CBSE Class 10 Maths (Standard) Question Paper 2025 Solved Code 30-1-3

SECTION A

1. What is the mode of a data if median and mean of the same data are 9·6 and 10·5, respectively ? [1 Mark]
(A) 7·8
(B) 12·3
(C) 8·4
(D) 7

Answer: (A) 7·8

Teacher's Note:
a) Use the empirical formula Mode \( = 3(\text{Median}) - 2(\text{Mean}) \).
b) Substitute values carefully: \( 3(9.6) - 2(10.5) = 28.8 - 21 = 7.8 \).

2. The value of \( (\tan A \, \text{cosec}\, A)^2 - (\sin A \sec A)^2 \) is : [1 Mark]
(A) 0
(B) 1
(C) -1
(D) 2

Answer: (B) 1

Teacher's Note:
a) Simplify each term: \( \tan A \, \text{cosec}\, A = \sec A \) and \( \sin A \sec A = \tan A \).
b) Then use the identity \( \sec^2 A - \tan^2 A = 1 \).

3. A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of \( 30^{\circ} \) to the horizontal. The length of the string is : [1 Mark]
(A) \( 100\sqrt{3} \) m
(B) 300 m
(C) \( 150\sqrt{2} \) m
(D) \( 150\sqrt{3} \) m

Answer: (B) 300 m

Teacher's Note:
a) Use \( \sin\theta = \dfrac{\text{height}}{\text{string length}} \).
b) \( \sin 30^{\circ} = \dfrac{1}{2} \), so length \( = \dfrac{150}{1/2} = 300 \) m.

4. In triangles ABC and DEF, \( \angle B = \angle E \), \( \angle F = \angle C \) and AB \( = 3 \) DE. Then, the two triangles are : [1 Mark]
(A) congruent but not similar
(B) congruent as well as similar
(C) neither congruent nor similar
(D) similar but not congruent

Answer: (D) similar but not congruent

Teacher's Note:
a) Equal corresponding angles give similarity by AA criterion.
b) Since sides are not equal (AB = 3DE), triangles cannot be congruent.

5. If \( \theta \) is an acute angle and \( 7 + 4\sin\theta = 9 \), then the value of \( \theta \) is : [1 Mark]
(A) \( 90^{\circ} \)
(B) \( 30^{\circ} \)
(C) \( 45^{\circ} \)
(D) \( 60^{\circ} \)

Answer: (B) \( 30^{\circ} \)

Teacher's Note:
a) Solve \( \sin\theta = \dfrac{2}{4} = \dfrac{1}{2} \).
b) Recall \( \sin 30^{\circ} = \dfrac{1}{2} \).

6. Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is : [1 Mark]
(A) 3
(B) 5
(C) 2
(D) 4

[Figure: A graph showing two curves - one appears to be an oval/closed curve intersecting the x-axis at two points, and another curve (an upward and downward opening parabola pair) crossing the x-axis. The curves intersect the x-axis at a total of 2 distinct points.]

Answer: (C) 2

Teacher's Note:
a) Zeroes of a polynomial are the points where its graph cuts the x-axis.
b) Count only the distinct x-axis crossing points shown in the figure.

7. In the given figure, PA is a tangent from an external point P to a circle with centre O. If \( \angle POB = 115^{\circ} \), then \( \angle APO \) is equal to : [1 Mark]
(A) \( 25^{\circ} \)
(B) \( 65^{\circ} \)
(C) \( 90^{\circ} \)
(D) \( 35^{\circ} \)

[Figure: A circle with centre O. P is an external point with tangent PA touching the circle at A. B is a point on the circle such that O, B and the line from P pass through, with angle POB = 115 degrees marked at the centre.]

Answer: (A) \( 25^{\circ} \)

Teacher's Note:
a) \( \angle POA = 180^{\circ} - 115^{\circ} = 65^{\circ} \) since A, O, B are collinear through the diameter.
b) In right triangle OAP, \( \angle OAP = 90^{\circ} \), so \( \angle APO = 90^{\circ} - 65^{\circ} = 25^{\circ} \).

