CBSE Class 10 Maths (Standard) Question Paper 2026 Solved Code 30-1-2

SECTION A

1. The natural number 2 is : [1 Mark]
(a) a prime number
(b) a composite number
(c) prime as well as composite
(d) neither prime nor composite

Answer: (a) a prime number

Teacher's Note:
a) 2 is the only even prime number.
b) Remember a prime number has exactly two factors, 1 and itself.

2. For any natural number \( n \), \( 6^{n} \) ends with the digit : [1 Mark]
(a) 0
(b) 6
(c) 3
(d) 2

Answer: (b) 6

Teacher's Note:
a) \( 6^{n} = 2^{n} \times 3^{n} \), and since it has no factor 5, it can never end in 0.
b) Check the unit digit of \( 6^{1}, 6^{2} \) - it is always 6.

3. The HCF of 960 and 432 is : [1 Mark]
(a) 48
(b) 54
(c) 72
(d) 36

Answer: (a) 48

Teacher's Note:
a) Use Euclid's division algorithm or prime factorisation to find HCF.
b) \( 960 = 2^{6} \times 3 \times 5 \), \( 432 = 2^{4} \times 3^{3} \); HCF \( = 2^{4} \times 3 = 48 \).

4. For an event E, \( P(E) + P(\overline{E}) = x \), then the value of \( x^{2} - 3 \) is : [1 Mark]
(a) \( -2 \)
(b) 2
(c) 1
(d) \( -1 \)

Answer: (a) \( -2 \)

Teacher's Note:
a) \( P(E) + P(\overline{E}) = 1 \) always, so \( x = 1 \).
b) Substitute \( x = 1 \) to get \( 1 - 3 = -2 \).

5. The graph of \( y = f(x) \) is given. The number of zeroes of \( f(x) \) is : [1 Mark]
(a) 0
(b) 1
(c) 2
(d) 4

[Figure: A smooth hump-shaped curve lies entirely above the X-axis between two arms going upward on both sides; the curve does not touch or cross the X-axis at any point. X'OX is the x-axis and YOY' is the y-axis.]

Answer: (a) 0

Teacher's Note:
a) The zeroes of \( f(x) \) are the x-coordinates of points where the graph cuts the x-axis.
b) Since the curve never touches the x-axis, it has no zeroes.

6. If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has : [1 Mark]
(a) a unique solution
(b) two solutions
(c) no solution
(d) an infinite number of solutions

Answer: (d) an infinite number of solutions

Teacher's Note:
a) Coincident lines means every point on one line lies on the other.
b) This happens when \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).

7. The first term of an AP is \( p \) and the common difference is \( q \), then its 10th term is : [1 Mark]
(a) \( q - 9p \)
(b) \( p - 9q \)
(c) \( p + 9q \)
(d) \( 2p + 9q \)

Answer: (c) \( p + 9q \)

Teacher's Note:
a) Use \( a_n = a + (n-1)d \).
b) Here \( a_{10} = p + 9q \).

8. Shown below are three triangles. The measures of two adjacent sides and included angle are given for each triangle : Which of these triangles are similar ? [1 Mark]
(a) \( \Delta \) RPQ and \( \Delta \) XZY
(b) \( \Delta \) RPQ and \( \Delta \) MNL
(c) \( \Delta \) XZY and \( \Delta \) MNL
(d) \( \Delta \) RPQ, \( \Delta \) XZY and \( \Delta \) MNL are similar to one another

[Figure: Triangle RPQ has PR = 6 cm, PQ = 4 cm and included angle P = 60 degrees. Triangle XYZ has XZ = 9 cm, YZ = 6 cm and included angle Z = 60 degrees. Triangle LMN has LN = 3 cm, MN = 4 cm and included angle N = 60 degrees.]

Answer: (a) \( \Delta \) RPQ and \( \Delta \) XZY

Teacher's Note:
a) Use SAS similarity: check if the ratio of two adjacent sides is equal with the same included angle.
b) Here \( \frac{PR}{ZX} = \frac{PQ}{ZY} = \frac{2}{3} \) with equal included angles, so they are similar.

9. The mid-point of the line segment joining the points \( (5, -4) \) and \( (6, 4) \) lies on : [1 Mark]
(a) \( x \)-axis
(b) \( y \)-axis
(c) origin
(d) neither \( x \)-axis nor \( y \)-axis

Answer: (a) \( x \)-axis

Teacher's Note:
a) Midpoint formula: \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \).
b) Here midpoint is \( (5.5, 0) \), which lies on the x-axis since y-coordinate is 0.

