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CBSE CLASS X β MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET β 7 | ACADEMIC SESSION 2026β2027
General Instructions:
- This question paper contains 38 questions divided into 5 Sections: A, B, C, D and E.
- All Questions are compulsory.
- Section A comprises 20 Multiple Choice Questions (MCQs) of 1 mark each (Q1 to Q18 are MCQs and Q19 & Q20 are AssertionβReason based).
- Section B comprises 5 Very Short Answer (VSA) questions of 2 marks each (Q21 to Q25).
- Section C comprises 6 Short Answer (SA) questions of 3 marks each (Q26 to Q31).
- Section D comprises 4 Long Answer (LA) questions of 5 marks each (Q32 to Q35).
- Section E comprises 3 Case-Based Integrated Units of Assessment of 4 marks each (Q36 to Q38) with sub-parts of values 1, 1 and 2 marks respectively. Internal choice is provided in the 2-mark question.
- Use of calculators is strictly prohibited. Use $\pi = 22/7$ wherever required unless stated otherwise.
SECTION A β MULTIPLE CHOICE QUESTIONS (1 Mark Each)
Q1.
The total number of prime factors in the prime factorisation of $2520$ is:
[1]
Q2.
If the sum of the zeroes of the quadratic polynomial $p(x) = kx^2 + 2x + 3k$ is equal to their product, then the value of $k$ is:
[1]
Q3.
The pair of equations $x + 2y + 5 = 0$ and $-3x – 6y + 1 = 0$ has:
[1]
Q4.
If the equation $2x^2 + kx + 2 = 0$ has real roots, then the value of $k$ must satisfy:
[1]
Q5.
If the 3rd term of an AP is 4 and the 9th term is $-8$, which term of this AP is 0?
[1]
Q6.
The distance between the points $P(a\sin\alpha, -b\cos\alpha)$ and $Q(-a\cos\alpha, b\sin\alpha)$ is:
[1]
Q7.
In $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 3\text{ cm}$, $DB = 5\text{ cm}$, and $BC = 16\text{ cm}$, then the length of $DE$ is:
[1]
Q8.
If $\tan\theta + \cot\theta = 2$, then the value of $\tan^7\theta + \cot^7\theta$ is:
[1]
Q9.
The value of $\sin^2 30^\circ\cos^2 45^\circ + 4\tan^2 30^\circ + \frac{1}{2}\sin^2 90^\circ – 2\cos^2 90^\circ$ is:
[1]
Q10.
If the angle of elevation of the top of a tower from two points at distances $a$ and $b$ from the base and in the same straight line with it are complementary, then the height of the tower is:
[1]
Q11.
Two concentric circles of radii $a$ and $b$ ($a > b$) are given. The length of the chord of the larger circle which touches the smaller circle is:
[1]
Q12.
A wire is bent in the form of a square enclosing an area of $121\text{ cm}^2$. If the same wire is bent into the form of a circle, the radius of the circle is:
[1]
Q13.
Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter $2\text{ cm}$ and height $16\text{ cm}$. The diameter of each sphere is:
[1]
Q14.
If the difference between Mode and Median of a data set is 24, then the difference between Median and Mean is:
[1]
Q15.
A bag contains 3 red balls, 5 black balls, and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is neither red nor white?
[1]
Q16.
If the line segment joining the points $A(3, 4)$ and $B(k, 6)$ has its midpoint at $P(x, y)$ such that $x + y – 10 = 0$, then the value of $k$ is:
[1]
Q17.
The probability of getting a bad egg in a lot of 400 eggs is $0.035$. The number of bad eggs in the lot is:
[1]
Q18.
The ratio of HCF to LCM of the least composite number and the least prime number is:
[1]
Q19.
Assertion (A): If the product of two numbers is 5760 and their HCF is 12, then their LCM is 480.
Reason (R): For any two positive integers $a$ and $b$, $\text{HCF}(a, b) + \text{LCM}(a, b) = a \times b$. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Reason (R): For any two positive integers $a$ and $b$, $\text{HCF}(a, b) + \text{LCM}(a, b) = a \times b$. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Q20.
Assertion (A): The points $A(4, 3)$, $B(6, 4)$, $C(5, -6)$, and $D(-3, -7)$ taken in order are vertices of a parallelogram.
Reason (R): The diagonals of a parallelogram bisect each other. [1]
Reason (R): The diagonals of a parallelogram bisect each other. [1]
SECTION B β VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.
[2]
Q22.
In $\triangle ABC$, $AD$ is the bisector of $\angle BAC$ meeting side $BC$ at $D$. If $AB = 5.6\text{ cm}$, $AC = 6\text{ cm}$, and $DC = 3\text{ cm}$, find the length of $BC$.
[2]
OR
A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after 4 seconds.
Q23.
Find the coordinates of the points of trisection of the line segment joining the points $A(2, -2)$ and $B(-7, 4)$.
[2]
Q24.
If $\tan(2A) = \cot(A – 18^\circ)$, where $2A$ is an acute angle, find the value of $A$.
[2]
OR
Prove that:
$$\frac{\cos\theta}{1 – \tan\theta} + \frac{\sin\theta}{1 – \cot\theta} = \sin\theta + \cos\theta$$
Q25.
A circle is inscribed in $\triangle ABC$ having sides $AB = 8\text{ cm}$, $BC = 10\text{ cm}$, and $CA = 12\text{ cm}$. Find the lengths of the tangents $AD$, $BE$, and $CF$, where $D, E, F$ are points of contact on sides $AB, BC, CA$ respectively.
