CBSE Class 10 Mathematics Basic Sample Paper 2026 27 with Solutions PDF Download

SECTION A

1. If HCF (2520, 6600) = 40 and LCM (2520, 6600) = \( 252 \times k \), then the value of \( k \) is [1 Mark]
(A) 165
(B) 1600
(C) 1625
(D) 1650

Answer: (D) 1650

Teacher's Note:
a) Use HCF \( \times \) LCM = product of the two numbers: \( 40 \times 252 \times k = 2520 \times 6600 \).
b) \( k = \frac{2520 \times 6600}{40 \times 252} = \frac{16632000}{10080} = 1650 \).

2. If \( \alpha \) and \( \beta \) are the Zeroes of the polynomial \( 2x^{2} - 4x - 5 \), the value of \( (\alpha - \beta)^{2} \) is [1 Mark]
(A) 4
(B) 6
(C) 14
(D) 56

Answer: (C) 14

Teacher's Note:
a) \( \alpha + \beta = -\frac{b}{a} = 2 \) and \( \alpha\beta = \frac{c}{a} = -\frac{5}{2} \).
b) \( (\alpha - \beta)^{2} = (\alpha + \beta)^{2} - 4\alpha\beta = 4 - 4\left(-\frac{5}{2}\right) = 4 + 10 = 14 \).

3. For what value of \( p \) does the pair of Linear equations \( 4x + py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \) has a unique solution [1 Mark]
(A) \( p = 4 \) only
(B) \( p = 2 \) only
(C) \( p \neq 4 \)
(D) \( p \neq 2 \)

Answer: (C) \( p \neq 4 \)

Teacher's Note:
a) For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
b) \( \frac{4}{2} \neq \frac{p}{2} \Rightarrow p \neq 4 \).

4. The sum of the numerator and denominator of a fraction is 11. If the denominator is increased by 1, the fraction becomes \( \frac{1}{2} \), then the fraction is [1 Mark]
(A) \( \frac{2}{9} \)
(B) \( \frac{3}{8} \)
(C) \( \frac{4}{7} \)
(D) \( \frac{5}{6} \)

Answer: (C) \( \frac{4}{7} \)

Teacher's Note:
a) Let the fraction be \( \frac{x}{y} \): then \( x + y = 11 \) and \( \frac{x}{y + 1} = \frac{1}{2} \), so \( 2x = y + 1 \).
b) Solving gives \( x = 4, y = 7 \). Check: \( \frac{4}{8} = \frac{1}{2} \).

5. If one root of the quadratic equation \( ax^{2} + bx + c = 0 \) is the reciprocal of the other, then [1 Mark]
(A) \( b = c \)
(B) \( a = b \)
(C) \( ac = 1 \)
(D) \( a = c \)

Answer: (D) \( a = c \)

Teacher's Note:
a) Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \); their product is 1.
b) Product of roots \( = \frac{c}{a} \), so \( \frac{c}{a} = 1 \Rightarrow a = c \).

6. The first term of AP is \( p \) and the common difference is \( q \), then its 10th term is [1 Mark]
(A) \( q + 10p \)
(B) \( p - 9q \)
(C) \( p + 9q \)
(D) \( p + 10q \)

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

Teacher's Note:
a) Use \( a_n = a + (n - 1)d \), so \( a_{10} = p + (10 - 1)q = p + 9q \).
b) A common mistake is to take \( n \) instead of \( n - 1 \), which gives option (D).

7. Which term of the AP: 21, 42, 63, 84 ... is 210? [1 Mark]
(A) 9th
(B) 10th
(C) 11th
(D) 12th

Answer: (B) 10th

Teacher's Note:
a) Here \( a = 21 \) and \( d = 21 \): \( 210 = 21 + (n - 1)21 \Rightarrow n - 1 = 9 \Rightarrow n = 10 \).
b) Quick check: each term is \( 21 \times n \), and \( 21 \times 10 = 210 \).

8. If the point P (5, 2) divides the line segment joining A( 8, 5) and B( 4, \( y \)) in the ratio 3 : 1, then the value of \( y \) is [1 Mark]
(A) 4
(B) 3
(C) 2
(D) 1

Answer: (D) 1

Teacher's Note:
a) By section formula, y-coordinate of P \( = \frac{3y + 1 \times 5}{3 + 1} = 2 \).
b) \( 3y + 5 = 8 \Rightarrow y = 1 \).

9. A letter from the word INDEPENDENCE is selected at random. What is the probability that the letter selected is a vowel which occurs the maximum number of times in the given word? [1 Mark]
(A) \( \frac{1}{12} \)
(B) \( \frac{1}{4} \)
(C) \( \frac{1}{3} \)
(D) \( \frac{5}{12} \)

Answer: (C) \( \frac{1}{3} \)

Teacher's Note:
a) The word has 12 letters; the vowel 'I' occurs once and 'E' occurs 4 times.
b) P(E) \( = \frac{4}{12} = \frac{1}{3} \).

10. For the following distribution
Class: 0-5 | 5-10 | 10-15 | 15-20 | 20-25
Frequency: 10 | 15 | 12 | 20 | 9
The lower limit of the median class is [1 Mark]
(A) 5
(B) 10
(C) 15
(D) 20

Answer: (B) 10

Teacher's Note:
a) \( N = 66 \), so \( \frac{N}{2} = 33 \). Cumulative frequencies are 10, 25, 37, 57, 66.
b) 33 lies in the class 10-15, so the median class is 10-15 and its lower limit is 10.

