CBSE Class 9 · Mathematics (NCERT ‘Ganita Manjari’) · Max Marks 80 · Time 3 hrs · HOTS / analytical
CBSE Class 9 Maths — Model Question Paper (Set 2 · HOTS) · Max Marks 80 · Time 3 hrs
A fully higher-order-thinking paper — assertion-reason, reasoning, application and case-study across all 8 chapters. Answers in the ✍️ Answers tab.
General Instructions: Five sections A–E, 38 compulsory questions. A (Q1–20) 1 mark; B (Q21–25) 2 marks; C (Q26–31) 3 marks; D (Q32–35) 5 marks; E (Q36–38) 4 marks. Internal choices given. Use π = 22/7 unless stated.
SECTION A — Objective (HOTS) [20 × 1 = 20]
Q1. A point equidistant from both axes with a negative x and positive y lies on the line y = −x in quadrant:
- (i) II
- (ii) I
- (iii) III
- (iv) IV
Q2. If (x, y) = (y, x) for a point, then necessarily:
- (i) x = y
- (ii) x = −y
- (iii) x = 0
- (iv) y = 0
Q3. The decimal 0.͖9 (0.999…) is exactly equal to:
- (i) 1
- (ii) 0.9
- (iii) 0.99
- (iv) just below 1
Q4. Which number is irrational?
- (i) √7
- (ii) 0.25
- (iii) 3/4
- (iv) √16
Q5. If a + b = 7 and ab = 10, then a² + b² equals:
- (i) 29
- (ii) 49
- (iii) 19
- (iv) 9
Q6. A line y = ax + b with a fixed and b changing gives lines that are:
- (i) parallel
- (ii) perpendicular
- (iii) intersecting at origin
- (iv) the same line
Q7. Two chords of a circle are equal iff they are:
- (i) equidistant from the centre
- (ii) both diameters
- (iii) perpendicular
- (iv) on the same side
Q8. Of two unequal chords, the longer one is:
- (i) nearer the centre
- (ii) farther from the centre
- (iii) a tangent
- (iv) equal to the radius
Q9. Heron’s formula is a special case of Brahmagupta’s formula when one side:
- (i) is zero
- (ii) equals the others
- (iii) is the diameter
- (iv) is doubled
Q10. For the same perimeter, the shape with the greatest area is a:
- (i) circle
- (ii) square
- (iii) triangle
- (iv) rectangle
Q11. After 6 heads in a row, P(head) on the next fair toss is:
- (i) 1/2
- (ii) 1
- (iii) 0
- (iv) less than 1/2
Q12. The square numbers 1, 4, 9, 16, … form:
- (i) neither an AP nor a GP
- (ii) an AP
- (iii) a GP
- (iv) a constant sequence
Q13. If the n-th term of a sequence is 3n − 2, the sequence is an AP with common difference:
- (i) 3
- (ii) 2
- (iii) −2
- (iv) 1
Q14. Doubling the radius of a circle multiplies its area by:
- (i) 4
- (ii) 2
- (iii) 8
- (iv) √2
Q15. (x + a)(x + b) = x² + (a + b)x + :
- (i) ab
- (ii) a + b
- (iii) a − b
- (iv) a/b
Q16. A cyclic parallelogram must be a:
- (i) rectangle
- (ii) rhombus
- (iii) kite
- (iv) trapezium
Q17–20 (Assertion–Reason): (a) Both true, R explains A; (b) Both true, R not the explanation; (c) A true, R false; (d) A false, R true.
Q17. Assertion: A line parallel to the y-axis has all points with the same x-coordinate. Reason: Such a line is vertical.
Q18. Assertion: π = 22/7 exactly. Reason: π is an irrational number.
Q19. Assertion: The points (0,0), (3,4), (6,8) are collinear. Reason: The slope between each pair of these points is the same.
Q20. Assertion: n³ − n is always divisible by 6. Reason: n³ − n = (n−1)·n·(n+1), a product of three consecutive integers.
