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

SECTION A

1. 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 prime factorisation or Euclid's division algorithm to find HCF quickly.
b) \(960 = 2^{6} \times 3 \times 5\) and \(432 = 2^{4} \times 3^{3}\), so HCF \( = 2^{4} \times 3 = 48\).

2. 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.

3. 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) \(6^{n} = (2 \times 3)^{n}\), which never ends in 0 as 5 is not a factor.
b) Check the unit digit pattern: \(6^1=6, 6^2=36, 6^3=216\), always ending in 6.

4. 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: Graph of \(y = f(x)\) showing a curve with two upward humps lying entirely above the x-axis; the curve does not touch or cross the x-axis anywhere.]

Answer: (a) 0

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

5. 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 mean the same line is drawn twice, so every point on it satisfies both equations.
b) Compare ratios: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) gives infinite solutions.

6. The common difference of the AP : \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, ..... \) is : [1 Mark]
(a) \( \sqrt{2} \)
(b) 1
(c) \( 2\sqrt{2} \)
(d) \( -\sqrt{2} \)

Answer: (a) \( \sqrt{2} \)

Teacher's Note:
a) Common difference \(d = a_2 - a_1\).
b) \(d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}\).

7. If \( \triangle ABC \) and \( \triangle DEF \) are similar such that \(2 AB = DE\) and \(BC = 8\) cm, then EF is equal to : [1 Mark]
(a) 4 cm
(b) 8 cm
(c) 12 cm
(d) 16 cm

Answer: (d) 16 cm

Teacher's Note:
a) In similar triangles, corresponding sides are proportional: \( \frac{AB}{DE} = \frac{BC}{EF} \).
b) Since \(DE = 2AB\), the ratio \( \frac{AB}{DE} = \frac{1}{2}\), so \(EF = 2 \times BC = 16\) cm.

8. 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) Midpoint is \(\left(5.5, 0\right)\); since y-coordinate is 0, it lies on the x-axis.

9. Given that \( \sin \theta = \frac{a}{b} \), then \( \cos \theta \) is equal to : [1 Mark]
(a) \( \frac{b}{\sqrt{b^2-a^2}} \)
(b) \( \frac{b}{a} \)
(c) \( \frac{\sqrt{b^2-a^2}}{b} \)
(d) \( \frac{a}{\sqrt{b^2-a^2}} \)

Answer: (c) \( \frac{\sqrt{b^2-a^2}}{b} \)

Teacher's Note:
a) Use a right triangle with perpendicular \(a\) and hypotenuse \(b\).
b) Base \( = \sqrt{b^2-a^2}\), so \( \cos \theta = \frac{\text{base}}{\text{hypotenuse}} \).

10. 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) \(1 + \frac{1}{4} = \frac{5}{4}\).

11. 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) Use \( \tan \theta = \frac{\text{height}}{\text{distance}} \).
b) \( \tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3} \Rightarrow \theta = 60^{\circ} \).

12. 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 forms a quadrilateral where \( \angle OPT = \angle OQT = 90^{\circ} \).
b) Sum of angles of quadrilateral is \(360^{\circ}\), so \( \angle PTQ = 360 - 90 - 90 - 120 = 60^{\circ} \).

13. 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 external point P touching the circle at A. B is another point on the circle such that angle POB = 125 degrees, with P, A, O, B forming the given configuration.]

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

Teacher's Note:
a) Since A, O, B are collinear on the diameter side, \( \angle POA = 180 - 125 = 55^{\circ} \).
b) In right triangle OAP, \( \angle OAP = 90^{\circ} \), so \( \angle APO = 90 - 55 = 35^{\circ} \).

14. 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.

15. 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 \( = 70 \times 0.5 = 35^{\circ}\).

16. 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) TSA of a hemisphere of radius \(r\) is \(3 \pi r^2\).
b) Here radius \(r = d\), so TSA \( = 3 \pi d^2\).

17. 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) \(21 = 3 \times \text{Median} - 24 \Rightarrow \text{Median} = 15\).

18. A die is thrown once. Probability of getting a number other than 3 is : [1 Mark]
(a) \( \frac{1}{6} \)
(b) \( \frac{3}{6} \)
(c) \( \frac{5}{6} \)
(d) 1

Answer: (c) \( \frac{5}{6} \)

Teacher's Note:
a) There are 5 favourable outcomes (1,2,4,5,6) out of 6.
b) Probability \( = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{5}{6}\).

