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

SECTION A

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

Answer: (D) 4

Teacher's Note:
a) Use \( \alpha+\beta = -\dfrac{6}{3} = -2 \) and \( \alpha\beta = \dfrac{k}{3} \).
b) Substitute in the given condition to solve for k quickly.

2. 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 directly into both equations to find a and b.
b) Then check which option is satisfied by the values obtained.

3. 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 formula gives \( \left(0, \dfrac{11}{2}\right) \).
b) Any point with x-coordinate 0 lies on the y-axis.

4. 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 degrees

Teacher's Note:
a) Simplify to get \( \sin\theta = \dfrac{1}{2} \).
b) Recall standard angle values of sine to identify \( \theta \).

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

Answer: (C) -1

Teacher's Note:
a) Note that \( \dfrac{1}{\cos\theta}\times\sec\theta = \sec^2\theta \).
b) Use the identity \( \tan^2\theta - \sec^2\theta = -1 \).

6. 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) Compute \( n - 7m = 196 - 98 = 98 \).

7. 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) Both tangents are perpendicular to the same diameter, so they never meet.
b) Two lines perpendicular to the same line are parallel to each other.

8. 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 make the triangles similar (AA test).
b) Since corresponding sides are not equal (AB = 3DE), they cannot be congruent.

9. 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 we need \( (-1)^n = -1 \).
b) This happens only when n is odd.

10. 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 intersecting curves, one an ellipse-like closed curve and the other two intersecting straight lines forming an X shape crossing the x-axis, all meeting the x-axis at two common points.]

Answer: (C) 2

Teacher's Note:
a) Zeroes of a polynomial are the x-coordinates where its graph cuts the x-axis.
b) Count only the distinct points common to both graphs on the x-axis.

11. If the sum of first m terms of an AP is \( 2m^2 + 3m \), then its second term is : [1 Mark]
(A) 10
(B) 9
(C) 12
(D) 4

Answer: (B) 9

Teacher's Note:
a) \( a_1 = S_1 = 5 \) and \( S_2 = 14 \).
b) Second term \( = S_2 - S_1 = 9 \).

12. 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 the empirical relation Mode = 3 Median - 2 Mean.
b) Substitute values to get Median = 17x.

13. 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) \( \dfrac{3}{13} \)
(B) \( \dfrac{2}{13} \)
(C) \( \dfrac{1}{2} \)
(D) \( \dfrac{3}{26} \)

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

Teacher's Note:
a) There are 6 red face cards (King, Queen, Jack of hearts and diamonds).
b) Probability = favourable outcomes divided by total outcomes = \( \dfrac{6}{52} \).

14. 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 that lies strictly between these two values.

15. 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) Area of sector \( = \dfrac{\theta}{360^{\circ}} \times \pi r^2 \).
b) Substitute values to get \( r^2 = 200 \), so \( r = 10\sqrt{2} \).

16. 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. AB is a diameter (A at top, B at bottom). P is an external point to the left with PA drawn as a tangent to the circle at A, and PO drawn to the centre. The angle POB at the centre is marked as 115 degrees.]

Answer: (A) 25 degrees

Teacher's Note:
a) Since AB is a straight line through O, \( \angle AOP = 180^{\circ} - 115^{\circ} = 65^{\circ} \).
b) In right triangle OAP, \( \angle OAP = 90^{\circ} \), so \( \angle APO = 180^{\circ} - 90^{\circ} - 65^{\circ} \).

17. 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 \( \sin 30^{\circ} = \dfrac{\text{height}}{\text{string length}} \).
b) Substitute to get string length \( = \dfrac{150}{1/2} = 300 \) m.

18. 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 degrees

Teacher's Note:
a) Arc length = radius \( \times \) angle (in radians).
b) Solve \( 20 = \dfrac{60}{\pi} \times \theta \) to get \( \theta = \dfrac{\pi}{3} \) radians = 60 degrees.

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 one number out of 1 to 20 is a sure event since one number will definitely be selected.
b) A sure event always has probability exactly equal to 1.

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 of the same radius along their bases does give a sphere, so the Assertion is correct.
b) The correct total surface area of a sphere is \( 4\pi r^2 \), not \( 3\pi r^2 \), so the Reason is false.

SECTION B

21. (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 the standard values: \( x\left(\dfrac{1}{2}\right) + y(1) + \dfrac{1}{2} - 1 = 5 \).
2. Simplify to get \( \dfrac{x}{2} + y = \dfrac{11}{2} \), so \( x + 2y = 11 \).