SECTION B — Very Short Answer (HOTS) [5 × 2 = 10]
Q21. Without full multiplication, evaluate 998² using a suitable identity. [2]
Q22. Show, using algebra, that 0.͖47 (0.4747…) is a rational number and find it. [2]
Q23. A chord of a circle of radius 13 cm is 5 cm from the centre; another is 12 cm from the centre. Which is longer and by how much? [2]
Q24. Find a point on the x-axis equidistant from A(2, −5) and B(−2, 9). [2]
Q25. The n-th term of a sequence is 4n + 3. Which term equals 59, and what is the constant difference? [2]
SECTION C — Short Answer (HOTS) [6 × 3 = 18]
Q26. If a + b = 5 and ab = 6, find a² + b² and a³ + b³ without finding a and b. [3]
Q27. Prove that the sum of a rational number and an irrational number is always irrational. [3]
Q28. A, B, C, D lie on a circle; AB subtends 50° at C. A point P (same side as C) has ∠APB = 50°. Is P on the circle? Justify. [3]
Q29. A wire bent into a circle of radius 28 cm is re-bent into a square. Find the square’s side and state which shape encloses more area. [3]
Q30. Two mobile plans cost y = 15x + 100 and y = 20x + 50 (x = GB). For what usage are they equal, and which is cheaper beyond that? [3]
Q31. Find all ways of writing 100 as a sum of consecutive natural numbers. [3]
SECTION D — Long Answer (HOTS) [4 × 5 = 20]
Q32. (a) Factorise x⁴ + x² + 1 completely. (b) If x + y = −4, find the value of x³ + y³ − 12xy + 64. [5]
OR — (a) Simplify (x² − 7x + 12)/(5x² + 5x − 100). (b) Prove that a³ + b³ + c³ − 3abc = −25 when a + b + c = 5 and ab + bc + ca = 10.
Q33. Prove that a rectangle is the only parallelogram that can be inscribed in a circle, and that its diagonals meet at the centre. [5]
OR — Prove that the opposite angles of a cyclic quadrilateral are supplementary, and use it to find the angles of a cyclic quadrilateral in which ∠A = (2x+10)° and ∠C = (3x−20)°.
Q34. Two circles of radius r pass through each other’s centres. (a) Show each ‘lens’ arc subtends 120° at its centre. (b) Find the total length of the boundary of the figure in terms of r. [5]
OR — Semicircles are drawn on the three sides of a right triangle; prove that the area on the hypotenuse equals the sum of the areas on the two legs.
Q35. The sum of the first three terms of a GP is 26 and the sum of their squares is 364. Find the terms. [5]
OR — The 4th and 8th terms of an AP add to 24, and the 6th and 10th add to 44. Find the first three terms, and the smallest n for which the sum 1 + 2 + … + n exceeds 1000.
SECTION E — Case / Source-based (HOTS) [3 × 4 = 12]
Q36. A drone starts at O(0, 0) and visits A(6, 0), B(6, 8) and returns to O along straight lines.
- (i) Find OA and AB. [1]
- (ii) Find OB (the return leg). [1]
- (iii) What type of triangle is OAB? [1]
- (iv) Find the total distance flown (the perimeter). [1]
Q37. A tiled floor grows in stages: 1, 5, 9, 13, … tiles (Stage 1, 2, 3, …).
- (i) Is this an AP or a GP? Give the common difference/ratio. [1]
- (ii) Write the rule for the number of tiles at stage n. [1]
- (iii) How many tiles are needed at stage 20? [1]
- (iv) Which stage uses exactly 101 tiles? [1]
Q38. A bag has 4 red, 3 green and 2 blue balls. One ball is drawn at random.
- (i) Find P(red). [1]
- (ii) Find P(not blue). [1]
- (iii) If two balls are drawn without replacement, find P(both different colours). [1]
- (iv) What is the probability of drawing a yellow ball? [1]
Model Answers & Marking Scheme (Set 2 · HOTS)
Value-points; reward correct reasoning.
Section A (1 mark each)
1 — (i) II | 2 — (i) x = y | 3 — (i) 1 | 4 — (i) √7 | 5 — (i) 29 | 6 — (i) parallel | 7 — (i) equidistant from the centre | 8 — (i) nearer the centre | 9 — (i) is zero | 10 — (i) circle | 11 — (i) 1/2 | 12 — (i) neither an AP nor a GP | 13 — (i) 3 | 14 — (i) 4 | 15 — (i) ab | 16 — (i) rectangle
17 — (a) both true, R explains A. 18 — (d) A false (π ≠ 22/7 exactly), R true. 19 — (a) both true, R explains A. 20 — (a) both true, R explains A.
Section B (2 marks each)
21. 998² = (1000 − 2)² = 1000000 − 4000 + 4 = 996004.
22. x = 0.4747… ⇒ 100x = 47.4747… ⇒ 99x = 47 ⇒ x = 47/99 — a ratio of integers, hence rational.
23. Chord at 5 cm: 2√(13²−25) = 2√144 = 24 cm. Chord at 12 cm: 2√(169−144) = 2√25 = 10 cm. The one nearer the centre (5 cm) is longer, by 14 cm.