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

20. Assertion (A) : The polynomial \( p(y) = y^2 + 4y + 3 \) has two zeroes.
Reason (R) : A quadratic polynomial can have at most two zeroes. [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: (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) \(p(y) = (y+1)(y+3)\), giving zeroes \(-1\) and \(-3\), so the Assertion is true.
b) The Reason is a general true fact but does not specifically explain why this particular polynomial has exactly two zeroes.

SECTION B

21. If \( \alpha, \beta \) are the zeroes of the polynomial \( p(x) = x^2 - 3x - 1 \), then find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \). [2 Marks]

Answer:
1. Here \( \alpha + \beta = 3 \) and \( \alpha \beta = -1 \).
2. \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{3}{-1} = -3 \).

Teacher's Note:
a) Use sum and product of zeroes from \( p(x) = ax^2+bx+c \): sum \( = -\frac{b}{a}\), product \( = \frac{c}{a}\).
b) Combine the fraction before substituting values to avoid errors.

22. (A) In \( \triangle ABC \), 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, \( \frac{AD}{DB} = \frac{AE}{EC} \).
2. \( \frac{x}{x-2} = \frac{x+2}{x-1} \Rightarrow x(x-1) = (x+2)(x-2) \Rightarrow x^2 - x = x^2 - 4 \Rightarrow x = 4 \).

Teacher's Note:
a) BPT applies only when the line is parallel to one side of the triangle.
b) Cross-multiply carefully to avoid sign errors while simplifying.

OR

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

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

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

Teacher's Note:
a) Match corresponding vertices carefully: A↔X, B↔Y, C↔Z.
b) Write all three ratios equal before solving for the unknowns.

23. 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 point on circle equals radius: \( \sqrt{(x-7+9)^2+(2x-11)^2} = 5\sqrt{2} \).
2. \( (x+2)^2 + (2x-11)^2 = 50 \Rightarrow 5x^2 - 40x + 75 = 0 \Rightarrow x^2 - 8x + 15 = 0 \Rightarrow (x-5)(x-3)=0 \).
3. So \(x = 3\) or \(x = 5\).

Teacher's Note:
a) Use the distance formula between the centre and a point on the circle equal to the radius.
b) Both solutions of the quadratic are valid unless the question restricts \(x\) further.

24. (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 perpendicular = 24, base = 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 Pythagoras theorem before writing sin and cos.
b) Keep the same right triangle for both ratios to avoid inconsistency.

OR

(B) If \( \cot \theta = \frac{7}{8} \) is, 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. \( \cot^2\theta = \left( \frac{7}{8} \right)^2 = \frac{49}{64} \).

Teacher's Note:
a) Use the identity \((1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta = \cos^2\theta\).
b) Recognise the final simplified ratio directly as \(\cot^2\theta\) to save time.

25. Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle. [2 Marks]

Answer:
1. The perpendicular from the centre O to the chord AB (which is tangent to the smaller circle) meets AB at M, with OM = 4 cm (radius of smaller circle) and OA = 5 cm.
2. By Pythagoras theorem, AM \( = \sqrt{5^2-4^2} = 3 \) cm, so AB \( = 2 \times 3 = 6 \) cm.

Teacher's Note:
a) The perpendicular from the centre bisects the chord.
b) Use the radius of the smaller circle as the perpendicular distance from the centre to the chord.

SECTION C

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

Answer:
1. Let \( \sqrt{3} \) be rational, so \( \sqrt{3} = \frac{p}{q} \), where \(q \neq 0\) and \(p, q\) are coprime integers.
2. Squaring, \(3q^2 = p^2\), so \(p^2\) is divisible by 3, which means \(p\) is divisible by 3. Let \(p = 3a\).
3. Then \(9a^2 = 3q^2 \Rightarrow q^2 = 3a^2\), so \(q\) is also divisible by 3.
4. This contradicts the assumption that \(p, q\) are coprime. Hence \( \sqrt{3} \) is irrational.

