Assertion-Reason Questions Class 10 Maths: Chapter-Wise with Answers
Chapter-wise assertion-reason questions for CBSE Class 10 Maths with answers and explanations for all 14 chapters, plus a method to solve any A-R question.
Assertion-reason questions give two statements: an Assertion (A) and a Reason (R). Check whether each is true on its own, then check whether R correctly explains A. Choose (a) both true and R explains A, (b) both true but R does not explain A, (c) A true and R false, or (d) A false and R true. Below are chapter-wise practice questions with answers.
Assertion-reason questions look tricky because they test two statements at once. Once you know the method, they become some of the quickest marks in the paper. This page explains the method and gives you chapter-wise assertion-reason questions for all 14 chapters of CBSE Class 10 Maths, each with the answer and a short explanation. All questions are written by the Intellicamp team and follow the current rationalised NCERT syllabus.
How to solve assertion-reason questions
Every question uses the same four options:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Use this three-step method:
- Judge A alone. Is the assertion true? Work it out if needed.
- Judge R alone. Is the reason a true mathematical statement?
- Only if both are true: ask "Does R tell me why A is true?" If yes, (a). If R is true but about something else, (b).
Tip: if A is false, the answer must be (d). If R is false (and A true), it must be (c). Try each question before opening the answer.
Chapter 1: Real Numbers
Q1. A: HCF(12, 18) × LCM(12, 18) = 216.
R: For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.
Show answer
(a). HCF = 6 and LCM = 36, so the product is 216, which equals 12 × 18. R is the rule that explains it.
Q2. A: 6n can never end with the digit 0 for any natural number n.
R: A number ends with 0 only if its prime factorisation contains both 2 and 5.
Show answer
(a). 6n = 2n × 3n, which has no factor 5, so it cannot end in 0. R explains exactly why.
Q3. A: 7 × 11 × 13 + 13 is a composite number.
R: Every composite number can be written as a product of primes.
Show answer
(b). 7 × 11 × 13 + 13 = 13 × (77 + 1) = 13 × 78, so it has factors other than 1 and itself: A is true. R (the Fundamental Theorem of Arithmetic) is true, but it does not show why this particular number is composite; taking out the common factor 13 does.
Q4. A: 5 + √3 is an irrational number.
R: The sum of a rational number and an irrational number is irrational.
Show answer
(a). 5 is rational and √3 is irrational, so their sum is irrational, as R states.
Chapter 2: Polynomials
Q1. A: The sum of the zeroes of 2x² − 8x + 6 is 4.
R: The sum of the zeroes of ax² + bx + c is −b/a.
Show answer
(a). −b/a = −(−8)/2 = 4. (The zeroes are 1 and 3.)
Q2. A: The graph of y = x² − 4 cuts the x-axis at two points.
R: A quadratic polynomial can have at most two zeroes.
Show answer
(b). x² − 4 = 0 gives x = 2 and x = −2, so A is true. R is true, but "at most two" does not explain why there are exactly two here.
Q3. A: If the graph of y = p(x) does not meet the x-axis, p(x) has no real zeroes.
R: The zeroes of p(x) are the x-coordinates of the points where the graph of y = p(x) meets the x-axis.
Show answer
(a). No meeting points means no zeroes, which follows directly from R.
Q4. A: x³ − 2x² + x has exactly three distinct zeroes.
R: A polynomial of degree n has at most n zeroes.
Show answer
(d). x³ − 2x² + x = x(x − 1)², whose zeroes are 0, 1 and 1: only two distinct zeroes. R is true.
Chapter 3: Pair of Linear Equations in Two Variables
Q1. A: The pair x + 2y = 3 and 2x + 4y = 6 has infinitely many solutions.
R: If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines coincide.
Show answer
(a). The ratios are 1/2, 2/4 and 3/6, all equal to 1/2, so the lines coincide.
Q2. A: The pair 3x − y = 2 and 6x − 2y = 7 is inconsistent.
R: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
Show answer
(a). 3/6 = 1/2 and (−1)/(−2) = 1/2, but 2/7 ≠ 1/2. Parallel lines never meet, so there is no solution.
Q3. A: x = 2, y = 1 is a solution of 2x + 3y = 7.
R: Every linear equation in two variables has a unique solution.
Show answer
(c). 2(2) + 3(1) = 7, so A is true. R is false: a single linear equation in two variables has infinitely many solutions.
Chapter 4: Quadratic Equations
Q1. A: x² + 4x + 5 = 0 has no real roots.
R: A quadratic equation has no real roots when b² − 4ac < 0.
Show answer
(a). b² − 4ac = 16 − 20 = −4, which is negative.
Q2. A: The roots of x² − 6x + 9 = 0 are real and equal.
R: If the discriminant of a quadratic equation is zero, its roots are real and equal.
Show answer
(a). b² − 4ac = 36 − 36 = 0. Both roots are 3.
