Class IX Mathematics (041)Annual Examination Solutions 2024-25 (ENGLISH MEDIUM)

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Class IX Mathematics (041)Annual Examination Solutions 2024-25

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SECTION – A (Multiple Choice Questions)

1. If (1, 2) is a solution of the equation x + 3y = p, then the value of p is:
Substitute x = 1 and y = 2 into the equation:
1 + 3(2) = p
1 + 6 = 7
p = 7

Answer: (c) 7
2. The coefficient of x2 in the polynomial 2x2 + 2x – 4 is:
The term containing x2 is 2x2.
The coefficient (number multiplying the variable) is 2.

Answer: (a) 2
3. If two given triangles ABC and XYZ are congruent (ΔABC ≅ ΔXYZ), which of the following statements is NOT correct?
In congruence ΔABC ≅ ΔXYZ, corresponding parts match: A↔X, B↔Y, C↔Z.
BC = YZ (Correct)
AC = XZ (Correct)
∠B = ∠Y (Correct)
AC = XY (Incorrect, because side AC corresponds to side XZ, not XY).

Answer: (d) AC = XY
4. √3 is a ______ number.

The square root of a non-perfect square (like 3) is irrational.

Answer: (d) an irrational
5. If the perimeter of an equilateral triangle is 180 cm, then its area will be: [cite: 175]
Side (a) = Perimeter / 3 = 180 / 3 = 60 cm.
Area = (√3 / 4) × a2 = (√3 / 4) × 602
= (√3 / 4) × 3600 = 900√3 cm2.

Answer: (b) 900√3 cm2
6. If an angle is equal to its supplement, then the angle is: [cite: 185]
Let the angle be x. Supplement is 180 – x.
x = 180 – x → 2x = 180 → x = 90°.

Answer: (b) 90°
7. The point (9, -7) lies in which of the following quadrants: [cite: 262]
x is positive (+), y is negative (-).
This sign combination (+, -) corresponds to the 4th Quadrant.

Answer: (d) IV
8. In ΔPRS and ΔQPT, the common angle is: [cite: 273]
Looking at the notation, vertex P is present in both ΔPRS and ΔQPT.
Geometrically, they share the angle at vertex P.

Answer: (b) ∠P
9. The number of Euclid’s axioms is/are: [cite: 281]

Euclid listed 7 axioms (common notions).

Answer: (b) 7
10.For polynomial P(x) = x3 + 1, the value of P(-1) is: [cite: 289]

P(-1) = (-1)3 + 1 = -1 + 1 = 0.

Answer: (d) 0
11. The sides of a triangle are in the ratio 2:3:5 and its perimeter is 800 cm. [cite_start]The length of the shortest side is: [cite: 297]
2x + 3x + 5x = 800 → 10x = 800 → x = 80.
Shortest side = 2x = 2(80) = 160 cm.

Answer: (c) 160 cm
12. [cite_start]The quadrilateral where diagonals are not equal but bisect each other at right angles is: [cite: 307]

A square has equal diagonals. A rhombus has unequal diagonals that bisect at 90°.

Answer: (c) Rhombus
13. Two similar conical birthday caps of equal volume have ratio of their radii as 1:2. [cite_start]The ratio of their heights is: [cite: 316]
Volume V is constant. r1 : r2 = 1 : 2 → r2 = 2r1.
(1/3)πr12h1 = (1/3)πr22h2
r12h1 = (2r1)2h2 → r12h1 = 4r12h2
h1 = 4h2 → h1 : h2 = 4:1.

Answer: (b) 4:1
14. [cite_start]In the class intervals 15-25, 25-35, 35-45 the number 35 is included in: [cite: 388]
In exclusive intervals, the upper limit is excluded.
35 is the upper limit of 25-35 and lower limit of 35-45. It belongs to 35-45.

Answer: (a) 35-45
15. (210)2/5 × [cite_start]5 is equal to: [cite: 397]
Power rule: 2(10 × 2/5) = 24 = 16.
Multiply by 5: 16 × 5 = 80.

Answer: (a) 80
16. [cite_start]The angle subtended by the diameter of a semicircle at the circumference is: [cite: 406]

The angle in a semicircle is a right angle.

Answer: (d) 90°
17. [cite_start]The number of zero/zeroes of the polynomial 7x + 14 is/are: [cite: 415]

This is a linear polynomial (degree 1), so it has exactly 1 zero.

