PA-1 ka teesra paper Class 9 Maths ke naye syllabus (Ganita Manjari Part 1, 2026-27) par: Chapter 1 (Orienting Yourself: The Use of Coordinates) aur Chapter 2 (Introduction to Linear Polynomials). Note: is saal Class 9 Maths ki poori textbook badal gayi hai, isliye yeh paper sirf pehle 2 chapters cover karta hai (aage ke chapters abhi content mein nahi hain).
Periodic Assessment pattern — school pattern se thoda alag ho sakta hai. Har question ke saath uska chapter tag [Ch 1] ya [Ch 2] diya hai.
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 compulsory hain. Internal choice nahi hai.
Section A — Multiple Choice Questions (5 × 1 = 5 marks)
Q1. [Ch 1] Point (−5, −8) kis quadrant mein hai?
(A) I (B) II (C) III (D) IV
Q2. [Ch 1] Point (0, 7) kahaan hai?
(A) x-axis par (B) y-axis par (C) origin par (D) inmein se koi nahi
Q3. [Ch 1] Point (3, −9) ko y-axis mein reflect karne par kya milega?
(A) (3, 9) (B) (−3, −9) (C) (−3, 9) (D) (3, −9)
Q4. [Ch 2] Polynomial p(x) = 7x − 2 ki degree kya hai?
(A) 0 (B) 1 (C) 2 (D) 7
Q5. [Ch 2] Inmein se kaunsa ek linear polynomial hai?
(A) x² + 1 (B) 5 (C) 3x − 8 (D) x³
Section B — Very Short Answer (4 × 2 = 8 marks)
Q6. [Ch 1] Point (−2, 5) kis quadrant mein hai? Ise x-axis mein reflect karne par kya milega?
Q7. [Ch 1] Point P ki abscissa (a − 2) aur ordinate (2a + 1) hai. Agar P y-axis par hai, to a ka maan aur P ke coordinates nikaliye.
Q8. [Ch 2] Polynomial p(x) = 9x − 27 ka zero nikaliye.
Q9. [Ch 2] Ek taxi ka fixed charge ₹50 hai aur har km ke liye ₹15 extra lagte hain. (i) x km ke liye fare y ka linear polynomial likhiye. (ii) 8 km ke liye fare nikaliye.
Section C — Short Answer (3 × 3 = 9 marks)
Q10. [Ch 1] Point P(−6, 4) ko pehle x-axis mein reflect kijiye, phir us naye point ko y-axis mein reflect kijiye. Final point Q ke coordinates aur uska quadrant bataiye.
Q11. [Ch 2] Ek paudhe (plant) ki height h(x) = 2x + 10 (cm) se di gayi hai, jahan x hafton ki sankhya hai. (i) Shuru mein (x = 0) height kya thi? (ii) Kitne hafton mein height 30 cm ho jaayegi? (iii) Kya height kabhi poore hafte par 25 cm hogi? Reason dijiye.
Q12. [Ch 2] p(x) = (2k − 3)x + 5 ek linear polynomial hai basharte k ka kaunsa maan na ho? Us case mein p(x) ka zero k ke terms mein likhiye.
Section D — Case Study / Application (1 × 3 = 3 marks)
Q13. [Ch 1 + Ch 2] Aman ek seedhe raste (straight path) par chal raha hai. Uski position ko coordinate plane par (x, y) se represent kiya jaata hai, jahan uska path is polynomial se model hota hai: y = 2x + 1 (x = ghar se horizontal distance, positive maana gaya hai).
- (i) Jab x = 3, uska y-coordinate kya hoga? Yeh point kis quadrant mein hai? (1)
- (ii) Is polynomial ka zero (woh x jahan y = 0 hota hai) nikaliye. (1)
- (iii) Agar x = 3 wale point ko x-axis mein reflect kar diya jaaye, to naya y-coordinate kya hoga? (1)
Show the Full Answer Key
Section A
Q1 — (C) III. x = −5 (negative), y = −8 (negative) — dono negative hone par point Quadrant III mein hota hai.
Q2 — (B) y-axis par. x-coordinate 0 hai, isliye point y-axis par hai.
Q3 — (B) (−3, −9). y-axis mein reflect karne par (x, y) → (−x, y) — sirf x ka sign badalta hai, y same rehta hai. To (3, −9) → (−3, −9).
Q4 — (B) 1. x ki sabse badi power 1 hai, isliye degree 1 hai (linear polynomial).
Q5 — (C) 3x − 8. Yeh degree-1 polynomial hai (a = 3 ≠ 0). Option (A) degree 2 hai, (B) ek constant hai (degree 0), (D) degree 3 hai.
Section B
Q6. (−2, 5): x negative, y positive → Quadrant II. x-axis mein reflect karne par (x, y) → (x, −y): (−2, 5) → (−2, −5).
Q7. y-axis par hone ka matlab abscissa (x) = 0: a − 2 = 0 ⇒ a = 2. Ordinate = 2(2) + 1 = 5. Isliye 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) x = 8 par: y = 15(8) + 50 = 120 + 50 = ₹170.
Section C
Q10. P(−6, 4) ko x-axis mein reflect kijiye: (x, y) → (x, −y) ⇒ (−6, −4).
Ab is naye point (−6, −4) ko y-axis mein reflect kijiye: (x, y) → (−x, y) ⇒ (6, −4).
Isliye Q = (6, −4), jo x positive aur y negative hone ki wajah se Quadrant IV mein hai.
Q11. (i) x = 0 par: h(0) = 2(0) + 10 = 10 cm.
(ii) 30 = 2x + 10 ⇒ 2x = 20 ⇒ x = 10 hafte.
(iii) 25 = 2x + 10 ⇒ 2x = 15 ⇒ x = 7.5. Kyunki x poora number nahi hai (7.5 hafte, jo whole week nahi hai), 25 cm height kisi bhi poore hafte ke ant mein nahi hogi.
Q12. Linear polynomial ke liye x ka coefficient zero nahi ho sakta: 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. Formula se: −5/(2(2)−3) = −5/1 = −5 ✓.
Section D
Q13 (i). x = 3 par: y = 2(3) + 1 = 7. Point (3, 7): x positive, y positive → Quadrant I.
Q13 (ii). Zero: 0 = 2x + 1 ⇒ 2x = −1 ⇒ x = −1/2.
Q13 (iii). Point (3, 7) ko x-axis mein reflect karne par (x, y) → (x, −y): naya y-coordinate = −7.
Apna Score Kaise Padhein
- 20–25: dono chapters strong hain.
- 14–19: reflection ya zero nikalne mein practice chahiye.
- 13 se kam: chapter notes dobara padhiye, phir 3 din baad yeh paper dobara dijiye.
Chapter notes yahan hain: Chapter 1 — Coordinates aur Chapter 2 — Linear Polynomials. Aur bhi papers yahan: Unit Test & Practice Paper Hub.
Kaizen: aaj sirf wo ek question dobara solve kijiye jo galat hua tha — bina answer dekhe. Roz ek chhoti si sudhaar, aur PA-1 tak farak saaf dikhega.