CBSE Class 10 Maths (Standard) Question Paper 2025 Solved Code 30-1-2

SECTION A

1. 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. PB is a line through O meeting the circle at B, with \( \angle POB = 115^{\circ} \) marked at the centre.]

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

Teacher's Note:
a) Since PA is tangent, \( \angle OAP = 90^{\circ} \).
b) \( \angle AOP = 180^{\circ} - 115^{\circ} = 65^{\circ} \), so \( \angle APO = 180^{\circ} - 90^{\circ} - 65^{\circ} = 25^{\circ} \).

2. A piece of wire 20 cm long is bent into the form of an arc of a circle of radius \( \frac{60}{\pi} \) 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 \( l = \frac{\theta}{360^{\circ}} \times 2\pi r \).
b) Substituting values gives \( \theta = 60^{\circ} \).

3. Three numbers in AP have the sum 30. What is its middle term ? [1 Mark]
(A) 4
(B) 10
(C) 16
(D) 8

Answer: (B) 10

Teacher's Note:
a) For three terms in AP, sum = \( 3 \times \) middle term.
b) So middle term \( = \frac{30}{3} = 10 \).

4. An arc of a circle is of length \( 5\pi \) cm and the sector it bounds has an area of \( 20\pi \) cm². Its radius is : [1 Mark]
(A) 10 cm
(B) 1 cm
(C) 5 cm
(D) 8 cm

Answer: (D) 8 cm

Teacher's Note:
a) Use Area \( = \frac{1}{2} \times r \times l \).
b) \( 20\pi = \frac{1}{2} \times r \times 5\pi \Rightarrow r = 8 \) cm.

5. 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 to get a=4, b=8.
b) Check each option; 2a=b holds true (2×4=8).

6. Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is : [1 Mark]

[Figure: A graph showing two curves, one an oval-like shape and another an X-shaped pair of lines, intersecting the x-axis at two common points.]
(A) 3
(B) 5
(C) 2
(D) 4

Answer: (C) 2

Teacher's Note:
a) Zeroes are the points where the graphs cut the x-axis.
b) Count only distinct points common to both curves on the x-axis.

7. If \( \alpha + \beta = 90^{\circ} \) and \( \alpha = 2\beta \), then \( \cos^2 \alpha + \sin^2 \beta \) is equal to : [1 Mark]
(A) 0
(B) \( \frac{1}{2} \)
(C) 1
(D) 2

Answer: (B) \( \frac{1}{2} \)

Teacher's Note:
a) Solve \( \alpha = 60^{\circ}, \beta = 30^{\circ} \).
b) Substitute to get \( \cos^2 60^{\circ} + \sin^2 30^{\circ} = \frac{1}{4}+\frac{1}{4} = \frac{1}{2} \).

8. A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is : [1 Mark]
(A) \( \frac{3}{13} \)
(B) \( \frac{2}{13} \)
(C) \( \frac{1}{2} \)
(D) \( \frac{3}{26} \)

Answer: (D) \( \frac{3}{26} \)

Teacher's Note:
a) Red face cards = 6 (J, Q, K of Hearts and Diamonds).
b) Probability \( = \frac{6}{52} = \frac{3}{26} \).

9. If \( \alpha \) and \( \beta \) are the zeroes of polynomial \( 3x^2 + 6x + k \) such that \( \alpha + \beta + \alpha\beta = -\frac{2}{3} \), then the value of k is : [1 Mark]
(A) -8
(B) 8
(C) -4
(D) 4

Answer: (D) 4

Teacher's Note:
a) \( \alpha+\beta = -2 \), \( \alpha\beta = \frac{k}{3} \).
b) \( -2 + \frac{k}{3} = -\frac{2}{3} \Rightarrow k = 4 \).

10. The value of \( \tan^2 \theta - \left( \frac{1}{\cos \theta} \times \sec \theta \right) \) is : [1 Mark]
(A) 1
(B) 0
(C) -1
(D) 2

Answer: (C) -1

Teacher's Note:
a) This simplifies to \( \tan^2\theta - \sec^2\theta \).
b) Use identity \( \sec^2\theta - \tan^2\theta = 1 \), so the value is -1.

11. Which of the following is a rational number between \( \sqrt{3} \) and \( \sqrt{5} \) ? [1 Mark]
(A) 1.4142387954012....
(B) \( 2.3\overline{26} \)
(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 in this range and is terminating.

