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

SECTION - A (20 x 1 = 20)

This section comprises of 18 multiple choice questions and two questions of assertion and reasoning type of 1 mark each.

1. The greatest number which divides both 134 and 188, leaving remainders 4 and 6 respectively, is: [1 Mark]
(A) 13
(B) 26
(C) 39
(D) 65

Answer: (B) 26

Teacher's Note:
a) First subtract the remainders: \(134 - 4 = 130\) and \(188 - 6 = 182\).
b) The required number is \( \text{HCF}(130, 182) = 26 \), since \(130 = 2 \times 5 \times 13\) and \(182 = 2 \times 7 \times 13\).
c) Check: 26 must be greater than both remainders, and it is.

2. If \( f(x) = px^2 + qx + r \), \( p \neq 0 \) and \( p + r = q \), then one of the zeroes of \( f(x) \) is: [1 Mark]
(A) \( \frac{q}{p} \)
(B) \( \frac{r}{p} \)
(C) \( -\frac{r}{p} \)
(D) \( -\frac{q}{p} \)

Answer: (C) \( -\frac{r}{p} \)

Teacher's Note:
a) Put \( q = p + r \): \( f(x) = px^2 + (p + r)x + r = (x + 1)(px + r) \).
b) So the zeroes are \( -1 \) and \( -\frac{r}{p} \); only \( -\frac{r}{p} \) is among the options.

3. Tarun correctly solved a pair of linear equations in two variables and found their only point of intersection as \( (5, -1) \). One of the lines was \( x - y = 6 \).
Which of the following could have been the other line?
I: \( 3x - 3y = 18 \)
II: \( 2x - 3y = 13 \)
III: \( 2x - 3y = 16 \) [1 Mark]
(A) I only
(B) II only
(C) I and II
(D) II and III

Answer: (B) II only

Teacher's Note:
a) Line I is \( 3(x - y) = 18 \), the same line as \( x - y = 6 \), so it gives infinitely many solutions, not only one.
b) Line II: \( 2(5) - 3(-1) = 13 \), so it passes through \( (5, -1) \) and meets \( x - y = 6 \) at only one point.
c) Line III: \( 2(5) - 3(-1) = 13 \neq 16 \), so \( (5, -1) \) is not on it.

4. If \( (1 - p) \) is a root of the quadratic equation \( x^2 + px + 1 - p = 0 \), then its roots are: [1 Mark]
(A) 0, 1
(B) \( -1, 1 \)
(C) \( 0, -1 \)
(D) \( -1, 2 \)

Answer: (C) \( 0, -1 \)

Teacher's Note:
a) Substitute \( x = 1 - p \): \( (1 - p)^2 + p(1 - p) + 1 - p = 0 \Rightarrow 2(1 - p) = 0 \Rightarrow p = 1 \).
b) The equation becomes \( x^2 + x = 0 \), i.e. \( x(x + 1) = 0 \), so the roots are 0 and \( -1 \).

5. The middle term of the A.P.: \( 10, 7, 4, \ldots, -62 \) is: [1 Mark]
(A) \( -26 \)
(B) \( -29 \)
(C) \( -32 \)
(D) \( -35 \)

Answer: (A) \( -26 \)

Teacher's Note:
a) \( -62 = 10 + (n - 1)(-3) \Rightarrow n = 25 \), so the middle term is the \( \frac{25 + 1}{2} = 13^{\text{th}} \) term.
b) \( a_{13} = 10 + 12(-3) = -26 \).

6. The perimeters of two similar triangles are 56 cm and 70 cm respectively. If one side of the first triangle is 14 cm, then the corresponding side of the second triangle (in cm) is: [1 Mark]
(A) 5
(B) 7.5
(C) 10
(D) 17.5

Answer: (D) 17.5

Teacher's Note:
a) Ratio of perimeters of similar triangles = ratio of corresponding sides.
b) \( \frac{56}{70} = \frac{14}{x} \Rightarrow x = \frac{14 \times 70}{56} = 17.5 \) cm.

7. The point which lies on the perpendicular bisector of the line segment joining the points A \( (-3, -4) \) and B \( (3, 4) \) is: [1 Mark]
(A) \( (0, 0) \)
(B) \( (0, 3) \)
(C) \( (3, 0) \)
(D) \( (0, 4) \)

Answer: (A) \( (0, 0) \)

Teacher's Note:
a) The mid-point of AB always lies on the perpendicular bisector of AB.
b) Mid-point \( = \left( \frac{-3 + 3}{2}, \frac{-4 + 4}{2} \right) = (0, 0) \).

8. If \( \tan (A + B) = \sqrt{3} \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \), \( 0^{\circ} \lt A + B \lt 90^{\circ} \), \( A \gt B \), then the value of A and B respectively are: [1 Mark]
(A) \( 60^{\circ}, 30^{\circ} \)
(B) \( 60^{\circ}, 45^{\circ} \)
(C) \( 45^{\circ}, 15^{\circ} \)
(D) \( 60^{\circ}, 15^{\circ} \)

Answer: (C) \( 45^{\circ}, 15^{\circ} \)

Teacher's Note:
a) \( \tan (A + B) = \sqrt{3} \Rightarrow A + B = 60^{\circ} \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30^{\circ} \).
b) Adding: \( 2A = 90^{\circ} \Rightarrow A = 45^{\circ} \), so \( B = 15^{\circ} \).