8. A piece of wire 20 cm long is bent into the form of an arc of a circle of radius \( \dfrac{60}{\pi} \) cm. The angle subtended by the arc at the centre of the circle is : [1 Mark]
(A) \( 30^{\circ} \)
(B) \( 60^{\circ} \)
(C) \( 90^{\circ} \)
(D) \( 50^{\circ} \)

Answer: (B) \( 60^{\circ} \)

Teacher's Note:
a) Use arc length formula \( l = \dfrac{\theta}{360^{\circ}} \times 2\pi r \).
b) Substituting values gives \( \theta = 60^{\circ} \).

9. If HCF(98, 28) = m and LCM(98, 28) = n, then the value of \( n - 7m \) is : [1 Mark]
(A) 0
(B) 28
(C) 98
(D) 198

Answer: (C) 98

Teacher's Note:
a) HCF(98,28) = 14, LCM(98,28) = 196.
b) \( n - 7m = 196 - 7(14) = 196 - 98 = 98 \).

10. Which of the following is a rational number between \( \sqrt{3} \) and \( \sqrt{5} \) ? [1 Mark]
(A) 1·4142387954012 ....
(B) \( 2.32\overline{6} \)
(C) \( \pi \)
(D) 1·857142

Answer: (D) 1·857142

Teacher's Note:
a) \( \sqrt{3} \approx 1.732 \) and \( \sqrt{5} \approx 2.236 \).
b) A rational number is a terminating or repeating decimal; 1.857142 lies between and is rational.

11. The sum of the zeroes of the polynomial \( p(x) = 5x - 7x^2 + 3 \) is : [1 Mark]
(A) \( \dfrac{-7}{5} \)
(B) \( \dfrac{7}{5} \)
(C) \( \dfrac{5}{7} \)
(D) \( \dfrac{-5}{7} \)

Answer: (C) \( \dfrac{5}{7} \)

Teacher's Note:
a) Write in standard form: \( -7x^2 + 5x + 3 \).
b) Sum of zeroes \( = \dfrac{-b}{a} = \dfrac{-5}{-7} = \dfrac{5}{7} \).

12. If x = 1 and y = 2 is a solution of the pair of linear equations \( 2x - 3y + a = 0 \) and \( 2x + 3y - b = 0 \), then : [1 Mark]
(A) a = 2b
(B) 2a = b
(C) a + 2b = 0
(D) 2a + b = 0

Answer: (B) 2a = b

Teacher's Note:
a) Substitute x=1, y=2 in both equations to get a=4 and b=8.
b) Check the options; \( 2a = 8 = b \) holds true.

13. If a sector of a circle has an area of \( 40\pi \) sq. units and a central angle of \( 72^{\circ} \), the radius of the circle is : [1 Mark]
(A) 200 units
(B) 100 units
(C) 20 units
(D) \( 10\sqrt{2} \) units

Answer: (D) \( 10\sqrt{2} \) units

Teacher's Note:
a) Use \( \text{Area} = \dfrac{\theta}{360^{\circ}} \times \pi r^2 \).
b) Solve \( 40\pi = \dfrac{72}{360}\pi r^2 \Rightarrow r^2 = 200 \Rightarrow r = 10\sqrt{2} \).

14. The tangents drawn at the extremities of the diameter of a circle are always : [1 Mark]
(A) parallel
(B) perpendicular
(C) equal
(D) intersecting

Answer: (A) parallel

Teacher's Note:
a) Tangent at any point is perpendicular to the radius at that point.
b) Since the diameter's two radii are collinear (opposite directions), the tangents are parallel.

15. If \( (-1)^n + (-1)^8 = 0 \), then n is : [1 Mark]
(A) any positive integer
(B) any negative integer
(C) any odd number
(D) any even number

Answer: (C) any odd number

Teacher's Note:
a) \( (-1)^8 = 1 \), so \( (-1)^n = -1 \).
b) This is true only when n is odd.