12. 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 = 14, LCM = 196.
b) \( n - 7m = 196 - 98 = 98 \).

13. If the length of a chord of a circle is equal to its radius, then the angle subtended by chord at the centre is : [1 Mark]
(A) \( 60^{\circ} \)
(B) \( 30^{\circ} \)
(C) \( 120^{\circ} \)
(D) \( 90^{\circ} \)

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

Teacher's Note:
a) When chord equals radius, the triangle formed with the centre is equilateral.
b) All angles of an equilateral triangle are \( 60^{\circ} \).

14. The greatest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is : [1 Mark]
(A) 13
(B) 65
(C) 875
(D) 1750

Answer: (A) 13

Teacher's Note:
a) Required number is HCF of (70-5) and (125-8), that is HCF(65, 117).
b) HCF(65, 117) = 13.

15. A ladder 14 m long leans against a wall. If the foot of the ladder is 7 m from the wall, then the angle of elevation of the top of the wall is : [1 Mark]
(A) \( 15^{\circ} \)
(B) \( 30^{\circ} \)
(C) \( 45^{\circ} \)
(D) \( 60^{\circ} \)

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

Teacher's Note:
a) \( \cos\theta = \frac{7}{14} = \frac{1}{2} \).
b) So \( \theta = 60^{\circ} \).

16. 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) Two equal angle pairs give similarity by AA criterion.
b) Since sides are not equal (AB = 3DE), the triangles cannot be congruent.

17. The mid-point of the line segment joining the points P(-4, 5) and Q(4, 6) lies on : [1 Mark]
(A) x-axis
(B) y-axis
(C) origin
(D) neither x-axis nor y-axis

Answer: (B) y-axis

Teacher's Note:
a) Midpoint \( = \left( \frac{-4+4}{2}, \frac{5+6}{2} \right) = (0, 5.5) \).
b) A point with x-coordinate 0 lies on the y-axis.

18. Mode and Mean of a data are 15x and 18x, respectively. Then the median of the data is : [1 Mark]
(A) x
(B) 11x
(C) 17x
(D) 34x

Answer: (C) 17x

Teacher's Note:
a) Use empirical relation: Mode = 3 Median - 2 Mean.
b) \( 15x = 3M - 36x \Rightarrow M = 17x \).

19. 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 base does give a sphere, so Assertion is correct.
b) TSA of a sphere is \( 4\pi r^2 \), not \( 3\pi r^2 \), so Reason is false.

20. 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 one number out of the given 20 numbers is certain to happen, so its probability is 1.
b) This matches the definition of a sure event given in the Reason.

SECTION B

21. If the zeroes of the polynomial \( x^2 + ax + b \) are in the ratio 3 : 4, then prove that \( 12a^2 = 49b \). [2 Marks]

Answer:
1. Let the zeroes be \( 3\alpha \) and \( 4\alpha \). Then sum of zeroes \( 3\alpha + 4\alpha = -a \Rightarrow 7\alpha = -a \).
2. Product of zeroes \( 12\alpha^2 = b \).
3. Now \( 12a^2 = 12(-7\alpha)^2 = 12 \times 49\alpha^2 = 49 \times 12\alpha^2 = 49b \). Hence proved.

Teacher's Note:
a) Use sum of zeroes \( = -\frac{b_1}{a_1} \) and product \( = \frac{c_1}{a_1} \) for a monic quadratic.
b) Express everything in terms of a single variable \( \alpha \) before substituting.

22. 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. P is an external point 26 m from O. PA and PB are tangents of length 10 m each, touching the circle at A and B.]

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 radius = 24 m.

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

23. (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 triangles are similar, \( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \), that is \( \frac{6}{PQ} = \frac{4}{QR} = \frac{8}{6} \).
2. From \( \frac{6}{PQ} = \frac{8}{6} \), \( PQ = \frac{36}{8} = 4.5 \) cm; from \( \frac{4}{QR} = \frac{8}{6} \), \( QR = 3 \) cm.
3. So \( PQ + QR = 4.5 + 3 = 7.5 \) cm.

Teacher's Note:
a) Corresponding sides of similar triangles are proportional; match vertices carefully (A-P, B-Q, C-R).
b) Use the known ratio AC/PR to find the scale factor first.

OR

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

[Figure: Triangle QTR with P on QT and S on QR such that lines PS and PR are drawn; angle 1 at Q (angle PQS) and angle 2 at R (angle PRS marked as 2) are shown equal.]