9. T-shirts marked with numbers 4 to 99 are placed in a box. Gunika is fond of numbers. She randomly takes out a T-shirt from this box. The probability that she gets a T-shirt marked with a number that is either a perfect square or a perfect cube is: [1 Mark]
(A) \( \frac{1}{12} \)
(B) \( \frac{1}{32} \)
(C) \( \frac{11}{96} \)
(D) \( \frac{5}{48} \)

Answer: (D) \( \frac{5}{48} \)

Teacher's Note:
a) Total outcomes from 4 to 99 \( = 99 - 4 + 1 = 96 \).
b) Perfect squares: 4, 9, 16, 25, 36, 49, 64, 81; perfect cubes: 8, 27, 64. Count 64 only once, so favourable outcomes = 10.
c) Probability \( = \frac{10}{96} = \frac{5}{48} \).

10. The diameter of a car wheel is 21 cm. The number of complete revolutions it will make in moving 66 km is: [1 Mark]
(A) \( 10^{4} \)
(B) \( 10^{5} \)
(C) \( 10^{6} \)
(D) \( 10^{7} \)

Answer: (B) \( 10^{5} \)

Teacher's Note:
a) Circumference \( = \pi d = \frac{22}{7} \times 21 = 66 \) cm, and 66 km \( = 66 \times 10^{5} \) cm.
b) Revolutions \( = \frac{66 \times 10^{5}}{66} = 10^{5} \).
c) Convert km to cm before dividing; this is the most common mistake.

11. Two cubes each of volume 64 cm³ are joined end to end to form a solid. The surface area of the resultant cuboid is: [1 Mark]
(A) 192 cm²
(B) 160 cm²
(C) 96 cm²
(D) 80 cm²

Answer: (B) 160 cm²

Teacher's Note:
a) Side of each cube \( = \sqrt[3]{64} = 4 \) cm, so the cuboid is \( 8 \times 4 \times 4 \) cm.
b) Surface area \( = 2(8 \times 4 + 4 \times 4 + 8 \times 4) = 160 \) cm².
c) Do not simply double the surface area of one cube; two faces get hidden where the cubes join.

12. The mean age of a combined group of men and women is 35 years. If the mean ages of the men and women are 38 years and 30 years respectively, then the percentage of women in the group is: [1 Mark]
(A) 15
(B) 25.5
(C) 35
(D) 37.5

Answer: (D) 37.5

Teacher's Note:
a) Let there be \( x \) men and \( y \) women: \( \frac{38x + 30y}{x + y} = 35 \Rightarrow 3x = 5y \Rightarrow \frac{x}{y} = \frac{5}{3} \).
b) Percentage of women \( = \frac{3}{8} \times 100 = 37.5 \).

13. Two dice are rolled simultaneously. The probability of getting number less than 4 on each die is: [1 Mark]
(A) \( \frac{1}{4} \)
(B) \( \frac{1}{9} \)
(C) \( \frac{1}{36} \)
(D) \( \frac{1}{6} \)

Answer: (A) \( \frac{1}{4} \)

Teacher's Note:
a) Total outcomes = 36. Each die must show 1, 2 or 3, so favourable outcomes \( = 3 \times 3 = 9 \).
b) Probability \( = \frac{9}{36} = \frac{1}{4} \).

14. PM is a median of \( \Delta \) PQR with vertices P \( (5, -6) \), Q \( (6, 4) \) and R \( (0, 0) \). The length of PM is: [1 Mark]
(A) \( 2\sqrt{17} \) units
(B) \( 2\sqrt{15} \) units
(C) \( \sqrt{101} \) units
(D) 10 units

Answer: (A) \( 2\sqrt{17} \) units

Teacher's Note:
a) M is the mid-point of QR: \( M = \left( \frac{6 + 0}{2}, \frac{4 + 0}{2} \right) = (3, 2) \).
b) \( PM = \sqrt{(5 - 3)^2 + (-6 - 2)^2} = \sqrt{68} = 2\sqrt{17} \) units.

15. The arc of a circle is of length \( 6\pi \) cm and the sector it bound has an area of \( 24\pi \) cm². The radius of the circle is: [1 Mark]
(A) 4 cm
(B) 8 cm
(C) 16 cm
(D) 18 cm

Answer: (B) 8 cm

Teacher's Note:
a) Use area of sector \( = \frac{1}{2} \times l \times r \).
b) \( 24\pi = \frac{1}{2} \times 6\pi \times r \Rightarrow r = 8 \) cm.

16. The mean and median of the data are 35.5 and 32 respectively. The value of mode for this data is: [1 Mark]
(A) 23
(B) 26
(C) 25
(D) 30

Answer: (C) 25

Teacher's Note:
a) Use the empirical relation: Mode \( = 3 \) Median \( - 2 \) Mean.
b) Mode \( = 3(32) - 2(35.5) = 96 - 71 = 25 \).

17. A flying kite is tied to a point on the ground with the help of a string. The string makes an angle \( \theta \) with the ground level such that \( \tan \theta = \frac{12}{5} \). If the length of the string is 52 m, then the height of the kite above the ground is: [1 Mark]
(A) 40 m
(B) 45.5 m
(C) 48 m
(D) 50 m

Answer: (C) 48 m

Teacher's Note:
a) \( \tan \theta = \frac{12}{5} \) gives hypotenuse \( = \sqrt{12^2 + 5^2} = 13 \), so \( \sin \theta = \frac{12}{13} \).
b) The string is the hypotenuse: \( \frac{h}{52} = \frac{12}{13} \Rightarrow h = 48 \) m.

18. 2 cards of diamonds and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from this pack. The probability of getting a card of heart is: [1 Mark]
(A) \( \frac{13}{52} \)
(B) \( \frac{13}{46} \)
(C) \( \frac{11}{52} \)
(D) \( \frac{11}{46} \)

Answer: (B) \( \frac{13}{46} \)

Teacher's Note:
a) Cards left \( = 52 - (2 + 4) = 46 \); all 13 hearts are still in the pack.
b) P(heart) \( = \frac{13}{46} \). Remember to reduce the total, not the number of hearts.