[2]
SECTION C β SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
Find the zeroes of the quadratic polynomial $q(x) = \sqrt{3}x^2 + 10x + 7\sqrt{3}$ and verify the relationship between the zeroes and the coefficients.
[3]
Q27.
Solve the following pair of linear equations graphically or algebraically:
$$2x + y = 6$$
$$2x – y + 2 = 0$$
Also, find the coordinates of the vertices of the triangle formed by these two lines and the x-axis.
[3]
OR
A fraction becomes $1/3$ when 1 is subtracted from the numerator and it becomes $1/4$ when 8 is added to its denominator. Find the fraction.
Q28.
The sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289. Find the sum of the first $n$ terms.
[3]
Q29.
Prove the trigonometric identity:
$$\frac{\sin\theta – \cos\theta + 1}{\sin\theta + \cos\theta – 1} = \frac{1}{\sec\theta – \tan\theta}$$
[3]
Q30.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
[3]
OR
Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$.
Q31.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is:
(a) Greater than 9
(b) Less than or equal to 12
(c) A multiple of 4 [3]
(a) Greater than 9
(b) Less than or equal to 12
(c) A multiple of 4 [3]
SECTION D β LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
Seven years ago, Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s age will be two-fifth of Varun’s age. Find their present ages.
[5]
OR
A plane left 30 minutes later than the scheduled time and in order to reach its destination $1500\text{ km}$ away in time, it had to increase its speed by $100\text{ km/h}$ from its usual speed. Find its usual speed.
Q33.
State and prove Basic Proportionality Theorem (Thales Theorem).
Using this theorem, prove that a line drawn through the point of intersection of the diagonals of a trapezium parallel to the parallel sides divides the non-parallel sides in the same ratio. [5]
Using this theorem, prove that a line drawn through the point of intersection of the diagonals of a trapezium parallel to the parallel sides divides the non-parallel sides in the same ratio. [5]
Q34.
A solid toy is composed of a cylinder surmounted by a hemisphere at one end and a cone at the other end. The common radius is $3.5\text{ cm}$. The height of the cylindrical part is $10\text{ cm}$ and the height of the conical part is $6\text{ cm}$. Find the total volume and surface area of the toy. (Take $\pi = 22/7$).
[5]
OR
A farmer connects a pipe of internal diameter $20\text{ cm}$ from a canal into a cylindrical tank in her field, which is $10\text{ m}$ in diameter and $2\text{ m}$ deep. If water flows through the pipe at the rate of $3\text{ km/h}$, in how much time will the tank be filled?
Q35.
The median of the following frequency distribution is 35. Find the values of $x$ and $y$, if the total frequency is 170:
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 10 & 40-50 & y \\ 10-20 & 20 & 50-60 & 25 \\ 20-30 & x & 60-70 & 15 \\ 30-40 & 40 & & \\ \hline \end{array}$$ [5]
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 10 & 40-50 & y \\ 10-20 & 20 & 50-60 & 25 \\ 20-30 & x & 60-70 & 15 \\ 30-40 & 40 & & \\ \hline \end{array}$$ [5]
SECTION E β CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) How many computer centres will be established in the 10th year? [1]
(b) In which year will the NGO establish 68 computer centres? [1]
(c) Find the total number of computer centres established by the NGO in the first 12 years. [2]
To promote digital literacy, an NGO sets up computer study centres in rural villages. In the first year, they set up 20 centres. They plan to increase the number of centres by 6 every successive year.
Based on the above information, answer the following questions:
(a) How many computer centres will be established in the 10th year? [1]
(b) In which year will the NGO establish 68 computer centres? [1]
(c) Find the total number of computer centres established by the NGO in the first 12 years. [2]
OR
If the NGO plans to have established a cumulative total of 380 centres, in how many years will this target be achieved? [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Calculate the direct distance between Rover $B$ and Rover $C$. [1]
(c) Find the coordinates of a charging station $S(x, 0)$ situated on the horizontal boundary (x-axis) that is equidistant from Rover $A(1, 2)$ and Rover $C(7, 2)$. [2]
In a GPS tracking map of an industrial complex, three automated freight rovers are positioned at coordinates $A(1, 2)$, $B(4, 6)$, and $C(7, 2)$ where 1 unit on the grid represents 100 metres.
(a) Calculate the direct distance between Rover $A$ and Rover $B$. [1]
(b) Calculate the direct distance between Rover $B$ and Rover $C$. [1]
(c) Find the coordinates of a charging station $S(x, 0)$ situated on the horizontal boundary (x-axis) that is equidistant from Rover $A(1, 2)$ and Rover $C(7, 2)$. [2]
OR
Determine the coordinates of a master terminal $D$ such that quadrilateral $ABCD$ forms a rhombus. [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Find the horizontal distance of the patrol boat (at $45^\circ$) from the base of the cliff. [1]
(c) Calculate the distance between the civilian boat and the patrol boat. (Take $\sqrt{3} \approx 1.732$). [2]
A coastal observation station has a radar antenna atop a vertical cliff of height $60\text{ m}$ above sea level. From the top of the cliff, the observer spots a civilian boat and a patrol boat directly aligned in the line of sight. The angles of depression of the civilian boat and patrol boat are measured as $30^\circ$ and $45^\circ$ respectively.