10. If \( \cos y = 0 \), then what is the value of \( \frac{1}{2} \cos \frac{y}{2} \) ? [1 Mark]
(a) 0
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{\sqrt{2}} \)
(d) \( \frac{1}{2\sqrt{2}} \)

Answer: (d) \( \frac{1}{2\sqrt{2}} \)

Teacher's Note:
a) \( \cos y = 0 \) gives \( y = 90^{\circ} \), so \( \frac{y}{2} = 45^{\circ} \).
b) \( \frac{1}{2} \cos 45^{\circ} = \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}} \).

11. If \( \cos A = \frac{1}{2} \), then the value of \( \sin^{2} A + 2 \cos^{2} A \) is : [1 Mark]
(a) \( \frac{3}{2} \)
(b) \( \frac{5}{4} \)
(c) \( -1 \)
(d) \( \frac{1}{2} \)

Answer: (b) \( \frac{5}{4} \)

Teacher's Note:
a) Rewrite as \( \sin^{2}A + \cos^{2}A + \cos^{2}A = 1 + \cos^{2}A \).
b) Substitute \( \cos^{2}A = \frac{1}{4} \) to get \( \frac{5}{4} \).

12. A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is \( 10\sqrt{3} \) m away from the base of the tower, is : [1 Mark]
(a) \( 30^{\circ} \)
(b) \( 45^{\circ} \)
(c) \( 90^{\circ} \)
(d) \( 60^{\circ} \)

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

Teacher's Note:
a) \( \tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3} \).
b) \( \sqrt{3} = \tan 60^{\circ} \), so the angle is \( 60^{\circ} \).

13. If TP and TQ are two tangents to a circle with centre O from an external point T so that \( \angle \) POQ \( = 120^{\circ} \), then \( \angle \) PTQ is equal to : [1 Mark]
(a) \( 60^{\circ} \)
(b) \( 70^{\circ} \)
(c) \( 80^{\circ} \)
(d) \( 90^{\circ} \)

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

Teacher's Note:
a) OPTQ is a cyclic quadrilateral with two right angles at P and Q.
b) \( \angle PTQ = 180^{\circ} - \angle POQ = 60^{\circ} \).

14. In the given figure, PA is a tangent from an external point P to a circle with centre O. If \( \angle \) POB \( = 125^{\circ} \), 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. PA is a tangent from an external point P touching the circle at A. B is another point on the circle such that A, O and B are collinear (AB is a diameter). Angle POB = 125 degrees is marked at O.]

Answer: (d) \( 35^{\circ} \)

Teacher's Note:
a) Since AOB is a straight line, \( \angle AOP = 180^{\circ} - 125^{\circ} = 55^{\circ} \).
b) In triangle OAP, \( \angle OAP = 90^{\circ} \) (tangent perpendicular to radius), so \( \angle APO = 180^{\circ} - 90^{\circ} - 55^{\circ} = 35^{\circ} \).

15. The length of the arc of the sector of a circle with radius 21 cm and of central angle \( 60^{\circ} \), is : [1 Mark]
(a) 22 cm
(b) 44 cm
(c) 88 cm
(d) 11 cm

Answer: (a) 22 cm

Teacher's Note:
a) Arc length \( = \frac{\theta}{360} \times 2\pi r \).
b) \( = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22 \) cm.

16. The hour hand of a clock is 7 cm long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is : [1 Mark]
(a) \( \left( \frac{35}{4} \right)^{\circ} \)
(b) \( \left( \frac{35}{2} \right)^{\circ} \)
(c) \( 35^{\circ} \)
(d) \( 70^{\circ} \)

Answer: (c) \( 35^{\circ} \)

Teacher's Note:
a) The hour hand sweeps \( 0.5^{\circ} \) per minute.
b) Time elapsed is 70 minutes, so angle \( = 70 \times 0.5 = 35^{\circ} \).