Assertion-Reason Based Questions

Directions: Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer from the options (A), (B), (C) and (D) as given below:

19. Assertion (A): If three vertices of a parallelogram taken in order are \( (-1, -6) \), \( (2, -5) \) and \( (7, 2) \), then its fourth vertex is \( (4, 1) \).
Reason (R): Diagonals of a parallelogram bisect each other. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (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 Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) Mid-point of the diagonal joining \( (-1, -6) \) and \( (7, 2) \) is \( (3, -2) \).
b) The fourth vertex \( (x, y) \) must satisfy \( \frac{2 + x}{2} = 3 \) and \( \frac{-5 + y}{2} = -2 \), giving \( (4, 1) \). This uses the Reason directly.

20. Assertion (A): If zeroes of the polynomial \( (2k - 1)x^2 + 4x - 3 \) are reciprocal of each other, then \( k = -1 \).
Reason (R): If \( a = c \), then zeroes of the polynomial \( ax^2 + bx + c \), \( a \neq 0 \) are reciprocal of each other. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but Reason (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 Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) Zeroes are reciprocals when their product \( \frac{c}{a} = 1 \), i.e. \( a = c \).
b) So \( 2k - 1 = -3 \Rightarrow k = -1 \), which is exactly what the Reason explains.

SECTION - B (5 x 2 = 10)

This section comprises of 5 Very Short Answer (VSA) type questions of 2 marks each.

21. (A) A hall has a length of 9.75 m, breadth of 6.75 m and height of 5.25 m. What is the length of the longest unmarked ruler that can exactly measure the dimensions of the hall? [2 Marks]

Answer:
1. Length = 975 cm, breadth = 675 cm, height = 525 cm. The longest ruler is \( \text{HCF}(975, 675, 525) \).
2. \( 975 = 3 \times 5^2 \times 13 \), \( 675 = 3^3 \times 5^2 \), \( 525 = 3 \times 5^2 \times 7 \).
3. \( \text{HCF} = 3 \times 5^2 = 75 \).
Hence, the length of the longest unmarked ruler is 75 cm or 0.75 m.

Teacher's Note:
a) Convert metres to centimetres first so that all numbers are whole numbers.
b) "Longest length that measures exactly" means HCF; take the lowest power of each common prime.

OR

(B) There are three bells placed at different swings in a park, which toll at intervals of 5, 6 and 8 minutes, respectively. They all toll together when the park opens for the visitors at 10:00 a.m.. How many more times do they all toll together till the park is closed at 7:00 p.m.? [2 Marks]

Answer:
1. \( \text{LCM}(5, 6, 8) = 2^3 \times 3 \times 5 = 120 \). So the bells toll together every 120 minutes, i.e. every 2 hours.
2. After 10:00 a.m., they toll together at 12:00 noon, 2:00 p.m., 4:00 p.m. and 6:00 p.m.
Hence, the bells toll together 4 more times till the park closes at 7:00 p.m.

Teacher's Note:
a) "Toll together again" means LCM, not HCF.
b) Do not count the 10:00 a.m. toll; the question asks for "more times".
c) 8:00 p.m. is after closing, so it is not counted.

22. (A) ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that \( \frac{AO}{BO} = \frac{CO}{DO} \). [2 Marks]

Answer:
1. In \( \Delta AOB \) and \( \Delta COD \): \( \angle OAB = \angle OCD \) and \( \angle OBA = \angle ODC \) (alternate interior angles, as \( AB \parallel DC \)).
2. So \( \Delta AOB \sim \Delta COD \) (AA criterion).
3. Therefore \( \frac{AO}{CO} = \frac{BO}{DO} \), which gives \( \frac{AO}{BO} = \frac{CO}{DO} \).

Teacher's Note:
a) Always write the reason (alternate interior angles) next to each pair of equal angles.
b) Write the vertices of similar triangles in the correct order: A with C, B with D, O with O.

OR

(B) In the given figure, if \( \angle PQR = \angle QSP \), PQ = 6 cm and PS = 3 cm, then find the length of PR. [2 Marks]

[Figure: Triangle PQR with base PR. Point S lies on PR between P and R, and the segment QS is drawn. Arcs mark \( \angle PQR \) at Q and \( \angle QSP \) at S.]

Answer:
1. In \( \Delta PRQ \) and \( \Delta PQS \): \( \angle PQR = \angle QSP \) (given) and \( \angle QPR = \angle QPS \) (common).
2. So \( \Delta PRQ \sim \Delta PQS \) (AA criterion).
3. \( \frac{PQ}{PS} = \frac{PR}{PQ} \Rightarrow \frac{6}{3} = \frac{PR}{6} \Rightarrow PR = 12 \) cm.

Teacher's Note:
a) Match the equal angles to fix the order of vertices: Q in \( \Delta PRQ \) matches S in \( \Delta PQS \).
b) This gives \( PQ^2 = PS \times PR \), a quick way to check: \( 36 = 3 \times 12 \).

For Visually Impaired Candidates (in lieu of Q. 22 (B))

In a right-angled triangle ABC, right-angled at B, a perpendicular BD is drawn to the hypotenuse AC. Prove that the triangle BCD is similar to the triangle ACB. [2 Marks]

Answer:
1. In \( \Delta BCD \) and \( \Delta ACB \): \( \angle BDC = \angle ABC \) (each \( 90^{\circ} \)).
2. \( \angle BCD = \angle ACB \) (common angle).
3. Therefore \( \Delta BCD \sim \Delta ACB \) (AA criterion).

Teacher's Note:
a) Two pairs of equal angles are enough for similarity (AA criterion).
b) Keep the correct order: D matches B, and C matches C.