16. The end points of a diameter of circle are (2, 4) and (-3, -1). The length of its radius is : [1 Mark]
(A) \( \dfrac{5\sqrt{2}}{2} \) units
(B) \( 5\sqrt{2} \) units
(C) \( 3\sqrt{2} \) units
(D) \( \pm\dfrac{5\sqrt{2}}{2} \) units

Answer: (A) \( \dfrac{5\sqrt{2}}{2} \) units

Teacher's Note:
a) Find diameter using distance formula: \( \sqrt{(2-(-3))^2+(4-(-1))^2} = \sqrt{50} = 5\sqrt{2} \).
b) Radius is half of diameter: \( \dfrac{5\sqrt{2}}{2} \).

17. The 11^th and 13^th term of an AP are 39 and 45, respectively. What is the common difference of the AP ? [1 Mark]
(A) 42
(B) 21
(C) 6
(D) 3

Answer: (D) 3

Teacher's Note:
a) \( a_{13} - a_{11} = 2d \).
b) \( 45 - 39 = 6 = 2d \Rightarrow d = 3 \).

18. A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a spade or a king ? [1 Mark]
(A) \( \dfrac{1}{13} \)
(B) \( \dfrac{2}{13} \)
(C) \( \dfrac{4}{13} \)
(D) \( \dfrac{9}{13} \)

Answer: (C) \( \dfrac{4}{13} \)

Teacher's Note:
a) Number of spades = 13, kings = 4, but king of spades is common (counted once).
b) Favourable outcomes = \( 13+4-1=16 \), so probability \( = \dfrac{16}{52} = \dfrac{4}{13} \).

19. Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1.
Reason (R) : For any event E, if P(E) = 1, then E is called a sure event. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) Selecting any number from a defined finite set always has probability 1 (sure event).
b) The Reason correctly explains why the Assertion is true.

20. Assertion (A) : If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason (R) : Total Surface Area of a sphere of radius r is \( 3\pi r^2 \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Joining two hemispheres along the same base does form a sphere.
b) The correct TSA of a sphere is \( 4\pi r^2 \), not \( 3\pi r^2 \), so Reason is false.

SECTION B

21. (a) If \( \triangle ABC \sim \triangle PQR \) in which AB = 6 cm, BC = 4 cm, AC = 8 cm and PR = 6 cm, then find the length of (PQ + QR). [2 Marks]

Answer:
1. Since \( \triangle ABC \sim \triangle PQR \), \( \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} \), i.e. \( \dfrac{6}{PQ} = \dfrac{4}{QR} = \dfrac{8}{6} \).
2. Solving gives \( PQ = 4.5 \) cm and \( QR = 3 \) cm, so \( PQ + QR = 7.5 \) cm.

Teacher's Note:
a) Write the correct correspondence of vertices before setting up ratios.
b) Cross-multiply carefully to avoid calculation errors.

OR

(b) In the given figure, \( \dfrac{QR}{QS} = \dfrac{QT}{PR} \) and \( \angle 1 = \angle 2 \), show that \( \triangle PQS \sim \triangle TQR \). [2 Marks]

[Figure: Triangle QPR with T above P, S on QR between Q and R. Angle 1 is at Q (angle PQS) and angle 2 is at R (angle TRQ, i.e. angle 2 marked near R).]

Answer:
1. In \( \triangle PQR \), since \( \angle 1 = \angle 2 \), PR = PQ (sides opposite equal angles).
2. Given \( \dfrac{QR}{QS} = \dfrac{QT}{PR} \), substituting PR = PQ gives \( \dfrac{QR}{QS} = \dfrac{QT}{PQ} \).
3. Also \( \angle PQS = \angle TQR \) (common angle), so by SAS similarity, \( \triangle PQS \sim \triangle TQR \).

Teacher's Note:
a) First establish PR = PQ using the isosceles triangle property.
b) Use SAS similarity criterion with the common angle.