17. The total surface area of a solid hemisphere of diameter '2d' is : [1 Mark]
(a) \( 3 \pi d^{2} \)
(b) \( 2 \pi d^{2} \)
(c) \( \frac{1}{2} \pi d^{2} \)
(d) \( \frac{3}{4} \pi d^{2} \)

Answer: (a) \( 3 \pi d^{2} \)

Teacher's Note:
a) Total surface area of a hemisphere of radius \( r \) is \( 3\pi r^{2} \).
b) Here radius \( = d \), so TSA \( = 3\pi d^{2} \).

18. If the mean and mode of a data are 12 and 21 respectively, then its median is : [1 Mark]
(a) 6
(b) 13.5
(c) 15
(d) 14

Answer: (c) 15

Teacher's Note:
a) Use the empirical relation Mode \( = 3 \) Median \( - 2 \) Mean.
b) \( 21 = 3 \times \text{Median} - 24 \Rightarrow \text{Median} = 15 \).

19. Assertion (A) : The probability that a leap year has 53 Mondays is \( \frac{2}{7} \). [1 Mark]
Reason (R) : The probability that a non-leap year has 53 Mondays is \( \frac{5}{7} \).
(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) A leap year has 366 days = 52 weeks + 2 extra days, giving P(53 Mondays) \( = \frac{2}{7} \), so Assertion is true.
b) A non-leap year has 365 days = 52 weeks + 1 extra day, so P(53 Mondays) \( = \frac{1}{7} \), not \( \frac{5}{7} \); Reason is false.

20. Assertion (A) : The polynomial \( p(y) = y^{2} + 4y + 3 \) has two zeroes. [1 Mark]
Reason (R) : A quadratic polynomial can have at most two zeroes.
(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: (b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) \( y^{2}+4y+3=(y+1)(y+3) \), giving zeroes \( -1 \) and \( -3 \), so Assertion is true.
b) The Reason is a general true fact about quadratics but does not specifically explain why this polynomial has exactly two zeroes.

SECTION B

21. (A) In \( \Delta \) ABC, DE \( \parallel \) BC. If AD \( = x \), DB \( = x - 2 \), AE \( = x + 2 \) and EC \( = x - 1 \), then find the value of \( x \). [2 Marks]

Answer:
1. Since DE \( \parallel \) BC, by Basic Proportionality Theorem, \( \frac{AD}{DB} = \frac{AE}{EC} \), so \( \frac{x}{x-2} = \frac{x+2}{x-1} \).
2. Cross-multiplying and simplifying gives \( x = 4 \).

Teacher's Note:
a) BPT applies only when the line is parallel to one side of the triangle.
b) Always cross check by substituting back the value of x to ensure all lengths are positive.

OR

(B) In the figure given above, \( \Delta \) ABC \( \sim \) \( \Delta \) XYZ, then find the values of \( x \) and \( y \). [2 Marks]

[Figure: Triangle ABC with AB = 4 cm, BC = 6 cm, AC = y; Triangle XYZ with XY = x, YZ = 7.2 cm, XZ = 6 cm.]

Answer:
1. Since \( \Delta ABC \sim \Delta XYZ \), \( \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ} \Rightarrow \frac{4}{x} = \frac{6}{7.2} = \frac{y}{6} \).
2. Solving, \( x = 4.8 \) cm and \( y = 5 \) cm.

Teacher's Note:
a) Write the ratio of corresponding sides in the same order as the similarity statement.
b) Simplify \( \frac{6}{7.2} = \frac{5}{6} \) first to make calculations easier.

22. The coordinates of the centre of a circle are \( (x - 7, 2x) \). Find the value(s) of 'x', if the circle passes through the point \( (-9, 11) \) and has radius \( 5\sqrt{2} \) units. [2 Marks]

Answer:
1. Distance from centre to the given point equals radius: \( \sqrt{(x-7+9)^{2}+(2x-11)^{2}} = 5\sqrt{2} \).
2. Squaring gives \( 5x^{2} - 40x + 75 = 0 \Rightarrow x^{2} - 8x + 15 = 0 \Rightarrow (x-5)(x-3) = 0 \).
3. Therefore \( x = 3 \) or \( x = 5 \).

Teacher's Note:
a) Use the distance formula and equate it to the given radius.
b) Both values of x should be checked as valid solutions unless a restriction is given.