Teacher's Note:
a) Remember \( \cos 0^{\circ} = 1 \) and \( \cot 45^{\circ} = 1 \), a common slip point.
b) Multiply through by 2 at the end to match the required expression x + 2y.

OR

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

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

Teacher's Note:
a) Note that \( \sin 60^{\circ} = \cos 30^{\circ} = \dfrac{\sqrt{3}}{2} \), so both terms are equal.
b) Simplify the denominator before dividing to avoid arithmetic errors.

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

Answer:
1. Multiply by 3: \( 3x^2 + 4x - 4 = 0 \), which factors as \( (3x-2)(x+2) = 0 \).
2. So the zeroes are \( x = \dfrac{2}{3} \) and \( x = -2 \).

Teacher's Note:
a) Clearing fractions first makes factorisation much easier.
b) Always verify zeroes by substituting back into the original polynomial.

23. 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 \( = 5\sqrt{2} \), so \( (2a-11)^2 + (a-7+9)^2 = 50 \).
2. Simplifying gives \( a^2 - 8a + 15 = 0 \), which factors as \( (a-5)(a-3)=0 \), so \( a = 5 \) or \( a = 3 \).

Teacher's Note:
a) Use the distance formula between the centre and a point on the circle equal to the radius.
b) Do not forget that diameter divided by 2 gives the radius before squaring.

24. (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. Using \( \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} \), we get \( \dfrac{6}{PQ} = \dfrac{4}{QR} = \dfrac{8}{6} \).
2. Solving gives \( PQ = 4.5 \) cm and \( QR = 3 \) cm, so \( PQ + QR = 7.5 \) cm.

Teacher's Note:
a) Match corresponding vertices correctly: A to P, B to Q, C to R.
b) Set up one common ratio using the known pair of sides (AC and PR) first.

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: A triangle QTR with point P on side QT and point S on side QR, forming triangle PQS inside. Angle 1 is marked at Q between QP and QS, and angle 2 is marked at R between RS and RT.]

Answer:
1. In triangle PQR, since \( \angle 1 = \angle 2 \), the sides opposite equal angles are equal, so PR = PQ.
2. Then \( \dfrac{QR}{QS} = \dfrac{QT}{PQ} \), and with the common angle \( \angle Q = \angle Q \), by SAS similarity \( \triangle PQS \sim \triangle TQR \).

Teacher's Note:
a) Converting the given angle condition into a side equality (PR = PQ) is the key step.
b) Identify the included angle carefully to apply the SAS similarity criterion correctly.

25. 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. An external point P is joined to O, and to two points A and B on the circle by tangents PA and PB, each of length 10 m. OP = 26 m.]

Answer:
1. Since OA is a radius and PA is a tangent, \( \angle OAP = 90^{\circ} \).
2. In right triangle OAP, \( 26^2 = OA^2 + 10^2 \), giving \( OA = \sqrt{576} = 24 \) m, so the radius is 24 m.

Teacher's Note:
a) The radius drawn to the point of tangency is always perpendicular to the tangent.
b) Apply the Pythagoras theorem in the right triangle formed by the radius, tangent and OP.

SECTION C

26. (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 at top right, and OA is drawn. OC is drawn as a dashed vertical line from O to C on the circle, where BCD is a tangent line touching the circle at C. P is another point on the circle on the tangent side, and A is joined to both B (on the line) and C.]

Answer:
1. In triangle OAC, since OA = OC (radii), \( \angle OCA = \angle OAC \).
2. Since OC is perpendicular to the tangent BCD, \( \angle OCD = 90^{\circ} \), so \( \angle OCA + \angle ACD = 90^{\circ} \).
3. Replacing \( \angle OCA \) with the equal angle \( \angle OAC \) (which is \( \angle BAC \)), we get \( \angle BAC + \angle ACD = 90^{\circ} \).

Teacher's Note:
a) The isosceles triangle property (equal radii give equal base angles) is the key starting step.
b) Remember that a 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, and join O to all four points of contact and to A, B, C, D.
2. Using congruent triangles formed by equal tangent segments from each vertex, angles at O on each side of the four points of contact are equal in pairs, say \( \angle 1 = \angle 2 \), \( \angle 3 = \angle 4 \), \( \angle 5 = \angle 6 \), \( \angle 7 = \angle 8 \).
3. Since the sum of all eight angles around O is \( 360^{\circ} \), we get \( 2(\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^{\circ} \), which gives \( \angle AOB + \angle COD = 180^{\circ} \), and similarly \( \angle BOC + \angle AOD = 180^{\circ} \).