22. (a) If \( x\cos 60^{\circ} + y\cos 0^{\circa} + \sin 30^{\circ} - \cot 45^{\circ} = 5 \), then find the value of \( x + 2y \). [2 Marks]

Answer:
1. Substitute values: \( x\left(\dfrac{1}{2}\right) + y(1) + \dfrac{1}{2} - 1 = 5 \).
2. Simplify: \( \dfrac{x}{2} + y - \dfrac{1}{2} = 5 \Rightarrow x + 2y = 11 \).

Teacher's Note:
a) Memorise standard trigonometric ratio values for \( 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ} \).
b) Multiply through by 2 to clear the fraction before solving.

OR

(b) Evaluate : \( \dfrac{\tan^2 60^{\circ}}{\sin^2 60^{\circ} + \cos^2 30^{\circ}} \) [2 Marks]

Answer:
1. \( \tan 60^{\circ} = \sqrt{3} \), \( \sin 60^{\circ} = \cos 30^{\circ} = \dfrac{\sqrt{3}}{2} \).
2. So expression \( = \dfrac{3}{\frac{3}{4}+\frac{3}{4}} = \dfrac{3}{\frac{3}{2}} = 2 \).

Teacher's Note:
a) Square each trigonometric value carefully before adding.
b) Simplify the fraction step by step to avoid sign errors.

23. A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground. [2 Marks]

[Figure: A circle with centre O, external point P at distance 26 m from O. Two tangent lines PA and PB, each 10 m long, touch the circle at points A and B respectively.]

Answer:
1. Since PA is a tangent, \( \angle OAP = 90^{\circ} \).
2. In right \( \triangle OAP \): \( OP^2 = OA^2 + PA^2 \Rightarrow 26^2 = OA^2 + 10^2 \).
3. \( OA^2 = 676 - 100 = 576 \Rightarrow OA = 24 \) m. So the radius is 24 m.

Teacher's Note:
a) Radius is always perpendicular to tangent at the point of contact.
b) Apply the Pythagoras theorem in the right triangle formed.

24. Find the zeroes of the polynomial \( p(x) = x^2 + \dfrac{4}{3}x - \dfrac{4}{3} \). [2 Marks]

Answer:
1. Multiply through by 3: \( \dfrac{1}{3}(3x^2+4x-4) = \dfrac{1}{3}(3x-2)(x+2) \).
2. Setting each factor to zero gives zeroes \( \dfrac{2}{3} \) and \( -2 \).

Teacher's Note:
a) Clear fractions first to simplify factorisation.
b) Verify zeroes by substituting back into the original polynomial.

25. Find the length of the median through the vertex B of \( \triangle ABC \) with vertices A(9, -2), B(-3, 7) and C(-1, 10). [2 Marks]

Answer:
1. Midpoint of AC \( = \left(\dfrac{9-1}{2}, \dfrac{-2+10}{2}\right) = (4, 4) \).
2. Length of median \( = \sqrt{(4-(-3))^2+(4-7)^2} = \sqrt{49+9} = \sqrt{58} \) units.

Teacher's Note:
a) A median connects a vertex to the midpoint of the opposite side.
b) Use the distance formula carefully with correct sign handling.

SECTION C

26. Prove that \( \sqrt{5} \) is an irrational number. [3 Marks]

Answer:
1. Assume \( \sqrt{5} \) is rational, so \( \sqrt{5} = \dfrac{p}{q} \), where p and q are co-prime integers, \( q \neq 0 \).
2. Then \( 5q^2 = p^2 \), so \( p^2 \) is divisible by 5, hence p is divisible by 5. Let \( p = 5a \).
3. Substituting gives \( 25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \), so q is also divisible by 5, contradicting that p, q are co-prime. Hence \( \sqrt{5} \) is irrational.

Teacher's Note:
a) This is a proof by contradiction; state the assumption clearly.
b) Show both p and q share a common factor 5 to reach the contradiction.