Answer:
1. In \( \triangle PQR \), \( \angle 1 = \angle 2 \Rightarrow PR = PQ \) (sides opposite equal angles).
2. Given \( \frac{QR}{QS} = \frac{QT}{PR} \), substitute PR = PQ to get \( \frac{QR}{QS} = \frac{QT}{PQ} \).
3. Also \( \angle Q = \angle Q \) (common angle in both triangles, since \( \angle 1 = \angle 1 \)). By SAS similarity, \( \triangle PQS \sim \triangle TQR \).

Teacher's Note:
a) First convert PR into PQ using the isosceles triangle property.
b) Then apply the SAS similarity criterion using the common angle at Q.

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

Answer:
1. Substitute values: \( x \left(\frac{1}{2}\right) + y(1) + \frac{1}{2} - 1 = 5 \).
2. Simplify: \( \frac{x}{2} + y - \frac{1}{2} = 5 \Rightarrow \frac{x}{2} + y = 5.5 \).
3. Multiply by 2: \( x + 2y = 11 \).

Teacher's Note:
a) Memorise standard trigonometric values: \( \cos 60^{\circ}=\frac{1}{2}, \cos 0^{\circ}=1, \sin 30^{\circ}=\frac{1}{2}, \cot 45^{\circ}=1 \).
b) Simplify step by step to avoid sign errors.

OR

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

Answer:
1. \( \tan 60^{\circ} = \sqrt{3} \), so \( \tan^2 60^{\circ} = 3 \).
2. \( \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), so denominator \( = \frac{3}{4} + \frac{3}{4} = \frac{3}{2} \).
3. Value \( = \frac{3}{3/2} = 2 \).

Teacher's Note:
a) Convert all terms to standard angle values before simplifying.
b) Double-check squaring of \( \frac{\sqrt{3}}{2} \) gives \( \frac{3}{4} \), not \( \frac{3}{2} \).

25. The coordinates of the centre of a circle are (2a, a - 7). Find the value(s) of 'a' if the circle passes through the point (11, -9) and has diameter \( 10\sqrt{2} \) units. [2 Marks]

Answer:
1. Radius \( = 5\sqrt{2} \) units, so distance from centre to (11,-9) equals \( 5\sqrt{2} \).
2. \( (2a-11)^2 + (a-7+9)^2 = 50 \Rightarrow (2a-11)^2+(a+2)^2=50 \).
3. Expanding: \( 5a^2 - 40a + 125 = 50 \Rightarrow a^2 - 8a + 15 = 0 \Rightarrow (a-5)(a-3)=0 \).
4. So \( a = 5 \) or \( a = 3 \).

Teacher's Note:
a) Use distance formula between centre and a point on the circle equal to the radius.
b) Remember diameter/2 = radius before substituting.

SECTION C

26. If the radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3, find the ratio of their volumes. [3 Marks]

Answer:
1. Let radii be \( r_1 : r_2 = 3:4 \) and heights \( h_1 : h_2 = 2:3 \) for cylinder and cone respectively.
2. \( \dfrac{\text{Volume of cylinder}}{\text{Volume of cone}} = \dfrac{\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = 3 \times \left(\dfrac{r_1}{r_2}\right)^2 \times \dfrac{h_1}{h_2} \).
3. Substituting: \( = 3 \times \left(\dfrac{3}{4}\right)^2 \times \dfrac{2}{3} = \dfrac{9}{8} \). So the ratio of volumes is 9 : 8.

Teacher's Note:
a) Remember the factor of 3 comes from cancelling the \( \frac{1}{3} \) in the cone's volume formula.
b) Keep ratios as fractions throughout to avoid decimal errors.

27. Three sets of Physics, Chemistry and Mathematics books have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The number of Physics books is 144, the number of Chemistry books is 180 and the number of Mathematics books is 192. Assuming that the books are of same thickness, determine the number of stacks of Physics, Chemistry and Mathematics books. [3 Marks]

Answer:
1. \( 144 = 2^4 \times 3^2 \), \( 180 = 2^2 \times 3^2 \times 5 \), \( 192 = 2^6 \times 3 \).
2. HCF \( = 2^2 \times 3 = 12 \), which gives the greatest number of books per stack.
3. Number of stacks: Physics \( = \frac{144}{12}=12 \), Chemistry \( = \frac{180}{12}=15 \), Mathematics \( = \frac{192}{12}=16 \).