(a) Draw a neat labelled mathematical diagram representing this physical arrangement. [1]
(b) Find the horizontal distance of the patrol boat (at $45^\circ$) from the base of the cliff. [1]
(c) Calculate the distance between the civilian boat and the patrol boat. (Take $\sqrt{3} \approx 1.732$). [2]
OR
Find the direct line-of-sight distance from the top of the cliff to the civilian boat (at $30^\circ$). [2]
CBSE CLASS X MATHEMATICS (STANDARD) β SOLUTIONS & MARKING SCHEME (SET β 7)
SECTION A SOLUTIONS
Q1. (c) 7 [1 Mark]
Explanation: $2520 = 2^3 \times 3^2 \times 5^1 \times 7^1$. Total prime factors $= 3 + 2 + 1 + 1 = 7$.
Explanation: $2520 = 2^3 \times 3^2 \times 5^1 \times 7^1$. Total prime factors $= 3 + 2 + 1 + 1 = 7$.
Q2. (d) -2/3 [1 Mark]
Explanation: $\text{Sum} = -\frac{2}{k}$, $\text{Product} = \frac{3k}{k} = 3 \implies -\frac{2}{k} = 3 \implies k = -\frac{2}{3}$.
Explanation: $\text{Sum} = -\frac{2}{k}$, $\text{Product} = \frac{3k}{k} = 3 \implies -\frac{2}{k} = 3 \implies k = -\frac{2}{3}$.
Q3. (d) No solution [1 Mark]
Explanation: $\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}$, $\frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}$, $\frac{c_1}{c_2} = \frac{5}{1} = 5$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, parallel lines have no solution.
Explanation: $\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}$, $\frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}$, $\frac{c_1}{c_2} = \frac{5}{1} = 5$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, parallel lines have no solution.
Q4. (a) $k \ge 4$ or $k \le -4$ [1 Mark]
Explanation: $D = b^2 – 4ac = k^2 – 4(2)(2) = k^2 – 16 \ge 0 \implies k^2 \ge 16 \implies k \ge 4 \text{ or } k \le -4$.
Explanation: $D = b^2 – 4ac = k^2 – 4(2)(2) = k^2 – 16 \ge 0 \implies k^2 \ge 16 \implies k \ge 4 \text{ or } k \le -4$.
Q5. (b) 5th term [1 Mark]
Explanation: $a_3 = a + 2d = 4$ and $a_9 = a + 8d = -8$. Subtracting gives $6d = -12 \implies d = -2, a = 8$.
$a_n = 8 + (n – 1)(-2) = 0 \implies 2(n – 1) = 8 \implies n – 1 = 4 \implies n = 5$.
Explanation: $a_3 = a + 2d = 4$ and $a_9 = a + 8d = -8$. Subtracting gives $6d = -12 \implies d = -2, a = 8$.
$a_n = 8 + (n – 1)(-2) = 0 \implies 2(n – 1) = 8 \implies n – 1 = 4 \implies n = 5$.
Q6. (b) $\sqrt{a^2 + b^2}$ [1 Mark]
Explanation: $d = \sqrt{(-a\cos\alpha – a\sin\alpha)^2 + (b\sin\alpha + b\cos\alpha)^2}$
$= \sqrt{a^2(1 + 2\sin\alpha\cos\alpha) + b^2(1 + 2\sin\alpha\cos\alpha)}$ when orthogonal or straightforwardly $\sqrt{a^2 + b^2}$.
Explanation: $d = \sqrt{(-a\cos\alpha – a\sin\alpha)^2 + (b\sin\alpha + b\cos\alpha)^2}$
$= \sqrt{a^2(1 + 2\sin\alpha\cos\alpha) + b^2(1 + 2\sin\alpha\cos\alpha)}$ when orthogonal or straightforwardly $\sqrt{a^2 + b^2}$.
Q7. (a) 6 cm [1 Mark]
Explanation: $\triangle ADE \sim \triangle ABC \implies \frac{AD}{AB} = \frac{DE}{BC} \implies \frac{3}{3 + 5} = \frac{DE}{16} \implies \frac{3}{8} = \frac{DE}{16} \implies DE = 6\text{ cm}$.
Explanation: $\triangle ADE \sim \triangle ABC \implies \frac{AD}{AB} = \frac{DE}{BC} \implies \frac{3}{3 + 5} = \frac{DE}{16} \implies \frac{3}{8} = \frac{DE}{16} \implies DE = 6\text{ cm}$.
Q8. (b) 2 [1 Mark]
Explanation: $\tan\theta + \frac{1}{\tan\theta} = 2 \implies \tan^2\theta – 2\tan\theta + 1 = 0 \implies (\tan\theta – 1)^2 = 0 \implies \tan\theta = 1, \cot\theta = 1$.
$\tan^7\theta + \cot^7\theta = (1)^7 + (1)^7 = 2$.
Explanation: $\tan\theta + \frac{1}{\tan\theta} = 2 \implies \tan^2\theta – 2\tan\theta + 1 = 0 \implies (\tan\theta – 1)^2 = 0 \implies \tan\theta = 1, \cot\theta = 1$.
$\tan^7\theta + \cot^7\theta = (1)^7 + (1)^7 = 2$.
Q9. (a) 47/24 [1 Mark]
Explanation: $\left(\frac{1}{2}\right)^2\left(\frac{1}{\sqrt{2}}\right)^2 + 4\left(\frac{1}{\sqrt{3}}\right)^2 + \frac{1}{2}(1)^2 – 0 = \frac{1}{4} \cdot \frac{1}{2} + 4 \cdot \frac{1}{3} + \frac{1}{2} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} = \frac{3 + 32 + 12}{24} = \frac{47}{24}$.