23. If \( \tan \theta = \frac{1}{\sqrt{5}} \), then find the value of \( \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} \). [2 Marks]

Answer:
1. \( \tan \theta = \frac{1}{\sqrt{5}} \Rightarrow \cot \theta = \sqrt{5} \).
2. \( \sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{1}{5} = \frac{6}{5} \).
3. \( \text{cosec}^2 \theta = 1 + \cot^2 \theta = 1 + 5 = 6 \).
4. \( \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} = \frac{6 - \frac{6}{5}}{6 + \frac{6}{5}} = \frac{\frac{24}{5}}{\frac{36}{5}} = \frac{2}{3} \).

Teacher's Note:
a) Use the identities \( \sec^2 \theta = 1 + \tan^2 \theta \) and \( \text{cosec}^2 \theta = 1 + \cot^2 \theta \).
b) Multiply numerator and denominator by 5 to clear fractions quickly: \( \frac{30 - 6}{30 + 6} = \frac{24}{36} \).

24. Find the area of a sector of a circle with diameter 28 cm, If the length of the corresponding arc is 22 cm. [2 Marks]

Answer:
1. Radius \( r = \frac{28}{2} = 14 \) cm.
2. Area of sector \( = \frac{1}{2} \times l \times r = \frac{1}{2} \times 22 \times 14 \).
3. Area of sector = 154 cm².

Teacher's Note:
a) When the arc length is given, use \( \frac{1}{2} l r \); there is no need to find the angle.
b) Use the radius, not the diameter, in the formula.

25. Two dice are thrown simultaneously. Find the probability that the product of the numbers appearing on them is a prime number. [2 Marks]

Answer:
1. Total number of outcomes = 36.
2. Favourable outcomes: (1, 2), (1, 3), (1, 5), (2, 1), (3, 1), (5, 1), i.e. 6 outcomes.
3. P(prime product) \( = \frac{6}{36} = \frac{1}{6} \).

Teacher's Note:
a) A product is prime only when one die shows 1 and the other shows a prime (2, 3 or 5).
b) Count both orders, such as (1, 2) and (2, 1).

SECTION - C (6 x 3 = 18)

This section comprises of 6 Short Answer (SA) type questions of 3 marks each.

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

Answer:
1. Let us assume that \( 2 + 3\sqrt{3} \) is a rational number.
2. Then \( 2 + 3\sqrt{3} = \frac{a}{b} \), where a and b are integers and \( b \neq 0 \).
3. So \( \sqrt{3} = \frac{a - 2b}{3b} \).
4. Since a and b are integers, \( \frac{a - 2b}{3b} \) is a rational number, so \( \sqrt{3} \) would be rational.
5. This contradicts the given fact that \( \sqrt{3} \) is irrational. Hence our assumption is wrong, and \( 2 + 3\sqrt{3} \) is an irrational number.

Teacher's Note:
a) This is proof by contradiction: start by assuming the opposite.
b) Mention clearly that a and b are integers with \( b \neq 0 \).
c) End by stating the contradiction and the conclusion; this line carries a mark.

27. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( 2x^2 - 8x + 5 \), then find the value of \( \left( \alpha + \frac{1}{\beta} \right) \times \left( \beta + \frac{1}{\alpha} \right) \). [3 Marks]

Answer:
1. \( \alpha + \beta = -\frac{-8}{2} = 4 \) and \( \alpha \beta = \frac{5}{2} \).
2. \( \left( \alpha + \frac{1}{\beta} \right) \times \left( \beta + \frac{1}{\alpha} \right) = \frac{\alpha \beta + 1}{\beta} \times \frac{\alpha \beta + 1}{\alpha} = \frac{(\alpha \beta + 1)^2}{\alpha \beta} \).
3. \( = \frac{\left( \frac{5}{2} + 1 \right)^2}{\frac{5}{2}} = \frac{\frac{49}{4}}{\frac{5}{2}} = \frac{49}{10} \).

Teacher's Note:
a) Sum of zeroes \( = -\frac{b}{a} \) and product \( = \frac{c}{a} \); take care of the sign of b.
b) Combine each bracket into a single fraction before substituting.
c) The final answer \( \frac{49}{10} \) can also be written as 4.9.

28. (A) Ridhi drew a polygon with n sides. The smallest exterior angle is \( 8^{\circ} \) and each subsequent exterior angle is \( 4^{\circ} \) more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn. [3 Marks]

Answer:
1. The exterior angles form an A.P. with \( a = 8 \) and \( d = 4 \). The sum of all exterior angles of a polygon is \( 360^{\circ} \), so \( S_n = 360 \).
2. \( \frac{n}{2}[2 \times 8 + (n - 1) \times 4] = 360 \Rightarrow n(2n + 6) = 360 \Rightarrow n^2 + 3n - 180 = 0 \).
3. \( (n + 15)(n - 12) = 0 \Rightarrow n = -15 \) or \( n = 12 \).
4. n cannot be negative, so \( n = 12 \). The polygon has 12 sides.

Teacher's Note:
a) The key fact is that the exterior angles of any polygon add up to \( 360^{\circ} \).
b) Reject the negative root with a reason; this step carries marks.

OR

(B) Find the sum of integers between 1 and 400 that are multiples of 4 as well as of 5. [3 Marks]

Answer:
1. Multiples of both 4 and 5 are multiples of 20. Required A.P.: 20, 40, 60, ..., 380, with \( a = 20 \), \( d = 20 \).
2. \( a_n = 380 \Rightarrow 20 + (n - 1) \times 20 = 380 \Rightarrow n = 19 \).
3. \( S_{19} = \frac{19}{2}[20 + 380] = \frac{19}{2} \times 400 = 3800 \).

Teacher's Note:
a) "Multiple of 4 as well as of 5" means a multiple of \( \text{LCM}(4, 5) = 20 \).
b) 400 is not included because the question says "between 1 and 400".
c) Use \( S_n = \frac{n}{2}(a + l) \) when the last term is known.