Teacher's Note:
a) Equal height of each stack means dividing by the HCF of the three numbers.
b) Show prime factorisation clearly to earn full marks.

28. Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2. [3 Marks]

Answer:
1. Total possible outcomes when two dice are thrown \( = 36 \).
2. Favourable outcomes (difference = 2): (1,3), (3,1), (4,2), (2,4), (5,3), (3,5), (4,6), (6,4), i.e., 8 outcomes.
3. Required probability \( = \dfrac{8}{36} = \dfrac{2}{9} \).

Teacher's Note:
a) List all pairs carefully to avoid missing or repeating an outcome.
b) Always simplify the fraction to its lowest terms.

29. (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. Write \( \tan\theta=\frac{\sin\theta}{\cos\theta} \) and \( \cot\theta=\frac{\cos\theta}{\sin\theta} \), then LHS \( = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)} \).
2. Combine over a common denominator: \( = \dfrac{1}{\sin\theta-\cos\theta}\left[\dfrac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}\right] \).
3. Factor \( \sin^3\theta - \cos^3\theta = (\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta) \), cancel to get \( \dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta} = 1+\sec\theta\,\text{cosec}\,\theta \) = RHS.

Teacher's Note:
a) Use the identity for difference of cubes to simplify the numerator.
b) Keep \( \sin\theta\cos\theta \) as a common factor in the denominator throughout.

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. Combine LHS over a common denominator: \( \dfrac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \).
2. Numerator \( = 2\sin^2A + 2\cos^2A = 2 \); denominator \( = \sin^2A-\cos^2A \).
3. Write \( \cos^2A = 1-\sin^2A \), so denominator \( = 2\sin^2A - 1 \). LHS \( = \dfrac{2}{2\sin^2A-1} \) = RHS.

Teacher's Note:
a) Use \( (a+b)^2+(a-b)^2=2a^2+2b^2 \) to simplify the numerator quickly.
b) Replace \( \cos^2A \) with \( 1-\sin^2A \) to match the RHS form.

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. A is a point on the circle, and BCD is a straight line tangent to the circle at C. P is another point on the circle, and OC is joined to O.]

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

Teacher's Note:
a) The radius is always perpendicular to the tangent at the point of contact.
b) Use the isosceles triangle property (equal radii) to replace one angle with another.

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. Let ABCD be a quadrilateral circumscribing a circle with centre O, touching the sides at P, Q, R, S. Join O to A, B, C, D and to the points of contact.
2. By congruency of triangles OAP and OAS (tangents from A are equal, OP=OS, OA common), \( \angle 1 = \angle 2 \); similarly \( \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8 \).
3. Since \( \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ} \), 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 figure with the circle, quadrilateral, and lines joining centre to vertices and points of tangency.
b) Use congruent triangles formed by equal tangent lengths to prove 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 the section formula for x-coordinate: \( 0 = \dfrac{-k+5}{k+1} \Rightarrow k = 5 \).
2. So the ratio is 5:1.
3. Now \( y = \dfrac{-4(5)-6}{5+1} = \dfrac{-26}{6} = -\dfrac{13}{3} \). So the point of intersection is \( \left(0, -\dfrac{13}{3}\right) \).

Teacher's Note:
a) Since the y-axis divides the segment, the x-coordinate of the section formula must be zero.
b) Once k is found, substitute back to find the y-coordinate of the intersection point.

SECTION D

32. (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 AE intersects BD at F.]

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

Teacher's Note:
a) Use alternate angles from the parallel sides AD and BC.
b) Similar triangles give proportional sides; cross-multiply carefully to get the required product.

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 between B and C.]

Answer:
1. Given \( AD^2 = BD \times DC \), so \( \dfrac{AD}{DC} = \dfrac{BD}{AD} \).
2. Also \( \angle ADB = \angle ADC = 90^{\circ} \).
3. By SAS similarity, \( \triangle DBA \sim \triangle DAC \), so \( \angle DBA = \angle DAC \) and \( \angle BAD = \angle DCA \).
4. Adding: \( \angle DBA + \angle DCA = \angle DAC + \angle BAD \).
5. Hence \( \angle BAC = \angle DAC + \angle BAD = \angle DBA + \angle DCA = 90^{\circ} \) (since angles of triangle ABC sum to 180° and the remaining two angles equal this sum).