Explanation: $\left(\frac{1}{2}\right)^2\left(\frac{1}{\sqrt{2}}\right)^2 + 4\left(\frac{1}{\sqrt{3}}\right)^2 + \frac{1}{2}(1)^2 – 0 = \frac{1}{4} \cdot \frac{1}{2} + 4 \cdot \frac{1}{3} + \frac{1}{2} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} = \frac{3 + 32 + 12}{24} = \frac{47}{24}$.
Q10. (a) $\sqrt{ab}$ [1 Mark]
Explanation: $\tan\theta = \frac{h}{a}$ and $\tan(90^\circ – \theta) = \cot\theta = \frac{h}{b}$. Multiplying gives $\tan\theta\cot\theta = \frac{h^2}{ab} \implies 1 = \frac{h^2}{ab} \implies h = \sqrt{ab}$.
Explanation: $\tan\theta = \frac{h}{a}$ and $\tan(90^\circ – \theta) = \cot\theta = \frac{h}{b}$. Multiplying gives $\tan\theta\cot\theta = \frac{h^2}{ab} \implies 1 = \frac{h^2}{ab} \implies h = \sqrt{ab}$.
Q11. (b) $2\sqrt{a^2 – b^2}$ [1 Mark]
Explanation: Half-chord length is $\sqrt{a^2 – b^2}$. Total chord length is $2\sqrt{a^2 – b^2}$.
Explanation: Half-chord length is $\sqrt{a^2 – b^2}$. Total chord length is $2\sqrt{a^2 – b^2}$.
Q12. (a) 7 cm [1 Mark]
Explanation: Side of square $a = \sqrt{121} = 11\text{ cm}$. Perimeter $= 4 \times 11 = 44\text{ cm}$.
$2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\text{ cm}$.
Explanation: Side of square $a = \sqrt{121} = 11\text{ cm}$. Perimeter $= 4 \times 11 = 44\text{ cm}$.
$2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\text{ cm}$.
Q13. (c) 2 cm [1 Mark]
Explanation: Cylinder volume $= \pi (1)^2(16) = 16\pi$. Volume of 12 spheres $= 12 \times \frac{4}{3}\pi r^3 = 16\pi r^3$.
$16\pi r^3 = 16\pi \implies r = 1\text{ cm} \implies \text{Diameter} = 2\text{ cm}$.
Explanation: Cylinder volume $= \pi (1)^2(16) = 16\pi$. Volume of 12 spheres $= 12 \times \frac{4}{3}\pi r^3 = 16\pi r^3$.
$16\pi r^3 = 16\pi \implies r = 1\text{ cm} \implies \text{Diameter} = 2\text{ cm}$.
Q14. (b) 12 [1 Mark]
Explanation: $\text{Mode} – \text{Mean} = 3(\text{Median} – \text{Mean})$.
$\text{Mode} – \text{Median} = 2(\text{Median} – \text{Mean}) \implies 24 = 2(\text{Median} – \text{Mean}) \implies \text{Median} – \text{Mean} = 12$.
Explanation: $\text{Mode} – \text{Mean} = 3(\text{Median} – \text{Mean})$.
$\text{Mode} – \text{Median} = 2(\text{Median} – \text{Mean}) \implies 24 = 2(\text{Median} – \text{Mean}) \implies \text{Median} – \text{Mean} = 12$.
Q15. (a) 5/12 [1 Mark]
Explanation: Total balls $= 3 + 5 + 4 = 12$. Neither red nor white means black balls $= 5$. $P = \frac{5}{12}$.
Explanation: Total balls $= 3 + 5 + 4 = 12$. Neither red nor white means black balls $= 5$. $P = \frac{5}{12}$.
Q16. (d) 11 [1 Mark]
Explanation: Midpoint $P = \left(\frac{3 + k}{2}, \frac{4 + 6}{2}\right) = \left(\frac{3 + k}{2}, 5\right)$.
$\frac{3 + k}{2} + 5 – 10 = 0 \implies \frac{3 + k}{2} = 5 \implies 3 + k = 10 \implies k = 7$ (check equation $x + y – 10 = 0 \implies \frac{3+k}{2} + 5 = 10 \implies k = 7$).
Explanation: Midpoint $P = \left(\frac{3 + k}{2}, \frac{4 + 6}{2}\right) = \left(\frac{3 + k}{2}, 5\right)$.
$\frac{3 + k}{2} + 5 – 10 = 0 \implies \frac{3 + k}{2} = 5 \implies 3 + k = 10 \implies k = 7$ (check equation $x + y – 10 = 0 \implies \frac{3+k}{2} + 5 = 10 \implies k = 7$).
Q17. (b) 14 [1 Mark]
Explanation: $\text{Number of bad eggs} = 400 \times 0.035 = 14$.
Explanation: $\text{Number of bad eggs} = 400 \times 0.035 = 14$.
Q18. (a) 1 : 2 [1 Mark]
Explanation: Least composite number $= 4$, least prime number $= 2$. $\text{HCF}(4, 2) = 2$, $\text{LCM}(4, 2) = 4$. Ratio $= 2 : 4 = 1 : 2$.
Explanation: Least composite number $= 4$, least prime number $= 2$. $\text{HCF}(4, 2) = 2$, $\text{LCM}(4, 2) = 4$. Ratio $= 2 : 4 = 1 : 2$.
Q19. (c) Assertion (A) is true but Reason (R) is false. [1 Mark]
Explanation: $\text{LCM} = \frac{5760}{12} = 480$ is true. But the formula is $\text{HCF} \times \text{LCM} = a \times b$, not $+$.
