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CBSE CLASS X – MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET – 1 | ACADEMIC SESSION 2026–2027
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 Assertion–Reason 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 – MULTIPLE CHOICE QUESTIONS (1 Mark Each)
Q1.
If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x, y$ are prime numbers, then $\text{HCF}(a, b)$ is:
[1]
Q2.
If one zero of the quadratic polynomial $f(x) = 4x^2 – 8kx – 9$ is negative of the other, then the value of $k$ is:
[1]
Q3.
The pair of linear equations $2x + 3y = 7$ and $4x + 6y = 12$ represents two lines which are:
[1]
Q4.
The discriminant of the quadratic equation $2x^2 – 4x + 3 = 0$ is:
[1]
Q5.
In an Arithmetic Progression, if $a = 3.5$, $d = 0$, and $n = 101$, then $a_n$ will be:
[1]
Q6.
The distance of the point $P(-6, 8)$ from the origin is:
[1]
Q7.
If $\triangle ABC \sim \triangle PQR$ with $\frac{AB}{PQ} = \frac{1}{3}$, and the perimeter of $\triangle ABC$ is $15\text{ cm}$, then the perimeter of $\triangle PQR$ is:
[1]
Q8.
If $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to:
[1]
Q9.
The value of $(\sec^2 45^\circ – \tan^2 45^\circ) + (\sin^2 30^\circ + \cos^2 30^\circ)$ is:
[1]
Q10.
If a vertical pole 6 m high casts a shadow $2\sqrt{3}\text{ m}$ long on the ground, then the angle of elevation of the sun is:
[1]
Q11.
From a point $Q$, the length of the tangent to a circle is 24 cm and the distance of $Q$ from the centre is 25 cm. The radius of the circle is:
[1]
Q12.
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is:
[1]
Q13.
A solid metallic sphere of radius 6 cm is melted and recast into the shape of a solid cylinder of radius 6 cm. The height of the cylinder is:
[1]
Q14.
For the following distribution, the modal class is:
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{No. of Students} & 3 & 12 & 20 & 15 & 5 \\ \hline \end{array}$$ [1]
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{No. of Students} & 3 & 12 & 20 & 15 & 5 \\ \hline \end{array}$$ [1]
Q15.
Which of the following numbers cannot be the probability of an event?
[1]
Q16.
If the mean of the observations $x, x+3, x+5, x+7,$ and $x+10$ is 9, then the mean of the last three observations is:
[1]
Q17.
A card is drawn at random from a well-shuffled deck of 52 playing cards. The probability of getting a black face card is:
[1]
Q18.
The point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ is:
[1]
Q19.
Assertion (A): The HCF of two numbers is 5 and their product is 150, then their LCM is 30.
Reason (R): For any two positive integers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$. [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 $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$. [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 line segment joining the points $A(3, -4)$ and $B(1, 2)$ is trisected at the points $P(7/3, -2)$ and $Q(5/3, 0)$.
Reason (R): The midpoint of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. [1]
Reason (R): The midpoint of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. [1]
SECTION B – VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Prove that $\sqrt{5}$ is an irrational number.
[2]
Q22.
In $\triangle ABC$, $DE \parallel BC$ such that $AD = x$, $DB = x – 2$, $AE = x + 2$, and $EC = x – 1$. Find the value of $x$.
[2]
OR
$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
Q23.
Find the coordinates of the point which divides the line segment joining $(4, -3)$ and $(8, 5)$ in the ratio $3 : 1$ internally.
[2]
Q24.
Evaluate:
$$\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$$
[2]
OR
If $\sin(A – B) = \frac{1}{2}$ and $\cos(A + B) = \frac{1}{2}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the acute angles $A$ and $B$.
Q25.
A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
[2]
SECTION C – SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
Find the zeroes of the quadratic polynomial $p(x) = 6x^2 – 3 – 7x$ and verify the relationship between the zeroes and the coefficients.
[3]
Q27.
Solve the following pair of linear equations algebraically by elimination method:
$$2x + 3y = 11$$
$$2x – 4y = -24$$
Hence, find the value of ‘$m$’ for which $y = mx + 3$.
[3]
OR
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Q28.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
[3]
Q29.
Prove the trigonometric identity:
$$\frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta} = \tan \theta$$
[3]
Q30.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
[3]
OR
In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle of radius 3 cm. Find the radius of the larger circle.