Teacher's Note:
a) Convert the given product relation into a ratio to use in similarity.
b) Use the two pairs of equal angles from similar triangles to build up \( \angle BAC \).

33. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the mean and mode of the data :
Monthly Consumption (in units): 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205
Number of Consumers: 4 | 5 | 13 | 20 | 14 | 8 | 4 [5 Marks]

Answer:
1. Taking assumed mean a = 135 and h = 20, the deviations \( u_i = \frac{x_i-135}{20} \) give \( \Sigma f_iu_i = 7 \) and \( \Sigma f_i = 68 \).
2. Mean \( = 135 + \dfrac{7}{68}\times 20 = 135 + 2.06 = 137.06 \) units.
3. The modal class is 125-145 (highest frequency 20). Mode \( = 125 + \left(\dfrac{20-13}{2(20)-13-14}\right)\times 20 = 125 + \dfrac{7}{13}\times 20 = 125+10.77 = 135.77 \) units.

Teacher's Note:
a) Use the step-deviation method to simplify calculations with large class marks.
b) The modal class is always the one with the highest frequency; apply the mode formula carefully with f1, f0, f2.

34. Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received Rs. 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received Rs. 20 more as annual interest. How much money did he invest in each scheme ? [5 Marks]

Answer:
1. Let Vijay invest Rs. x in scheme A (8%) and Rs. y in scheme B (9%).
2. From the given condition: \( \dfrac{8x}{100} + \dfrac{9y}{100} = 1860 \Rightarrow 8x + 9y = 186000 \) ... (i)
3. After interchanging: \( \dfrac{9x}{100} + \dfrac{8y}{100} = 1880 \Rightarrow 9x + 8y = 188000 \) ... (ii)
4. Solving (i) and (ii) simultaneously gives \( x = 12000 \) and \( y = 10000 \).
5. So Vijay invested Rs. 12,000 in scheme A and Rs. 10,000 in scheme B.

Teacher's Note:
a) Form two linear equations carefully based on original and interchanged interest amounts.
b) Solve using elimination method; check the answer by substituting back.

35. (a) A two-digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number. [5 Marks]

Answer:
1. Let the ten's digit be x and unit's digit be y, so the number is \( 10x+y \), with \( xy = 12 \) ... (i)
2. Given \( 10x+y+36 = 10y+x \Rightarrow x-y+4=0 \) ... (ii)
3. From (i) and (ii): \( y = x+4 \), substitute in \( xy=12 \) to get \( x^2+4x-12=0 \Rightarrow (x+6)(x-2)=0 \Rightarrow x=2 \).
4. So \( y = 6 \), and the number is 26.

Teacher's Note:
a) Represent the two-digit number correctly as \( 10x+y \), not \( xy \).
b) Reject the negative value of x since a digit cannot be negative.

OR

(b) A student scored a total of 32 marks in class tests in Mathematics and Science. Had he scored 2 marks less in Science and 4 marks more in Mathematics, the product of his marks would have been 253. Find his marks in the two subjects. [5 Marks]

Answer:
1. Let marks in Mathematics be x and in Science be y, so \( x+y=32 \) ... (i)
2. Given \( (x+4)(y-2) = 253 \) ... (ii)
3. From (i), \( y = 32-x \); substituting in (ii) gives \( x^2 - 26x + 133 = 0 \Rightarrow (x-19)(x-7)=0 \Rightarrow x=19 \) or \( x=7 \).
4. If \( x=19 \), \( y=13 \); if \( x=7 \), \( y=25 \). So the marks are 19 and 13, or 7 and 25.

Teacher's Note:
a) Convert the word problem into two equations using the given conditions.
b) Both solutions of the quadratic are valid here; state both possibilities.

SECTION E

36. 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.

A collage of decorative circular mandala-style brooch designs in various colours, followed by a circular brooch diagram with centre point, 5 diameters dividing it into 10 equal sectors, with points A, B and C marked near the top right of the circle.

Based on the above given information, answer the following questions :

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

Answer: The central angle of each sector \( = \dfrac{360^{\circ}}{10} = 36^{\circ} \).

Teacher's Note:
a) Total angle at the centre of a circle is always \( 360^{\circ} \).
b) Divide by the number of equal sectors to get each central angle.

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

Answer: Length of arc ACB \( = \dfrac{1}{10} \times 2 \times \dfrac{22}{7} \times \dfrac{35}{2} = 11 \) mm.