11. The length of a tangent drawn from a point at a distance of 10 cm from the centre of the circle is 8 cm. The radius of the circle is [1 Mark]
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 7 cm

Answer: (C) 6 cm

Teacher's Note:
a) The radius is perpendicular to the tangent, so (distance)\(^{2}\) = (tangent)\(^{2}\) + (radius)\(^{2}\).
b) \( r = \sqrt{10^{2} - 8^{2}} = \sqrt{36} = 6 \) cm.

12. Mean and median of certain data are 32 and 30 respectively. Using empirical formula, the value of mode is [1 Mark]
(A) 36
(B) 26
(C) 30
(D) 20

Answer: (B) 26

Teacher's Note:
a) Empirical formula: Mode = 3 Median - 2 Mean.
b) Mode \( = 3 \times 30 - 2 \times 32 = 90 - 64 = 26 \).

13. A ladder 15 m long reaches a window 12 m above the ground. The distance of the foot of the ladder from the base of the wall is [1 Mark]
(A) 8 m
(B) 9 m
(C) 10 m
(D) 13 m

Answer: (B) 9 m

Teacher's Note:
a) The ladder is the hypotenuse: \( 15^{2} = (\text{Base})^{2} + 12^{2} \).
b) Base \( = \sqrt{225 - 144} = \sqrt{81} = 9 \) m.

14. The value of \( \sin^{2} 60^{\circ} - 2\tan^{2} 45^{\circ} - \cos^{2} 30^{\circ} \) is [1 Mark]
(A) \( -2 \)
(B) \( -1 \)
(C) 1
(D) 2

Answer: (A) \( -2 \)

Teacher's Note:
a) \( \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) and \( \tan 45^{\circ} = 1 \).
b) \( \frac{3}{4} - 2(1) - \frac{3}{4} = -2 \).

15. The perimeter of two similar triangles is 28 cm and 35 cm respectively. If one side of the first triangle is 8cm, then the corresponding side of the second triangle is [1 Mark]
(A) 10 cm
(B) 12cm
(C) 14cm
(D) 16cm

Answer: (A) 10 cm

Teacher's Note:
a) In similar triangles, ratio of perimeters = ratio of corresponding sides.
b) \( \frac{28}{35} = \frac{8}{x} \Rightarrow x = \frac{8 \times 35}{28} = 10 \) cm.

16. If in \( \triangle ABC \) and \( \triangle PQR \), \( \angle B = \angle Q \), \( \angle R = \angle C \) and AB = 2PQ, then the two triangles are [1 Mark]
(A) Congruent but not similar
(B) Similar but not congruent
(C) Neither congruent nor similar
(D) Congruent as well as similar

Answer: (B) Similar but not congruent

Teacher's Note:
a) Two pairs of equal angles give \( \triangle ABC \sim \triangle PQR \) by AA similarity.
b) Since \( \frac{AB}{PQ} = \frac{2}{1} \), AB is not equal to PQ, so the triangles are not congruent.

17. If tangents PA and PB from a point P to a circle with centre O and respective point of contacts as A and B, are inclined to each other at angle of \( 80^{\circ} \), then \( \angle AOB \) is equal to [1 Mark]
(A) \( 60^{\circ} \)
(B) \( 70^{\circ} \)
(C) \( 80^{\circ} \)
(D) \( 100^{\circ} \)

Answer: (D) \( 100^{\circ} \)

Teacher's Note:
a) \( \angle OAP = \angle OBP = 90^{\circ} \) because the radius is perpendicular to the tangent at the point of contact.
b) In quadrilateral AOBP: \( \angle AOB = 360^{\circ} - 90^{\circ} - 90^{\circ} - 80^{\circ} = 100^{\circ} \).

18. Two cubes each of volume \( 64 \text{ cm}^{3} \) are joined end to end to form a cuboid. The total surface area of the resulting cuboid is [1 Mark]
(A) \( 128 \text{ cm}^{2} \)
(B) \( 160 \text{ cm}^{2} \)
(C) \( 176 \text{ cm}^{2} \)
(D) \( 192 \text{ cm}^{2} \)

Answer: (B) \( 160 \text{ cm}^{2} \)

Teacher's Note:
a) \( a^{3} = 64 \Rightarrow a = 4 \) cm, so the cuboid is \( 8 \text{ cm} \times 4 \text{ cm} \times 4 \text{ cm} \).
b) TSA \( = 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(80) = 160 \text{ cm}^{2} \).

Assertion-Reason Based Questions

19. ASSERTION (A): The probability of getting number 8 on rolling a die is zero (0)
REASON (R): The probability of an impossible event is zero (0) [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of 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 (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) A die has only the numbers 1 to 6, so getting 8 is an impossible event: P(8) \( = \frac{0}{6} = 0 \).
b) The Reason states exactly why the Assertion is true, so (R) explains (A).

20. ASSERTION (A): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
REASON (R): Line drawn from midpoint of one side of triangle parallel to another will bisect the third side. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of 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 (R) is not the correct explanation of Assertion (A).

Teacher's Note:
a) The Assertion is the converse of the Basic Proportionality Theorem, so it is true.
b) The Reason is the converse of the mid-point theorem; it is true but is only a special case, so it does not explain (A).

SECTION B

21. (A) Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. [2 Marks]

Answer:
1. \( 520 = 2^{3} \times 5 \times 13 \) and \( 468 = 2^{2} \times 3^{2} \times 13 \).
2. LCM \( = 2^{3} \times 3^{2} \times 5 \times 13 = 4680 \).
3. The number increased by 17 must equal the LCM, so the required number \( = 4680 - 17 = 4663 \).

Teacher's Note:
a) "Exactly divisible by both" means the smallest such value is the LCM.
b) Remember to subtract 17 at the end; many students stop at 4680.