Q3. A: (x + 1)² = x² + 2 is a quadratic equation.
R: A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0.
Show answer
(d). Expanding gives x² + 2x + 1 = x² + 2, so 2x − 1 = 0. The x² terms cancel; it is linear, not quadratic. R is true.
Chapter 5: Arithmetic Progressions
Q1. A: 3, 7, 11, 15, … is an arithmetic progression.
R: In an AP, the difference between consecutive terms is constant.
Show answer
(a). Each term is 4 more than the one before.
Q2. A: The 10th term of the AP 2, 5, 8, … is 29.
R: The nth term of an AP is aₙ = a + (n − 1)d.
Show answer
(a). a = 2, d = 3: a₁₀ = 2 + 9 × 3 = 29.
Q3. A: The sum of the first 20 natural numbers is 210.
R: The sum of the first n terms of an AP is Sₙ = n/2 [2a + (n − 1)d].
Show answer
(a). S₂₀ = 20/2 × [2(1) + 19(1)] = 10 × 21 = 210.
Q4. A: The common difference of the AP −5, −2, 1, 4, … is −3.
R: The common difference of an AP is d = a₂ − a₁.
Show answer
(d). d = −2 − (−5) = 3, not −3. R is true.
Chapter 6: Triangles
Q1. A: All equilateral triangles are similar.
R: Two triangles are similar if their corresponding angles are equal.
Show answer
(a). Every angle of an equilateral triangle is 60°, so corresponding angles are always equal.
Q2. A: In ΔABC, if D and E lie on AB and AC with DE ∥ BC, then AD/DB = AE/EC.
R: A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
Show answer
(a). R is the Basic Proportionality Theorem, which gives A directly.
Q3. A: All isosceles triangles are similar.
R: In an isosceles triangle, the angles opposite the equal sides are equal.
Show answer
(d). A triangle with angles 40°, 70°, 70° and one with 80°, 50°, 50° are both isosceles but not similar. R is true.
Chapter 7: Coordinate Geometry
Q1. A: The distance between (0, 0) and (3, 4) is 5 units.
R: The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²].
Show answer
(a). √(9 + 16) = √25 = 5.
Q2. A: The midpoint of (2, 4) and (6, 8) is (4, 6).
R: The midpoint of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).
Show answer
(a). ((2 + 6)/2, (4 + 8)/2) = (4, 6).
Q3. A: The point (−3, 0) lies on the y-axis.
R: Every point on the x-axis has y-coordinate 0.
Show answer
(d). A y-coordinate of 0 puts (−3, 0) on the x-axis, not the y-axis. R is true.
Chapter 8: Introduction to Trigonometry
Q1. A: sin²θ + cos²θ = 1 for every acute angle θ.
R: In a right triangle, (perpendicular)² + (base)² = (hypotenuse)².
Show answer
(a). Dividing the Pythagoras relation by (hypotenuse)² gives sin²θ + cos²θ = 1.
Q2. A: sin θ = 1.5 for some angle θ.
R: In a right triangle, the hypotenuse is the longest side.
Show answer
(d). sin θ = perpendicular / hypotenuse, and the perpendicular is shorter than the hypotenuse, so sin θ cannot exceed 1. R is true.
Q3. A: tan 45° = 1.
R: In a right triangle with one angle of 45°, the two sides containing the right angle are equal.
Show answer
(a). The other acute angle is also 45°, so the legs are equal and tan 45° = perpendicular/base = 1.
Chapter 9: Some Applications of Trigonometry
Q1. A: As the sun rises higher in the sky, the shadow of a vertical pole becomes shorter.
R: Shadow length = height of pole ÷ tan θ, where θ is the sun's angle of elevation, and tan θ increases as θ increases.
Show answer
(a). A larger θ gives a larger tan θ, so the shadow length becomes smaller.
Q2. A: A 10 m tall tower casts a 10 m long shadow when the sun's angle of elevation is 45°.
R: tan 45° = 1.
Show answer
(a). tan 45° = height / shadow = 10/10 = 1.
Q3. A: The angle of elevation of the top of a tower from a point P equals the angle of depression of P from the top of the tower.
R: The angle of elevation is always greater than the angle of depression.
Show answer
(c). The two angles are alternate angles between parallel horizontal lines, so they are equal. R is false.
Chapter 10: Circles
Q1. A: If PA and PB are tangents from an external point P to a circle, then PA = PB.
R: The lengths of tangents drawn from an external point to a circle are equal.
Show answer
(a). R is the theorem that gives A.
Q2. A: A point P is 13 cm from the centre of a circle of radius 5 cm. The tangent from P to the circle is 12 cm long.
R: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Show answer
(a). The radius, tangent and line OP form a right triangle, so tangent = √(13² − 5²) = √144 = 12 cm.
Q3. A: Two tangents can be drawn to a circle from a point inside it.
R: A tangent meets a circle at exactly one point.
