeroes: Product of zeroes = c/a.

• Polynomial from Zeroes: A quadratic polynomial with zeroes α and β can be written as k[x² − (α + β)x + αβ].

Pair of Linear Equations

• Unique Solution: a₁/a₂ ≠ b₁/b₂.

• Infinitely Many Solutions: a₁/a₂ = b₁/b₂ = c₁/c₂.

• No Solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Quadratic Equations

• Discriminant: D = b² − 4ac.

• Nature of Roots: D > 0 gives real and distinct roots, D = 0 gives real and equal roots, and D < 0 gives no real roots.

• Quadratic Formula: x = [−b ± √D]/2a.

Trigonometry

• Identity 1: sin²θ + cos²θ = 1.

• Identity 2: 1 + tan²θ = sec²θ.

• Identity 3: 1 + cot²θ = cosec²θ.

Arithmetic Progressions

• nth Term: aₙ = a + (n − 1)d.

• Sum of n Terms: Sₙ = n/2[2a + (n − 1)d].

• Sum Using Last Term: Sₙ = n/2(a + l).

• nth Term from the End: l − (n − 1)d.

Triangles

• Basic Proportionality Theorem: A line parallel to one side of a triangle divides the other two sides in the same ratio.

• Similarity Criteria: Triangles can be proved similar using AA, AAA, SSS, or SAS criteria.

Coordinate Geometry

• Distance Formula: √[(x₂ − x₁)² + (y₂ − y₁)²].

• Section Formula: For the ratio m:n, x = (mx₂ + nx₁)/(m + n) and y = (my₂ + ny₁)/(m + n).

• Midpoint Formula: ((x₁ + x₂)/2, (y₁ + y₂)/2).

ICSE Class 10 Maths: 3, 5 & 7-Mark Important Questions

 Most Important 3-Marker Questions

1. Using the prime factorisation method, find the HCF and LCM of 26, 65 and 117.

Solution:

Prime factorisation:

26 = 2 × 13
65 = 5 × 13
117 = 3² × 13

The common prime factor is 13.

Therefore,

HCF = 13

For LCM, take the greatest power of every prime factor:

LCM = 2 × 3² × 5 × 13

LCM = 1170

Answer: HCF = 13, LCM = 1170

2. Find the zeroes of the quadratic polynomial 21y² − 11y − 2 and verify the relationship between the zeroes and the coefficients.

Solution:

21y² − 11y − 2
= 21y² − 14y + 3y − 2
= 7y(3y − 2) + 1(3y − 2)
= (3y − 2)(7y + 1)

Therefore,

3y − 2 = 0 or 7y + 1 = 0

So, the zeroes are:

y = 2/3 and y = −1/7

Verification:

Sum of zeroes:

2/3 − 1/7
= (14 − 3)/21
= 11/21

For 21y² − 11y − 2:

−b/a = −(−11)/21 = 11/21

Therefore, sum of zeroes = −b/a.

Product of zeroes:

(2/3)(−1/7) = −2/21

c/a = −2/21

Therefore, product of zeroes = c/a.

Hence, the relationship is verified.

3. Without solving the equation, find the nature of the roots of 2x² − 4x + 3 = 0.

Solution:

Here,

a = 2, b = −4 and c = 3

Discriminant:

D = b² − 4ac
= (−4)² − 4(2)(3)
= 16 − 24
= −8

Since D < 0, the equation has no real roots.

Answer: The equation has no real roots.

4. Is −100 a term of the AP −7, −12, −17, …? Justify your answer.

Solution:

Here,

a = −7

d = −12 − (−7) = −5

Let −100 be the nth term.

Using:

aₙ = a + (n − 1)d

−100 = −7 + (n − 1)(−5)

−100 = −7 − 5n + 5

−100 = −5n − 2

−98 = −5n

n = 98/5

n = 19.6

Since n is not a natural number, −100 is not a term of the given AP.

5. Prove that 6ⁿ can never end with the digit 0 for any natural number n.

Solution:

We know:

6ⁿ = (2 × 3)ⁿ

Therefore,

6ⁿ = 2ⁿ × 3ⁿ

For a number to end in 0, it must be divisible by 10.

Since:

10 = 2 × 5

the prime factorisation of the number must contain both 2 and 5 as factors.

However, 6ⁿ contains only the prime factors 2 and 3. It does not contain 5.

Therefore, 6ⁿ can never end with the digit 0 for any natural number n.