OR

(B) Show that \( (25)^{n} \) will never end with digit zero for any natural number \( n \). [2 Marks]

Answer:
1. \( (25)^{n} = (5^{2})^{n} = 5^{2n} \). So the prime factorisation of \( (25)^{n} \) contains only the prime 5, and this factorisation is unique.
2. For a number to end with digit 0, its prime factorisation must contain both 2 and 5.
3. But 2 is not a factor of \( (25)^{n} \).
4. Hence, \( (25)^{n} \) can never end with digit zero for any natural number \( n \).

Teacher's Note:
a) Quote the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic).
b) The key step is stating that the factor 2 is missing.

22. If \( 3\sin\theta = 4\cos\theta \), find the value of \( \sin\theta + \cos^{2}\theta - 1 \); \( 0^{\circ} \lt \theta \lt 90^{\circ} \) [2 Marks]

Answer:
1. \( 3\sin\theta = 4\cos\theta \Rightarrow \frac{\sin\theta}{\cos\theta} = \frac{4}{3} \Rightarrow \tan\theta = \frac{4}{3} \).
2. Take perpendicular = 4k and base = 3k, so hypotenuse \( = \sqrt{16k^{2} + 9k^{2}} = 5k \).
3. So \( \sin\theta = \frac{4}{5} \) and \( \cos\theta = \frac{3}{5} \).
4. \( \sin\theta + \cos^{2}\theta - 1 = \frac{4}{5} + \frac{9}{25} - 1 = \frac{20 + 9 - 25}{25} = \frac{4}{25} \).

Teacher's Note:
a) Getting \( \tan\theta = \frac{4}{3} \) carries the first mark.
b) Since \( \theta \) is acute, both \( \sin\theta \) and \( \cos\theta \) are positive.

23. If one zero of the quadratic polynomial \( 2x^{2} - 3x + p \) is 3, find the value of \( p \). Also, find the other zero. [2 Marks]

Answer:
1. Since 3 is a zero of \( p(x) = 2x^{2} - 3x + p \), \( p(3) = 0 \).
2. \( 2(3)^{2} - 3(3) + p = 0 \Rightarrow 18 - 9 + p = 0 \Rightarrow p = -9 \).
3. Let the other zero be \( \beta \). Sum of zeroes: \( 3 + \beta = -\frac{(-3)}{2} = \frac{3}{2} \).
4. \( \beta = \frac{3}{2} - 3 = -\frac{3}{2} \).

Teacher's Note:
a) Substitute the given zero first; this gives \( p \) directly.
b) Check with the product of zeroes: \( 3 \times \left(-\frac{3}{2}\right) = -\frac{9}{2} = \frac{p}{2} \).

24. In the given figure, PA is a common tangent and QB and PC are the tangents from Q and P to the smaller and larger circle respectively. If QB = 5cm and PC = 9cm, then evaluate length of PQ. [2 Marks]

[Figure: Two circles touching each other internally at point A, with the smaller circle (centre O marked) inside the larger one. The common tangent at A passes through points Q and P on a straight line. QB is a tangent from Q touching the smaller circle at B, marked 5cm. PC is a tangent from P touching the larger circle at C, marked 9cm.]

Answer:
1. Lengths of tangents drawn from an external point to a circle are equal.
2. From Q to the smaller circle: QA = QB = 5 cm.
3. From P to the larger circle: PA = PC = 9 cm.
4. PQ = PA - QA = 9 cm - 5 cm = 4 cm.

Teacher's Note:
a) PA is a tangent to both circles, so it helps to link the two given tangents.
b) Write the reason "tangents from an external point are equal" to get full marks.

For Visually Impaired Candidates (in lieu of Q. 24)

Two concentric circles are of radii 5cm and 3cm. Find the length of the chord of the larger circle which touches the smaller circle. [2 Marks]

Answer:
1. Let O be the common centre and AB be the chord of the larger circle touching the smaller circle at C. Join OC and OA.
2. \( \angle ACO = 90^{\circ} \) (radius through the point of contact is perpendicular to the tangent).
3. In right \( \triangle ACO \): \( AC = \sqrt{OA^{2} - OC^{2}} = \sqrt{5^{2} - 3^{2}} = \sqrt{16} = 4 \) cm.
4. The perpendicular from the centre bisects the chord, so \( AB = 2AC = 2 \times 4 = 8 \) cm.

Teacher's Note:
a) OA is the radius of the larger circle (5 cm) and OC is the radius of the smaller circle (3 cm).
b) Do not forget to double AC; the chord length is 8 cm, not 4 cm.

25. (A) Points A (3,1), B (5, 1), C (a, b) and D (4, 3) are vertices of a parallelogram ABCD. Find the values of a and b. [2 Marks]

Answer:
1. Diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD.
2. \( \left(\frac{3 + a}{2}, \frac{1 + b}{2}\right) = \left(\frac{5 + 4}{2}, \frac{1 + 3}{2}\right) \).
3. \( \frac{3 + a}{2} = \frac{9}{2} \Rightarrow a = 6 \) and \( \frac{1 + b}{2} = \frac{4}{2} \Rightarrow b = 3 \).

Teacher's Note:
a) Pair the opposite vertices correctly: A with C and B with D.
b) Check: C(6, 3) makes AB = DC = 2 units, both horizontal.

OR

(B) Find a linear relation between x and y such that P (x, y) is equidistant from the points A (1, 4) and B (\( -1 \), 2). [2 Marks]

Answer:
1. PA = PB \( \Rightarrow \sqrt{(x - 1)^{2} + (y - 4)^{2}} = \sqrt{(x + 1)^{2} + (y - 2)^{2}} \).
2. Squaring: \( x^{2} - 2x + 1 + y^{2} - 8y + 16 = x^{2} + 2x + 1 + y^{2} - 4y + 4 \).
3. \( -4x - 4y + 12 = 0 \Rightarrow 4x + 4y = 12 \Rightarrow x + y = 3 \).