Answer: (d) 1
18. A hemisphere dome of radius 3.5 m is to be painted outside at a rate of Rs 600/m2. What is the cost? [cite_start][cite: 424]
CSA = 2πr2 = 2 × (22/7) × 3.5 × 3.5 = 77 m2.
Cost = 77 × 600 = 46,200.

Answer: (a) 46200
19. Assertion (A): The line segment joining any two points on the circumference of a circle is called chord.
Reason (R): Radius is the longest chord. [cite_start][cite: 2283]
Assertion (A) is True.
Reason (R) is False (Diameter is the longest chord, not radius).

Answer: (c) A is true but R is false.
20. Assertion (A): If height is 24 cm and base diameter is 14 cm, slant height is 15 cm.
Reason (R): Height, radius, and slant height form a right triangle. [cite_start][cite: 2288]
r = 7, h = 24. l = √(72 + 242) = 25 cm.
Assertion says 15 cm, which is False. Reason is True.

Answer: (d) A is false but R is true.

SECTION – B (Very Short Answer)

21. If the ratio of the volume of two spheres is 8:27, find ratio of surface areas. [cite_start][cite: 2294]
Vol ratio = (r1/r2)3 = 8/27 → r1/r2 = 2/3.
Area ratio = (r1/r2)2 = (2/3)2 = 4/9.

Ratio = 4:9

OR: CSA and TSA of hemisphere with r=21 cm. [cite_start][cite: 2298]
CSA = 2πr2 = 2 × (22/7) × 21 × 21 = 2772 cm2.
TSA = 3πr2 = 4158 cm2.

22. [cite_start]Give one example each of a binomial of degree 3. [cite: 468]
Example: x3 + 2x
23. If ΔABC ≅ ΔDEF then: (i) ∠C = ? (ii) DF = ? [cite_start][cite: 470]
(i) ∠C = ∠F (CPCT)
(ii) DF = AC (CPCT)
24. [cite_start]Check whether -2 and 2 are zeroes of the polynomial x + 2. [cite: 478]
P(-2) = -2 + 2 = 0 (Yes).
P(2) = 2 + 2 = 4 (No).
25. Radius of cone = 8 cm, Height = 15 cm. Find Curved Surface Area (π = 3.14). [cite_start][cite: 479]
Slant height (l) = √(82 + 152) = √289 = 17 cm.
CSA = πrl = 3.14 × 8 × 17 = 427.04 cm2.

SECTION – C (Short Answer)

26. [cite_start]Simplify: (x + y + z)2 – (x – y – z)2 [cite: 483]
Use a2 – b2 = (a-b)(a+b).
Let A = (x+y+z) and B = (x-y-z).
A + B = 2x.
A – B = 2y + 2z.

Result: (2x)(2y + 2z) = 4xy + 4xz
27. Lines PQ and RS intersect at O. If ∠POR : ∠ROQ = 5:7, find all angles. [cite_start][cite: 484]
Sum = 180°. Ratio 5:7 means 5x + 7x = 180 → 12x = 180 → x = 15.
∠POR = 5(15) = 75°.
∠ROQ = 7(15) = 105°.

Angles are 75°, 105°, 75°, 105°.
28. (i) Name of horizontal and vertical lines in Cartesian plane? (ii) Name of intersection point? [cite_start][cite: 489]
(i) Horizontal: x-axis, Vertical: y-axis.
(ii) Intersection: Origin
OR (Graph Questions):
(i) Point (-3, -5) is E.
(ii) Ordinate of G is -4.
(iii) Abscissa of D is 6.
29. Prove that the diagonal divides a parallelogram into two congruent triangles. [cite_start][cite: 555]
In parallelogram ABCD with diagonal AC:
1. AB || DC → Alt. Interior angles equal.
2. AD || BC → Alt. Interior angles equal.
3. AC is common.

By ASA rule, ΔABC ≅ ΔCDA.
30. Find three rational numbers between 3/5 and 4/5. [cite_start][cite: 560]
Multiply numerator and denominator by 4.
3/5 = 12/20 and 4/5 = 16/20.

Numbers: 13/20, 14/20, 15/20.
31. Prove that equal chords of a circle subtend equal angles at the center. [cite_start][cite: 562]
Given chords AB = CD. Center O.
In ΔAOB and ΔCOD:
AO = CO (Radii), BO = DO (Radii), AB = CD (Given).

By SSS rule, ΔAOB ≅ ΔCOD. Thus, ∠AOB = ∠COD (CPCT).