Q31.
Two dice are thrown simultaneously. What is the probability that:
(a) The sum of the two numbers appearing on the top is 8?
(b) 5 will not come up on either die?
(c) The two numbers appearing on top are equal (a doublet)? [3]
(a) The sum of the two numbers appearing on the top is 8?
(b) 5 will not come up on either die?
(c) The two numbers appearing on top are equal (a doublet)? [3]
SECTION D – LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
A motor boat whose speed is $18\text{ km/h}$ in still water takes 1 hour more to go $24\text{ km}$ upstream than to return downstream to the same spot. Find the speed of the stream.
[5]
OR
Two water taps together can fill a tank in $9\frac{3}{8}\text{ hours}$. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Q33.
State and prove Basic Proportionality Theorem (Thales Theorem).
[5]
Q34.
A solid toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (Take $\pi = 3.14$).
Further, if a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. [5]
Further, if a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. [5]
OR
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the conical top is $2.8\text{ m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of $₹500\text{ per m}^2$. (Note that the base of the tent is not covered with canvas).
Q35.
The median of the following frequency distribution is 525. Find the values of $x$ and $y$, if the total frequency is 100:
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-100 & 2 & 500-600 & 20 \\ 100-200 & 5 & 600-700 & y \\ 200-300 & x & 700-800 & 9 \\ 300-400 & 12 & 800-900 & 7 \\ 400-500 & 17 & 900-1000 & 4 \\ \hline \end{array}$$ [5]
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-100 & 2 & 500-600 & 20 \\ 100-200 & 5 & 600-700 & y \\ 200-300 & x & 700-800 & 9 \\ 300-400 & 12 & 800-900 & 7 \\ 400-500 & 17 & 900-1000 & 4 \\ \hline \end{array}$$ [5]
SECTION E – CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) Find the production of radio sets in the 1st year ($a$). [1]
(b) Find the fixed annual increase in production ($d$). [1]
(c) Find the total production of radio sets in the first 10 years. [2]
A production manager in a radio manufacturing factory plans to increase the production of radio sets uniformly every year. The factory produced 600 sets in the 3rd year and 700 sets in the 7th year.
Based on the above information, answer the following questions:
(a) Find the production of radio sets in the 1st year ($a$). [1]
(b) Find the fixed annual increase in production ($d$). [1]
(c) Find the total production of radio sets in the first 10 years. [2]
OR
In which year will the total annual production reach 1000 radio sets? [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Write the coordinates of the red flag posted by Preet. [1]
(c) Find the shortest distance between both the flags. [2]
To conduct Sports Day activities in a rectangular school ground $ABCD$, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots are placed at a distance of 1 m from each other along $AD$. Niharika runs $1/4\text{th}$ the distance $AD$ on the 2nd line and posts a green flag. Preet runs $1/5\text{th}$ the distance $AD$ on the 8th line and posts a red flag.
(a) Write the coordinates of the green flag posted by Niharika. [1]
(b) Write the coordinates of the red flag posted by Preet. [1]
(c) Find the shortest distance between both the flags. [2]
OR
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, at what distance and on which line should she post her flag? [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Find the height of the temple up to the roof. (Use $\sqrt{3} \approx 1.732$). [1]
(c) Calculate the height of the flagstaff standing on the roof of the temple. [2]
A straight vertical flagstaff stands on the top of a temple roof. From a point on the horizontal ground at a distance of 100 m from the base of the temple, the angles of elevation of the bottom and top of the flagstaff are observed to be $30^\circ$ and $45^\circ$ respectively.
(a) Draw a neat labelled mathematical diagram representing the given situation. [1]
(b) Find the height of the temple up to the roof. (Use $\sqrt{3} \approx 1.732$). [1]
(c) Calculate the height of the flagstaff standing on the roof of the temple. [2]
OR
Find the distance of the top of the flagstaff from the point of observation on the ground. [2]
CBSE CLASS X MATHEMATICS (STANDARD) – SOLUTIONS & MARKING SCHEME (SET – 1)
SECTION A SOLUTIONS
Q1. (b) $x y^2$ [1 Mark]
Explanation: $\text{HCF}$ is the product of the smallest powers of common prime factors: $x^{\min(3,1)} \cdot y^{\min(2,3)} = x^1 y^2$.