Teacher's Note:
a) Square both sides to remove the roots; the \( x^{2} \) and \( y^{2} \) terms cancel.
b) Check: the midpoint (0, 3) of AB satisfies \( x + y = 3 \).

SECTION C

26. Given that \( \sqrt{5} \) is irrational, prove that \( 2 + 3\sqrt{5} \) is irrational. [3 Marks]

Answer:
1. Let us assume, to the contrary, that \( 2 + 3\sqrt{5} \) is rational.
2. Then \( 2 + 3\sqrt{5} = \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
3. So \( \sqrt{5} = \frac{p - 2q}{3q} \).
4. Since \( p \) and \( q \) are integers, \( \frac{p - 2q}{3q} \) is a rational number, so \( \sqrt{5} \) would be rational.
5. But \( \sqrt{5} \) is irrational. A rational number cannot equal an irrational number, so this is a contradiction.
6. Hence, our assumption is incorrect. Therefore, \( 2 + 3\sqrt{5} \) is irrational.

Teacher's Note:
a) Clearly state "let us assume the contrary" at the start of a proof by contradiction.
b) Write the condition \( q \neq 0 \) and the final conclusion; both carry marks.

27. (A) In the given figure, AX, AY and BC are tangents to the circle with P, R and Q respective point of contacts. Prove that AP = \( \frac{1}{2} \)(AB + BC + CA) [3 Marks]

[Figure: A circle with an external point A above it. Tangents AX and AY from A touch the circle at P and R respectively. A third tangent BC touches the circle at Q, with B on AX and C on AY, so that triangle ABC lies outside the circle.]

Answer:
1. AB + BC + CA = AB + BQ + CQ + AC (since BC = BQ + QC).
2. Lengths of tangents from an external point to a circle are equal, so BQ = BP and CQ = CR.
3. So AB + BC + CA = AB + BP + CR + AC = AP + AR.
4. Also AP = AR (tangents from A), so AB + BC + CA = 2AP.
5. Hence, AP = \( \frac{1}{2} \)(AB + BC + CA).

Teacher's Note:
a) Split BC at Q, then replace each part by an equal tangent.
b) Mention the equal-tangent property each time you use it.

OR

(B) Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that \( \angle APB = 2\angle OAB \) [3 Marks]

Answer:
1. Figure: a circle with centre O; tangents PA and PB from external point P touch the circle at A and B; join OA and AB.
2. PA = PB (tangents from an external point), so \( \angle BAP = \angle ABP \).
3. In \( \triangle APB \): \( \angle APB + \angle BAP + \angle ABP = 180^{\circ} \Rightarrow \angle APB + 2\angle BAP = 180^{\circ} \) ... (i)
4. \( \angle OAP = 90^{\circ} \) (tangent is perpendicular to the radius at the point of contact), so \( \angle OAB + \angle BAP = 90^{\circ} \Rightarrow \angle BAP = 90^{\circ} - \angle OAB \) ... (ii)
5. Substituting (ii) in (i): \( \angle APB + 2(90^{\circ} - \angle OAB) = 180^{\circ} \).
6. Hence, \( \angle APB = 2\angle OAB \).

Teacher's Note:
a) A correct figure carries half a mark, so draw and label it.
b) The two key facts are PA = PB and \( OA \perp PA \).

For Visually Impaired Candidates (in lieu of Q. 27 (A))

Prove that a parallelogram circumscribing a circle is a rhombus. [3 Marks]

Answer:
1. Let parallelogram ABCD circumscribe a circle, touching AB, BC, CD and DA at P, Q, R and S respectively.
2. Lengths of tangents drawn from an external point are equal: AP = AS, PB = BQ, CR = CQ, DR = DS.
3. Adding: (AP + PB) + (CR + RD) = (AS + DS) + (CQ + QB), so AB + CD = AD + BC.
4. Opposite sides of a parallelogram are equal (CD = AB, AD = BC), so 2AB = 2BC, i.e. AB = BC.
5. A parallelogram with a pair of adjacent sides equal is a rhombus. Hence, ABCD is a rhombus.

Teacher's Note:
a) Write all four pairs of equal tangents; this step carries 1 mark.
b) End by stating why equal adjacent sides make it a rhombus.

28. Determine the ratio in which the point (\( -6 \), y) divides the line segment joining the points A (\( -3 \), \( -1 \)) and B (\( -8 \), 9). Also, find the value of y. [3 Marks]

Answer:
1. Let the required ratio be \( k : 1 \).
2. By section formula, the point is \( \left(\frac{-8k - 3}{k + 1}, \frac{9k - 1}{k + 1}\right) \).
3. \( \frac{-8k - 3}{k + 1} = -6 \Rightarrow -8k - 3 = -6k - 6 \Rightarrow 2k = 3 \Rightarrow k = \frac{3}{2} \). So the required ratio is 3 : 2.
4. \( y = \frac{9\left(\frac{3}{2}\right) - 1}{\frac{3}{2} + 1} = \frac{\frac{25}{2}}{\frac{5}{2}} = 5 \).

Teacher's Note:
a) Taking the ratio as \( k : 1 \) keeps the algebra to one unknown.
b) Use the x-coordinate to find \( k \), then the y-coordinate to find \( y \).