Teacher's Note:
a) Draw a clear, correctly labelled figure since one mark is reserved just for this.
b) The congruence of the triangles formed at each vertex (SSS or RHS) is the main working step.

27. (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 \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \) and \( \cot\theta = \dfrac{\cos\theta}{\sin\theta} \), then simplify each term to get a common denominator of \( (\sin\theta - \cos\theta) \).
2. Combine the fractions to get \( \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)} \).
3. Factor using \( a^3-b^3 \) identity, cancel \( (\sin\theta-\cos\theta) \), and simplify to get \( 1 + \sec\theta\,\text{cosec}\,\theta \), which is the RHS.

Teacher's Note:
a) Converting to sine and cosine is the standard first step for such identities.
b) Use the identity \( a^3-b^3=(a-b)(a^2+ab+b^2) \) to simplify the numerator neatly.

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. Take LCM to combine both fractions: \( \dfrac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \).
2. Expand the numerator using \( (a+b)^2+(a-b)^2 = 2(a^2+b^2) \) to get \( \dfrac{2}{\sin^2 A - \cos^2 A} \).
3. Replace \( \cos^2 A = 1 - \sin^2 A \) to get \( \dfrac{2}{2\sin^2 A - 1} \), which is the RHS.

Teacher's Note:
a) The expansion identity \( (a+b)^2+(a-b)^2=2(a^2+b^2) \) saves a lot of time here.
b) Use \( \sin^2A+\cos^2A=1 \) in the last step to match the given RHS form.

28. 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, so the x-coordinate of the point of intersection is \( \dfrac{-k+5}{k+1} = 0 \), giving \( k = 5 \).
2. Hence the ratio is 5:1.
3. Substituting k = 5 in the y-coordinate formula: \( y = \dfrac{-4(5)+(-6)}{5+1} = \dfrac{-26}{6} = -\dfrac{13}{3} \), so the point of intersection is \( \left(0, -\dfrac{13}{3}\right) \).

Teacher's Note:
a) Since the point lies on the y-axis, its x-coordinate must be 0; use this to find the ratio directly.
b) After finding the ratio, use the section formula again to get the y-coordinate.

29. Prove that \( \dfrac{1}{\sqrt{5}} \) is an irrational number. [3 Marks]

Answer:
1. Assume, to the contrary, that \( \dfrac{1}{\sqrt{5}} \) is rational, so \( \dfrac{1}{\sqrt{5}} = \dfrac{p}{q} \), where p and q are coprime integers and \( q \neq 0 \).
2. This gives \( 5p^2 = q^2 \), so \( q^2 \) is divisible by 5, hence q is divisible by 5. Let q = 5a for some integer a.
3. Then \( 25a^2 = 5p^2 \), so \( p^2 = 5a^2 \), meaning p is also divisible by 5. This contradicts that p and q are coprime, so \( \dfrac{1}{\sqrt{5}} \) is irrational.

Teacher's Note:
a) This is the standard proof by contradiction used for all such irrationality proofs.
b) State clearly the coprime assumption at the start, as this earns dedicated marks.

30. 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 the common radius be r and the height of the cylinder be 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. Setting this equal to \( \dfrac{1408}{21} \) gives \( r^3 = 8 \), so r = 2 m, and hence h = 4 m.

Teacher's Note:
a) Express both the cylinder height and hemisphere radius in terms of a single variable r first.
b) Add the volume of the cylinder and the hemisphere carefully, not the full sphere.

31. 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 number of outcomes when two dice are thrown = 36.
2. Favourable outcomes where the difference is 2 are: (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4), which are 8 in number.
3. Required probability \( = \dfrac{8}{36} = \dfrac{2}{9} \).

Teacher's Note:
a) List the outcomes systematically to avoid missing any pair.
b) Always simplify the final probability fraction to its lowest terms.

SECTION D

32. 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. According to the given condition: \( \dfrac{8x}{100} + \dfrac{9y}{100} = 1860 \), i.e., \( 8x + 9y = 186000 \).
3. After interchanging: \( \dfrac{9x}{100} + \dfrac{8y}{100} = 1880 \), i.e., \( 9x + 8y = 188000 \).
4. Solving these two equations simultaneously gives \( x = 12000 \) and \( y = 10000 \).
5. Hence, Vijay invested Rs. 12,000 in scheme A and Rs. 10,000 in scheme B.