27. Two dice are rolled together. Find the probability of getting :
(i) a multiple of 2 on one and a multiple of 3 on the other die.
(ii) the product of two numbers on the top of the two dice is a perfect square number. [3 Marks]

Answer:
1. Total outcomes when two dice are rolled = 36.
2. (i) Favourable outcomes: (2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6) = 11 outcomes, so \( P(E) = \dfrac{11}{36} \).
3. (ii) Favourable outcomes with perfect square product: (1,1),(2,2),(3,3),(1,4),(4,1),(4,4),(5,5),(6,6) = 8 outcomes, so \( P(E) = \dfrac{8}{36} = \dfrac{2}{9} \).

Teacher's Note:
a) List all favourable outcomes systematically to avoid missing cases.
b) Remember total outcomes for two dice is always 36.

28. (a) Prove that : \( \dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\, \text{cosec}\,\theta \) [3 Marks]

Answer:
1. LHS \( = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)} \).
2. Combine over common denominator: \( = \dfrac{1}{\sin\theta-\cos\theta}\left[\dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta}\right] \).
3. Using \( a^3-b^3=(a-b)(a^2+ab+b^2) \), this simplifies to \( \dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta} = 1 + \sec\theta\,\text{cosec}\,\theta \) = RHS.

Teacher's Note:
a) Convert tan and cot to sin/cos form first.
b) Use the identity for difference of cubes to simplify the numerator.

OR

(b) Prove that : \( \dfrac{\sin A + \cos A}{\sin A - \cos A} + \dfrac{\sin A - \cos A}{\sin A + \cos A} = \dfrac{2}{2\sin^2 A - 1} \) [3 Marks]

Answer:
1. LHS \( = \dfrac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \).
2. Numerator simplifies to \( 2\sin^2A+2\cos^2A = 2 \); denominator is \( \sin^2A-\cos^2A \).
3. So LHS \( = \dfrac{2}{\sin^2A-(1-\sin^2A)} = \dfrac{2}{2\sin^2A-1} \) = RHS.

Teacher's Note:
a) Expand squares using \( (a+b)^2 \) and \( (a-b)^2 \) formulas.
b) Replace \( \cos^2A \) with \( 1-\sin^2A \) to match the RHS form.

29. A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains \( \dfrac{1408}{21} \) m³ of air, find the height of the cylindrical part. (Use \( \pi = \dfrac{22}{7} \)). [3 Marks]

Answer:
1. Let radius = r, height of cylinder h = 2r.
2. Total volume \( = \dfrac{2}{3}\pi r^3 + \pi r^2 h = \dfrac{2}{3}\pi r^3 + 2\pi r^3 = \dfrac{8}{3}\pi r^3 \).
3. \( \dfrac{1408}{21} = \dfrac{8}{3}\times\dfrac{22}{7}\times r^3 \Rightarrow r^3 = 8 \Rightarrow r = 2 \) m, so \( h = 4 \) m.

Teacher's Note:
a) Express hemisphere radius in terms of cylinder height before setting up the volume equation.
b) Simplify the constants carefully to isolate \( r^3 \).

30. (a) In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that \( \angle BAC + \angle ACD = 90^{\circ} \). [3 Marks]

[Figure: A circle with centre O. Point A on the circle, chord AC drawn, with OA and OC as radii (dashed line from O to C). BCD is a straight tangent line touching the circle at C, with B to the left and D to the right of C. P is another point on the circle on line PA.]

Answer:
1. In \( \triangle OAC \), OA = OC (radii), so \( \angle OCA = \angle OAC \).
2. Since BCD is tangent at C, \( \angle OCD = 90^{\circ} \), so \( \angle OCA + \angle ACD = 90^{\circ} \).
3. Substituting \( \angle OCA = \angle OAC = \angle BAC \), we get \( \angle BAC + \angle ACD = 90^{\circ} \).

Teacher's Note:
a) Use the isosceles triangle property for equal radii to get equal base angles.
b) Remember tangent is always perpendicular to the radius at the point of contact.