Teacher's Note:
a) Use \( l = \dfrac{\theta}{360^{\circ}} \times 2\pi r \) with \( \theta = 36^{\circ} \).
b) Radius here is half the diameter, that is 17.5 mm.

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

Answer:
1. Area of each sector \( = \dfrac{1}{10} \times \dfrac{22}{7} \times \dfrac{35}{2} \times \dfrac{35}{2} \).
2. This gives \( \dfrac{385}{4} = 96.25 \) 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 (35/2) to avoid rounding errors.

OR

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

Answer:
1. Length of wire in the circle (circumference) \( = 2 \times \dfrac{22}{7} \times \dfrac{35}{2} = 110 \) mm.
2. Length of wire used for 5 diameters \( = 5 \times 35 = 175 \) mm.
3. Total length of silver wire \( = 110 + 175 = 285 \) mm.

Teacher's Note:
a) The wire is used both for the circle's circumference and for the 5 diameters.
b) Add both lengths together for the final answer.

37. 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 \( 45^{\circ} \).

A photograph of a white lighthouse on a rocky coastal cliff, followed by a diagram: point A on the ground, point B directly above A at height 40 m (observation deck), point E is the base of the lighthouse on the ground in line with A, point D is on the lighthouse at the same height as B, and point C is the top of the lighthouse with CD = h. The angle at B (angle DBC) is 45° and the angle at A (angle EAC) is 60°.

Based on the above given information, answer the following questions :

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

Answer: Since \( \angle DBC = 45^{\circ} \), \( \tan 45^{\circ} = \dfrac{CD}{BD} = 1 \), so BD = h metres.

Teacher's Note:
a) At \( 45^{\circ} \), the opposite and adjacent sides of the right triangle are equal.
b) BD and CD form the two legs of a right triangle with the 45° angle at B.

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

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

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

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

Answer:
1. Since AE = BD = h and DE = AB = 40, \( \tan 60^{\circ} = \dfrac{CE}{AE} = \dfrac{h+40}{h} = \sqrt{3} \).
2. Solving: \( h(\sqrt{3}-1) = 40 \Rightarrow h = \dfrac{40}{\sqrt{3}-1} = 20(\sqrt{3}+1) = 20 \times 2.73 = 54.6 \) m.
3. CE \( = h + 40 = 54.6 + 40 = 94.6 \) m.

Teacher's Note:
a) Use the fact that ABDE forms a rectangle, so AE = BD and DE = AB.
b) Rationalise the denominator before substituting the value of \( \sqrt{3} \).

OR

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

Answer:
1. In right triangle ACE, \( \cos 60^{\circ} = \dfrac{AE}{AC} \).
2. \( \dfrac{1}{2} = \dfrac{AE}{100} \Rightarrow AE = 50 \) m.

Teacher's Note:
a) Use the cosine ratio since AC is the hypotenuse and AE is the adjacent side.
b) \( \cos 60^{\circ} = \frac{1}{2} \) is a standard value to remember.

38. 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.

An illustration of colourful runner silhouettes arranged in a circle around a central ring, representing a charity run.

Based on the information given above, answer the following questions :

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

Answer: The AP is 300, 350, 400, .... with a = 300, d = 50. So \( a_4 = 450 \), \( a_5 = 500 \), \( a_6 = 550 \) metres.

Teacher's Note:
a) Use \( a_n = a + (n-1)d \) to find each term.
b) Double check by simply adding 50 to each previous term.

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

Answer: \( a_8 = 300 + 7 \times 50 = 650 \) metres.

Teacher's Note:
a) Substitute n = 8 in the nth term formula.
b) Keep track that (n-1) is used, not n, in the formula.

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

Answer:
1. \( S_{10} = \dfrac{10}{2} \times (2 \times 300 + 9 \times 50) \).
2. \( = 5 \times (600+450) = 5 \times 1050 = 5250 \) metres.

Teacher's Note:
a) Use the sum formula \( S_n = \dfrac{n}{2}[2a+(n-1)d] \).
b) Always substitute n, a and d correctly before simplifying.

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. \( S_6 = \dfrac{6}{2} \times (2 \times 300 + 5 \times 50) \).
2. \( = 3 \times (600+250) = 3 \times 850 = 2550 \) metres.

Teacher's Note:
a) Verify by directly adding the first six terms: 300+350+400+450+500+550 = 2550.
b) Use the sum formula as a quick check for such AP problems.

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