Explanation: $\text{LCM} = \frac{5760}{12} = 480$ is true. But the formula is $\text{HCF} \times \text{LCM} = a \times b$, not $+$.
Q20. (d) Assertion (A) is false but Reason (R) is true. [1 Mark]
Explanation: Midpoint of $AC = \left(\frac{9}{2}, -\frac{3}{2}\right) \neq \text{Midpoint of } BD = \left(\frac{3}{2}, -\frac{3}{2}\right)$, so $ABCD$ is not a parallelogram.
Explanation: Midpoint of $AC = \left(\frac{9}{2}, -\frac{3}{2}\right) \neq \text{Midpoint of } BD = \left(\frac{3}{2}, -\frac{3}{2}\right)$, so $ABCD$ is not a parallelogram.
SECTION B SOLUTIONS
Q21. Composite Numbers: [2 Marks]
- $7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13(78) = 13 \times 13 \times 6$, which has more than two factors, hence composite. [1 Mark]
- $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5(1008 + 1) = 5 \times 1009$, which has more than two factors, hence composite. [1 Mark]
Q22. Angle Bisector Property: [2 Marks]
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{5.6}{6} = \frac{BD}{3} \implies BD = \frac{5.6 \times 3}{6} = 2.8\text{ cm}$. [1 Mark]
$BC = BD + DC = 2.8 + 3 = \mathbf{5.8\text{ cm}}$. [1 Mark]
OR
Distance walked in $4\text{ s} = 1.2 \times 4 = 4.8\text{ m}$. Let shadow be $x\text{ m}$.
$\frac{3.6}{0.9} = \frac{4.8 + x}{x} \implies 4 = \frac{4.8 + x}{x} \implies 4x = 4.8 + x \implies 3x = 4.8 \implies \mathbf{x = 1.6\text{ m}}$. [2 Marks]
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{5.6}{6} = \frac{BD}{3} \implies BD = \frac{5.6 \times 3}{6} = 2.8\text{ cm}$. [1 Mark]
$BC = BD + DC = 2.8 + 3 = \mathbf{5.8\text{ cm}}$. [1 Mark]
OR
Distance walked in $4\text{ s} = 1.2 \times 4 = 4.8\text{ m}$. Let shadow be $x\text{ m}$.
$\frac{3.6}{0.9} = \frac{4.8 + x}{x} \implies 4 = \frac{4.8 + x}{x} \implies 4x = 4.8 + x \implies 3x = 4.8 \implies \mathbf{x = 1.6\text{ m}}$. [2 Marks]
Q23. Trisection Points: [2 Marks]
Point $P$ divides $AB$ in $1 : 2$: $P = \left(\frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2}\right) = \mathbf{(-1, 0)}$. [1 Mark]
Point $Q$ divides $AB$ in $2 : 1$: $Q = \left(\frac{2(-7) + 1(2)}{3}, \frac{2(4) + 1(-2)}{3}\right) = \mathbf{(-4, 2)}$. [1 Mark]
Point $P$ divides $AB$ in $1 : 2$: $P = \left(\frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2}\right) = \mathbf{(-1, 0)}$. [1 Mark]
Point $Q$ divides $AB$ in $2 : 1$: $Q = \left(\frac{2(-7) + 1(2)}{3}, \frac{2(4) + 1(-2)}{3}\right) = \mathbf{(-4, 2)}$. [1 Mark]
Q24. Trigonometric Value: [2 Marks]
$\tan(2A) = \cot(90^\circ – 2A) \implies 90^\circ – 2A = A – 18^\circ \implies 3A = 108^\circ \implies \mathbf{A = 36^\circ}$. [2 Marks]
OR
$\text{LHS} = \frac{\cos\theta}{1 – \sin\theta/\cos\theta} + \frac{\sin\theta}{1 – \cos\theta/\sin\theta} = \frac{\cos^2\theta}{\cos\theta – \sin\theta} – \frac{\sin^2\theta}{\cos\theta – \sin\theta}$
$= \frac{\cos^2\theta – \sin^2\theta}{\cos\theta – \sin\theta} = \frac{(\cos\theta – \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta – \sin\theta} = \mathbf{\sin\theta + \cos\theta} = \text{RHS}$. [2 Marks]
$\tan(2A) = \cot(90^\circ – 2A) \implies 90^\circ – 2A = A – 18^\circ \implies 3A = 108^\circ \implies \mathbf{A = 36^\circ}$. [2 Marks]
OR
$\text{LHS} = \frac{\cos\theta}{1 – \sin\theta/\cos\theta} + \frac{\sin\theta}{1 – \cos\theta/\sin\theta} = \frac{\cos^2\theta}{\cos\theta – \sin\theta} – \frac{\sin^2\theta}{\cos\theta – \sin\theta}$
$= \frac{\cos^2\theta – \sin^2\theta}{\cos\theta – \sin\theta} = \frac{(\cos\theta – \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta – \sin\theta} = \mathbf{\sin\theta + \cos\theta} = \text{RHS}$. [2 Marks]
Q25. Tangents on Inscribed Circle: [2 Marks]
Let $AD = AF = x, BD = BE = y, CE = CF = z$.
$x + y = 8$, $y + z = 10$, $z + x = 12$. Adding: $2(x + y + z) = 30 \implies x + y + z = 15$.