Explanation: $\text{HCF}$ is the product of the smallest powers of common prime factors: $x^{\min(3,1)} \cdot y^{\min(2,3)} = x^1 y^2$.
Q2. (a) 0 [1 Mark]
Explanation: If roots are $\alpha$ and $-\alpha$, $\text{Sum} = \alpha + (-\alpha) = 0 = -\frac{-8k}{4} = 2k \implies k = 0$.
Explanation: If roots are $\alpha$ and $-\alpha$, $\text{Sum} = \alpha + (-\alpha) = 0 = -\frac{-8k}{4} = 2k \implies k = 0$.
Q3. (c) Parallel [1 Mark]
Explanation: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{7}{12}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, lines are parallel.
Explanation: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{7}{12}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, lines are parallel.
Q4. (a) -8 [1 Mark]
Explanation: $D = b^2 – 4ac = (-4)^2 – 4(2)(3) = 16 – 24 = -8$.
Explanation: $D = b^2 – 4ac = (-4)^2 – 4(2)(3) = 16 – 24 = -8$.
Q5. (b) 3.5 [1 Mark]
Explanation: $a_n = a + (n – 1)d = 3.5 + (100)(0) = 3.5$.
Explanation: $a_n = a + (n – 1)d = 3.5 + (100)(0) = 3.5$.
Q6. (c) 10 units [1 Mark]
Explanation: $\text{Distance} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ units}$.
Explanation: $\text{Distance} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ units}$.
Q7. (c) 45 cm [1 Mark]
Explanation: $\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)} = \frac{AB}{PQ} \implies \frac{15}{\text{Perimeter}(\triangle PQR)} = \frac{1}{3} \implies \text{Perimeter}(\triangle PQR) = 45\text{ cm}$.
Explanation: $\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)} = \frac{AB}{PQ} \implies \frac{15}{\text{Perimeter}(\triangle PQR)} = \frac{1}{3} \implies \text{Perimeter}(\triangle PQR) = 45\text{ cm}$.
Q8. (b) $\frac{\sqrt{b^2 – a^2}}{b}$ [1 Mark]
Explanation: $\cos \theta = \sqrt{1 – \sin^2 \theta} = \sqrt{1 – \frac{a^2}{b^2}} = \frac{\sqrt{b^2 – a^2}}{b}$.
Explanation: $\cos \theta = \sqrt{1 – \sin^2 \theta} = \sqrt{1 – \frac{a^2}{b^2}} = \frac{\sqrt{b^2 – a^2}}{b}$.
Q9. (b) 2 [1 Mark]
Explanation: $(\sec^2 45^\circ – \tan^2 45^\circ) + (\sin^2 30^\circ + \cos^2 30^\circ) = 1 + 1 = 2$.
Explanation: $(\sec^2 45^\circ – \tan^2 45^\circ) + (\sin^2 30^\circ + \cos^2 30^\circ) = 1 + 1 = 2$.
Q10. (a) 60° [1 Mark]
Explanation: $\tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \implies \theta = 60^\circ$.
Explanation: $\tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \implies \theta = 60^\circ$.
Q11. (a) 7 cm [1 Mark]
Explanation: $r = \sqrt{OQ^2 – PQ^2} = \sqrt{25^2 – 24^2} = \sqrt{625 – 576} = \sqrt{49} = 7\text{ cm}$.
Explanation: $r = \sqrt{OQ^2 – PQ^2} = \sqrt{25^2 – 24^2} = \sqrt{625 – 576} = \sqrt{49} = 7\text{ cm}$.
Q12. (a) 2 units [1 Mark]
Explanation: $2\pi r = \pi r^2 \implies r = 2\text{ units}$.
Explanation: $2\pi r = \pi r^2 \implies r = 2\text{ units}$.
Q13. (b) 8 cm [1 Mark]
Explanation: $\pi r^2 h = \frac{4}{3}\pi r^3 \implies h = \frac{4}{3} r = \frac{4}{3}(6) = 8\text{ cm}$.
Explanation: $\pi r^2 h = \frac{4}{3}\pi r^3 \implies h = \frac{4}{3} r = \frac{4}{3}(6) = 8\text{ cm}$.
Q14. (b) 20–30 [1 Mark]
Explanation: Maximum frequency is 20, corresponding to class interval 20–30.
