PA-1 is the very first periodic test of the session, and the 10 marks of internal assessment are built on it. This 25-mark practice paper covers Chapters 1, 2 and 3 of CBSE Class 10 Maths: Real Numbers, Polynomials, and Pair of Linear Equations in Two Variables. All the questions are original, and every answer has been solved step by step.
Periodic Assessment pattern — school pattern se thoda alag ho sakta hai. Every question carries its chapter tag: [Ch 1], [Ch 2] or [Ch 3]. If your school has kept only Chapters 1 and 2 for PA-1, skip the [Ch 3] questions — the rest of the paper will work exactly as it is.
General Instructions
- Time: 40 minutes | Maximum Marks: 25
- Chapter coverage: Ch 1 Real Numbers (7 marks), Ch 2 Polynomials (9 marks), Ch 3 Pair of Linear Equations in Two Variables (9 marks).
- The paper has four sections — A (5 × 1 mark), B (4 × 2 marks), C (3 × 3 marks), D (1 case-study of 3 marks).
- All questions are compulsory. There is no internal choice.
- The use of a calculator is not allowed.
- Keep a separate page for rough work, and always write the working along with the final answer — in a PA you get marks for the steps as well.
Section A — Multiple Choice Questions (5 × 1 = 5 marks)
Q1. [Ch 1] Two positive integers a and b are written as a = p³q² and b = pq³, where p and q are two different prime numbers. What is HCF(a, b)?
(A) pq² (B) p³q³ (C) pq³ (D) p³q²
Q2. [Ch 1] If HCF(96, 404) = 4, then what is LCM(96, 404)?
(A) 9696 (B) 4848 (C) 19392 (D) 2424
Q3. [Ch 2] One zero of the quadratic polynomial x² − 5x + k is 2. Then the value of k is:
(A) 4 (B) 6 (C) −6 (D) 10
Q4. [Ch 2] The graph of a quadratic polynomial p(x) meets the x-axis at exactly one point. Then p(x) has:
(A) no zeroes (B) two different zeroes (C) two equal zeroes (D) three zeroes
Q5. [Ch 3] The pair of linear equations 2x + 3y = 7 and 4x + 6y = 5 has:
(A) exactly one solution (B) infinitely many solutions (C) no solution (D) exactly two solutions
Section B — Very Short Answer (4 × 2 = 8 marks)
Q6. [Ch 1] Using prime factorisation, find the HCF and the LCM of 90 and 144. Verify your answer using HCF × LCM = first number × second number.
Q7. [Ch 2] Find the zeroes of the polynomial p(x) = x² − 2x − 15 and verify the relationship between the zeroes and the coefficients.
Q8. [Ch 3] Solve by the substitution method: x + 2y = 8 and 2x − y = 1.
Q9. [Ch 2] If α and β are the zeroes of the polynomial 2x² − 7x + 3, find the value of 1/α + 1/β.
Section C — Short Answer (3 × 3 = 9 marks)
Q10. [Ch 1] Prove that 5 − 2√3 is an irrational number. (You may assume that √3 is irrational.)
Q11. [Ch 3] The sum of the digits of a two-digit number is 11. If the digits are interchanged, the new number is 27 more than the original number. Find the original number.
Q12. [Ch 2] Form a quadratic polynomial whose zeroes are 3 + √2 and 3 − √2. After that, also find the sum of the squares of its zeroes.
Section D — Case Study / Application (1 × 3 = 3 marks)
Q13. [Ch 3] A stationery shop sells notebooks and pens, both at a fixed rate. Riya paid ₹240 for 3 notebooks and 4 pens. At the same shop, Mohit paid ₹330 for 5 notebooks and 2 pens. Let the price of one notebook be ₹x and the price of one pen be ₹y.
- (i) Using the given information, write the pair of linear equations. (1)
- (ii) Find the price of one notebook and one pen. (1)
- (iii) What will be the total cost of 2 notebooks and 5 pens? (1)
Show the Full Answer Key
Section A
Q1 — (A) pq². For the HCF, the lowest power of every common prime is taken. a = p³q², b = p¹q³. The powers of p are 3 and 1 → the lower power is 1. The powers of q are 2 and 3 → the lower power is 2. Therefore HCF = p¹q² = pq².
Q2 — (A) 9696. For two numbers, HCF × LCM = first number × second number. So LCM = (96 × 404) ÷ HCF = 38784 ÷ 4 = 9696.
Q3 — (B) 6. Since 2 is a zero, p(2) = 0. (2)² − 5(2) + k = 0 → 4 − 10 + k = 0 → −6 + k = 0 → k = 6.