$\mathbf{AD = x = 15 – 10 = 5\text{ cm}}$, $\mathbf{BE = y = 15 – 12 = 3\text{ cm}}$, $\mathbf{CF = z = 15 – 8 = 7\text{ cm}}$. [2 Marks]
Let $AD = AF = x, BD = BE = y, CE = CF = z$.
$x + y = 8$, $y + z = 10$, $z + x = 12$. Adding: $2(x + y + z) = 30 \implies x + y + z = 15$.
$\mathbf{AD = x = 15 – 10 = 5\text{ cm}}$, $\mathbf{BE = y = 15 – 12 = 3\text{ cm}}$, $\mathbf{CF = z = 15 – 8 = 7\text{ cm}}$. [2 Marks]
SECTION C SOLUTIONS
Q26. Zeroes & Coefficients: [3 Marks]
$q(x) = \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = (x + \sqrt{3})(\sqrt{3}x + 7)$.
Zeroes are $\mathbf{-\sqrt{3}}$ and $\mathbf{-\frac{7}{\sqrt{3}}}$. [1.5 Marks]
$\alpha + \beta = -\sqrt{3} – \frac{7}{\sqrt{3}} = -\frac{10}{\sqrt{3}} = -\frac{b}{a}$; $\alpha\beta = (-\sqrt{3})\left(-\frac{7}{\sqrt{3}}\right) = 7 = \frac{7\sqrt{3}}{\sqrt{3}} = \frac{c}{a}$. (Verified). [1.5 Marks]
$q(x) = \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = (x + \sqrt{3})(\sqrt{3}x + 7)$.
Zeroes are $\mathbf{-\sqrt{3}}$ and $\mathbf{-\frac{7}{\sqrt{3}}}$. [1.5 Marks]
$\alpha + \beta = -\sqrt{3} – \frac{7}{\sqrt{3}} = -\frac{10}{\sqrt{3}} = -\frac{b}{a}$; $\alpha\beta = (-\sqrt{3})\left(-\frac{7}{\sqrt{3}}\right) = 7 = \frac{7\sqrt{3}}{\sqrt{3}} = \frac{c}{a}$. (Verified). [1.5 Marks]
Q27. Linear Equations Solution: [3 Marks]
Adding: $4x = 4 \implies \mathbf{x = 1}$, $y = 6 – 2(1) = \mathbf{4}$. Intersection is $(1, 4)$. [1.5 Marks]
For $2x + y = 6$, y=0 gives $(3, 0)$. For $2x – y + 2 = 0$, y=0 gives $(-1, 0)$.
Vertices of triangle formed with x-axis: $\mathbf{(1, 4), (3, 0), (-1, 0)}$. [1.5 Marks]
OR
Let fraction be $\frac{x}{y}$.
$\frac{x – 1}{y} = \frac{1}{3} \implies 3x – y = 3$ …(1); $\frac{x}{y + 8} = \frac{1}{4} \implies 4x – y = 8$ …(2)
Subtracting (1) from (2): $\mathbf{x = 5}$; $y = 3(5) – 3 = \mathbf{12}$. Fraction is $\mathbf{\frac{5}{12}}$. [3 Marks]
Adding: $4x = 4 \implies \mathbf{x = 1}$, $y = 6 – 2(1) = \mathbf{4}$. Intersection is $(1, 4)$. [1.5 Marks]
For $2x + y = 6$, y=0 gives $(3, 0)$. For $2x – y + 2 = 0$, y=0 gives $(-1, 0)$.
Vertices of triangle formed with x-axis: $\mathbf{(1, 4), (3, 0), (-1, 0)}$. [1.5 Marks]
OR
Let fraction be $\frac{x}{y}$.
$\frac{x – 1}{y} = \frac{1}{3} \implies 3x – y = 3$ …(1); $\frac{x}{y + 8} = \frac{1}{4} \implies 4x – y = 8$ …(2)
Subtracting (1) from (2): $\mathbf{x = 5}$; $y = 3(5) – 3 = \mathbf{12}$. Fraction is $\mathbf{\frac{5}{12}}$. [3 Marks]
Q28. Sum of $n$ terms of AP: [3 Marks]
$S_7 = \frac{7}{2}[2a + 6d] = 49 \implies a + 3d = 7$ …(1)
$S_{17} = \frac{17}{2}[2a + 16d] = 289 \implies a + 8d = 17$ …(2) [1.5 Marks]
Subtracting gives $5d = 10 \implies d = 2, a = 1$.
$S_n = \frac{n}{2}[2(1) + (n – 1)2] = \frac{n}{2}[2n] = \mathbf{n^2}$. [1.5 Marks]
$S_7 = \frac{7}{2}[2a + 6d] = 49 \implies a + 3d = 7$ …(1)
$S_{17} = \frac{17}{2}[2a + 16d] = 289 \implies a + 8d = 17$ …(2) [1.5 Marks]
Subtracting gives $5d = 10 \implies d = 2, a = 1$.