Explanation: Maximum frequency is 20, corresponding to class interval 20–30.
Q15. (b) -1.5 [1 Mark]
Explanation: Probability of an event cannot be negative ($0 \le P(E) \le 1$).
Explanation: Probability of an event cannot be negative ($0 \le P(E) \le 1$).
Q16. (b) 11.33 [1 Mark]
Explanation: $\text{Mean} = \frac{5x + 25}{5} = x + 5 = 9 \implies x = 4$. Last three observations: $9, 11, 14 \implies \text{Mean} = \frac{34}{3} \approx 11.33$.
Explanation: $\text{Mean} = \frac{5x + 25}{5} = x + 5 = 9 \implies x = 4$. Last three observations: $9, 11, 14 \implies \text{Mean} = \frac{34}{3} \approx 11.33$.
Q17. (b) 3/26 [1 Mark]
Explanation: Total black face cards $= 6 \implies P = \frac{6}{52} = \frac{3}{26}$.
Explanation: Total black face cards $= 6 \implies P = \frac{6}{52} = \frac{3}{26}$.
Q18. (b) $(-7, 0)$ [1 Mark]
Explanation: $(x – 2)^2 + (0 + 5)^2 = (x + 2)^2 + (0 – 9)^2 \implies -4x + 29 = 4x + 85 \implies 8x = -56 \implies x = -7$.
Explanation: $(x – 2)^2 + (0 + 5)^2 = (x + 2)^2 + (0 – 9)^2 \implies -4x + 29 = 4x + 85 \implies 8x = -56 \implies x = -7$.
Q19. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: $\text{LCM} = \frac{\text{Product}}{\text{HCF}} = \frac{150}{5} = 30$.
Explanation: $\text{LCM} = \frac{\text{Product}}{\text{HCF}} = \frac{150}{5} = 30$.
Q20. (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A). [1 Mark]
Explanation: Trisection points divide the line segment in ratios $1:2$ and $2:1$. Both statements are mathematically true.
Explanation: Trisection points divide the line segment in ratios $1:2$ and $2:1$. Both statements are mathematically true.
SECTION B SOLUTIONS
Q21. Proof of Irrationality of $\sqrt{5}$: [2 Marks]
- Assume $\sqrt{5} = \frac{a}{b}$, where $a, b$ are coprime integers and $b \neq 0$. [0.5 Mark]
- $5b^2 = a^2 \implies 5 \text{ divides } a^2 \implies 5 \text{ divides } a$. Let $a = 5c$. [0.5 Mark]
- $5b^2 = 25c^2 \implies b^2 = 5c^2 \implies 5 \text{ divides } b$. [0.5 Mark]
- Thus $a$ and $b$ share a common factor of 5, contradicting their coprime nature. Hence, $\sqrt{5}$ is irrational. [0.5 Mark]
Q22. Value of $x$: [2 Marks]
By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{x}{x-2} = \frac{x+2}{x-1}$ [0.5 Mark]
$x(x – 1) = (x – 2)(x + 2) \implies x^2 – x = x^2 – 4 \implies \mathbf{x = 4}$. [1.5 Marks]
OR
In $\triangle ABE$ and $\triangle CFB$: $\angle A = \angle C$ (opposite angles of $\|gm$) and $\angle AEB = \angle CBF$ (alt. interior angles). By AA similarity, $\mathbf{\triangle ABE \sim \triangle CFB}$. [2 Marks]
By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{x}{x-2} = \frac{x+2}{x-1}$ [0.5 Mark]
$x(x – 1) = (x – 2)(x + 2) \implies x^2 – x = x^2 – 4 \implies \mathbf{x = 4}$. [1.5 Marks]
OR
In $\triangle ABE$ and $\triangle CFB$: $\angle A = \angle C$ (opposite angles of $\|gm$) and $\angle AEB = \angle CBF$ (alt. interior angles). By AA similarity, $\mathbf{\triangle ABE \sim \triangle CFB}$. [2 Marks]
Q23. Section Formula: [2 Marks]
$x = \frac{3(8) + 1(4)}{3 + 1} = \frac{28}{4} = \mathbf{7}$; $y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{12}{4} = \mathbf{3}$. Coordinates are $\mathbf{(7, 3)}$. [2 Marks]
$x = \frac{3(8) + 1(4)}{3 + 1} = \frac{28}{4} = \mathbf{7}$; $y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{12}{4} = \mathbf{3}$. Coordinates are $\mathbf{(7, 3)}$. [2 Marks]
Q24. Evaluation: [2 Marks]
$\text{Numerator} = 5(1/2)^2 + 4(2/\sqrt{3})^2 – 1^2 = \frac{5}{4} + \frac{16}{3} – 1 = \frac{67}{12}$.