SECTION – D (Long Answer)

32. Frequency Polygons & Data Analysis

Comparison: Section A has more students in higher mark ranges (peaks later and sustains). Section B peaks early at low marks (10-20) and drops.

Conclusion: Section A is better.
33. If a = (√5 – √3)/(√5 + √3) and b = (√5 + √3)/(√5 – √3), find a2 + b2.
Rationalizing: a = 4 – √15, b = 4 + √15.
a + b = 8.
ab = 16 – 15 = 1.
a2 + b2 = (a+b)2 – 2ab = 64 – 2.

Answer: 62
34. (a) Factorize: x2 – 22x + 120. (b) Find remainder when (3x4 – 4x3 – 3x – 1) is divided by (x – 1).
(a) Factors of 120 summing to -22 are -12 and -10. Result: (x – 12)(x – 10).
(b) Remainder Theorem: Put x = 1.
3(1) – 4(1) – 3(1) – 1 = 3 – 4 – 3 – 1 = -5.

Remainder: -5
35. Geometry Proof (AB=AD, ∠1=∠2, ∠3=∠4) [cite: 650]
1. Prove ΔABC ≅ ΔADC (SAS) using AB=AD, common AC, and ∠BAC=∠DAC (sum of equal angles).
2. Get ∠B = ∠D by CPCT.
3. Prove ΔABP ≅ ΔADQ (ASA) using AB=AD, ∠1=∠2, ∠B=∠D.

Result: AP = AQ (CPCT).

SECTION – E (Case Study)

36. Rectangular Park (90m x 74m) with Triangles (5, 12, 13) at corners/center.
(i) Semi-perimeter = (5+12+13)/2 = 15 m.
(ii) Rectangular Area = 90 × 74 = 6660 m2.
(iii) Flower Area = 5 triangles × (1/2 × 5 × 12) = 150 m2.
Grass Area = 6660 – 150 = 6510 m2.

OR Cost: 150 × 95 = Rs 14,250.
37. STD Booths (Equilateral setup on Circle). [cite_start][cite: 715]
(i) Equilateral Triangle.
(ii) Chord: PQ. Arc: PQ.
(iii) Angle at center is double the angle at circumference (60°). ∠QOR = 120°.

OR: If OQ bisects ∠PQR (60°), then ∠OQR = 30°.
38. Sweets Distribution (2 per child, 1 per adult).
(i) Equation: y = 2x + 8.
(ii) A linear equation in two variables has infinitely many solutions.
(iii) If y = 108: 108 = 2x + 8 → 100 = 2x → x = 50 children.

OR: Solutions for 3x + 2y = 9: (1, 3) and (3, 0).

 

 

Here are the solutions for the Optional/OR questions from your question paper in English:

Section B

[cite_start]

Question 21 (OR): Find the curved surface area and the total surface area of a hemisphere of radius 21 cm. [cite: 825]

  • Solution:
    • Given: Radius (r) = 21 cm.
    • Curved Surface Area (CSA):
      CSA = 2πr2
      = 2 × (22/7) × 21 × 21
      = 44 × 3 × 21
      = 2772 cm2
    • Total Surface Area (TSA):
      TSA = 3πr2
      = 3 × (22/7) × 21 × 21
      = 66 × 3 × 21
      = 4158 cm2

[cite_start]

Question 23 (OR): Two circles of the same _______ are congruent and two squares of the same _______ are congruent. [cite: 473, 476, 477]

  • Solution:
    • Two circles of the same radii are congruent.
    • Two squares of the same sides are congruent.

Section C

[cite_start]

Question 28 (OR): Based on the given graph answer the following questions [cite: 537]

 

[Image of cartesian plane graph with points]

  • Solution:
      1. [cite_start]

      1. Name the point represented by (-3, -5): [cite: 540]
        On the graph, the point at x = -3 and y = -5 is E.

    [cite_start]

      1. What is the ordinate of point G? [cite: 540]
        The coordinates of point G are (2, -4). The ordinate (y-value) is -4.

    [cite_start]

    1. What is the abscissa of point D? [cite: 541]
      The coordinates of point D are (6, 2). The abscissa (x-value) is 6.