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

Answer:
Given: ABCD is a parallelogram circumscribing a circle. The circle touches AB, BC, CD and DA at S, P, Q and R respectively.
To prove: ABCD is a rhombus.
Proof:
1. Tangents drawn from an external point to a circle are equal, so AS = AR ...(i), BS = BP ...(ii), CQ = CP ...(iii), DQ = DR ...(iv).
2. Adding (i), (ii), (iii) and (iv): AS + BS + CQ + DQ = AR + BP + CP + DR, so AB + CD = AD + BC.
3. In a parallelogram, AB = CD and AD = BC. So 2AB = 2AD, i.e. AB = AD.
4. Since adjacent sides of parallelogram ABCD are equal, ABCD is a rhombus.

Teacher's Note:
a) Draw the figure and name the points of contact before writing the proof.
b) Quote the reason "tangents from an external point are equal" for each pair.
c) Use opposite sides of a parallelogram being equal to reach AB = AD.

30. In what ratio does the x-axis divide the line segment joining the points \( (-4, -6) \) and \( (-1, 7) \)? Also, find the coordinates of the point of division. [3 Marks]

Answer:
1. Let the x-axis divide the segment joining A \( (-4, -6) \) and B \( (-1, 7) \) at P \( (x, 0) \) in the ratio \( k : 1 \).
2. By the section formula, P \( = \left( \frac{-k - 4}{k + 1}, \frac{7k - 6}{k + 1} \right) \).
3. P lies on the x-axis, so \( \frac{7k - 6}{k + 1} = 0 \Rightarrow k = \frac{6}{7} \). The ratio is \( 6 : 7 \).
4. \( x = \frac{-\frac{6}{7} - 4}{\frac{6}{7} + 1} = \frac{-34}{13} \). So the point of division is \( \left( -\frac{34}{13}, 0 \right) \).

Teacher's Note:
a) Any point on the x-axis has y-coordinate 0; use this to find k.
b) Taking the ratio as \( k : 1 \) keeps the working short.
c) Check: the ratio \( 6 : 7 \) also equals \( |-6| : |7| \), the distances of the points from the x-axis.

31. (A) During a math class, Ms. Isha wrote the expression given below on the board and asked the students to simplify it.
\( \frac{\cos \theta}{1 - \sin \theta} + \frac{1 - \sin \theta}{\cos \theta} \)
Jyoti solved it in her note book as follows:
\( \frac{\cos \theta}{1 - \sin \theta} + \frac{1 - \sin \theta}{\cos \theta} \)
\( = \frac{\cos^2 \theta + (1 - \sin \theta)^2}{(1 - \sin \theta) \times \cos \theta} \) .... (step 1)
\( = \frac{\cos^2 \theta + \cos^2 \theta}{(1 - \sin \theta) \times \cos \theta} \) .... (step 2)
\( = \frac{2\cos^2 \theta}{(1 - \sin \theta) \times \cos \theta} \) .... (step 3)
\( = \frac{2\cos \theta}{(1 - \sin \theta)} \) .... (step 4)
Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer. [3 Marks]

Answer:
Jyoti made an error in step 2: \( (1 - \sin \theta)^2 \) is not equal to \( \cos^2 \theta \).
The correct solution is:
1. \( \frac{\cos \theta}{1 - \sin \theta} + \frac{1 - \sin \theta}{\cos \theta} = \frac{\cos^2 \theta + (1 - \sin \theta)^2}{(1 - \sin \theta) \times \cos \theta} \)
2. \( = \frac{\cos^2 \theta + 1 + \sin^2 \theta - 2\sin \theta}{(1 - \sin \theta) \times \cos \theta} \)
3. \( = \frac{2 - 2\sin \theta}{(1 - \sin \theta) \times \cos \theta} \) (using \( \sin^2 \theta + \cos^2 \theta = 1 \))
4. \( = \frac{2(1 - \sin \theta)}{(1 - \sin \theta) \times \cos \theta} = \frac{2}{\cos \theta} = 2\sec \theta \)

Teacher's Note:
a) Naming the wrong step (step 2) earns the first mark; say why it is wrong.
b) Expand \( (1 - \sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta \) carefully.
c) Take out 2 as a common factor so that \( (1 - \sin \theta) \) cancels.

OR

(B) If \( \frac{1}{\sin x - \cos x} = \frac{\text{cosec}\, x}{\sqrt{2}} \), then prove that \( \left( \frac{1}{\sin x + \cos x} \right)^2 = \frac{\sec^2 x}{2} \) [3 Marks]

Answer:
1. Given \( \frac{1}{\sin x - \cos x} = \frac{1}{\sqrt{2} \sin x} \Rightarrow \sin x - \cos x = \sqrt{2} \sin x \).
2. Squaring both sides: \( \sin^2 x + \cos^2 x - 2\sin x \cos x = 2\sin^2 x \Rightarrow 2\sin x \cos x = 1 - 2\sin^2 x \) ...(i)
3. LHS \( = \left( \frac{1}{\sin x + \cos x} \right)^2 = \frac{1}{\sin^2 x + \cos^2 x + 2\sin x \cos x} = \frac{1}{1 + 2\sin x \cos x} \)
4. Using (i): \( = \frac{1}{1 + (1 - 2\sin^2 x)} = \frac{1}{2 - 2\sin^2 x} = \frac{1}{2\cos^2 x} \)
5. \( = \frac{\sec^2 x}{2} = \) RHS. Hence proved.

Teacher's Note:
a) Rewrite \( \text{cosec}\, x \) as \( \frac{1}{\sin x} \) to simplify the given condition.
b) Squaring and using \( \sin^2 x + \cos^2 x = 1 \) gives relation (i), which is the key step.
c) Use \( 1 - \sin^2 x = \cos^2 x \) to reach the RHS.