23. (A) If \( \tan \theta = \frac{24}{7} \), then find the value of \( \sin \theta + \cos \theta \). [2 Marks]

Answer:
1. Using a right triangle with opposite = 24, adjacent = 7, hypotenuse \( = \sqrt{24^{2}+7^{2}} = 25 \).
2. \( \sin \theta = \frac{24}{25} \), \( \cos \theta = \frac{7}{25} \), so \( \sin \theta + \cos \theta = \frac{31}{25} \).

Teacher's Note:
a) Always find the hypotenuse using the Pythagoras theorem before finding sin and cos.
b) Keep the same right triangle for both ratios to avoid sign errors.

OR

(B) If \( \cot \theta = \frac{7}{8} \), then find the value of \( \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} \). [2 Marks]

Answer:
1. \( \frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^{2}\theta}{1-\cos^{2}\theta} = \frac{\cos^{2}\theta}{\sin^{2}\theta} = \cot^{2}\theta \).
2. Substituting \( \cot \theta = \frac{7}{8} \) gives \( \cot^{2}\theta = \frac{49}{64} \).

Teacher's Note:
a) Recognise the identity \( (1+\sin\theta)(1-\sin\theta) = 1-\sin^{2}\theta = \cos^{2}\theta \).
b) The final expression simplifies neatly to \( \cot^{2}\theta \), so avoid unnecessary expansion.

24. In the given figure, a circle with centre O is inscribed inside \( \Delta \) LMN. A and B are the points of tangency. Find \( \angle \)ANB. [2 Marks]

[Figure: Triangle LMN with an inscribed circle centred at O. A is the point of tangency on side LN, and B is the point of tangency on side MN. The reflex angle AOB is marked as 240 degrees.]

Answer:
1. \( \angle AOB = 360^{\circ} - 240^{\circ} = 120^{\circ} \) (non-reflex angle).
2. Since \( \angle OAN = \angle OBN = 90^{\circ} \) (tangent perpendicular to radius), \( \angle ANB = 360^{\circ} - (90^{\circ}+90^{\circ}+120^{\circ}) = 60^{\circ} \).

Teacher's Note:
a) The tangent is always perpendicular to the radius at the point of contact.
b) Use the angle sum property of a quadrilateral (360°) formed by O, A, N, B.

25. Find the zeroes of the quadratic polynomial \( x^{2} + 7x + 10 \), and verify the relationship between the zeroes and its coefficients. [2 Marks]

Answer:
1. \( x^{2}+7x+10 = (x+2)(x+5) \), so the zeroes are \( -2 \) and \( -5 \).
2. Sum of zeroes \( = -7 = -\frac{\text{coefficient of } x}{\text{coefficient of } x^{2}} \); product of zeroes \( = 10 = \frac{\text{constant term}}{\text{coefficient of } x^{2}} \), both verified.

Teacher's Note:
a) Factorise by splitting the middle term to find the zeroes quickly.
b) Remember the sign convention: sum of zeroes is negative of the ratio of coefficients.

SECTION C

26. Prove that \( \sqrt{2} \) is an irrational number. [3 Marks]

Answer:
1. Assume, to the contrary, that \( \sqrt{2} \) is rational, so \( \sqrt{2} = \frac{p}{q} \), where \( q \neq 0 \) and \( p, q \) are coprime integers.
2. Squaring gives \( 2q^{2} = p^{2} \), so \( p^{2} \) is divisible by 2, hence \( p \) is divisible by 2. Let \( p = 2a \).
3. Substituting, \( 4a^{2} = 2q^{2} \Rightarrow q^{2} = 2a^{2} \), so \( q \) is also divisible by 2, contradicting that \( p, q \) are coprime. Hence \( \sqrt{2} \) is irrational.

Teacher's Note:
a) This is a proof by contradiction; state the assumption clearly at the start.
b) Both steps showing p and q divisible by 2 must be shown to reach the contradiction.

27. (A) If \( x = h + a \cos \theta \), \( y = k + b \sin \theta \), then prove that : \( \left( \frac{x-h}{a} \right)^{2} + \left( \frac{y-k}{b} \right)^{2} = 1 \) [3 Marks]

Answer:
1. From \( x = h + a\cos\theta \), \( \frac{x-h}{a} = \cos\theta \).
2. From \( y = k + b\sin\theta \), \( \frac{y-k}{b} = \sin\theta \).
3. So \( \left(\frac{x-h}{a}\right)^{2} + \left(\frac{y-k}{b}\right)^{2} = \cos^{2}\theta + \sin^{2}\theta = 1 \).