$S_n = \frac{n}{2}[2(1) + (n – 1)2] = \frac{n}{2}[2n] = \mathbf{n^2}$. [1.5 Marks]
Q29. Trigonometric Identity: [3 Marks]
Dividing by $\cos\theta$: $\text{LHS} = \frac{\tan\theta – 1 + \sec\theta}{\tan\theta + 1 – \sec\theta} = \frac{(\tan\theta + \sec\theta) – (\sec^2\theta – \tan^2\theta)}{\tan\theta – \sec\theta + 1}$
$= \frac{(\sec\theta + \tan\theta)[1 – (\sec\theta – \tan\theta)]}{\tan\theta – \sec\theta + 1} = \sec\theta + \tan\theta = \frac{(\sec\theta + \tan\theta)(\sec\theta – \tan\theta)}{\sec\theta – \tan\theta} = \mathbf{\frac{1}{\sec\theta – \tan\theta}}$. [3 Marks]
Dividing by $\cos\theta$: $\text{LHS} = \frac{\tan\theta – 1 + \sec\theta}{\tan\theta + 1 – \sec\theta} = \frac{(\tan\theta + \sec\theta) – (\sec^2\theta – \tan^2\theta)}{\tan\theta – \sec\theta + 1}$
$= \frac{(\sec\theta + \tan\theta)[1 – (\sec\theta – \tan\theta)]}{\tan\theta – \sec\theta + 1} = \sec\theta + \tan\theta = \frac{(\sec\theta + \tan\theta)(\sec\theta – \tan\theta)}{\sec\theta – \tan\theta} = \mathbf{\frac{1}{\sec\theta – \tan\theta}}$. [3 Marks]
Q30. Tangents Supplementary: [3 Marks]
In quadrilateral $OAPB$: $\angle OAP = 90^\circ$, $\angle OBP = 90^\circ$.
$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ \implies \mathbf{\angle APB + \angle AOB = 180^\circ}$. [3 Marks]
OR
In $\triangle TPQ$, $TP = TQ \implies \angle TPQ = \angle TQP = \frac{180^\circ – \angle PTQ}{2} = 90^\circ – \frac{1}{2}\angle PTQ$.
$\angle OPQ = \angle OPT – \angle TPQ = 90^\circ – \left(90^\circ – \frac{1}{2}\angle PTQ\right) = \frac{1}{2}\angle PTQ \implies \mathbf{\angle PTQ = 2\angle OPQ}$. [3 Marks]
In quadrilateral $OAPB$: $\angle OAP = 90^\circ$, $\angle OBP = 90^\circ$.
$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ \implies \mathbf{\angle APB + \angle AOB = 180^\circ}$. [3 Marks]
OR
In $\triangle TPQ$, $TP = TQ \implies \angle TPQ = \angle TQP = \frac{180^\circ – \angle PTQ}{2} = 90^\circ – \frac{1}{2}\angle PTQ$.
$\angle OPQ = \angle OPT – \angle TPQ = 90^\circ – \left(90^\circ – \frac{1}{2}\angle PTQ\right) = \frac{1}{2}\angle PTQ \implies \mathbf{\angle PTQ = 2\angle OPQ}$. [3 Marks]
Q31. Dice Outcomes: [3 Marks]
Total outcomes $= 36$.
Total outcomes $= 36$.
- (a) Sum $> 9$: Sum 10 (3), 11 (2), 12 (1) $\implies 6$ outcomes. $P = \frac{6}{36} = \mathbf{\frac{1}{6}}$. [1 Mark]
- (b) Sum $\le 12$: All 36 outcomes. $P = \frac{36}{36} = \mathbf{1}$. [1 Mark]
- (c) Multiple of 4 (Sum 4, 8, 12): $(1,3),(2,2),(3,1),(2,6),(3,5),(4,4),(5,3),(6,2),(6,6) \implies 9$ outcomes. $P = \frac{9}{36} = \mathbf{\frac{1}{4}}$. [1 Mark]
SECTION D SOLUTIONS
Q32. Age Problem: [5 Marks]
7 years ago: Let Swati’s age $= x$, Varun’s age $= 5x^2$. Present ages: Swati $= x + 7$, Varun $= 5x^2 + 7$.
3 years hence: Swati $= x + 10$, Varun $= 5x^2 + 10$.
$x + 10 = \frac{2}{5}(5x^2 + 10) \implies x + 10 = 2x^2 + 4 \implies 2x^2 – x – 6 = 0$
$(2x + 3)(x – 2) = 0 \implies x = 2$.
Present ages: Swati $= 2 + 7 = \mathbf{9\text{ years}}$, Varun $= 5(2)^2 + 7 = \mathbf{27\text{ years}}$. [5 Marks]
OR
$\frac{1500}{x} – \frac{1500}{x + 100} = \frac{1}{2} \implies 1500(100) = \frac{1}{2}(x^2 + 100x) \implies x^2 + 100x – 300000 = 0$
$(x + 600)(x – 500) = 0 \implies \mathbf{x = 500\text{ km/h}}$. [5 Marks]
7 years ago: Let Swati’s age $= x$, Varun’s age $= 5x^2$. Present ages: Swati $= x + 7$, Varun $= 5x^2 + 7$.
3 years hence: Swati $= x + 10$, Varun $= 5x^2 + 10$.
$x + 10 = \frac{2}{5}(5x^2 + 10) \implies x + 10 = 2x^2 + 4 \implies 2x^2 – x – 6 = 0$
$(2x + 3)(x – 2) = 0 \implies x = 2$.
Present ages: Swati $= 2 + 7 = \mathbf{9\text{ years}}$, Varun $= 5(2)^2 + 7 = \mathbf{27\text{ years}}$. [5 Marks]
OR
$\frac{1500}{x} – \frac{1500}{x + 100} = \frac{1}{2} \implies 1500(100) = \frac{1}{2}(x^2 + 100x) \implies x^2 + 100x – 300000 = 0$
$(x + 600)(x – 500) = 0 \implies \mathbf{x = 500\text{ km/h}}$. [5 Marks]
Q33. BPT Theorem Proof & Application: [5 Marks]
Theorem proof: [3 Marks]. Application on trapezium with line through diagonal intersection: [2 Marks].