29. Consider \( \triangle \) ACB, right-angled at C, in which AB = 29 units, BC = 21 units and \( \angle ABC = \theta \). Determine the values of
i. \( 1 + \tan^{2}\theta \)
ii. \( \cos^{2}\theta - \sin^{2}\theta \) [3 Marks]

Answer:
1. \( AC = \sqrt{AB^{2} - BC^{2}} = \sqrt{29^{2} - 21^{2}} = \sqrt{841 - 441} = \sqrt{400} = 20 \) units.
2. \( \sin\theta = \frac{AC}{AB} = \frac{20}{29} \), \( \cos\theta = \frac{BC}{AB} = \frac{21}{29} \), \( \tan\theta = \frac{AC}{BC} = \frac{20}{21} \).
3. (i) \( 1 + \tan^{2}\theta = 1 + \left(\frac{20}{21}\right)^{2} = 1 + \frac{400}{441} = \frac{841}{441} \).
4. (ii) \( \cos^{2}\theta - \sin^{2}\theta = \left(\frac{21}{29}\right)^{2} - \left(\frac{20}{29}\right)^{2} = \frac{441 - 400}{841} = \frac{41}{841} \).

Teacher's Note:
a) For angle B, the opposite side is AC and the adjacent side is BC.
b) Check (i) with the identity \( 1 + \tan^{2}\theta = \sec^{2}\theta = \left(\frac{29}{21}\right)^{2} = \frac{841}{441} \).

30. (A) The mean of the following distribution is 48 and sum of all the frequencies is 50. Find the missing frequencies x and y.
Class: 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70
Frequency: 8 | 6 | x | 11 | y [3 Marks]

Answer:
1. Take assumed mean \( A = 45 \), class width \( h = 10 \), \( u_i = \frac{x_i - 45}{10} \).
2. Mid-values \( x_i \): 25, 35, 45, 55, 65; \( u_i \): \( -2, -1, 0, 1, 2 \); \( f_i u_i \): \( -16, -6, 0, 11, 2y \).
3. \( \sum f_i = 25 + x + y = 50 \Rightarrow x + y = 25 \) ... (1), and \( \sum f_i u_i = 2y - 11 \).
4. Mean \( = A + \frac{\sum f_i u_i}{\sum f_i} \times h \Rightarrow 48 = 45 + \frac{2y - 11}{50} \times 10 = 45 + \frac{2y - 11}{5} \).
5. \( 2y - 11 = 15 \Rightarrow y = 13 \). From (1), \( x = 25 - 13 = 12 \).

Teacher's Note:
a) The correct table with \( f_i u_i \) carries 1 mark, so show it.
b) Check: \( \sum f_i x_i = 200 + 210 + 540 + 605 + 845 = 2400 \) and \( \frac{2400}{50} = 48 \).

OR

(B) The distribution below gives the weight of 50 students of class X. Find the Modal weight of the students.
Weight ( in kg ): 35 - 45 | 45 - 55 | 55 - 65 | 65 - 75 | 75 - 85
Number of students: 5 | 10 | 20 | 12 | 3 [3 Marks]

Answer:
1. The highest frequency is 20, so the modal class is 55 - 65.
2. \( l = 55 \), \( h = 10 \), \( f_1 = 20 \), \( f_0 = 10 \), \( f_2 = 12 \).
3. Mode \( = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 55 + \left(\frac{20 - 10}{40 - 10 - 12}\right) \times 10 = 55 + \frac{100}{18} = 60.555... \)
4. Hence, the modal weight of the students is 60.56 kg (approx.).

Teacher's Note:
a) \( f_0 \) is the frequency of the class before the modal class and \( f_2 \) the one after it.
b) The mode must lie inside the modal class (55 to 65), which is a quick check.

31. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. [3 Marks]

Answer:
1. Let the ten's digit be \( x \) and the unit's digit be \( y \). Number \( = 10x + y \); reversed number \( = 10y + x \).
2. According to the question: \( x + y = 9 \) ... (1) and \( 9(10x + y) = 2(10y + x) \) ... (2).
3. From (2): \( 90x + 9y = 20y + 2x \Rightarrow 88x = 11y \Rightarrow y = 8x \).
4. Substituting in (1): \( x + 8x = 9 \Rightarrow x = 1 \), \( y = 8 \).
5. Required number \( = 18 \).

Teacher's Note:
a) Write the number as \( 10x + y \), not \( xy \).
b) Check: \( 9 \times 18 = 162 \) and \( 2 \times 81 = 162 \).

SECTION D

32. Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. [5 Marks]

Answer:
1. Figure: \( \triangle ABC \) with a line DE parallel to BC, meeting AB at D and AC at E. Join BE and CD, and draw \( DM \perp AC \) and \( EN \perp AB \).
2. Given: In \( \triangle ABC \), \( DE \parallel BC \). To prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).
3. \( \text{ar}(ADE) = \frac{1}{2} \times AD \times EN \) and \( \text{ar}(BDE) = \frac{1}{2} \times DB \times EN \), so \( \frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{AD}{DB} \) ... (1)
4. \( \text{ar}(ADE) = \frac{1}{2} \times AE \times DM \) and \( \text{ar}(DEC) = \frac{1}{2} \times EC \times DM \), so \( \frac{\text{ar}(ADE)}{\text{ar}(DEC)} = \frac{AE}{EC} \) ... (2)
5. \( \triangle BDE \) and \( \triangle DEC \) are on the same base DE and between the same parallels BC and DE, so \( \text{ar}(BDE) = \text{ar}(DEC) \) ... (3)
6. From (1), (2) and (3): \( \frac{AD}{DB} = \frac{AE}{EC} \). Hence proved.

Teacher's Note:
a) Figure, Given, To prove and Construction carry half a mark each (2 marks in all).
b) The key step is \( \text{ar}(BDE) = \text{ar}(DEC) \); give its reason in full.
c) This is the Basic Proportionality Theorem (Thales theorem).