Teacher's Note:
a) Express cos θ and sin θ separately before squaring and adding.
b) Use the fundamental identity \( \sin^{2}\theta+\cos^{2}\theta=1 \) to complete the proof.

OR

(B) Prove that : \( \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A} = 2 \operatorname{cosec} A \) [3 Marks]

Answer:
1. LHS \( = \frac{\sin A/\cos A}{1+1/\cos A} - \frac{\sin A/\cos A}{1-1/\cos A} = \frac{\sin A}{\cos A+1} - \frac{\sin A}{\cos A-1} \).
2. Combining over a common denominator: \( \sin A \left( \frac{-2}{-\sin^{2}A} \right) \).
3. This simplifies to \( \frac{2}{\sin A} = 2 \operatorname{cosec} A = \) RHS.

Teacher's Note:
a) Convert tan A and sec A to sin A and cos A first.
b) Use \( \cos^{2}A - 1 = -\sin^{2}A \) to simplify the denominator.

28. (A) In the given figure, \( \Delta \) ABC is a right triangle in which \( \angle \) B \( = 90^{\circ} \), AB \( = 4 \) cm and BC \( = 3 \) cm. Find the radius of the circle inscribed in the triangle ABC. [3 Marks]

[Figure: Right triangle ABC with the right angle at B, AB = 4 cm, BC = 3 cm, and a circle inscribed touching all three sides.]

Answer:
1. \( AC = \sqrt{3^{2}+4^{2}} = 5 \) cm (by Pythagoras theorem).
2. Area of \( \Delta ABC = \frac{1}{2} \times 4 \times 3 = 6 \) cm\(^{2}\).
3. Using Area \( = \frac{1}{2} r (AB+BC+AC) \Rightarrow 6 = \frac{1}{2} r (4+3+5) \Rightarrow r = 1 \) cm.

Teacher's Note:
a) Alternatively, use \( r = \frac{AB+BC-AC}{2} \) for a right triangle's inradius.
b) Always find the hypotenuse first using Pythagoras theorem.

OR

(B) In the given figure, if a circle touches the side QR of \( \Delta \) PQR at S and extended sides PQ and PR at M and N respectively, then prove that : \( PM = \frac{1}{2} (PQ + QR + PR) \) [3 Marks]

[Figure: Triangle PQR with a circle touching side QR at S, and touching the extended sides PQ and PR at M and N respectively.]

Answer:
1. Since tangents from an external point are equal: \( PM = PN \), \( QS = QM \), \( RS = RN \).
2. \( PM + PN = PQ + QM + PR + RN \Rightarrow 2PM = PQ + QS + PR + RS \).
3. This simplifies to \( 2PM = PQ + QR + PR \), so \( PM = \frac{1}{2}(PQ+QR+PR) \).

Teacher's Note:
a) The key property used is that tangent segments from an external point are equal in length.
b) Carefully replace QM with QS and RN with RS to get QR in the final expression.

29. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use \( \pi = \frac{22}{7} \)) [3 Marks]

Answer:
1. Radius of cylinder = radius of hemisphere \( = \frac{7}{2} \) cm.
2. Height of cylinder \( = 20 - \frac{7}{2} - \frac{7}{2} = 13 \) cm.
3. Total volume \( = \pi r^{2} h + 2 \times \frac{2}{3}\pi r^{3} = \frac{22}{7} \times \left(\frac{7}{2}\right)^{2} \times \left(13+\frac{14}{3}\right) = \frac{4081}{6} \approx 680.17 \) cm\(^{3}\).

Teacher's Note:
a) Height of the cylindrical part is total height minus radii of both hemispherical ends.
b) Combine the cylinder and hemisphere volume formulas by taking \( \pi r^{2} \) common.

30. Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :
(i) same number appears on both the dice ?
(ii) different number appears on both the dice ? [3 Marks]

Answer:
1. Total possible outcomes = 36, being all ordered pairs (1,1), (1,2), ..., (6,6).
2. P(same number on both dice) \( = \frac{6}{36} = \frac{1}{6} \).
3. P(different numbers on both dice) \( = 1 - \frac{1}{6} = \frac{5}{6} \).

Teacher's Note:
a) Listing all 36 outcomes systematically avoids missing any pair.
b) Same number outcomes are the 6 pairs (1,1) to (6,6).