 Most Important 5-Marker Questions

1. Two cars start simultaneously from two places 100 km apart. If they travel towards each other, they meet in 1 hour. If they travel in the same direction, they meet in 5 hours. Find the speed of each car.

Solution:

Let the speeds of the two cars be x km/h and y km/h, where x > y.

When they travel towards each other:

x + y = 100/1

x + y = 100 …(i)

When they travel in the same direction:

x − y = 100/5

x − y = 20 …(ii)

Adding (i) and (ii):

2x = 120

x = 60

Substituting x = 60 in (i):

60 + y = 100

y = 40

Answer: The speeds of the two cars are 60 km/h and 40 km/h.

2. A boat covers 30 km upstream and 42 km downstream, each in 3 hours. Find the speed of the boat in still water and the speed of the stream.

Solution:

Let the speed of the boat in still water be x km/h.

Let the speed of the stream be y km/h.

Upstream speed = x − y

Downstream speed = x + y

For the upstream journey:

x − y = 30/3

x − y = 10 …(i)

For the downstream journey:

x + y = 42/3

x + y = 14 …(ii)

Adding (i) and (ii):

2x = 24

x = 12

Substituting in (i):

12 − y = 10

y = 2

Answer:

• Speed of boat in still water = 12 km/h

• Speed of stream = 2 km/h

3. The longest side of a right triangle is 4 m more than its shortest side, and the third side is 2 m less than the longest side. Find the length of each side.

Solution:

Let the shortest side be x m.

Longest side = x + 4 m

Third side = x + 4 − 2 = x + 2 m

Since the triangle is right-angled, by Pythagoras' theorem:

(x + 4)² = x² + (x + 2)²

x² + 8x + 16 = x² + x² + 4x + 4

x² − 4x − 12 = 0

Factorising:

(x − 6)(x + 2) = 0

Therefore:

x = 6 or x = −2

Since length cannot be negative:

x = 6

Therefore:

Shortest side = 6 m

Third side = 8 m

Longest side = 10 m

Answer: The three sides are 6 m, 8 m and 10 m.

4. In a test of 120 MCQs, 1 mark is awarded for a correct answer and ¼ mark is deducted for a wrong answer. Sarita knew the answers to some questions and guessed the rest. All her guesses were wrong. If she scored 95 marks, how many questions did she answer correctly?

Solution:

Let the number of questions answered correctly be x.

Then the number of wrong answers is:

120 − x

Marks obtained for correct answers = x

Marks deducted for wrong answers:

(120 − x)/4

According to the question:

x − (120 − x)/4 = 95

Multiplying by 4:

4x − 120 + x = 380

5x = 500

x = 100

Answer: Sarita answered 100 questions correctly.

5. State and prove the Basic Proportionality Theorem (Thales' Theorem).

Solution:

Statement: If a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those two sides in the same ratio.

Given: In ΔABC, DE ∥ BC, where D lies on AB and E lies on AC.

To prove:

AD/DB = AE/EC

Construction: Join BE and CD. Draw EN perpendicular to AB and DM perpendicular to AC.

Proof:

Area of ΔADE:

ar(ΔADE) = 1/2 × AD × EN

Area of ΔBDE:

ar(ΔBDE) = 1/2 × DB × EN

Therefore,

ar(ΔADE)/ar(ΔBDE) = AD/DB …(i)

Similarly,

ar(ΔADE) = 1/2 × AE × DM

and

ar(ΔCDE) = 1/2 × EC × DM

Therefore,

ar(ΔADE)/ar(ΔCDE) = AE/EC …(ii)

Now, ΔBDE and ΔCDE lie on the same base DE and between the same parallels DE and BC.

Therefore,

ar(ΔBDE) = ar(ΔCDE) …(iii)

Using (i), (ii) and (iii):

AD/DB = AE/EC

Hence proved.

Most Important 7-Marker Questions

1. If 2p = q + r, prove that the roots of the equation (p − q)x² + (q − r)x + (r − p) = 0 are equal.

Solution:

For the equation:

(p − q)x² + (q − r)x + (r − p) = 0

we have:

a = p − q

b = q − r

c = r − p

The discriminant is:

D = (q − r)² − 4(p − q)(r − p)

Expanding:

D = q² − 2qr + r² − 4pr + 4p² + 4qr − 4pq

D = q² + 2qr + r² + 4p² − 4p(q + r)

D = (q + r)² + 4p² − 4p(q + r)

Given:

2p = q + r

Therefore:

D = (2p)² + 4p² − 4p(2p)

D = 4p² + 4p² − 8p²

D = 0

Since the discriminant is zero, the equation has real and equal roots.

