This is an original PA 1 practice paper for CBSE Class 10 Mathematics (Standard), session 2026-27, built for schools whose first Periodic Assessment covers Chapters 1 to 4 — Real Numbers, Polynomials, Pair of Linear Equations in Two Variables and Quadratic Equations. It carries 25 marks and is designed to be finished in 40 minutes.
Periodic Assessment pattern — school pattern se thoda alag ho sakta hai. Every question below carries a chapter tag such as [Ch 3]. If your school has only covered up to Chapter 2 or Chapter 3, skip the questions tagged beyond your syllabus and adjust your total accordingly.
Before you start
- Maximum marks: 25
- Time allowed: 40 minutes
- Chapters covered: Chapter 1 (Real Numbers), Chapter 2 (Polynomials), Chapter 3 (Pair of Linear Equations in Two Variables), Chapter 4 (Quadratic Equations)
- Sections: A — 5 MCQs of 1 mark; B — 4 questions of 2 marks; C — 3 questions of 3 marks; D — 1 case-based question of 3 marks
- All questions are compulsory. Calculators are not allowed. Sit with a timer and write the full working, exactly as you would in the real assessment.
Section A — Multiple Choice Questions (5 × 1 = 5 marks)
Q1. [Ch 1] Two positive integers are written as a = x4y2 and b = x2y3, where x and y are distinct prime numbers. The LCM of a and b is:
(A) x2y2 (B) x4y3 (C) x6y5 (D) x2y3
Q2. [Ch 2] If 3 is a zero of the polynomial p(x) = x2 − 7x + k, then the value of k is:
(A) 10 (B) 12 (C) −12 (D) 4
Q3. [Ch 3] The pair of linear equations 3x − y = 5 and 6x − 2y = 11 has:
(A) a unique solution (B) exactly two solutions (C) infinitely many solutions (D) no solution
Q4. [Ch 4] The quadratic equation 4x2 − kx + 9 = 0 has two equal real roots when k equals:
(A) 6 only (B) 12 only (C) ±12 (D) ±6
Q5. [Ch 1] The HCF of two numbers is 9 and their LCM is 360. If one of the numbers is 45, the other number is:
(A) 72 (B) 80 (C) 90 (D) 63
Section B — Very Short Answer (4 × 2 = 8 marks)
Q6. [Ch 1] Prove that 5 + 2√3 is an irrational number. (You may assume that √3 is irrational.)
Q7. [Ch 2] Write a quadratic polynomial whose sum of zeroes is −5 and product of zeroes is 6. Also find its zeroes.
Q8. [Ch 3] Solve the following pair of equations by the elimination method: 2x + 3y = 13 and 3x − y = 3.
Q9. [Ch 4] Find the roots of the quadratic equation 2x2 − 7x + 3 = 0 by the factorisation method.
Section C — Short Answer (3 × 3 = 9 marks)
Q10. [Ch 1] Three bells in a school ring at intervals of 9 minutes, 12 minutes and 15 minutes respectively. All three ring together at 7:00 a.m. At what time will they next ring together? Show your working.
Q11. [Ch 2] If α and β are the zeroes of the polynomial p(x) = 2x2 − 5x + 1, find the value of (i) α + β, (ii) αβ, and (iii) 1/α + 1/β.
Q12. [Ch 3] The sum of the present ages of a father and his son is 45 years. Five years ago, the father was six times as old as his son. Find their present ages.
Section D — Case-Based / Application (1 × 3 = 3 marks)
Q13. [Ch 4] A school is planning a rectangular vegetable garden in one corner of its playground. The gardener is told that the length of the plot must be 3 metres more than twice its breadth, and the plot must have an area of 90 square metres. Let the breadth of the plot be x metres.
(i) Form a quadratic equation in x from the information given. (1)
(ii) Solve the equation and state the breadth and the length of the plot in metres. (2)
Show the Full Answer Key
Section A
Q1 — (B) x4y3. For the LCM of numbers written as products of primes, take the highest power of each prime that appears. Here a = x4y2 and b = x2y3. The highest power of x is 4 and the highest power of y is 3, so the LCM is x4y3.
Q2 — (B) 12. If 3 is a zero, then p(3) = 0. So (3)2 − 7(3) + k = 0, that is 9 − 21 + k = 0, which gives −12 + k = 0 and therefore k = 12.
Q3 — (D) no solution. Compare the coefficients. a1/a2 = 3/6 = 1/2 and b1/b2 = (−1)/(−2) = 1/2, but c1/c2 = 5/11. Since a1/a2 = b1/b2 but this is not equal to c1/c2, the two lines are parallel and the pair is inconsistent. There is no solution.
Q4 — (C) ±12. Equal roots means the discriminant is zero. D = b2 − 4ac = (−k)2 − 4(4)(9) = k2 − 144. Setting k2 − 144 = 0 gives k2 = 144, so k = 12 or k = −12. Both values work, so the answer is ±12.