31. Parthi and Alisha found a treasure that is exactly on the straight line joining their locations. Parthi's location is at point \( (-6, -5) \) and Alisha's location is at point \( (10, 11) \). The distance from the treasure to Parthi's location is three times that of the distance to Alisha's location. Find the coordinates of the location of the treasure. [3 Marks]

Answer:
1. Let the treasure T divide PA in the ratio \( PT:TA = 3:1 \), where P \( = (-6,-5) \) and A \( = (10,11) \).
2. Using the section formula: \( T = \left( \frac{3 \times 10 + 1 \times (-6)}{3+1}, \frac{3 \times 11 + 1 \times (-5)}{3+1} \right) \).
3. This gives \( T = (6, 7) \).

Teacher's Note:
a) Identify the correct ratio direction: treasure is closer to Alisha, so it divides PA in ratio 3:1 from P.
b) Apply the section formula carefully, keeping numerator and denominator terms consistent.

SECTION D

32. (A) A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains. [5 Marks]

Answer:
1. Let the speed of the faster train be \( x \) km/h, so the speed of the slower train \( = (x-10) \) km/h.
2. Time difference equation: \( \frac{200}{x-10} - \frac{200}{x} = 1 \).
3. Simplifying gives \( x^{2} - 10x - 2000 = 0 \).
4. Factorising: \( (x-50)(x+40) = 0 \), so \( x = 50 \) (rejecting negative value \( x = -40 \)).
5. Hence, speed of faster train \( = 50 \) km/h and speed of slower train \( = 40 \) km/h.

Teacher's Note:
a) Always reject the negative root since speed cannot be negative.
b) Set up the time equation carefully: slower train takes more time, so it is subtracted second.

OR

(B) The sum of the areas of two squares is 640 m\(^{2}\). If the difference in their perimeters is 64 m, find the sides of the two squares. [5 Marks]

Answer:
1. Let the sides of the two squares be \( x \) m and \( y \) m, with \( x \gt y \).
2. \( x^{2} + y^{2} = 640 \) and \( 4x - 4y = 64 \Rightarrow y = x - 16 \).
3. Substituting: \( x^{2} + (x-16)^{2} = 640 \Rightarrow x^{2} - 16x - 192 = 0 \).
4. Factorising: \( (x-24)(x+8) = 0 \), so \( x = 24 \) (rejecting \( x = -8 \)).
5. Hence \( y = 24 - 16 = 8 \). The sides of the two squares are 24 m and 8 m.

Teacher's Note:
a) Convert the perimeter condition into a relation between the sides before substituting.
b) Reject negative side lengths since sides of a square must be positive.

33. (A) State and prove Basic Proportionality Theorem. [5 Marks]

Answer:
1. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. Given: In \( \Delta ABC \), DE \( \parallel \) BC, with D on AB and E on AC. To prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).
3. Construction: Join BE and CD, and draw \( DM \perp AC \) and \( EN \perp AB \).
4. Proof: Compare areas of triangles ADE, BDE and ADE, CDE using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \); since \( \Delta BDE \) and \( \Delta CDE \) have equal areas (same base DE, between the same parallels), we get \( \frac{AD}{DB} = \frac{AE}{EC} \).
5. Hence the theorem is proved.

Teacher's Note:
a) A clear, labelled figure with the construction lines earns key marks.
b) The area ratio argument, ar(BDE) = ar(CDE), is the crucial step of the proof.

OR

(B) In the given figure, CM and RN are respectively the medians of \( \Delta \) ABC and \( \Delta \) PQR. If \( \Delta \) ABC \( \sim \) \( \Delta \) PQR, then prove that :
(i) \( \Delta \) AMC \( \sim \) \( \Delta \) PNR
(ii) \( \Delta \) CMB \( \sim \) \( \Delta \) RNQ [5 Marks]

[Figure: Triangle ABC with median CM (M is the midpoint of AB), and triangle PQR with median RN (N is the midpoint of PQ).]