33. (A) A train travels a distance of 360 km at a uniform speed. If the speed had been 5 km/h more, then it would have taken 1 hour less 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. Then the increased speed is \( (x + 5) \) km/h.
2. According to the question: \( \frac{360}{x} - \frac{360}{x + 5} = 1 \).
3. \( 360(x + 5) - 360x = x(x + 5) \Rightarrow 1800 = x^{2} + 5x \Rightarrow x^{2} + 5x - 1800 = 0 \).
4. \( x = \frac{-5 \pm \sqrt{5^{2} + 4(1800)}}{2} = \frac{-5 \pm \sqrt{7225}}{2} = \frac{-5 \pm 85}{2} \).
5. \( x = 40 \) or \( x = -45 \). Speed cannot be negative, so \( x = 40 \).
6. Speed of the train \( = 40 \) km/h.

Teacher's Note:
a) Time = distance \( \div \) speed; the slower speed takes more time, so subtract in that order.
b) Reject the negative root with a reason.
c) Check: \( \frac{360}{40} - \frac{360}{45} = 9 - 8 = 1 \) hour.

OR

(B) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find the number of marbles they had with them in the beginning. [5 Marks]

Answer:
1. Let John have \( x \) marbles in the beginning. Then Jivanti had \( 45 - x \) marbles.
2. After losing 5 each: \( (x - 5)(45 - x - 5) = 124 \Rightarrow (x - 5)(40 - x) = 124 \).
3. \( 40x - x^{2} - 200 + 5x = 124 \Rightarrow x^{2} - 45x + 324 = 0 \).
4. \( x = \frac{45 \pm \sqrt{45^{2} - 4(324)}}{2} = \frac{45 \pm \sqrt{729}}{2} = \frac{45 \pm 27}{2} \), so \( x = 36 \) or \( x = 9 \).
5. Case 1: \( x = 36 \): John had 36 marbles and Jivanti had \( 45 - 36 = 9 \) marbles.
6. Case 2: \( x = 9 \): John had 9 marbles and Jivanti had \( 45 - 9 = 36 \) marbles.

Teacher's Note:
a) Both roots are valid here, so write both cases to get the last mark.
b) Check: \( (36 - 5)(9 - 5) = 31 \times 4 = 124 \).

34. The largest possible hemisphere is drilled out of a wooden cubical block of side 21cm such that the base of the hemisphere is on one of the faces of the cube. Then, find
i. the volume of the wood left in the block.
ii. the total surface area of the remaining solid. [5 Marks]

Answer:
1. Side of cube \( a = 21 \) cm. Radius of the largest hemisphere \( r = \frac{21}{2} \) cm.
2. (i) Volume of wood left \( = a^{3} - \frac{2}{3}\pi r^{3} = 21^{3} - \frac{2}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{2 \times 2 \times 2} \).
3. \( = 9261 - 2425.5 = 6835.5 \text{ cm}^{3} \).
4. (ii) Surface area of remaining solid \( = 6a^{2} - \pi r^{2} + 2\pi r^{2} = 6a^{2} + \pi r^{2} \).
5. \( = 6(21)^{2} + \frac{22}{7} \times \left(\frac{21}{2}\right)^{2} = 2646 + 346.5 = 2992.5 \text{ cm}^{2} \).

Teacher's Note:
a) The largest hemisphere has diameter equal to the side of the cube.
b) For surface area, remove the circular top (\( \pi r^{2} \)) and add the curved inner surface (\( 2\pi r^{2} \)).

35. (A) To assist an ambulance, a drone is hovering over an accident site at a constant height on a highway. The angle of depression of the ambulance, approaching the accident site, observed from drone is \( 30^{\circ} \). Twelve minutes later, the angle of depression of the ambulance is found to be \( 60^{\circ} \). If the ambulance is moving at a constant speed then, find the total time taken by the ambulance to reach the site of accident. [5 Marks]

Answer:
1. Figure: drone at A, vertically above the accident site B, with AB = \( h \) m. The ambulance is first at D (angle of depression \( 30^{\circ} \)) and after 12 minutes at C (angle of depression \( 60^{\circ} \)), with D, C, B on the highway.
2. Let the speed of the ambulance be \( x \) m/min. Then CD \( = 12x \) m.
3. In right \( \triangle ABC \): \( \tan 60^{\circ} = \frac{AB}{BC} \Rightarrow BC = \frac{h}{\sqrt{3}} \).
4. In right \( \triangle ABD \): \( \tan 30^{\circ} = \frac{AB}{BD} \Rightarrow BD = h\sqrt{3} \).
5. BD = BC + CD: \( \frac{h}{\sqrt{3}} + 12x = h\sqrt{3} \Rightarrow 12x = \frac{2h}{\sqrt{3}} \Rightarrow h = 6\sqrt{3}x \). So \( BC = \frac{6\sqrt{3}x}{\sqrt{3}} = 6x \).
6. Time from C to B \( = \frac{6x}{x} = 6 \) minutes. Total time taken \( = 12 + 6 = 18 \) minutes.

Teacher's Note:
a) A correct figure carries 1 mark; the angle of depression equals the angle of elevation from the ground (alternate angles).
b) The speed \( x \) cancels out, so the answer does not need the actual speed.

OR

(B) A statue standing on a pedestal, is 1.6 m tall. From a point on the ground, the angles of elevation of the top and bottom of the statue are \( 60^{\circ} \) and \( 45^{\circ} \) respectively. Find the height of the pedestal. [5 Marks]

Answer:
1. Figure: pedestal BC of height \( h \) m on the ground at B, statue CD = 1.6 m on top of it, and A the point on the ground.
2. In right \( \triangle ABC \): \( \tan 45^{\circ} = \frac{BC}{AB} \Rightarrow 1 = \frac{h}{AB} \Rightarrow AB = h \) ... (i)
3. In right \( \triangle ABD \): \( \tan 60^{\circ} = \frac{BD}{AB} \Rightarrow \sqrt{3} = \frac{h + 1.6}{AB} \Rightarrow AB = \frac{h + 1.6}{\sqrt{3}} \) ... (ii)
4. From (i) and (ii): \( h = \frac{h + 1.6}{\sqrt{3}} \Rightarrow (\sqrt{3} - 1)h = 1.6 \).
5. \( h = \frac{1.6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \) m.
6. Height of the pedestal \( = 0.8(\sqrt{3} + 1) \) m (about 2.19 m).