SECTION - D (4 x 5 = 20)

This section comprises of 4 Long Answer (LA) type questions of 5 marks each.

32. In a rectangular park measuring 50 m \( \times \) 40 m, the gram panchayat plans to construct a rectangular swimming pool in the middle, surrounded by a uniform-width grass strip throughout the park. Find the dimensions of the swimming pool, if the area of the grassy strip is 1184 m². [5 Marks]

Answer:
Let ABCD be the rectangular park and PQRS the swimming pool. Let x m be the uniform width of the grass strip.
1. Length of the pool, PQ \( = (50 - 2x) \) m and breadth of the pool, QR \( = (40 - 2x) \) m.
2. Area of park \( - \) Area of pool \( = \) Area of grass strip: \( 50 \times 40 - (50 - 2x)(40 - 2x) = 1184 \).
3. \( 2000 - (2000 - 180x + 4x^2) = 1184 \Rightarrow 4x^2 - 180x + 1184 = 0 \Rightarrow x^2 - 45x + 296 = 0 \).
4. \( (x - 8)(x - 37) = 0 \Rightarrow x = 8 \) or \( x = 37 \).
5. \( x = 37 \) is rejected, because then \( 50 - 2x \) and \( 40 - 2x \) would be negative. So \( x = 8 \).
Length of the pool \( = 50 - 16 = 34 \) m and breadth of the pool \( = 40 - 16 = 24 \) m.

Teacher's Note:
a) The strip is on both sides, so subtract \( 2x \) from each dimension, not \( x \).
b) Give a reason for rejecting \( x = 37 \); length and breadth cannot be negative.
c) Check: \( 2000 - 34 \times 24 = 2000 - 816 = 1184 \) m².

33. Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, divides the other two sides in the same ratio.
Using the above theorem solve the following:
PQRS is a trapezium with PQ || SR. X and Y are points on non-parallel sides PS and QR respectively such that XY || PQ. Show that \( \frac{PX}{XS} = \frac{QY}{YR} \). [5 Marks]

Answer:
Theorem (Basic Proportionality Theorem):
Given: In \( \Delta ABC \), a line DE \( \parallel \) BC meets AB at D and AC at E.
To prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).
Construction: Join BE and CD. Draw DM \( \perp \) AC and EN \( \perp \) AB.
Proof:
1. \( \text{ar}(ADE) = \frac{1}{2} \times AD \times EN \) and \( \text{ar}(BDE) = \frac{1}{2} \times DB \times EN \), so \( \frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{AD}{DB} \) ...(1)
2. \( \text{ar}(ADE) = \frac{1}{2} \times AE \times DM \) and \( \text{ar}(DEC) = \frac{1}{2} \times EC \times DM \), so \( \frac{\text{ar}(ADE)}{\text{ar}(DEC)} = \frac{AE}{EC} \) ...(2)
3. \( \Delta BDE \) and \( \Delta DEC \) are on the same base DE and between the same parallels DE and BC, so \( \text{ar}(BDE) = \text{ar}(DEC) \) ...(3)
4. From (1), (2) and (3): \( \frac{AD}{DB} = \frac{AE}{EC} \). Hence proved.
Application:
5. Join PR, intersecting XY at M.
6. In \( \Delta PSR \), XM \( \parallel \) SR (as XY \( \parallel \) PQ \( \parallel \) SR). By the theorem, \( \frac{PX}{XS} = \frac{PM}{MR} \) ...(i)
7. In \( \Delta PQR \), MY \( \parallel \) PQ. By the theorem, \( \frac{QY}{YR} = \frac{PM}{MR} \) ...(ii)
8. From (i) and (ii), \( \frac{PX}{XS} = \frac{QY}{YR} \). Hence shown.

Teacher's Note:
a) The theorem carries marks for the figure, given, to prove and construction, so write each heading.
b) In the application, joining the diagonal PR is the key step; it creates two triangles for the theorem.
c) Both ratios share \( \frac{PM}{MR} \), which links the two sides of the trapezium.

34. (A) A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, then find the area of the canvas required for making the tent, keeping a provision of 26 m² of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of Rs. 250 per m². [5 Marks]

Answer:
1. Height of conical part \( = 13.5 - 3 = 10.5 \) m; radius \( r = 14 \) m.
2. Slant height \( l = \sqrt{(14)^2 + (10.5)^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5 \) m.
3. CSA of conical part \( = \pi r l = \frac{22}{7} \times 14 \times 17.5 = 770 \) m².
4. CSA of cylindrical part \( = 2\pi r h = 2 \times \frac{22}{7} \times 14 \times 3 = 264 \) m².
5. Total canvas required \( = 770 + 264 + 26 = 1060 \) m².
Cost of canvas \( = 250 \times 1060 = \) Rs. 2,65,000.

Teacher's Note:
a) A tent has no base, so use only the curved surface areas of the cone and cylinder.
b) Find the height of the cone first by subtracting the cylinder's height from the total height.
c) Do not forget to add the 26 m² for stitching and wastage before finding the cost.

OR

(B) A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 3.5 cm and the total wood used in making the toy is \( 166\frac{5}{6} \) cm³. Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of Rs. 15 per cm². [5 Marks]

Answer:
Let h cm be the height of the conical part; \( r = 3.5 \) cm.
1. Volume of wood = Volume of cone + Volume of hemisphere: \( 166\frac{5}{6} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \).
2. \( \frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h + \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 \Rightarrow \frac{1001}{6} = \frac{77}{6}(h + 7) \).
3. \( h + 7 = 13 \Rightarrow h = 6 \) cm.
4. CSA of hemispherical part \( = 2\pi r^2 = 2 \times \frac{22}{7} \times (3.5)^2 = 77 \) cm².
5. Cost of painting \( = 15 \times 77 = \) Rs. 1155.