Answer:
1. Since \( \Delta ABC \sim \Delta PQR \), \( \angle A = \angle P \) and \( \frac{AB}{PQ} = \frac{AC}{PR} \).
2. Since M, N are midpoints, \( AB = 2AM \) and \( PQ = 2PN \), so \( \frac{AM}{PN} = \frac{AC}{PR} \). With \( \angle A = \angle P \), by SAS, \( \Delta AMC \sim \Delta PNR \).
3. Similarly, \( \angle B = \angle Q \) and \( \frac{BC}{QR} = \frac{2MB}{2NQ} \), giving \( \frac{MB}{NQ} = \frac{BC}{QR} \).
4. With \( \angle B = \angle Q \), by SAS, \( \Delta CMB \sim \Delta RNQ \).

Teacher's Note:
a) Use the fact that a median divides the opposite side into two equal halves.
b) Apply the SAS similarity criterion using the given similarity of the original triangles.

34. The monthly expenditure on fruits in 200 families of a Housing Society is given below. Find the value of \( x \) and also find the mode and mean expenditure on fruits.
Monthly Expenditure (in Rs.): 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000
No. of Families: 24 | 40 | 33 | 28 | x | 22 | 16 | 7 [5 Marks]

Answer:
1. Total families: \( 24+40+33+28+x+22+16+7 = 200 \Rightarrow 170+x = 200 \Rightarrow x = 30 \).
2. The modal class is 1500-2000 (highest frequency, 40). Using Mode \( = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \):
Mode \( = 1500 + \frac{40-24}{2(40)-24-33} \times 500 = 1500 + 347.83 = 1847.83 \) (approx.).
3. Using the direct method with midpoints \( x_i \), \( \sum f_i x_i = 532500 \) and \( \sum f_i = 200 \).
4. Mean \( = \frac{532500}{200} = 2662.50 \).
5. Hence, \( x = 30 \), modal expenditure \( \approx \) Rs. 1847.83 and mean expenditure = Rs. 2662.50.

Teacher's Note:
a) Find x first using the total frequency before calculating mode and mean.
b) Identify the modal class correctly as the one with the highest frequency.
c) The assumed mean method with step deviation can also be used to simplify mean calculation.

35. Two sections, A and B, of class X contributed a total of Rs. 1500 for the Uttarakhand flood victims. The contribution from X-A was Rs. 100 less than that of X-B. Graphically, find the amounts contributed by both sections. [5 Marks]

Answer:
1. Let the amounts contributed by X-A and X-B be Rs. \( x \) and Rs. \( y \) respectively.
2. The equations are \( x + y = 1500 \) and \( y - x = 100 \).
3. Plotting both lines on a graph, they intersect at the point \( (700, 800) \).
4. Hence, amount contributed by class X-A = Rs. 700 and by class X-B = Rs. 800.

Teacher's Note:
a) Plot at least three points for each line for an accurate graph.
b) The point of intersection of the two lines gives the required solution.

SECTION E

36. Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure). Distance between the base of the tower and point 'O' is 6 m. From point 'O', the angle of elevation of the top of the section 'B' is 30° and the angle of elevation of the top of section 'A' is 60°.

[Figure: A vertical tower PA with an intermediate point B on it. P is the base on the ground, O is a point on the ground 6 m from P. From O, the angle of elevation to B is 30 degrees and to A is 60 degrees.]

Based on the above information, answer the following questions :

(i) Find the length of the wire from the point 'O' to the top of section 'B'. [1 Mark]

Answer: Using \( \cos 30^{\circ} = \frac{6}{OB} \), we get \( OB = \frac{12}{\sqrt{3}} = 4\sqrt{3} \) m.

Teacher's Note:
a) OB is the hypotenuse of the right triangle formed with the base OP = 6 m.
b) Alternatively, use \( \sin 30^{\circ} = \frac{BP}{OB} \) after finding BP.

(ii) Find the length of the wire from the point 'O' to the top of section 'A'. [1 Mark]

Answer: Using \( \cos 60^{\circ} = \frac{6}{OA} \), we get \( OA = 12 \) m.

Teacher's Note:
a) OA is the hypotenuse for the 60 degree angle of elevation.
b) A larger elevation angle gives a shorter horizontal reach but a longer or comparable hypotenuse depending on the base.

(iii) (a) Find the distance AB. [2 Marks]

Answer:
1. \( \tan 30^{\circ} = \frac{BP}{6} \Rightarrow BP = 6 \tan 30^{\circ} = 2\sqrt{3} \) m.
2. \( \tan 60^{\circ} = \frac{AP}{6} \Rightarrow AP = 6\sqrt{3} \) m.
3. \( AB = AP - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \) m.