Teacher's Note:
a) The bottom of the statue is the top of the pedestal, so use \( 45^{\circ} \) for BC and \( 60^{\circ} \) for BD.
b) Rationalise the denominator to write the answer in its simplest form.

SECTION E

36. An Art and craft teacher prepared a fan which looked like the sector of a circle using ice-cream sticks and black sheet as shown in the given figure:

[Figure: An open hand fan shaped like the sector of a circle. The outer band is labelled "Black sheet" and the inner part near the centre is labelled "Ice-cream sticks". The full radius of the fan is marked 10 cm and the ice-cream stick part near the centre is marked 4 cm.]

(i) The angle subtended at the centre of the sector is \( 60^{\circ} \). Find the area of one face of the fan. [1 Mark]

Answer:
1. \( \theta = 60^{\circ} \), \( R = 10 \) cm.
2. Area of one face \( = \frac{\theta}{360} \pi R^{2} = \frac{60}{360} \times \frac{22}{7} \times 10^{2} = \frac{1100}{21} = 52.38 \text{ cm}^{2} \) (approx.).

Teacher's Note:
a) One face of the fan is the full sector of radius 10 cm.
b) \( \frac{1100}{21} \) is also an acceptable final form.

(ii) With the given sector angle of \( 60^{\circ} \) at the centre, find area of black sheet used in the fan. [1 Mark]

Answer:
1. The black sheet is the region between radii \( r = 4 \) cm and \( R = 10 \) cm.
2. Area \( = \frac{\theta}{360} \pi (R^{2} - r^{2}) = \frac{60}{360} \times \frac{22}{7} \times (10^{2} - 4^{2}) = \frac{1}{6} \times \frac{22}{7} \times 84 = 44 \text{ cm}^{2} \).

Teacher's Note:
a) Subtract the small sector (ice-cream stick part) from the big sector.
b) \( 100 - 16 = 84 \) is divisible by 7, so the answer comes out exact.

(iii)(A) The boundary of the black sheet region on one face is to be covered by a designer tape. What length of tape is required? [2 Marks]

Answer:
1. Boundary = two straight edges + outer arc + inner arc \( = 2(R - r) + \frac{\theta}{360} \times 2\pi (R + r) \).
2. \( = 2(10 - 4) + \frac{60}{360} \times 2 \times \frac{22}{7} \times (10 + 4) \).
3. \( = 12 + \frac{44}{3} = \frac{80}{3} \) cm (about 26.67 cm).

Teacher's Note:
a) Include both arcs and both straight edges of length \( 10 - 4 = 6 \) cm.
b) Adding the arcs first (\( R + r = 14 \)) saves time.

OR

(iii)(B) If 29.6 cm of designer tape is required to cover the boundary of the black sheet region on one face of the fan, then evaluate the central angle of the sector. [2 Marks]

Answer:
1. \( 2(10 - 4) + \frac{\theta}{360} \times 2 \times \frac{22}{7} \times (10 + 4) = 29.6 \).
2. \( 12 + \frac{\theta}{360} \times 88 = 29.6 \Rightarrow \frac{\theta}{360} \times 88 = 17.6 \).
3. \( \theta = \frac{17.6 \times 360}{88} = 72^{\circ} \).

Teacher's Note:
a) Use the same boundary formula as in part (iii)(A), with \( \theta \) unknown.
b) Subtract the straight edges (12 cm) before solving for \( \theta \).

For Visually Impaired Candidates (in lieu of Q. 36)

A motion sensor (or motion detector) is an electronic device that is designed to detect and measure movement. The motion detector can detect movement over a sector of angle \( \theta = 70^{\circ} \) to a distance of 24m.

(i) Derive a relation between length of arc(\( l \)) and the area of the sector(\( A \)) enclosed by it? [1 Mark]

Answer:
1. \( \frac{A}{l} = \frac{\frac{\theta}{360}\pi r^{2}}{\frac{\theta}{360} \times 2\pi r} = \frac{r}{2} \).
2. So \( A = \frac{1}{2} l r \).

Teacher's Note:
a) Divide the sector area formula by the arc length formula; \( \theta \) and \( \pi \) cancel.
b) This relation is like the area of a triangle: half of "base" \( l \) times "height" \( r \).

(ii) How much area is monitored by motion detector? [1 Mark]

Answer:
1. \( r = 24 \) m, \( \theta = 70^{\circ} \).
2. Area monitored \( = \frac{\theta}{360}\pi r^{2} = \frac{70}{360} \times \frac{22}{7} \times 24^{2} = 352 \text{ m}^{2} \).

Teacher's Note:
a) The range of distance is the radius of the sector.
b) Cancel 7 with 70 and 576 with 360 early to simplify.

(iii) (A) To increase the area to be monitored by 50% with same range of distance covered, what should be the angle? [2 Marks]

Answer:
1. New area \( = \frac{150}{100} \times 352 = 528 \text{ m}^{2} \).
2. \( \frac{\theta}{360}\pi (24)^{2} = \frac{150}{100} \times 352 \Rightarrow \theta = \frac{3}{2} \times \frac{7 \times 360}{22 \times 24^{2}} \times 352 \).
3. \( \theta = 105^{\circ} \).