24. Let P(x, 0): (x−2)² + 25 = (x+2)² + 81 ⇒ −4x + 29 = 4x + 85 ⇒ −8x = 56 ⇒ x = −7. So P(−7, 0).
25. 4n + 3 = 59 ⇒ 4n = 56 ⇒ n = 14 (the 14th term); the constant difference is the coefficient of n = 4.
Section C (3 marks each)
26. a² + b² = (a+b)² − 2ab = 25 − 12 = 13. a³ + b³ = (a+b)³ − 3ab(a+b) = 125 − 3(6)(5) = 125 − 90 = 35.
27. Suppose rational r + irrational s = t were rational. Then s = t − r is a difference of rationals, hence rational — contradicting s irrational. So r + s is irrational.
28. Yes. AB subtends equal angles (50°) at C and P on the same side of AB, so by the concyclicity theorem A, B, C, P lie on one circle — P is on the circle through A, B, C.
29. Wire = 2πr = 2(22/7)(28) = 176 cm = square perimeter ⇒ side = 44 cm. Circle area = (22/7)(784) = 2464 cm²; square area = 44² = 1936 cm². The circle encloses more area.
30. 15x + 100 = 20x + 50 ⇒ 50 = 5x ⇒ x = 10 GB (both ₹250). Beyond 10 GB, Plan A (slope 15) grows slower, so Plan A is cheaper for more than 10 GB.
31. k consecutive numbers from m: k(2m + k − 1) = 200. Integer solutions give 18+19+20+21+22 (5 terms) and 9+10+…+16 (8 terms).
Section D (5 marks each)
32. (a) x⁴ + x² + 1 = (x² + 1)² − x² = (x² − x + 1)(x² + x + 1). (b) x³ + y³ − 12xy + 64 = x³ + y³ + 4³ − 3·x·y·4 = (x + y + 4)(…); since x + y = −4, the factor (x+y+4) = 0, so the value is 0. OR (a) (x−3)(x−4)/[5(x−4)(x+5)] = (x−3)/[5(x+5)]; (b) a²+b²+c² = 25 − 20 = 5, so a³+b³+c³−3abc = 5(5−10) = −25.
33. A cyclic parallelogram has opposite angles equal (parallelogram) and supplementary (cyclic) ⇒ each = 90° ⇒ rectangle; each diagonal subtends 90° so each is a diameter, and the diagonals (which bisect each other) meet at the centre. OR — opposite angles each = half the two arcs that make the whole circle ⇒ sum 180°; then (2x+10)+(3x−20)=180 ⇒ x = 38 ⇒ ∠A = 86°, ∠C = 94° (and ∠B, ∠D also sum to 180°).
34. (a) Joining the two centres and an intersection point gives a triangle with all sides = r (equilateral), so each half-angle is 60° and the arc of one circle between the intersection points subtends 2×60° = 120° at its centre. (b) Each outer arc is the remaining 360−120 = 240° = ⅔ of a circle, so boundary = 2 × (⅔ × 2πr) = (8πr)/3. OR — a semicircle on a side of length x has area πx²/8 ∝ x²; since a² + b² = c², πa²/8 + πb²/8 = πc²/8, proving the hypotenuse semicircle equals the sum of the other two.
35. Terms a, ar, ar²: a(1+r+r²)=26, a²(1+r²+r⁴)=364. Using 1+r²+r⁴=(1+r+r²)(1−r+r²): ar = 6 and a(1+r²)=20 ⇒ 3r²−10r+3=0 ⇒ r = 3 (or 1/3), a = 2 → terms 2, 6, 18. OR — a+5d=12 and a+7d=22 ⇒ d=5, a=−13 → first three −13, −8, −3; for the sum, n(n+1)/2 > 1000 ⇒ n(n+1) > 2000 ⇒ n = 45 (45×46 = 2070).
Section E (4 marks each)
36. (i) OA = 6, AB = 8. (ii) OB = √(6² + 8²) = 10. (iii) Since OA ⊥ AB, ΔOAB is a right-angled triangle (right angle at A). (iv) Perimeter = 6 + 8 + 10 = 24 units.
37. (i) AP, common difference 4. (ii) Number of tiles = 1 + (n−1)×4 = 4n − 3. (iii) Stage 20 = 4×20 − 3 = 77 tiles. (iv) 4n − 3 = 101 ⇒ 4n = 104 ⇒ n = 26 (stage 26).
38. 9 balls total. (i) P(red) = 4/9. (ii) P(not blue) = 7/9. (iii) different colours (no replacement): 1 − same-colour = 1 − [C(4,2)+C(3,2)+C(2,2)]/C(9,2) = 1 − 10/36 = 26/36 = 13/18. (iv) No yellow balls → P(yellow) = 0.