OR

(b) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. [3 Marks]

Answer:
1. Draw a quadrilateral ABCD circumscribing a circle with centre O, touching the sides at P, Q, R, S. Join O to each vertex and to each point of contact.
2. By congruent triangles (e.g. \( \triangle OAP \cong \triangle OAS \)), pairs of angles at O are equal: \( \angle 1 = \angle 2 \), \( \angle 3 = \angle 4 \), \( \angle 5 = \angle 6 \), \( \angle 7 = \angle 8 \).
3. Since all 8 angles sum to \( 360^{\circ} \), we get \( 2(\angle 1+\angle 4+\angle 5+\angle 8) = 360^{\circ} \), so \( \angle AOB + \angle COD = 180^{\circ} \), and similarly \( \angle BOC + \angle AOD = 180^{\circ} \).

Teacher's Note:
a) Draw a clear, labelled figure with all points of contact marked.
b) Use tangent-length congruent triangles to establish equal angle pairs.

31. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection. [3 Marks]

Answer:
1. Let the ratio be k:1, and the point on the y-axis be P(0, y). Using section formula for x-coordinate: \( 0 = \dfrac{-k+5}{k+1} \Rightarrow k = 5 \).
2. So the ratio is 5:1.
3. y-coordinate: \( y = \dfrac{-4(5)-6}{5+1} = \dfrac{-26}{6} = -\dfrac{13}{3} \). Point of intersection is \( \left(0, -\dfrac{13}{3}\right) \).

Teacher's Note:
a) The point on the y-axis always has x-coordinate 0.
b) Apply the section formula for both coordinates separately.

SECTION D

32. (a) The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle. [5 Marks]

Answer:
1. Let the other two sides be x cm and y cm. Given \( x+y+25=60 \Rightarrow y=35-x \).
2. By Pythagoras theorem: \( x^2+y^2=625 \).
3. Substituting: \( x^2+(35-x)^2=625 \Rightarrow x^2-35x+300=0 \).
4. Factorising: \( (x-20)(x-15)=0 \Rightarrow x=20 \) or \( x=15 \).
5. So the sides are 15 cm and 20 cm.

Teacher's Note:
a) Form two equations: perimeter equation and Pythagoras equation.
b) Solve the resulting quadratic carefully by factorisation.

OR

(b) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train. [5 Marks]

Answer:
1. Let the speed of the train be x km/h. Reduced speed = (x-8) km/h.
2. Time equation: \( \dfrac{480}{x-8} - \dfrac{480}{x} = 3 \).
3. Simplify: \( 480x - 480(x-8) = 3x(x-8) \Rightarrow 3840 = 3x^2-24x \).
4. So \( x^2-8x-1280=0 \Rightarrow (x-40)(x+32)=0 \Rightarrow x=40 \).
5. Hence the speed of the train is 40 km/h.

Teacher's Note:
a) Set up the time difference equation carefully using \( \text{Time}=\dfrac{\text{Distance}}{\text{Speed}} \).
b) Reject the negative root as speed cannot be negative.

33. A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of 24. If three times the number of red balls exceeds the number of blue balls by 20, find the number of red and blue balls. [5 Marks]

Answer:
1. Let number of red balls = x and blue balls = y.
2. From the first condition: \( \dfrac{10x}{100}+\dfrac{20y}{100}=24 \Rightarrow x+2y=240 \).
3. From the second condition: \( 3x-y=20 \).
4. Solving these two equations simultaneously gives \( x=40 \) and \( y=100 \).
5. So there are 40 red balls and 100 blue balls.

Teacher's Note:
a) Translate percentages into fractions carefully before forming equations.
b) Use substitution or elimination method to solve the linear system.

34. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm): 118-126 | 127-135 | 136-144 | 145-153 | 154-162 | 163-171 | 172-180
Number of Leaves: 3 | 5 | 9 | 12 | 5 | 4 | 2
Find the median length of the leaves. [5 Marks]

Answer:
1. Convert to continuous class intervals (exclusive form): 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5, with cumulative frequencies 3, 8, 17, 29, 34, 38, 40.
2. \( n=40 \), so \( \dfrac{n}{2}=20 \); this lies in the class 144.5-153.5 (median class), where cf before = 17, f = 12, h = 9.
3. Median \( = 144.5 + \dfrac{20-17}{12}\times 9 = 144.5+2.25=146.75 \) mm.