Teacher's Note:
a) This is the standard proof by contradiction; state the assumption clearly at the start.
b) Both \(p\) and \(q\) being divisible by 3 is the key contradiction that earns full marks.

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

Answer:
1. Let the point P divide AB in the ratio \(k:1\), so \(P = \left( \frac{-4k-6}{k+1}, \frac{-k+5}{k+1} \right)\).
2. Since P lies on the x-axis, \( \frac{-k+5}{k+1} = 0 \Rightarrow k = 5\). So the required ratio is \(5:1\).
3. Substituting \(k=5\), coordinates of P are \(\left( \frac{-26}{6}, 0 \right) = \left( -\frac{13}{3}, 0 \right)\).

Teacher's Note:
a) The x-axis divides a segment where the y-coordinate of the section point becomes zero.
b) Use the section formula \( \left( \frac{kx_2+x_1}{k+1}, \frac{ky_2+y_1}{k+1} \right) \) carefully with correct order of points.

28. (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 \(x = h + a\cos\theta\), we get \( \frac{x-h}{a} = \cos\theta\).
2. From \(y = k + b\sin\theta\), we get \( \frac{y-k}{b} = \sin\theta\).
3. Squaring and adding: \( \left( \frac{x-h}{a} \right)^2 + \left( \frac{y-k}{b} \right)^2 = \cos^2\theta + \sin^2\theta = 1\).

Teacher's Note:
a) Isolate \(\cos\theta\) and \(\sin\theta\) first before squaring.
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. \( = \sin A \left( \frac{-2}{-\sin^2 A} \right) = \frac{2}{\sin A}\).
3. \( = 2 \operatorname{cosec} A = \) RHS.

Teacher's Note:
a) Multiply numerator and denominator by \(\cos A\) to simplify secant terms into sine and cosine.
b) Combine fractions carefully using \((\cos A+1)(\cos A-1)=\cos^2A-1=-\sin^2A\).

29. (A) In the given figure, \( \triangle ABC \) 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 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 (hypotenuse).
2. Let the inradius be \(r\); using tangent lengths, \(BD = BE = r\), \(AD = 4-r\), \(CE = 3-r\), and \(AD+CE = AC\).
3. \((4-r)+(3-r)=5 \Rightarrow r = 1\) cm.

Teacher's Note:
a) Alternatively use \(r = \frac{a+b-c}{2}\) for a right triangle, where \(c\) is the hypotenuse.
b) Tangent lengths from an external point to a circle are equal - this is the key property used here.

OR

(B) In the given figure, if a circle touches the side QR of \( \triangle PQR \) 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 points 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. \( = PQ + QR + PR\), so \(PM = \frac{1}{2}(PQ+QR+PR)\).

Teacher's Note:
a) Use the equal tangent-segment property from each external point (P, Q, R).
b) Note that \(QS+RS = QR\) is the key step linking the three tangent equalities.

30. 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. \( \left( \text{Use } \pi = \frac{22}{7} \right) \) [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 \( = \pi r^2 h + \frac{4}{3}\pi r^3 = \frac{22}{7} \times \left( \frac{7}{2} \right)^2 \times \left(13+\frac{14}{3}\right) = \frac{4081}{6} \approx 680.17\) cm³.

Teacher's Note:
a) Subtract both hemisphere radii from the total height to get the cylinder's height.
b) Combine the volume formulas of cylinder and two hemispheres (= one full sphere) for a quicker calculation.

31. 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, listed as ordered pairs (1,1) to (6,6).
2. (i) Same number on both dice: favourable outcomes = 6 (namely (1,1),(2,2),...,(6,6)), so probability \( = \frac{6}{36} = \frac{1}{6}\).
3. (ii) Different number on both dice: probability \( = 1 - \frac{1}{6} = \frac{5}{6}\).

Teacher's Note:
a) Always write the full sample space of 36 outcomes for two-dice problems.
b) Use the complement rule to quickly find the probability of "different number" once "same number" is known.

SECTION D

32. Determine graphically, the coordinates of vertices of a triangle whose equations are \( 2x-3y+6=0 \); \( 2x+3y-18=0 \) and \( x=0 \). Also, find the area of this triangle. [5 Marks]

[Figure: Graph showing three straight lines - \(2x-3y+6=0\), \(2x+3y-18=0\), and \(x=0\) (the y-axis) - intersecting to form a triangle with vertices at (0,2), (3,4) and (0,6).]