Teacher's Note:
a) Convert \( 166\frac{5}{6} \) to the improper fraction \( \frac{1001}{6} \) before solving.
b) Take \( \frac{1}{3}\pi r^2 \) common: volume \( = \frac{1}{3}\pi r^2 (h + 2r) \); this makes the arithmetic easy.
c) The curved surface area of a hemisphere is \( 2\pi r^2 \), not \( 3\pi r^2 \) (that is its total surface area).

35. (A) Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of \( 30^{\circ} \). Mahesh, standing on the roof of a 50 m high building on the opposite side of the kite, observes the same kite at an angle of elevation of \( 45^{\circ} \). Find the distance between the kite and Mahesh. [5 Marks]

Answer:
Let Manish stand at A and the kite be at B, with BE perpendicular to the ground. Let CD be the building (DC = 50 m), with Mahesh at D, and let DR be perpendicular to BE. So AB = 200 m, \( \angle BAE = 30^{\circ} \) and \( \angle BDR = 45^{\circ} \).
1. In right-angled \( \Delta AEB \): \( \frac{BE}{200} = \sin 30^{\circ} = \frac{1}{2} \Rightarrow BE = 100 \) m.
2. RE = DC = 50 m, so BR = BE \( - \) RE \( = 100 - 50 = 50 \) m.
3. In right-angled \( \Delta BRD \): \( \frac{BR}{BD} = \sin 45^{\circ} \Rightarrow \frac{50}{BD} = \frac{1}{\sqrt{2}} \).
4. \( BD = 50\sqrt{2} \) m.
Hence, the distance of the kite from Mahesh is \( 50\sqrt{2} \) m (about 70.7 m).

Teacher's Note:
a) A correct, labelled figure carries 1 mark; show the 50 m building and the horizontal line DR.
b) 200 m is the length of the line of sight (hypotenuse), so use sine, not tangent.
c) Subtract the building's height from the kite's height before using the \( 45^{\circ} \) triangle.

OR

(B) The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be \( 45^{\circ} \) and \( 30^{\circ} \). If the ships are 200m apart and one ship is exactly behind the other then find the height of lighthouse. [5 Marks]

Answer:
Let AB = h m be the lighthouse, with A at the top. Let C and D be the two ships, with CD = 200 m. The angles of elevation of A from C and D are \( 45^{\circ} \) and \( 30^{\circ} \) (equal to the angles of depression).
1. In right-angled \( \Delta ABC \): \( \frac{h}{BC} = \tan 45^{\circ} = 1 \Rightarrow BC = h \) ...(i)
2. In right-angled \( \Delta ABD \): \( \frac{h}{BD} = \tan 30^{\circ} \Rightarrow \frac{h}{h + 200} = \frac{1}{\sqrt{3}} \).
3. \( \sqrt{3}h = h + 200 \Rightarrow h(\sqrt{3} - 1) = 200 \Rightarrow h = \frac{200}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \).
4. \( h = \frac{200(\sqrt{3} + 1)}{2} = 100(\sqrt{3} + 1) \).
Hence, the height of the lighthouse is \( 100(\sqrt{3} + 1) \) m (about 273.2 m).

Teacher's Note:
a) Angle of depression from the top equals the angle of elevation from the ship (alternate angles).
b) The farther ship makes the smaller angle, so BD = h + 200.
c) Rationalise the denominator to get the answer in the standard form.

SECTION - E (3 x 4 = 12)

This section comprises of 3 case-study-based questions of 4 marks each with sub parts. Each case study questions has three sub parts (i), (ii), (iii) of marks 1, 1, 2 respectively.

36. A student entrepreneur started a company that manufactures sanitizers in two sizes - small and large. The cost of a small bottle of sanitizer is Rs. 10 and that of a large bottle is Rs. 15. In June, the company sold 1000 bottles and recorded a total sale of Rs. 12,750. Seeing the increased demand, the company decided to increase the price of both sanitizer bottles by Rs. 2 each. In the next month, the company sold 2500 bottles and recorded a total sale of Rs. 34,250.
Based on the above information, answer the following questions:

(i) Form a linear equation in two variables representing the sale for June. [1 Mark]

Answer: Let the number of small and large bottles be x and y respectively. Sale for June: \( 10x + 15y = 12750 \) or \( 2x + 3y = 2550 \).

Teacher's Note:
a) Always define the variables before writing the equation.
b) Total sale = price \( \times \) number of bottles, added for both sizes.

(ii) Form a linear equation in two variables representing the sale for July. [1 Mark]

Answer: In July the prices are Rs. 12 and Rs. 17. Sale for July: \( 12x + 17y = 34250 \).

Teacher's Note:
a) Add Rs. 2 to each price: \( 10 + 2 = 12 \) and \( 15 + 2 = 17 \).
b) Use the July total of Rs. 34,250, not the June total.

(iii) (A) How many sanitizer bottles of each type were actually sold in June? [2 Marks]

Answer:
1. \( x + y = 1000 \) ...(i) and \( 10x + 15y = 12750 \) ...(ii)
2. From (i), \( 10x + 10y = 10000 \). Subtracting from (ii): \( 5y = 2750 \Rightarrow y = 550 \), so \( x = 450 \).
Number of small and large bottles sold in June are 450 and 550 respectively.

Teacher's Note:
a) Use both facts for June: the number of bottles and the total sale.
b) Check: \( 10 \times 450 + 15 \times 550 = 4500 + 8250 = 12750 \).