Teacher's Note:
a) Always convert inclusive class intervals to exclusive form before finding cumulative frequency.
b) Identify the median class where cumulative frequency first exceeds \( n/2 \).

35. (a) The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF \( \times \) EF = FB \( \times \) FA. [5 Marks]

[Figure: Parallelogram ABCD with diagonal BD. E is a point on side BC, and line segment AE intersects BD at point F.]

Answer:
1. In \( \triangle ADF \) and \( \triangle EBF \): \( \angle DFA = \angle EFB \) (vertically opposite angles), and \( \angle ADF = \angle FBE \) (alternate interior angles, since AD parallel to BC).
2. By AA similarity, \( \triangle ADF \sim \triangle EBF \).
3. So \( \dfrac{DF}{FB} = \dfrac{FA}{EF} \), which gives \( DF \times EF = FB \times FA \).

Teacher's Note:
a) Use the property that opposite sides of a parallelogram are parallel to identify equal alternate angles.
b) Vertically opposite angles at F are always equal.

OR

(b) In \( \triangle ABC \), if AD \( \perp \) BC and \( AD^2 = BD \times DC \), then prove that \( \angle BAC = 90^{\circ} \). [5 Marks]

[Figure: Triangle ABC with AD perpendicular to BC, D lying on BC between B and C.]

Answer:
1. Given \( AD^2 = BD\times DC \), rewrite as \( \dfrac{AD}{DC} = \dfrac{BD}{AD} \).
2. Since \( \angle ADB = \angle ADC = 90^{\circ} \), by SAS similarity \( \triangle DBA \sim \triangle DAC \).
3. So \( \angle DBA = \angle DAC \) and \( \angle BAD = \angle DCA \).
4. Adding: \( \angle DBA + \angle DCA = \angle DAC + \angle BAD \), which gives \( \angle BAC = 90^{\circ} \) (since angles of triangle ABC sum to \( 180^{\circ} \)).

Teacher's Note:
a) Use the given relation to set up a ratio suitable for similarity.
b) Add the two pairs of equal angles to reach the required angle sum result.

SECTION E

Case Study - 1

36. Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be \( 60^{\circ} \). Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be \( 45^{\circ} \).

[Figure: A right-angled figure showing points A (ground level), B (40 m above A, observation deck base), E (base of lighthouse), D (point on lighthouse level with B), C (top of lighthouse). AB = 40 m (vertical), angle at B (angle DBC) = 45 degrees, angle at A (angle EAC) = 60 degrees. CD = h metres. BD and AE are horizontal.]

(i) If CD is h metres, find the distance BD in terms of 'h'. [1 Mark]

Answer: Using \( \tan 45^{\circ}=\dfrac{h}{BD}=1 \), we get BD = h metres.

Teacher's Note:
a) \( \tan 45^{\circ}=1 \), so the opposite and adjacent sides are equal.
b) This gives a simple direct relation between BD and h.

(ii) Find distance BC in terms of 'h'. [1 Mark]

Answer: Using \( \sin 45^{\circ}=\dfrac{h}{BC}=\dfrac{1}{\sqrt{2}} \), we get \( BC=\sqrt{2}\,h \) metres.

Teacher's Note:
a) BC is the hypotenuse of the right triangle BDC.
b) Use \( \sin 45^{\circ} = \dfrac{1}{\sqrt{2}} \) to relate BC and h.

(iii) (a) Find the height CE of the lighthouse [Use \( \sqrt{3} = 1.73 \)] [2 Marks]

Answer:
1. \( \tan 60^{\circ} = \dfrac{h+40}{h} = \sqrt{3} \Rightarrow h(\sqrt{3}-1)=40 \Rightarrow h=\dfrac{40}{\sqrt{3}-1}=20(\sqrt{3}+1) \).
2. \( h = 20\times 2.73 = 54.6 \) m, so \( CE = h+40 = 54.6+40 = 94.6 \) m.