Answer:
1. Line \(2x-3y+6=0\) meets \(x=0\) at \((0,2)\) and meets \(2x+3y-18=0\) at \((3,4)\).
2. Line \(2x+3y-18=0\) meets \(x=0\) at \((0,6)\).
3. So the vertices of the triangle are A(0,2), B(3,4) and C(0,6).
4. Base AC (on the y-axis) = 4 units; height from B to the y-axis = 3 units.
5. Area \( = \frac{1}{2} \times 4 \times 3 = 6\) sq. units.

Teacher's Note:
a) Find intersection points of each pair of lines algebraically to plot the graph accurately.
b) When one side lies on the y-axis, area can be found easily using base and height instead of the coordinate formula.

33. (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 speed of slower train \( = (x-10)\) km/h.
2. \( \frac{200}{x-10} - \frac{200}{x} = 1\).
3. Simplifying: \(x^2 - 10x - 2000 = 0 \Rightarrow (x-50)(x+40)=0\).
4. So \(x = 50\) (rejecting negative value \(x=-40\)).
5. Speed of faster train = 50 km/h, speed of slower train = 40 km/h.

Teacher's Note:
a) Set up the time difference equation using \( \text{time} = \frac{\text{distance}}{\text{speed}} \).
b) Always reject negative speed values as they are not physically meaningful.

OR

(B) The sum of the areas of two squares is 640 m². If the difference in their perimeters is 64 m, find the sides of the two squares. [5 Marks]

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

Teacher's Note:
a) Convert the perimeter difference into a side difference by dividing by 4.
b) Substitute \(y\) in terms of \(x\) to reduce to a single-variable quadratic equation.

34. (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 \( \triangle ABC\), DE \( \parallel \) BC, meeting AB at D and AC at E. To prove: \( \frac{AD}{DB} = \frac{AE}{EC}\).
3. Construction: Join BE and CD; draw \(DM \perp AC\) and \(EN \perp AB\).
4. Proof: \( \frac{ar(\triangle ADE)}{ar(\triangle DBE)} = \frac{AD}{DB}\) and \( \frac{ar(\triangle ADE)}{ar(\triangle DEC)} = \frac{AE}{EC}\).
5. But \( \triangle DBE\) and \( \triangle DEC\) are on the same base DE and between the same parallels DE and BC, so \( ar(\triangle DBE) = ar(\triangle DEC)\). Hence \( \frac{AD}{DB} = \frac{AE}{EC}\).

Teacher's Note:
a) Draw a clear, labelled figure with the parallel line before starting the proof.
b) The key idea is that triangles on the same base between the same parallels have equal areas.

OR

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

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

Answer:
1. Since \( \triangle ABC \sim \triangle PQR\), \( \angle A = \angle P\) and \( \frac{AB}{PQ} = \frac{AC}{PR}\).
2. As CM and RN are medians, \(AB = 2AM\) and \(PQ = 2PN\), so \( \frac{2AM}{2PN} = \frac{AC}{PR} \Rightarrow \frac{AM}{PN} = \frac{AC}{PR}\).
3. With \( \angle A = \angle P\), by SAS similarity, \( \triangle AMC \sim \triangle PNR\).
4. Similarly, \( \angle B = \angle Q\) and \( \frac{AB}{PQ} = \frac{BC}{QR} \Rightarrow \frac{2MB}{2NQ} = \frac{BC}{QR} \Rightarrow \frac{MB}{NQ} = \frac{BC}{QR}\).
5. With \( \angle B = \angle Q\), by SAS similarity, \( \triangle CMB \sim \triangle RNQ\).

Teacher's Note:
a) Medians divide the opposite side into two equal halves - use this to relate AM, PN, MB, NQ to full sides.
b) Apply SAS similarity criterion by matching the included angle with the two proportional sides.