Question 29 (OR): The diagonal AC of a parallelogram ABCD bisects ∠A. [cite_start]Show that it bisects ∠C also. [cite: 557, 558]

 

[Image of parallelogram diagonal bisecting angles]

  • Solution:
    • Given: ABCD is a parallelogram where AD || BC and AB || DC. Diagonal AC bisects ∠A, meaning ∠DAC = ∠BAC … (i)
    • To Prove: AC bisects ∠C (∠DCA = ∠BCA).
    • Proof:
      • Since AD || BC and AC is a transversal:
        ∠DAC = ∠BCA (Alternate Interior Angles) … (ii)
      • Since AB || DC and AC is a transversal:
        ∠BAC = ∠DCA (Alternate Interior Angles) … (iii)
      • From equation (i), we know ∠DAC = ∠BAC.
      • Therefore, using equations (ii) and (iii):
        ∠BCA = ∠DCA
    • Conclusion: Hence, diagonal AC bisects ∠C.

Section D

Question 32 (OR):
(i) [cite_start]Draw a histogram to represent the data. [cite: 641]
(ii) [cite_start]Find the difference of students spending lowest time and highest time on homework. [cite: 643]

  • Solution:
    • (i) Histogram: This needs to be drawn on graph paper. Take “Hours” on the x-axis (0-5, 5-10, etc.) and “Frequency” (Number of students) on the y-axis. Draw rectangles with heights corresponding to the table (0-5: 8, 5-10: 12, etc.).
    • (ii) Finding the difference:
        • [cite_start]

        • Students spending lowest time (0-5 hours) = 8 [cite: 591]

      [cite_start]

      • Students spending highest time (20-25 hours) = 5 [cite: 591]
      • Difference = 8 – 5 = 3

Question 35 (OR): In the given figure, ΔABC and ΔDEF are isosceles and ∠BAC = ∠EDF.
(i) Prove that ΔABC ≅ ΔDEF.
(ii) [cite_start]If ∠BAC = 30° then find ∠DFE. [cite: 660, 664, 665]

 

[Image of congruent isosceles triangles]

  • Solution:
    • (i) Proving Congruency:
      [cite_start]According to the markings in the figure[cite: 661]:

      • AB = DE (Given/Marked)
      • AC = DF (Given/Marked)
      • ∠BAC = ∠EDF (Given in question)
      • Therefore, by SAS (Side-Angle-Side) Congruence Rule:
        ΔABC ≅ ΔDEF
    • (ii) Value of ∠DFE:
      • Since ΔDEF is an isosceles triangle (DE = DF), angles opposite to equal sides are equal: ∠DEF = ∠DFE.
      • Given ∠BAC = 30° and triangles are congruent, then ∠EDF = 30°.
      • Using the Angle Sum Property of a triangle:
        ∠EDF + ∠DEF + ∠DFE = 180°
        30° + ∠DFE + ∠DFE = 180°
        2∠DFE = 150°
        ∠DFE = 75°

Section E (Case Study)

Question 36 (OR): The cost of planting the flowering plant is ₹ 95 per m2. [cite_start]Find the total cost of planting the flowering plants. [cite: 714]

  • Solution:
      [cite_start]

    • There are 4 corners and 1 center, so a total of 5 triangular areas for flowers[cite: 703].
    • Sides of the triangle: 5 m, 12 m, 13 m. This is a right-angled triangle (52 + 122 = 25 + 144 = 169 = 132).
    • Area of one triangle = (1/2) × base × height = (1/2) × 5 × 12 = 30 m2.
    • Total area for 5 triangles = 5 × 30 = 150 m2.
    • Total Cost = Area × Rate
      = 150 × 95
      = 14,250
    • Answer: ₹ 14,250

[cite_start]

Question 37 (OR): If OQ is the bisector of ∠PQR then find the value of ∠OQR. [cite: 746]

  • Solution:
      [cite_start]

    • According to the question, the three booths (P, Q, R) are equidistant, which means ΔPQR is an Equilateral Triangle[cite: 717].
    • Each angle of an equilateral triangle is 60°. Therefore, ∠PQR = 60°.
    • Given that OQ is the bisector of ∠PQR.
    • ∠OQR = (1/2) × ∠PQR
      = (1/2) × 60° = 30°
    • Answer: 30°

[cite_start]

Question 38 (OR): Find two solutions of the equation 3x + 2y = 9. [cite: 765]

  • Solution:
    1. Let x = 1:
      3(1) + 2y = 9 ⇒ 3 + 2y = 9 ⇒ 2y = 6 ⇒ y = 3
      First solution: (1, 3)
    2. Let y = 0:
      3x + 2(0) = 9 ⇒ 3x = 9 ⇒ x = 3
      Second solution: (3, 0)

 

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