OR

(iii) (B) How many sanitizer bottles of each type were sold in July? [2 Marks]

Answer:
1. \( x + y = 2500 \) ...(i) and \( 12x + 17y = 34250 \) ...(ii)
2. From (i), \( 12x + 12y = 30000 \). Subtracting from (ii): \( 5y = 4250 \Rightarrow y = 850 \), so \( x = 1650 \).
Number of small and large bottles sold in July are 1650 and 850 respectively.

Teacher's Note:
a) Use the new prices Rs. 12 and Rs. 17 with the July total of 2500 bottles.
b) Check: \( 12 \times 1650 + 17 \times 850 = 19800 + 14450 = 34250 \).

37. In a technology park in Hyderabad, a company installed a circular meditation garden for its employees as shown in the figure.

[Figure: A circle with centre O. P is a point outside the circle. Two tangents PA and PB are drawn from P, touching the circle at A and B, and the chord AB is drawn.]

Arjun, a designer, stood at a point P outside the garden and drew two tangents PA and PB to the circular boundary. The radius of the garden is 9 m, and the distance of point P from the centre O is 18 m. For preparing safety guidelines, Arjun needed the lengths of the tangents. During a demonstration to interns, he asked them to compute the angle formed between the two tangents using the properties of circle.
Based on the above information, answer the following questions:

(i) What is the measure of \( \angle OAP \)? [1 Mark]

Answer: \( \angle OAP = 90^{\circ} \), because the radius is perpendicular to the tangent at the point of contact.

Teacher's Note:
a) Always state the reason: radius \( \perp \) tangent at the point of contact.
b) This right angle is used in every other part of the case study.

(ii) Calculate the length of the tangent PA. [1 Mark]

Answer: In right-angled \( \Delta OAP \), \( OP^2 = OA^2 + AP^2 \Rightarrow (18)^2 = (9)^2 + AP^2 \Rightarrow AP^2 = 243 \Rightarrow AP = \sqrt{243} = 9\sqrt{3} \) m.

Teacher's Note:
a) OP is the hypotenuse because it is opposite the right angle at A.
b) Simplify the surd: \( \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3} \).

(iii) (A) Calculate the measure of \( \angle AOB \). [2 Marks]

Answer:
1. Let \( \angle AOP = \theta \). In right-angled \( \Delta OAP \), \( \cos \theta = \frac{OA}{OP} = \frac{9}{18} = \frac{1}{2} = \cos 60^{\circ} \Rightarrow \theta = 60^{\circ} \).
2. OP bisects \( \angle AOB \), so \( \angle AOB = 2\theta = 120^{\circ} \).

Teacher's Note:
a) \( \Delta OAP \cong \Delta OBP \), so \( \angle AOP = \angle BOP \).
b) Check: \( \angle APB = 180^{\circ} - 120^{\circ} = 60^{\circ} \), since \( \angle AOB \) and \( \angle APB \) are supplementary.

OR

(iii) (B) In the above design, if PA and PB were inclined at \( 60^{\circ} \), then what would be their length? [2 Marks]

Answer:
1. If the tangents are inclined at \( 60^{\circ} \), then \( \angle OPA = \frac{1}{2} \times 60^{\circ} = 30^{\circ} \).
2. In right-angled \( \Delta OPA \), \( \frac{OA}{AP} = \tan 30^{\circ} \Rightarrow \frac{9}{AP} = \frac{1}{\sqrt{3}} \Rightarrow AP = 9\sqrt{3} \) m.
So PA = PB = \( 9\sqrt{3} \) m.

Teacher's Note:
a) OP bisects the angle between the two tangents.
b) OA is opposite \( \angle OPA \) and AP is adjacent to it, so use tangent.

38. A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below:
Marks Obtained: 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50
Number of students: 3 | 6 | 12 | 15 | 14
The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.
Based on the above information, answer the following questions:

(i) Identify the class with the most common performance range. [1 Mark]

Answer: The class 30 - 40, as it has the highest frequency (15). This is the modal class.

Teacher's Note:
a) "Most common" means the class with the maximum frequency.
b) Do not confuse it with the class 40 - 50, whose frequency (14) is slightly lower.

(ii) Find the class interval containing the median. [1 Mark]

Answer: Cumulative frequencies are 3, 9, 21, 36, 50. Here N = 50, so \( \frac{N}{2} = 25 \). The first cumulative frequency greater than 25 is 36, so the median class is 30 - 40.

Teacher's Note:
a) Make the cumulative frequency column first.
b) The median class is the class whose cumulative frequency is just greater than \( \frac{N}{2} \).

(iii) (A) What is the average performance of the students? [2 Marks]

Answer:
1. Class marks \( x_i \): 5, 15, 25, 35, 45. Products \( f_i x_i \): 15, 90, 300, 525, 630.
2. \( \sum f_i = 50 \) and \( \sum f_i x_i = 1560 \).
3. Mean \( = \frac{\sum f_i x_i}{\sum f_i} = \frac{1560}{50} = 31.2 \).
The average performance of the students is 31.2 marks.

Teacher's Note:
a) A correct table of \( x_i \) and \( f_i x_i \) carries 1 mark.
b) Class mark \( = \frac{\text{lower limit} + \text{upper limit}}{2} \).

OR

(iii) (B) Find the mode of the data. [2 Marks]

Answer:
1. Modal class is 30 - 40: \( l = 30 \), \( h = 10 \), \( f_1 = 15 \), \( f_0 = 12 \), \( f_2 = 14 \).
2. Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{15 - 12}{2 \times 15 - 12 - 14} \right) \times 10 \).
3. Mode \( = 30 + \frac{3}{4} \times 10 = 30 + 7.5 = 37.5 \).

Teacher's Note:
a) \( f_0 \) is the frequency of the class before the modal class and \( f_2 \) of the class after it.
b) Write the formula first, then substitute; both steps earn marks.