Q4 — (C) two equal zeroes. The graph of a quadratic is a parabola. If it touches the x-axis at only one point, then the discriminant b² − 4ac = 0, which means the two zeroes are equal (coincident). For example: x² − 6x + 9 = (x − 3)², whose two zeroes are both 3.
Q5 — (C) no solution. Here a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = 3/6 = 1/2, but c₁/c₂ = 7/5. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel — the system is inconsistent and has no solution.
Section B
Q6. Prime factorisation: 90 = 2 × 3² × 5 and 144 = 2⁴ × 3².
HCF = the lower powers of the common primes = 2¹ × 3² = 18.
LCM = the higher powers of all the primes = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720.
Verification: HCF × LCM = 18 × 720 = 12960 and 90 × 144 = 12960. Both are equal — the answer is correct.
Q7. Factorise x² − 2x − 15: two numbers are needed whose product is −15 and whose sum is −2 → −5 and +3. So x² − 5x + 3x − 15 = x(x − 5) + 3(x − 5) = (x − 5)(x + 3). Zeroes: 5 and −3.
Verification: here a = 1, b = −2, c = −15. Sum of zeroes = 5 + (−3) = 2 and −b/a = −(−2)/1 = 2 ✓. Product = 5 × (−3) = −15 and c/a = −15/1 = −15 ✓.
Q8. From the first equation, x = 8 − 2y. Substitute this into the second one: 2(8 − 2y) − y = 1 → 16 − 4y − y = 1 → 16 − 5y = 1 → 5y = 15 → y = 3. Now x = 8 − 2(3) = x = 2.
Check: 2 + 2(3) = 8 ✓ and 2(2) − 3 = 1 ✓.
Q9. In 2x² − 7x + 3, a = 2, b = −7, c = 3. So α + β = −b/a = 7/2 and αβ = c/a = 3/2.
1/α + 1/β = (β + α)/(αβ) = (7/2) ÷ (3/2) = 7/2 × 2/3 = 7/3.
Check: the zeroes are in fact 3 and 1/2; 1/3 + 2 = 7/3 ✓.
Section C
Q10. Assume the opposite — that 5 − 2√3 is rational. Then it can be written as equal to some rational number r:
5 − 2√3 = r
⇒ 2√3 = 5 − r
⇒ √3 = (5 − r) ÷ 2.
Now 5 and r are both rational, and the difference of two rationals is rational; dividing that by 2 (a non-zero rational) also gives a rational. This means √3 turns out to be rational — but it is given that √3 is irrational. This is a contradiction.
Therefore the initial assumption was wrong, and 5 − 2√3 is irrational.
Q11. Let the tens digit be x and the units digit be y. Number = 10x + y.
Sum of the digits: x + y = 11 …(i)
On interchanging the digits the number becomes 10y + x, which is 27 more:
10y + x = (10x + y) + 27 → 9y − 9x = 27 → y − x = 3 …(ii)
(i) + (ii): 2y = 14 → y = 7; so x = 11 − 7 = 4.
Original number = 10(4) + 7 = 47.
Check: 4 + 7 = 11 ✓ and 74 − 47 = 27 ✓.
Q12. Zeroes α = 3 + √2, β = 3 − √2.
Sum = (3 + √2) + (3 − √2) = 6.
Product = (3)² − (√2)² = 9 − 2 = 7.
Polynomial = x² − (sum)x + (product) = x² − 6x + 7. (Any non-zero multiple of this is also correct.)
Sum of the squares of the zeroes: α² + β² = (α + β)² − 2αβ = (6)² − 2(7) = 36 − 14 = 22.
Section D
Q13 (i). Riya: 3x + 4y = 240. Mohit: 5x + 2y = 330.
Q13 (ii). Multiply the second equation by 2: 10x + 4y = 660. Now subtract the first equation:
(10x + 4y) − (3x + 4y) = 660 − 240 → 7x = 420 → x = 60.
5(60) + 2y = 330 → 300 + 2y = 330 → y = 15.
One notebook costs ₹60 and one pen costs ₹15. Check: 3(60) + 4(15) = 180 + 60 = 240 ✓.
Q13 (iii). 2 notebooks + 5 pens = 2(60) + 5(15) = 120 + 75 = ₹195.
Apna Score Kaise Padhein
- 20–25: all three chapters are strong. Now move ahead to the PA-2 chapters.
- 14–19: the concepts are fine, but calculation slips are happening. Solve every wrong question again, without looking at the answer.
- Below 13: first read the notes of the chapter where you lost the most marks, then take this paper again after 3 days.
More chapter-wise tests and unit test papers are available here: Unit Test & Practice Paper Hub.
Kaizen: the whole syllabus will not improve in one day. Today, solve again just that one question you got wrong — without looking at the answer. One small improvement every day, and by PA-1 the difference will be clearly visible.