Teacher's Note:
a) With the same radius, area is proportional to the angle, so \( 1.5 \times 70^{\circ} = 105^{\circ} \) is a quick check.
b) "Increase by 50%" means the new area is 150% of the old area.

OR

(iii) (B) For \( \theta = 91^{\circ} \), what range of distance is required for the detector to monitor 30% more area? [2 Marks]

Answer:
1. New area \( = \frac{130}{100} \times 352 \text{ m}^{2} \).
2. \( \frac{91}{360}\pi r^{2} = \frac{130}{100} \times 352 \Rightarrow r^{2} = \frac{13}{10} \times \frac{7 \times 360}{22 \times 91} \times 352 = 36 \times 16 = 576 \).
3. \( r = 24 \) m.

Teacher's Note:
a) "30% more area" means 130% of the area found in part (ii).
b) Since \( 91 = 1.3 \times 70 \), the angle alone gives the 30% increase, so the range stays 24 m.

37. Delhi's pollution is a severe, often seasonal issue caused by a combination of factors like high vehicular and industrial emissions, dust from construction, and crop burning in surrounding states. In a school, students of classes I to XII thought of planting trees in and around the school to reduce air pollution.
It was decided that the number of trees that each section of each class will plant be three more than the double of the class in which they are studying, e.g., a section of class IV will plant 11 trees. If in the given school, there are two sections of each class, then answer the following questions:

[Figure: Two pictures side by side - on the left, a field of dry crop stubble burning with flames and smoke; on the right, factory chimneys giving out smoke, a car, bare trees and a family of four walking.]

(i) If number of trees planted by students class-wise follow arithmetic progression, then find the common difference. [1 Mark]

Answer:
1. Trees planted by a class \( n \) (two sections) \( = 2\{2n + 3\} \).
2. Class I: \( 2\{2(1) + 3\} = 10 \); class II: \( 2\{2(2) + 3\} = 14 \); class III: \( 2\{2(3) + 3\} = 18 \). AP: 10, 14, 18, 22, ...
3. Common difference \( = 14 - 10 = 4 \).

Teacher's Note:
a) Multiply by 2 because each class has two sections.
b) A common mistake is to take \( d = 2 \), which is the difference for one section only.

(ii) What is the number of trees planted by students of class VI? [1 Mark]

Answer: Number of trees planted by class VI \( = 2\{2(6) + 3\} = 2 \times 15 = 30 \).

Teacher's Note:
a) Check using the AP: \( a_6 = 10 + 5 \times 4 = 30 \).
b) Remember to count both sections of class VI.

(iii) (A) What is the total number of trees planted by the students of given school? [2 Marks]

Answer:
1. \( a = 10 \), \( d = 4 \), \( n = 12 \) (classes I to XII).
2. \( S_n = \frac{n}{2}\{2a + (n - 1)d\} = \frac{12}{2}\{2(10) + (12 - 1)4\} \).
3. \( = 6 \times (20 + 44) = 6 \times 64 = 384 \).

Teacher's Note:
a) There are 12 classes, so \( n = 12 \).
b) Check: last term \( = 2\{2(12) + 3\} = 54 \) and \( S_{12} = \frac{12}{2}(10 + 54) = 384 \).

OR

(iii) (B) Identify the class that planted 34 trees. [2 Marks]

Answer:
1. Let the class that planted 34 trees be \( n \).
2. \( 2\{2(n) + 3\} = 34 \Rightarrow 2n + 3 = 17 \).
3. \( n = 7 \). So class VII planted 34 trees.

Teacher's Note:
a) Form the equation using the total for both sections.
b) Give the answer as a class (class VII), not just the number 7.

38. A labourer prints 400 T-shirts in a day. The supervisor checked the T-shirts and found that 312 prints were good, 54 prints were with minor defects and rest of the prints were of major defects.
Harish, a customer will buy a T-shirt only if it is good but a trader will buy, if it has no major defect. If a random T-shirt is picked, then

[Figure: Photograph of workers seated at long tables in a garment factory, working on T-shirts.]

(i) find the probability that Harish will buy a T-shirt. [1 Mark]

Answer: P(Harish buys T-shirt) \( = \frac{312}{400} = \frac{39}{50} \).

Teacher's Note:
a) Harish buys only good T-shirts, so the favourable outcomes are 312.
b) Simplify the fraction; \( \frac{312}{400} \) is also accepted.

(ii) find the probability that a trader will buy the T-shirt. [1 Mark]

Answer: P(Trader buys T-shirt) \( = \frac{312 + 54}{400} = \frac{366}{400} = \frac{183}{200} \).

Teacher's Note:
a) "No major defect" includes both good prints and prints with minor defects.
b) Number of T-shirts with major defects \( = 400 - 312 - 54 = 34 \).

(iii) (A) find the probability that the T-shirt is not good. [2 Marks]

Answer:
1. P(T-shirt is not good) \( = 1 - \) P(T-shirt is good).
2. \( = 1 - \frac{312}{400} = \frac{88}{400} = \frac{11}{50} \).

Teacher's Note:
a) Use the complement rule: P(not E) \( = 1 - \) P(E).
b) Check: not good = minor defects + major defects \( = 54 + 34 = 88 \).

OR

(iii) (B) find the probability that neither Harish nor trader buy the T-shirt. [2 Marks]

Answer:
1. Neither buys only when the T-shirt has a major defect: number \( = 400 - 54 - 312 = 34 \).
2. P(neither Harish nor Trader buys T-shirt) \( = \frac{34}{400} = \frac{17}{200} \).

Teacher's Note:
a) If the trader does not buy, Harish will not buy either, so only major-defect T-shirts count.
b) Check: \( 1 - \frac{183}{200} = \frac{17}{200} \).