$\text{Denominator} = \sin^2 30^\circ + \cos^2 30^\circ = 1$. $\text{Result} = \mathbf{\frac{67}{12}}$. [2 Marks]
OR
$A – B = 30^\circ$ and $A + B = 60^\circ \implies 2A = 90^\circ \implies \mathbf{A = 45^\circ}, \mathbf{B = 15^\circ}$. [2 Marks]
$\text{Numerator} = 5(1/2)^2 + 4(2/\sqrt{3})^2 – 1^2 = \frac{5}{4} + \frac{16}{3} – 1 = \frac{67}{12}$.
$\text{Denominator} = \sin^2 30^\circ + \cos^2 30^\circ = 1$. $\text{Result} = \mathbf{\frac{67}{12}}$. [2 Marks]
OR
$A – B = 30^\circ$ and $A + B = 60^\circ \implies 2A = 90^\circ \implies \mathbf{A = 45^\circ}, \mathbf{B = 15^\circ}$. [2 Marks]
Q25. Circumscribed Quadrilateral: [2 Marks]
Tangents from external points are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
Adding: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \implies \mathbf{AB + CD = AD + BC}$. [2 Marks]
Tangents from external points are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
Adding: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \implies \mathbf{AB + CD = AD + BC}$. [2 Marks]
SECTION C SOLUTIONS
Q26. Zeroes & Verification: [3 Marks]
$6x^2 – 7x – 3 = (2x – 3)(3x + 1) = 0 \implies \alpha = \mathbf{\frac{3}{2}}, \beta = \mathbf{-\frac{1}{3}}$. [1.5 Marks]
$\alpha + \beta = \frac{7}{6} = -\frac{b}{a} = \frac{7}{6}$; $\alpha\beta = -\frac{1}{2} = \frac{c}{a} = -\frac{1}{2}$. (Verified). [1.5 Marks]
$6x^2 – 7x – 3 = (2x – 3)(3x + 1) = 0 \implies \alpha = \mathbf{\frac{3}{2}}, \beta = \mathbf{-\frac{1}{3}}$. [1.5 Marks]
$\alpha + \beta = \frac{7}{6} = -\frac{b}{a} = \frac{7}{6}$; $\alpha\beta = -\frac{1}{2} = \frac{c}{a} = -\frac{1}{2}$. (Verified). [1.5 Marks]
Q27. Linear Equations: [3 Marks]
$2x + 3y = 11$ and $2x – 4y = -24 \implies 7y = 35 \implies \mathbf{y = 5}, \mathbf{x = -2}$. [2 Marks]
$y = mx + 3 \implies 5 = m(-2) + 3 \implies \mathbf{m = -1}$. [1 Mark]
OR
$(10x + y) + (10y + x) = 66 \implies x + y = 6$. Digit difference: $|x – y| = 2$.
Solving gives numbers $\mathbf{42}$ and $\mathbf{24}$ (Two such numbers). [3 Marks]
$2x + 3y = 11$ and $2x – 4y = -24 \implies 7y = 35 \implies \mathbf{y = 5}, \mathbf{x = -2}$. [2 Marks]
$y = mx + 3 \implies 5 = m(-2) + 3 \implies \mathbf{m = -1}$. [1 Mark]
OR
$(10x + y) + (10y + x) = 66 \implies x + y = 6$. Digit difference: $|x – y| = 2$.
Solving gives numbers $\mathbf{42}$ and $\mathbf{24}$ (Two such numbers). [3 Marks]
Q28. AP Terms: [3 Marks]
$a_4 + a_8 = 24 \implies a + 5d = 12$; $a_6 + a_{10} = 44 \implies a + 7d = 22$.