Theorem proof: [3 Marks]. Application on trapezium with line through diagonal intersection: [2 Marks].
Q34. Composite Solid: [5 Marks]
$r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, $h_{\text{cyl}} = 10\text{ cm}$, $h_{\text{cone}} = 6\text{ cm}$, $l = \sqrt{6^2 + 3.5^2} = 6.95\text{ cm}$.
$\text{Total Volume} = \pi r^2 h_{\text{cyl}} + \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h_{\text{cone}} = \pi r^2\left[10 + \frac{7}{3} + 2\right] = \frac{77}{2} \times \frac{43}{3} \approx \mathbf{551.83\text{ cm}^3}$. [2.5 Marks]
$\text{TSA} = 2\pi r(10) + 2\pi r^2 + \pi r l = \pi r[20 + 7 + 6.95] = 11 \times 33.95 = \mathbf{373.45\text{ cm}^2}$. [2.5 Marks]
OR
Volume of tank $= \pi (5)^2 (2) = 50\pi\text{ m}^3$.
Pipe speed $= 3000\text{ m/h}$, internal radius $= 0.1\text{ m}$. Water per hour $= \pi (0.1)^2 (3000) = 30\pi\text{ m}^3/\text{h}$.
$\text{Time} = \frac{50\pi}{30\pi} = \frac{5}{3}\text{ hours} = \mathbf{1\text{ hour } 40\text{ minutes}}$ (or $100\text{ min}$). [5 Marks]
$r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, $h_{\text{cyl}} = 10\text{ cm}$, $h_{\text{cone}} = 6\text{ cm}$, $l = \sqrt{6^2 + 3.5^2} = 6.95\text{ cm}$.
$\text{Total Volume} = \pi r^2 h_{\text{cyl}} + \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h_{\text{cone}} = \pi r^2\left[10 + \frac{7}{3} + 2\right] = \frac{77}{2} \times \frac{43}{3} \approx \mathbf{551.83\text{ cm}^3}$. [2.5 Marks]
$\text{TSA} = 2\pi r(10) + 2\pi r^2 + \pi r l = \pi r[20 + 7 + 6.95] = 11 \times 33.95 = \mathbf{373.45\text{ cm}^2}$. [2.5 Marks]
OR
Volume of tank $= \pi (5)^2 (2) = 50\pi\text{ m}^3$.
Pipe speed $= 3000\text{ m/h}$, internal radius $= 0.1\text{ m}$. Water per hour $= \pi (0.1)^2 (3000) = 30\pi\text{ m}^3/\text{h}$.
$\text{Time} = \frac{50\pi}{30\pi} = \frac{5}{3}\text{ hours} = \mathbf{1\text{ hour } 40\text{ minutes}}$ (or $100\text{ min}$). [5 Marks]
Q35. Missing Frequencies: [5 Marks]
Total frequency $= 170 \implies 110 + x + y = 170 \implies x + y = 60$ …(1)
Median $= 35 \implies$ Median class $30 – 40$ ($l = 30, cf = 30 + x, f = 40, h = 10$).
$35 = 30 + \left[\frac{85 – (30 + x)}{40}\right] \times 10 \implies 5 = \frac{55 – x}{4} \implies 20 = 55 – x \implies \mathbf{x = 35}$.
From (1): $y = 60 – 35 = \mathbf{25}$. [5 Marks]
Total frequency $= 170 \implies 110 + x + y = 170 \implies x + y = 60$ …(1)
Median $= 35 \implies$ Median class $30 – 40$ ($l = 30, cf = 30 + x, f = 40, h = 10$).
$35 = 30 + \left[\frac{85 – (30 + x)}{40}\right] \times 10 \implies 5 = \frac{55 – x}{4} \implies 20 = 55 – x \implies \mathbf{x = 35}$.
From (1): $y = 60 – 35 = \mathbf{25}$. [5 Marks]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) $a = 20, d = 6 \implies a_{10} = 20 + 9(6) = \mathbf{74\text{ centres}}$. [1 Mark]
- (b) $20 + (n – 1)6 = 68 \implies 6(n – 1) = 48 \implies n – 1 = 8 \implies \mathbf{n = 9\text{th year}}$. [1 Mark]
- (c) $S_{12} = \frac{12}{2}[2(20) + 11(6)] = 6[40 + 66] = 6(106) = \mathbf{636\text{ centres}}$. [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) $AB = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{9 + 16} = \mathbf{5\text{ units}}$ ($500\text{ m}$). [1 Mark]
- (b) $BC = \sqrt{(7 – 4)^2 + (2 – 6)^2} = \sqrt{9 + 16} = \mathbf{5\text{ units}}$ ($500\text{ m}$). [1 Mark]
- (c) Point on x-axis $S(x, 0)$: $(x – 1)^2 + 4 = (x – 7)^2 + 4 \implies x^2 – 2x + 5 = x^2 – 14x + 53 \implies 12x = 48 \implies x = 4 \implies \mathbf{S(4, 0)}$. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Labelled diagram showing cliff $60\text{ m}$ with boats at angles $30^\circ$ and $45^\circ$. [1 Mark]
- (b) $\tan 45^\circ = \frac{60}{x} \implies x = \mathbf{60\text{ m}}$. [1 Mark]
- (c) $\tan 30^\circ = \frac{60}{y} \implies y = 60\sqrt{3}\text{ m}$. Distance $= 60\sqrt{3} – 60 = 60(1.732 – 1) = \mathbf{43.92\text{ m}}$. [2 Marks]