Teacher's Note:
a) Convert the percentage interest conditions into linear equations carefully.
b) Solve by elimination, multiplying the equations suitably to cancel one variable.
c) Always verify both original conditions with the final values obtained.

33. (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 x EF = FB x FA. [5 Marks]

[Figure: Parallelogram ABCD with A at bottom left, B at bottom right, C at top right, D at top left. Diagonal BD is drawn, E is a point on side BC, and AE is drawn intersecting BD at F.]

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

Teacher's Note:
a) Use the property AD parallel to BC of a parallelogram to get equal alternate angles.
b) Correctly identifying vertically opposite angles at F is essential for the similarity proof.
c) A clearly labelled figure is worth separate marks in this question.

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 D on side BC such that AD is perpendicular to BC, drawn from vertex A down to BC.]

Answer:
1. Given \( AD^2 = BD \times DC \), this can be written as \( \dfrac{AD}{DC} = \dfrac{BD}{AD} \).
2. Since \( \angle ADB = \angle ADC = 90^{\circ} \), triangles DBA and DAC are similar (SAS similarity using the ratio and the right angle).
3. From this similarity, \( \angle DBA = \angle DAC \) and \( \angle BAD = \angle DCA \).
4. Adding these two angle equalities: \( \angle DBA + \angle DCA = \angle DAC + \angle BAD \), which simplifies to \( \angle BAC = 90^{\circ} \) using angle sum property of triangle ABC.

Teacher's Note:
a) Converting the given product relation into a ratio is the key first step for similarity.
b) Draw the perpendicular clearly and mark the right angles at D in the figure.
c) The final step uses the fact that angles of triangle ABC sum to 180 degrees.

34. (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 two other sides be x cm and y cm. Given \( x + y + 25 = 60 \), so \( y = 35 - x \).
2. By Pythagoras theorem: \( x^2 + y^2 = 25^2 = 625 \).
3. Substituting: \( x^2 + (35-x)^2 = 625 \), which simplifies to \( x^2 - 35x + 300 = 0 \).
4. Factoring: \( (x-20)(x-15) = 0 \), so \( x = 20 \) or \( x = 15 \).
5. Hence, the two sides are 15 cm and 20 cm.

Teacher's Note:
a) Use the perimeter condition first to express one variable in terms of the other.
b) Apply the Pythagoras theorem next and solve the resulting quadratic equation.
c) Both roots give a valid pair of sides, just swapped.

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, so the reduced speed is (x - 8) km/h.
2. According to the given condition: \( \dfrac{480}{x-8} - \dfrac{480}{x} = 3 \).
3. Simplifying gives \( x^2 - 8x - 1280 = 0 \).
4. Factoring: \( (x-40)(x+32) = 0 \), so \( x = 40 \) (rejecting the negative value).
5. Hence, the speed of the train is 40 km/h.

Teacher's Note:
a) Set up the time difference equation using time = distance divided by speed.
b) Reject the negative root of the quadratic since speed cannot be negative.
c) Always verify the answer by substituting back into the original condition.

35. Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode. [5 Marks]
Daily Allowance: 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25
Number of Children: 7 | 6 | 9 | 13 | f | 5 | 4

Answer:
1. Using class marks 12, 14, 16, 18, 20, 22, 24, the sum of frequencies is \( 44 + f \) and the sum of \( f_i x_i \) is \( 752 + 20f \).
2. Mean equation: \( 18 = \dfrac{752+20f}{44+f} \), which gives \( f = 20 \).
3. The modal class (highest frequency 13) is 19-21.
4. Using the mode formula: \( \text{Mode} = 19 + \dfrac{20-13}{40-13-5} \times 2 \), where class width is 2, giving mode approximately 19.95.

Teacher's Note:
a) Build the frequency table with class marks carefully before applying the mean formula.
b) Once f = 20 is found, identify the modal class as the one with highest frequency (17-19, since f=20 makes 19-21 highest, recheck carefully).
c) Substitute values step by step in the mode formula to avoid calculation errors.

SECTION E

Case Study 1

36. 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.

[Figure: An illustration of several runners arranged in a circular pattern around a circular track, representing rounds of a charity run.]

