This is the third PA-1 practice paper for Class 9 Maths, based on the new syllabus (Ganita Manjari Part 1, 2026-27): Chapter 1 (Orienting Yourself: The Use of Coordinates) and Chapter 2 (Introduction to Linear Polynomials). Note: the whole Class 9 Maths textbook has changed this year, so this paper covers only the first 2 chapters (the later chapters are not part of it yet).
Periodic Assessment pattern — school pattern se thoda alag ho sakta hai. Every question carries its chapter tag, either [Ch 1] or [Ch 2].
General Instructions
- Time: 40 minutes | Maximum Marks: 25
- Chapter coverage: Ch 1 Orienting Yourself: The Use of Coordinates, Ch 2 Introduction to Linear Polynomials (Ganita Manjari Part 1, new 2026-27 textbook).
- 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.
Section A — Multiple Choice Questions (5 × 1 = 5 marks)
Q1. [Ch 1] In which quadrant does the point (−5, −8) lie?
(A) I (B) II (C) III (D) IV
Q2. [Ch 1] Where does the point (0, 7) lie?
(A) On the x-axis (B) On the y-axis (C) At the origin (D) None of these
Q3. [Ch 1] What do you get when the point (3, −9) is reflected in the y-axis?
(A) (3, 9) (B) (−3, −9) (C) (−3, 9) (D) (3, −9)
Q4. [Ch 2] What is the degree of the polynomial p(x) = 7x − 2?
(A) 0 (B) 1 (C) 2 (D) 7
Q5. [Ch 2] Which one of the following is a linear polynomial?
(A) x² + 1 (B) 5 (C) 3x − 8 (D) x³
Section B — Very Short Answer (4 × 2 = 8 marks)
Q6. [Ch 1] In which quadrant does the point (−2, 5) lie? What do you get when it is reflected in the x-axis?
Q7. [Ch 1] The abscissa of a point P is (a − 2) and its ordinate is (2a + 1). If P lies on the y-axis, find the value of a and the coordinates of P.
Q8. [Ch 2] Find the zero of the polynomial p(x) = 9x − 27.
Q9. [Ch 2] A taxi charges a fixed fee of ₹50 plus ₹15 extra for every kilometre. (i) Write the linear polynomial for the fare y of a journey of x km. (ii) Find the fare for 8 km.
Section C — Short Answer (3 × 3 = 9 marks)
Q10. [Ch 1] First reflect the point P(−6, 4) in the x-axis, then reflect that new point in the y-axis. State the coordinates of the final point Q and the quadrant in which it lies.
Q11. [Ch 2] The height of a plant is given by h(x) = 2x + 10 (in cm), where x is the number of weeks. (i) What was the height at the start (x = 0)? (ii) After how many weeks will the height become 30 cm? (iii) Will the height ever be exactly 25 cm at the end of a whole week? Give a reason.
Q12. [Ch 2] p(x) = (2k − 3)x + 5 is a linear polynomial provided k does not take one particular value. Which value is that? In that case, write the zero of p(x) in terms of k.
Section D — Case Study / Application (1 × 3 = 3 marks)
Q13. [Ch 1 + Ch 2] Aman is walking along a straight path. His position on the coordinate plane is represented by (x, y), and his path is modelled by the polynomial: y = 2x + 1 (here x is the horizontal distance from his home, taken as positive).
- (i) When x = 3, what is his y-coordinate? In which quadrant does this point lie? (1)
- (ii) Find the zero of this polynomial (the value of x for which y = 0). (1)
- (iii) If the point at x = 3 is reflected in the x-axis, what is the new y-coordinate? (1)
Show the Full Answer Key
Section A
Q1 — (C) III. x = −5 (negative) and y = −8 (negative) — when both coordinates are negative, the point lies in Quadrant III.
Q2 — (B) On the y-axis. The x-coordinate is 0, so the point lies on the y-axis.
Q3 — (B) (−3, −9). Reflecting in the y-axis maps (x, y) → (−x, y) — only the sign of x changes, y stays the same. So (3, −9) → (−3, −9).
Q4 — (B) 1. The highest power of x is 1, so the degree is 1 (a linear polynomial).
Q5 — (C) 3x − 8. This is a degree-1 polynomial (a = 3 ≠ 0). Option (A) has degree 2, (B) is a constant (degree 0), and (D) has degree 3.
Section B
Q6. (−2, 5): x is negative and y is positive → Quadrant II. Reflecting in the x-axis maps (x, y) → (x, −y), so (−2, 5) → (−2, −5).
Q7. Lying on the y-axis means the abscissa (x) = 0: a − 2 = 0 ⇒ a = 2. Ordinate = 2(2) + 1 = 5. Therefore P = (0, 5).
Q8. 9x − 27 = 0 ⇒ 9x = 27 ⇒ x = 3. Check: p(3) = 9(3) − 27 = 27 − 27 = 0 ✓.
Q9. (i) y = 15x + 50. (ii) At x = 8: y = 15(8) + 50 = 120 + 50 = ₹170.
Section C
Q10. Reflect P(−6, 4) in the x-axis: (x, y) → (x, −y) ⇒ (−6, −4).
Now reflect this new point (−6, −4) in the y-axis: (x, y) → (−x, y) ⇒ (6, −4).
Therefore Q = (6, −4), and since x is positive and y is negative, it lies in Quadrant IV.
Q11. (i) At x = 0: h(0) = 2(0) + 10 = 10 cm.
(ii) 30 = 2x + 10 ⇒ 2x = 20 ⇒ x = 10 weeks.
(iii) 25 = 2x + 10 ⇒ 2x = 15 ⇒ x = 7.5. Since x is not a whole number (7.5 weeks is not a complete week), the height will never be 25 cm at the end of any whole week.
Q12. For a linear polynomial the coefficient of x cannot be zero: 2k − 3 ≠ 0 ⇒ k ≠ 3/2.
Zero: (2k − 3)x + 5 = 0 ⇒ x = −5 / (2k − 3).
Check (k = 2): coefficient = 2(2) − 3 = 1, polynomial = x + 5, zero = −5. From the formula: −5/(2(2)−3) = −5/1 = −5 ✓.
Section D
Q13 (i). At x = 3: y = 2(3) + 1 = 7. Point (3, 7): x is positive and y is positive → Quadrant I.
Q13 (ii). Zero: 0 = 2x + 1 ⇒ 2x = −1 ⇒ x = −1/2.
Q13 (iii). Reflecting the point (3, 7) in the x-axis maps (x, y) → (x, −y), so the new y-coordinate is −7.
Apna Score Kaise Padhein
- 20–25: both chapters are strong.
- 14–19: you need more practice with reflections or with finding zeroes.
- Below 13: read the chapter notes again, then attempt this paper once more after 3 days.
Chapter notes are here: Chapter 1 — Coordinates and Chapter 2 — Linear Polynomials. More practice papers are here: Unit Test & Practice Paper Hub.
Kaizen: today, solve just that one question you got wrong — without looking at the answer. One small improvement every day, and the difference will be clear by PA-1.