Hence proved.

2. A flight of 2800 km had its average speed reduced by 100 km/h due to bad weather. As a result, its duration increased by 30 minutes. Find the original duration of the flight.

Solution:

Let the original speed of the flight be x km/h.

Original time:

2800/x hours

Reduced speed:

(x − 100) km/h

New time:

2800/(x − 100) hours

The increase in time is 30 minutes = 1/2 hour.

Therefore:

2800/(x − 100) − 2800/x = 1/2

Taking the LCM:

2800[x − (x − 100)]/[x(x − 100)] = 1/2

280000/[x(x − 100)] = 1/2

Therefore:

x(x − 100) = 560000

x² − 100x − 560000 = 0

Using the quadratic formula:

x = [100 ± √(10000 + 2240000)]/2

x = [100 ± 1500]/2

Therefore:

x = 800 or x = −700

Speed cannot be negative, so:

x = 800 km/h

Original time:

2800/800 = 3.5 hours

Answer: The original duration of the flight was 3.5 hours, or 3 hours 30 minutes.

3. Three parallel lines are intersected by two transversals. They cut the first transversal at A, B, C and the second transversal at D, E, F, respectively. Using the Basic Proportionality Theorem, prove that AB/BC = DE/EF.

Solution:

Let the two transversals meet at O.

Since the three lines are parallel, the corresponding segments cut on the two transversals are proportional.

Consider the appropriate triangles formed by the transversals and the parallel lines.

Applying the Basic Proportionality Theorem to the first triangle:

AB/BC = AO/OF …(i)

Applying the Basic Proportionality Theorem to the second triangle:

DE/EF = AO/OF …(ii)

From (i) and (ii):

AB/BC = DE/EF

Therefore,

AB/BC = DE/EF

Hence proved.

4. A father's age is three times the sum of the ages of his two children. After 5 years, his age will be twice the sum of the ages of his two children. Find the present age of the father.

Solution:

Let the father's present age be x years.

Let the sum of the present ages of the two children be y years.

According to the question:

x = 3y …(i)

After 5 years, the father's age will be:

x + 5

Since there are two children, the sum of their ages will increase by 10 years.

Therefore, their combined age after 5 years will be:

y + 10

According to the question:

x + 5 = 2(y + 10)

x + 5 = 2y + 20

x = 2y + 15 …(ii)

From (i):

3y = 2y + 15

y = 15

Therefore:

x = 3 × 15

x = 45

Answer: The father's present age is 45 years.

5. A city's taxi fare consists of a fixed charge and a charge per kilometre travelled. A journey of 12 km costs Rs. 89, while a journey of 20 km costs Rs. 145. Find the fixed charge and the charge per kilometre.

Solution:

Let the fixed charge be Rs. x.

Let the charge per kilometre be Rs. y.

For a 12 km journey:

x + 12y = 89 …(i)

For a 20 km journey:

x + 20y = 145 …(ii)

Subtracting (i) from (ii):

8y = 56

y = 7

Substituting y = 7 in (i):

x + 12(7) = 89

x + 84 = 89

x = 5

Therefore:

Fixed charge = Rs. 5

Charge per kilometre = Rs. 7

Note: For more important questions and a deeper understanding of the concepts covered here, watch the PW Class 10 Mid-Term Marathon video and revise along with the explanations.

How Does the PW Marathon Bring Your Mid-Term Revision Together?

The PW Mid-Term Marathon lets you revise important Maths concepts, formulas, and question types in the same session. Instead of revisiting every chapter from the beginning, you can use the marathon to:

• Revisit Important Concepts and Formulas: Revise important concepts and formulas from the chapters covered in the marathon and refresh topics you may have forgotten.

• See Where Each Formula Is Used: Understand how formulas are applied in different questions and identify the right formula or concept needed for a problem.

• Practise Different Types of Maths Questions: Work through formula-based, numerical, proof-based, and application-based questions from the chapters covered in the marathon.

• Work on Proofs and Application-Based Questions: Practise important proofs such as BPT and trigonometric identities, along with word problems from Quadratic Equations, AP, and Coordinate Geometry.

• Find the Topics That Need Another Revision: Identify formulas, concepts, calculations, or proofs where you make mistakes and revisit those areas before your mid-term exam.

The PW Mid-Term Marathon gives you a way to revise Class 10 Maths without separating formula revision from question practice. Use the questions above to check which topics you can solve confidently and return to the marathon for concepts or questions that need another round of revision before your mid-term.