Show answer
(d). Every line through a point inside a circle cuts it at two points, so no tangent can be drawn. R is true.
Chapter 11: Areas Related to Circles
Q1. A: The area of a sector of angle 90° in a circle of radius 14 cm is 154 cm². (Take π = 22/7.)
R: The area of a sector of angle θ is (θ/360°) × πr².
Show answer
(a). (90/360) × (22/7) × 14 × 14 = (1/4) × 616 = 154 cm².
Q2. A: The arc of a sector of angle 60° in a circle of radius 21 cm is 22 cm long. (Take π = 22/7.)
R: The length of an arc of angle θ is (θ/360°) × 2πr.
Show answer
(a). (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm.
Q3. A: The area of a minor segment is greater than the area of the corresponding sector.
R: Area of minor segment = area of sector − area of the corresponding triangle.
Show answer
(d). R shows the segment is the sector minus a triangle, so it is smaller. A is false.
Chapter 12: Surface Areas and Volumes
Q1. A: A cone and a cylinder with the same base radius and height: the cone's volume is one-third of the cylinder's.
R: Volume of a cone = ⅓πr²h, and volume of a cylinder = πr²h.
Show answer
(a). Comparing the two formulas gives the ratio 1 : 3.
Q2. A: The total surface area of a solid hemisphere of radius r is 3πr².
R: The curved surface of a hemisphere is 2πr² and its flat base is a circle of area πr².
Show answer
(a). 2πr² + πr² = 3πr².
Q3. A: A cube of side 4 cm has a volume of 64 cm³.
R: The total surface area of a cube of side a is 4a².
Show answer
(c). 4³ = 64 cm³, so A is true. A cube has 6 faces, so its total surface area is 6a², not 4a².
Chapter 13: Statistics
Q1. A: The mode of 2, 3, 3, 5, 7, 3, 8 is 3.
R: The mode is the value that occurs most often.
Show answer
(a). 3 appears three times, more than any other value.
Q2. A: The median of 4, 8, 6, 10, 2 is 6.
R: For an odd number n of observations arranged in order, the median is the ((n + 1)/2)th observation.
Show answer
(a). In order: 2, 4, 6, 8, 10. The 3rd value is 6.
Q3. A: The mean, median and mode of any data set are always equal.
R: The median is the middle value of the data arranged in order.
Show answer
(d). For 2, 3, 3, 5, 7, 3, 8 above, the mode is 3 but the mean is 31/7 ≈ 4.43. R is true.
Chapter 14: Probability
Q1. A: The probability of getting a number greater than 6 on one throw of a die is 0.
R: The probability of an impossible event is 0.
Show answer
(a). A die shows 1 to 6, so a number greater than 6 is impossible.
Q2. A: For any event E, P(E) + P(not E) = 1.
R: An event and its complement cannot happen together, and between them they cover all possible outcomes.
Show answer
(a). R describes complementary events, which is why their probabilities add to 1.
Q3. A: The probability of an event can be 1.2.
R: For any event E, 0 ≤ P(E) ≤ 1.
Show answer
(d). R shows no probability can exceed 1, so A is false.
How to practise these well
- Time yourself: aim for under a minute per question.
- Say why: for every (b), explain what R would need to say to make it (a).
- Revisit mistakes: a wrong answer usually points to a definition you need to re-read in the NCERT textbook.
- Check the real paper: see how assertion-reason questions appear in CBSE's 2025-26 Class 10 sample papers, and practise more from the NCERT Exemplar problems.
Next, try our case study questions for every Class 10 Maths chapter, or read how to get good marks in Class 10 Maths. For Science, see assertion-reason questions for Class 10 Science. On Intellicamp you can take chapter-wise tests that mix assertion-reason with every other question type, marked instantly, with your weak topics shown after each test. See how it works for students.
- Judge A, then R, then whether R explains A.
- A false means (d); R false means (c); both true means (a) or (b).
- Practise across all 14 chapters and review every mistake against the NCERT text.
Frequently asked questions
How many marks are assertion-reason questions in the Class 10 Maths board paper?
They are usually 1-mark questions in Section A of the CBSE Class 10 Maths paper. Check the current sample paper for the exact number each year.
What is the fastest way to solve assertion-reason questions?
Judge A alone, then R alone. If either is false, the answer is (c) or (d) immediately. Only when both are true do you need to decide whether R explains A.
Are these questions from the latest rationalised syllabus?
Yes. They cover the 14 chapters of the current NCERT Class 10 Mathematics textbook and avoid topics removed in the rationalisation, such as Euclid's division lemma and frustums.
Where can I find more assertion-reason questions?
CBSE's official sample papers include assertion-reason questions every year. You can also practise chapter-wise tests with instant marking on Intellicamp.
What if both statements are false?
CBSE's standard four options do not include 'both false', so a well-written question always has at least one true statement.
Practise this chapter free
Take a short practice test on any chapter, get it marked instantly, and see exactly which topics are costing you marks.