35. The mean of the following frequency distribution is 35. Find the values of \(x\) and \(y\), if the sum of frequencies is 25 :
Class: 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70
Frequency: 1 | x | 5 | 7 | y | 3 | 1 [5 Marks]

Answer:
1. Sum of frequencies: \(1+x+5+7+y+3+1 = 25 \Rightarrow x+y = 8\) ... (i)
2. Using midpoints 5,15,25,35,45,55,65: \( \sum f_i x_i = 605+15x+45y\).
3. Mean \( = \frac{605+15x+45y}{25} = 35 \Rightarrow 15x+45y = 270 \Rightarrow x+3y = 18\) ... (ii)
4. Solving (i) and (ii): \(2y = 10 \Rightarrow y = 5\), and \(x = 3\).

Teacher's Note:
a) Build a table with class marks and \(f_i x_i\) to avoid calculation errors.
b) Form two linear equations (from total frequency and from mean) and solve simultaneously.

SECTION E

36. 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 :

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

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.
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: The distance covered is \(5+5 = 10\) m.

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

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

Answer: The distance covered is \(8+8 = 16\) m (since the second potato is at \(5+3=8\) m from the bucket).

Teacher's Note:
a) Each successive potato is 3 m farther, so the round trip increases by 6 m each time.
b) This confirms the AP has first term 10 and common difference 6.

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

Answer:
1. The distances form an AP with \(a=10\), \(d=6\), \(n=10\).
2. \(S_{10} = \frac{10}{2}[2(10)+9(6)] = 5 \times 74 = 370\) m.

Teacher's Note:
a) Use the sum of AP formula \(S_n = \frac{n}{2}[2a+(n-1)d]\).
b) Double-check that \(n=10\) 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 covered \(= 370\) m (using the AP sum as above).
2. Average time \( = \frac{370}{5} = 74\) seconds.

Teacher's Note:
a) First find total distance using the AP sum before applying speed-time-distance relation.
b) Time \( = \frac{\text{distance}}{\text{speed}}\); keep units consistent (metres and m/s).

37. 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).

[Figure: A vertical tower PA divided into two sections; P is the base, B is a point partway up the tower, and A is the top. A wire runs from a point O on the ground (6 m from P) to B making an angle of elevation of 30 degrees, and another wire from O to A making an angle of elevation of 60 degrees.]

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 \(30^{\circ}\) and the angle of elevation of the top of section 'A' is \(60^{\circ}\).
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 \( \cos 30^{\circ} = \frac{6}{OB}\), \(OB = \frac{12}{\sqrt{3}} = 4\sqrt{3}\) m.

Teacher's Note:
a) Use the cosine ratio since OP (base) and OB (hypotenuse) are involved.
b) Rationalise the denominator to express the answer as \(4\sqrt{3}\) m.

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

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

Teacher's Note:
a) \( \cos 60^{\circ} = \frac{1}{2}\), so \(OA\) is simply twice the base distance.
b) Check that the angle used corresponds correctly to the top of section A.

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

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

Teacher's Note:
a) Find the heights of both A and B from the ground separately using tan ratios.
b) AB is the difference between the two heights, not their sum.

OR

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

Answer:
1. \(BP = 2\sqrt{3}\) m (as found using \(\tan 30^{\circ}\)).
2. Area of \( \triangle OPB = \frac{1}{2} \times BP \times OP = \frac{1}{2} \times 2\sqrt{3} \times 6 = 6\sqrt{3}\) m².

Teacher's Note:
a) Triangle OPB is right-angled at P, so use the two legs BP and OP directly for the area.
b) Keep the answer in surd form unless a decimal value is asked.

38. 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 of diameter 35 cm with 5 diameters drawn through the centre, dividing the circle into 10 equal sectors.]

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 (\( \frac{35}{2}\)) for easier calculation in later parts.

(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 \(C = 2\pi r\) or equivalently \(\pi d\).
b) Using \(\pi = \frac{22}{7}\) with radius \(\frac{35}{2}\) simplifies nicely since 35 is a multiple of 7.

(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) Remember to add the circumference and all 5 diameters together, not just the sectors.
b) Each diameter equals 35 cm since it passes through the full circle.

OR

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

Answer:
1. Central angle of each of the 10 equal sectors \( = \frac{360}{10} = 36^{\circ}\).
2. Area of each sector \( = \frac{36}{360} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25\) cm².

Teacher's Note:
a) Use the sector area formula \( \frac{\theta}{360} \times \pi r^2\).
b) With 10 equal sectors, each central angle is exactly \(36^{\circ}\).

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