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

Answer: The AP is 300, 350, 400, ... with first term 300 and common difference 50, so the 4th, 5th and 6th terms are 450 m, 500 m and 550 m.

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

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

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

Teacher's Note:
a) Use n = 8 in the general term formula of an AP.
b) Keep the common difference sign consistent throughout.

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

Answer:
1. Use the sum formula \( S_n = \dfrac{n}{2}[2a + (n-1)d] \) with n = 10.
2. \( S_{10} = \dfrac{10}{2}[2(300) + 9(50)] = 5[600+450] = 5250 \) m.

Teacher's Note:
a) Substitute a = 300 and d = 50 carefully into the sum formula.
b) Keep track of units (metres) throughout the calculation.

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. Use \( S_6 = \dfrac{6}{2}[2(300) + 5(50)] \).
2. \( S_6 = 3[600+250] = 3 \times 850 = 2250 \) m.

Teacher's Note:
a) Use n = 6 instead of n = 10 in the same sum formula.
b) Recheck arithmetic inside the brackets before multiplying.

Case Study 2

37. 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 collage of 25 decorative circular brooch designs on top, and below, a circle divided by 5 diameters into 10 equal sectors, with points A, B and C marked near the top right sector boundary, each sector containing a small spiral pattern.]

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

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

Teacher's Note:
a) Divide the total angle at the centre (360 degrees) by the number of equal sectors.
b) This works because the 10 sectors are stated to be equal.

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

Answer: Arc ACB corresponds to one sector (36 degrees), so its length \( = \dfrac{1}{10} \times 2 \times \dfrac{22}{7} \times \dfrac{35}{2} = 11 \) mm.

Teacher's Note:
a) Use arc length \( = \dfrac{\theta}{360^{\circ}} \times 2\pi r \) with radius \( = \dfrac{35}{2} \) mm.
b) Since there are 10 equal sectors, one sector's arc is simply \( \dfrac{1}{10} \) of the circumference.

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

Answer:
1. Area of the full circle \( = \pi r^2 = \dfrac{22}{7} \times \left(\dfrac{35}{2}\right)^2 \).
2. Area of one sector (out of 10 equal sectors) \( = \dfrac{1}{10} \times \dfrac{22}{7} \times \dfrac{35}{2} \times \dfrac{35}{2} = \dfrac{385}{4} = 96.25 \) mm².

Teacher's Note:
a) Divide the total circle area by 10 since all sectors are equal.
b) Keep radius as a fraction (35/2) to avoid rounding errors mid-calculation.

OR

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

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

Teacher's Note:
a) Do not forget to add the wire used for the diameters, not just the circle's circumference.
b) Each diameter is a separate straight length of 35 mm, and there are 5 of them.

Case Study 3

38. 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 degrees. 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 degrees.

[Figure: A photograph of a lighthouse on a cliff, followed by a diagram: point A is at the bottom, E is at the base of the lighthouse on the ground, B is 40 m directly above A, D is directly above E at the same height as B, and C is the top of the lighthouse directly above D at height h from D. The angle at A (angle CAE) is 60 degrees and the angle at B (angle CBD) is 45 degrees.]

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

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

Teacher's Note:
a) Use the tangent ratio in right triangle BDC at the 45 degree angle.
b) Since tan 45 degrees equals 1, BD simply equals CD.

(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) Alternatively, use Pythagoras theorem with BD = h and CD = h since the triangle is isosceles right-angled.

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

Answer:
1. In right triangle AEC, \( \tan 60^{\circ} = \dfrac{EC}{AE} \), and AE = BD = h (since ABDE forms a rectangle), so \( \sqrt{3} = \dfrac{h+40}{h} \).
2. Solving: \( h = \dfrac{40}{\sqrt{3}-1} = 20(\sqrt{3}+1) = 20 \times 2.73 = 54.6 \) m.
3. Total height \( CE = h + 40 = 54.6 + 40 = 94.6 \) m.

Teacher's Note:
a) Recognise that AE equals BD since ABDE is a rectangle, both representing the same horizontal distance.
b) Rationalise the denominator carefully before substituting the value of root 3.

OR

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

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

Teacher's Note:
a) Use the cosine ratio since AC is the hypotenuse and AE is the adjacent side to the 60 degree angle.
b) A quick sanity check: AE must be less than AC, and 50 m satisfies this.

Please click the link below to download pdf file of CBSE Class 10 Maths (Standard) Question Paper 2025 Solved Code 30-1-1