Subtracting: $2d = 10 \implies \mathbf{d = 5} \implies \mathbf{a = -13}$. First 3 terms: $\mathbf{-13, -8, -3}$. [3 Marks]
$a_4 + a_8 = 24 \implies a + 5d = 12$; $a_6 + a_{10} = 44 \implies a + 7d = 22$.
Subtracting: $2d = 10 \implies \mathbf{d = 5} \implies \mathbf{a = -13}$. First 3 terms: $\mathbf{-13, -8, -3}$. [3 Marks]
Q29. Trigonometric Identity: [3 Marks]
$\text{LHS} = \frac{\sin\theta(1 – 2\sin^2\theta)}{\cos\theta(2\cos^2\theta – 1)} = \tan\theta \cdot \frac{1 – 2(1 – \cos^2\theta)}{2\cos^2\theta – 1} = \tan\theta \cdot \frac{2\cos^2\theta – 1}{2\cos^2\theta – 1} = \mathbf{\tan\theta} = \text{RHS}$. [3 Marks]
$\text{LHS} = \frac{\sin\theta(1 – 2\sin^2\theta)}{\cos\theta(2\cos^2\theta – 1)} = \tan\theta \cdot \frac{1 – 2(1 – \cos^2\theta)}{2\cos^2\theta – 1} = \tan\theta \cdot \frac{2\cos^2\theta – 1}{2\cos^2\theta – 1} = \mathbf{\tan\theta} = \text{RHS}$. [3 Marks]
Q30. Tangents at Diameter Ends: [3 Marks]
Radius $\perp$ tangent at contact point: $\angle OAP = 90^\circ$ and $\angle OBR = 90^\circ$.
$\angle PAB = \angle S BA = 90^\circ$ (Equal alternate interior angles) $\implies \mathbf{PQ \parallel RS}$. [3 Marks]
OR
$OP \perp AB \implies AP = 4\text{ cm}$. In right $\triangle OPA$: $OA = \sqrt{3^2 + 4^2} = \mathbf{5\text{ cm}}$. [3 Marks]
Radius $\perp$ tangent at contact point: $\angle OAP = 90^\circ$ and $\angle OBR = 90^\circ$.
$\angle PAB = \angle S BA = 90^\circ$ (Equal alternate interior angles) $\implies \mathbf{PQ \parallel RS}$. [3 Marks]
OR
$OP \perp AB \implies AP = 4\text{ cm}$. In right $\triangle OPA$: $OA = \sqrt{3^2 + 4^2} = \mathbf{5\text{ cm}}$. [3 Marks]
Q31. Two Dice Probability: [3 Marks]
Total outcomes $= 36$. (a) Sum = 8: $(2,6),(3,5),(4,4),(5,3),(6,2) \implies P = \mathbf{\frac{5}{36}}$. [1 Mark]
(b) 5 not on either: $P = 1 – \frac{11}{36} = \mathbf{\frac{25}{36}}$. [1 Mark]
(c) Doublet: $(1,1),\dots,(6,6) \implies P = \frac{6}{36} = \mathbf{\frac{1}{6}}$. [1 Mark]
Total outcomes $= 36$. (a) Sum = 8: $(2,6),(3,5),(4,4),(5,3),(6,2) \implies P = \mathbf{\frac{5}{36}}$. [1 Mark]
(b) 5 not on either: $P = 1 – \frac{11}{36} = \mathbf{\frac{25}{36}}$. [1 Mark]
(c) Doublet: $(1,1),\dots,(6,6) \implies P = \frac{6}{36} = \mathbf{\frac{1}{6}}$. [1 Mark]
SECTION D SOLUTIONS
Q32. Stream Speed: [5 Marks]
$\frac{24}{18 – x} – \frac{24}{18 + x} = 1 \implies 24(2x) = 324 – x^2 \implies x^2 + 48x – 324 = 0$.
$(x + 54)(x – 6) = 0 \implies \mathbf{x = 6\text{ km/h}}$. [5 Marks]
OR
$\frac{1}{x} + \frac{1}{x – 10} = \frac{8}{75} \implies 4x^2 – 115x + 375 = 0 \implies (4x – 15)(x – 25) = 0$.
Smaller tap $= \mathbf{25\text{ hours}}$, Larger tap $= \mathbf{15\text{ hours}}$. [5 Marks]
$\frac{24}{18 – x} – \frac{24}{18 + x} = 1 \implies 24(2x) = 324 – x^2 \implies x^2 + 48x – 324 = 0$.