Q5 — (A) 72. For two numbers, HCF × LCM = product of the numbers. So 9 × 360 = 45 × (other number), giving 3240 = 45 × (other number), and the other number = 3240 ÷ 45 = 72. Check: HCF(45, 72) = 9 and LCM(45, 72) = 360, so the answer is consistent.
Section B
Q6 (2 marks). Suppose, on the contrary, that 5 + 2√3 is rational. Then 5 + 2√3 = p/q for some integers p and q with q not equal to 0. Rearranging, 2√3 = p/q − 5 = (p − 5q)/q, so √3 = (p − 5q)/(2q). The right-hand side is a ratio of two integers with a non-zero denominator, so it is a rational number. That would make √3 rational, which contradicts the given fact that √3 is irrational. Hence the assumption is wrong, and 5 + 2√3 is irrational.
Q7 (2 marks). A quadratic polynomial with a given sum and product of zeroes is x2 − (sum)x + (product). Here the sum is −5 and the product is 6, so the polynomial is x2 − (−5)x + 6 = x2 + 5x + 6. Factorising, x2 + 5x + 6 = (x + 2)(x + 3), so the zeroes are x = −2 and x = −3. Check: (−2) + (−3) = −5 and (−2)(−3) = 6, as required.
Q8 (2 marks). The equations are 2x + 3y = 13 …(i) and 3x − y = 3 …(ii). Multiply equation (ii) by 3 so that the y terms match: 9x − 3y = 9 …(iii). Add (i) and (iii): 11x = 22, so x = 2. Substitute x = 2 in (ii): 3(2) − y = 3, so 6 − y = 3 and y = 3. The solution is x = 2, y = 3. Check in (i): 2(2) + 3(3) = 4 + 9 = 13. Correct.
Q9 (2 marks). Split the middle term of 2x2 − 7x + 3. Two numbers are needed whose product is 2 × 3 = 6 and whose sum is −7; these are −6 and −1. So 2x2 − 7x + 3 = 2x2 − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3). Setting each factor to zero gives x = 1/2 and x = 3. The roots are 1/2 and 3.
Section C
Q10 (3 marks). The bells ring together again after a time equal to the LCM of 9, 12 and 15 minutes. Write each number as a product of primes: 9 = 32, 12 = 22 × 3, and 15 = 3 × 5. Take the highest power of each prime: LCM = 22 × 32 × 5 = 4 × 9 × 5 = 180. So they ring together again after 180 minutes, which is 3 hours. Starting from 7:00 a.m., the next time all three ring together is 10:00 a.m.
Q11 (3 marks). For p(x) = 2x2 − 5x + 1 we have a = 2, b = −5 and c = 1. (i) α + β = −b/a = 5/2. (ii) αβ = c/a = 1/2. (iii) 1/α + 1/β = (β + α)/(αβ) = (5/2) ÷ (1/2) = 5. So the three required values are 5/2, 1/2 and 5.
Q12 (3 marks). Let the father’s present age be f years and the son’s present age be s years. The first condition gives f + s = 45 …(i). Five years ago their ages were (f − 5) and (s − 5), so the second condition gives f − 5 = 6(s − 5), that is f − 5 = 6s − 30, so f = 6s − 25 …(ii). Substituting (ii) into (i): (6s − 25) + s = 45, so 7s = 70 and s = 10. Then f = 45 − 10 = 35. The father is 35 years old and the son is 10 years old. Check: five years ago they were 30 and 5, and 30 = 6 × 5. Correct.
Section D
Q13 (3 marks). (i) Let the breadth be x metres. Then the length is (2x + 3) metres. Area = length × breadth, so x(2x + 3) = 90, which gives the quadratic equation 2x2 + 3x − 90 = 0. (ii) Here a = 2, b = 3 and c = −90. D = b2 − 4ac = 32 − 4(2)(−90) = 9 + 720 = 729, and √729 = 27. So x = (−b ± √D)/(2a) = (−3 ± 27)/4, which gives x = 24/4 = 6 or x = −30/4 = −7.5. A breadth cannot be negative, so x = 6 is taken. Breadth = 6 m and length = 2(6) + 3 = 15 m. Check: 6 × 15 = 90 square metres. Correct.
Marks distribution: Section A 5 × 1 = 5, Section B 4 × 2 = 8, Section C 3 × 3 = 9, Section D 1 × 3 = 3. Total = 25 marks.
Apna Score Kaise Padhein
- 21–25: Chapters 1 to 4 are solid. Move on to timed mixed practice.
- 16–20: Good base. Look at which chapter tag your lost marks came from and revise only that chapter.
- 11–15: The methods are known but the working is slipping. Redo Sections B and C without looking at the key.
- Below 11: Go back to the chapter notes first, then attempt this paper again after two days.
Looking for more assessment practice? The Unit Test and PA practice hub collects every chapter test and Periodic Assessment paper on the site, class by class.
Kaizen note: one paper attempted honestly under a timer teaches more than ten papers read with the answer key open. Improve by one mark at a time.