Teacher's Note:
a) Find the heights AP and BP separately using tangent ratios.
b) AB is simply the difference of the two heights since B lies on segment PA.

OR

(iii) (b) Find the area of \( \Delta \) OPB. [2 Marks]

Answer:
1. \( \tan 30^{\circ} = \frac{BP}{6} \Rightarrow BP = 2\sqrt{3} \) m.
2. Area of \( \Delta OPB = \frac{1}{2} \times BP \times OP = \frac{1}{2} \times 2\sqrt{3} \times 6 = 6\sqrt{3} \) m\(^{2}\).

Teacher's Note:
a) OPB is a right triangle with the right angle at P.
b) Use base = OP and height = BP for the area formula.

37. A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 cm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure.

[Figure: A circle divided into 10 equal sectors by 5 diameters, resembling a wheel with spokes.]

Based on the above information, answer the following questions :

(i) What is the radius of circle ? [1 Mark]

Answer: Radius \( = \frac{35}{2} = 17.5 \) cm.

Teacher's Note:
a) Radius is always half the diameter.
b) Keep the value as a fraction for accurate further calculations.

(ii) What is the circumference of the brooch ? [1 Mark]

Answer: Circumference \( = 2 \times \frac{22}{7} \times \frac{35}{2} = 110 \) cm.

Teacher's Note:
a) Use the formula \( 2\pi r \) for circumference.
b) \( \pi = \frac{22}{7} \) simplifies nicely with the radius 17.5 cm.

(iii) (a) What is the total length of silver wire required ? [2 Marks]

Answer:
1. Total wire length = circumference + length of 5 diameters \( = 110 + 5 \times 35 \).
2. \( = 110 + 175 = 285 \) cm.

Teacher's Note:
a) Don't forget to add the wire used for the diameters, not just the circumference.
b) Each diameter is 35 cm, and there are 5 of them.

OR

(iii) (b) What is the area of each sector of the brooch ? [2 Marks]

Answer:
1. Central angle of each sector \( = \frac{360^{\circ}}{10} = 36^{\circ} \).
2. Area of each sector \( = \frac{36}{360} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^{2} = \frac{385}{4} = 96.25 \) cm\(^{2}\).

Teacher's Note:
a) Divide 360 degrees by the number of equal sectors to get the central angle.
b) Use the sector area formula \( \frac{\theta}{360} \times \pi r^{2} \).

38. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes, as shown in the figure : A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.

[Figure: A bucket at the starting point. The first potato is 5 m from the bucket, and the remaining 9 potatoes are placed in a straight line, each 3 m apart from the previous one, making a total of 10 potatoes.]

Based on the above information, answer the following questions :

(i) What is the distance covered to pick up the first potato and drop it in bucket ? [1 Mark]

Answer: Distance \( = 5 + 5 = 10 \) m.

Teacher's Note:
a) The competitor runs to the potato and back, so distance is doubled.
b) This becomes the first term of an AP.

(ii) What is the distance covered to pick up the second potato and drop it in bucket ? [1 Mark]

Answer: Second potato is at \( 5+3=8 \) m from the bucket, so distance \( = 8+8 = 16 \) m.

Teacher's Note:
a) Each successive potato is 3 m farther, doubling the run distance by 6 m each time.
b) This confirms an AP with common difference 6.

(iii) (a) What is the total distance the competitor has to run ? [2 Marks]

Answer:
1. Distances covered form an AP with first term \( a = 10 \) and common difference \( d = 6 \), for \( n = 10 \) potatoes.
2. \( S_{10} = \frac{10}{2}[2(10)+9(6)] = 5 \times 74 = 370 \) m.

Teacher's Note:
a) Use the AP sum formula \( S_n = \frac{n}{2}[2a+(n-1)d] \).
b) There are exactly 10 terms since there are 10 potatoes.

OR

(iii) (b) If average speed of competitor is 5 m/s, then find the average time taken by competitor to put all the potatoes in the bucket. [2 Marks]

Answer:
1. Total distance \( S_{10} = 370 \) m (as calculated using the AP with \( a=10, d=6 \)).
2. Time \( = \frac{370}{5} = 74 \) seconds.

Teacher's Note:
a) First find the total distance using the AP sum formula before dividing by speed.
b) Time = Distance ÷ Speed, keeping units consistent (metres and m/s give seconds).

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