$(x + 54)(x – 6) = 0 \implies \mathbf{x = 6\text{ km/h}}$. [5 Marks]
OR
$\frac{1}{x} + \frac{1}{x – 10} = \frac{8}{75} \implies 4x^2 – 115x + 375 = 0 \implies (4x – 15)(x – 25) = 0$.
Smaller tap $= \mathbf{25\text{ hours}}$, Larger tap $= \mathbf{15\text{ hours}}$. [5 Marks]
Q33. BPT Theorem Proof: [5 Marks]
Statement [1 Mark], Given/Figure/Construction [1.5 Marks], Area ratio derivation and proof showing $\frac{AD}{DB} = \frac{AE}{EC}$ [2.5 Marks].
Statement [1 Mark], Given/Figure/Construction [1.5 Marks], Area ratio derivation and proof showing $\frac{AD}{DB} = \frac{AE}{EC}$ [2.5 Marks].
Q34. Volume of Toy: [5 Marks]
$\text{Toy Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}(3.14)(4)(6) = \mathbf{25.12\text{ cm}^3}$. [2.5 Marks]
$\text{Cylinder Volume} = \pi r^2 H = 3.14(4)(4) = 50.24\text{ cm}^3$.
$\text{Difference} = 50.24 – 25.12 = \mathbf{25.12\text{ cm}^3}$. [2.5 Marks]
OR
$\text{Canvas Area} = 2\pi rh + \pi rl = \frac{22}{7} \times 2 \times [4.2 + 2.8] = \mathbf{44\text{ m}^2}$. [3 Marks]
$\text{Cost} = 44 \times ₹500 = \mathbf{₹22,000}$. [2 Marks]
$\text{Toy Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}(3.14)(4)(6) = \mathbf{25.12\text{ cm}^3}$. [2.5 Marks]
$\text{Cylinder Volume} = \pi r^2 H = 3.14(4)(4) = 50.24\text{ cm}^3$.
$\text{Difference} = 50.24 – 25.12 = \mathbf{25.12\text{ cm}^3}$. [2.5 Marks]
OR
$\text{Canvas Area} = 2\pi rh + \pi rl = \frac{22}{7} \times 2 \times [4.2 + 2.8] = \mathbf{44\text{ m}^2}$. [3 Marks]
$\text{Cost} = 44 \times ₹500 = \mathbf{₹22,000}$. [2 Marks]
Q35. Missing Frequencies: [5 Marks]
$N = 100 \implies x + y = 24$. Median $= 525 \implies$ Median class $500-600$.
$525 = 500 + \left[\frac{50 – (36 + x)}{20}\right] \times 100 \implies 25 = 5(14 – x) \implies \mathbf{x = 9}$.
$y = 24 – 9 \implies \mathbf{y = 15}$. [5 Marks]
$N = 100 \implies x + y = 24$. Median $= 525 \implies$ Median class $500-600$.
$525 = 500 + \left[\frac{50 – (36 + x)}{20}\right] \times 100 \implies 25 = 5(14 – x) \implies \mathbf{x = 9}$.
$y = 24 – 9 \implies \mathbf{y = 15}$. [5 Marks]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) $a_3 = 600, a_7 = 700 \implies 4d = 100 \implies \mathbf{d = 25}$. [1 Mark]
- (b) $a = 600 – 2(25) = \mathbf{550\text{ sets}}$. [1 Mark]
- (c) $S_{10} = \frac{10}{2}[2(550) + 9(25)] = 5[1325] = \mathbf{6625\text{ sets}}$. [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) Green flag: $\mathbf{(2, 25)}$. [1 Mark]
- (b) Red flag: $\mathbf{(8, 20)}$. [1 Mark]
- (c) Distance $= \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \mathbf{\sqrt{61}\text{ m}} \approx 7.81\text{ m}$. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Schematic labelled diagram with ground distance 100 m, angles $30^\circ$ and $45^\circ$. [1 Mark]
- (b) Temple height: $\tan 30^\circ = \frac{h}{100} \implies h = \frac{100}{\sqrt{3}} \approx \mathbf{57.73\text{ m}}$. [1 Mark]
- (c) Flagstaff height $= 100 – 57.73 = \mathbf{42.27\text{ m}}$. [2 Marks]