Teacher's Note:
a) Rationalise the denominator to simplify the expression for h.
b) Remember CE is the full height including the 40 m already climbed.

OR

(iii) (b) Find distance AE, if AC = 100 m. [2 Marks]

Answer:
1. Using \( \cos 60^{\circ} = \dfrac{AE}{AC} = \dfrac{1}{2} \).
2. \( AE = \dfrac{100}{2} = 50 \) m.

Teacher's Note:
a) AC is the hypotenuse in right triangle AEC.
b) Use \( \cos 60^{\circ}=\dfrac{1}{2} \) directly for a quick solution.

Case Study - 2

37. A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.

(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. [1 Mark]

Answer: The AP is 300, 350, 400, ...; \( a_4 = 450 \) m, \( a_5 = 500 \) m, \( a_6 = 550 \) m.

Teacher's Note:
a) First term a = 300, common difference d = 50.
b) Use \( a_n = a+(n-1)d \) to find each term.

(ii) Determine the distance of the 8^th round. [1 Mark]

Answer: \( a_8 = 300+7\times50 = 650 \) m.

Teacher's Note:
a) Substitute n=8 in the nth term formula.
b) Double check by counting terms from the first round.

(iii) (a) Find the total distance run after completing all 10 rounds. [2 Marks]

Answer:
1. \( S_{10} = \dfrac{10}{2}\left[2(300)+9(50)\right] = 5(600+450) \).
2. \( S_{10} = 5\times1050 = 5250 \) m.

Teacher's Note:
a) Use the sum formula \( S_n=\dfrac{n}{2}[2a+(n-1)d] \).
b) Substitute n=10 carefully, not n=9.

OR

(iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? [2 Marks]

Answer:
1. \( S_6 = \dfrac{6}{2}\left[2(300)+5(50)\right] = 3(600+250) \).
2. \( S_6 = 3\times850 = 2550 \) m.

Teacher's Note:
a) Use n=6 in the sum formula for the first 6 rounds only.
b) Double-check arithmetic; small mistakes in multiplication are common here.

Case Study - 3

38. A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs.
One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure.

[Figure: A circle divided into 10 equal sectors by 5 diameters, each sector decorated with a spiral/swirl design. Points B, C, and A are marked close together near the top-right, indicating adjacent sector boundary points on the circle.]

(i) Find the central angle of each sector. [1 Mark]

Answer: Central angle \( = \dfrac{360^{\circ}}{10} = 36^{\circ} \).

Teacher's Note:
a) Equal sectors divide the full angle equally.
b) \( 360^{\circ} \) divided by the number of sectors gives the central angle.

(ii) Find the length of the arc ACB. [1 Mark]

Answer: Arc length \( = \dfrac{1}{10}\times 2\times\dfrac{22}{7}\times\dfrac{35}{2} = 11 \) mm.

Teacher's Note:
a) Use arc length formula \( l=\dfrac{\theta}{360^{\circ}}\times 2\pi r \).
b) Here the arc corresponds to one sector (\( 36^{\circ} \) out of \( 360^{\circ} \)).

(iii) (a) Find the area of each sector of the brooch. [2 Marks]

Answer:
1. Area of sector \( = \dfrac{1}{10}\times\dfrac{22}{7}\times\dfrac{35}{2}\times\dfrac{35}{2} \).
2. \( = \dfrac{385}{4} = 96.25 \) mm².

Teacher's Note:
a) Use \( \text{Area of sector} = \dfrac{\theta}{360^{\circ}}\times\pi r^2 \).
b) Keep the radius as a fraction to avoid rounding errors before the final step.

OR

(iii) (b) Find the total length of the silver wire used. [2 Marks]

Answer:
1. Circumference of circle \( = 2\times\dfrac{22}{7}\times\dfrac{35}{2} = 110 \) mm.
2. Length of 5 diameters \( = 5\times35 = 175 \) mm.
3. Total wire length \( = 110+175 = 285 \) mm.

Teacher's Note:
a) The wire forms both the circle's circumference and the 5 diameters.
b) Add both parts together for the total wire length.

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