ADEPT YOURSELF
www.adeptyourself.com β’ Academic Excellence Series
CBSE CLASS X β MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET β 3 | 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 exponent of 2 in the prime factorisation of 144 is:
[1]
Q2.
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 – 5x + 6$, then the value of $\frac{1}{\alpha} + \frac{1}{\beta} – 2\alpha\beta$ is:
[1]
Q3.
The pair of linear equations $ax + by = c$ and $lx + my = n$ has a unique solution if:
[1]
Q4.
The roots of the quadratic equation $x^2 – 0.04 = 0$ are:
[1]
Q5.
In an AP, if $a_{18} – a_{14} = 32$, then the common difference $d$ of the AP is:
[1]
Q6.
The distance between the points $(a\cos\theta + b\sin\theta, 0)$ and $(0, a\sin\theta – b\cos\theta)$ is:
[1]
Q7.
In $\triangle ABC$, $DE \parallel BC$ such that $\frac{AD}{DB} = \frac{2}{3}$. If $AE = 6\text{ cm}$, then the length of $EC$ is:
[1]
Q8.
If $\sqrt{3}\tan 2\theta – 3 = 0$, then the acute angle $\theta$ is equal to:
[1]
Q9.
The value of $\frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ}$ is equal to:
[1]
Q10.
The angle of elevation of the top of a $30\text{ m}$ high tower from a point on the horizontal ground at a distance of $30\text{ m}$ from the base of the tower is:
[1]
Q11.
Tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^\circ$. Then $\angle POA$ is equal to:
[1]
Q12.
If the radius of a circle is increased by 50%, then the percentage increase in its area is:
[1]
Q13.
The total surface area of a solid hemisphere of radius $r$ is:
[1]
Q14.
If $\sum f_i x_i = 1320$ and $\sum f_i = 40$, then the arithmetic mean of the distribution is:
[1]
Q15.
The probability that a non-leap year selected at random will contain 53 Sundays is:
[1]
Q16.
The coordinates of the mid-point of the line segment joining $A(-2, 8)$ and $B(-6, -4)$ are:
[1]
Q17.
If $P(E) = 0.05$, then the probability of ‘not $E$’ is:
[1]
Q18.
The LCM of the smallest two-digit composite number and the smallest composite number is:
[1]
Q19.
Assertion (A): $\sqrt{2}$ is an irrational number.
Reason (R): The square root of any prime number is always an irrational number. [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): The square root of any prime number is always an irrational number. [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 point $P(0, 4)$ lies on the y-axis.
Reason (R): The x-coordinate (abscissa) of any point lying on the y-axis is zero. [1]
Reason (R): The x-coordinate (abscissa) of any point lying on the y-axis is zero. [1]
SECTION B β VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Find the HCF and LCM of 72 and 120 using the prime factorisation method.
[2]
Q22.
In $\triangle ABC$, $XY \parallel BC$ and it divides $\triangle ABC$ into two regions of equal area. Find the ratio $\frac{AX}{AB}$.
[2]
OR
The diagonals of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
Q23.
Find a relation between $x$ and $y$ such that the point $P(x, y)$ is equidistant from the points $A(3, 6)$ and $B(-3, 4)$.
[2]
Q24.
If $\sin(A + B) = 1$ and $\cos(A – B) = \frac{\sqrt{3}}{2}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the values of $A$ and $B$.
[2]
OR
Evaluate:
$$\frac{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$$
Q25.
In two concentric circles of radii $13\text{ cm}$ and $5\text{ cm}$, find the length of the chord of the larger circle which touches the smaller circle.
[2]
SECTION C β SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
If the zeroes of the quadratic polynomial $f(x) = x^2 – px + q$ are in the ratio $2 : 3$, prove that $6p^2 = 25q$.
[3]
Q27.
Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. Find their present ages.
[3]
OR
Solve the following system of linear equations algebraically:
$$152x – 378y = -74$$
$$-378x + 152y = -604$$
Q28.
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
[3]
Q29.
Prove the trigonometric identity:
$$\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$$
[3]
Q30.
Prove that the parallelogram circumscribing a circle is a rhombus.
[3]
OR
In a circle with centre $O$, $PA$ and $PB$ are tangents drawn from an external point $P$. Prove that $\angle APB = 2\angle OAB$.
Q31.
A bag contains 24 marbles, some of which are green and others are blue. If a marble is drawn at random from the bag, the probability that it is green is $\frac{2}{3}$. Find the number of blue marbles in the bag.
[3]
SECTION D β LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
An aeroplane left 30 minutes later than its scheduled time and in order to reach its destination $1500\text{ km}$ away in time, it had to increase its speed by $250\text{ km/h}$ from its usual speed. Find its usual speed.
[5]
OR
Solve for $x$:
$$\frac{1}{x + 4} – \frac{1}{x – 7} = \frac{11}{30}, \quad x \neq -4, 7$$
Q33.
State Basic Proportionality Theorem.
Using this theorem, prove the following: In $\triangle ABC$, $DE \parallel BC$ and $CD \parallel EF$, where $D$ lies on $AB$, $E$ lies on $AC$, and $F$ lies on $AD$. Prove that: $$AD^2 = AB \times AF$$ [5]
Using this theorem, prove the following: In $\triangle ABC$, $DE \parallel BC$ and $CD \parallel EF$, where $D$ lies on $AB$, $E$ lies on $AC$, and $F$ lies on $AD$. Prove that: $$AD^2 = AB \times AF$$ [5]
Q34.
A solid right circular cylinder of diameter $12\text{ cm}$ and height $15\text{ cm}$ is full of ice cream. The ice cream is to be filled into cones of height $12\text{ cm}$ and diameter $6\text{ cm}$, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
[5]
OR
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere of the same base radius. If the radius of the hemisphere is $3.5\text{ cm}$ and the total height of the toy is $15.5\text{ cm}$, find the total surface area and volume of the toy. (Take $\pi = 22/7$).
Q35.
The arithmetic mean of the following frequency distribution is 50. Find the missing frequencies $f_1$ and $f_2$, given that the sum of all frequencies is 120:
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 \\ \hline \text{Frequency} & 17 & f_1 & 32 & f_2 & 19 \\ \hline \end{array}$$ [5]
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 \\ \hline \text{Frequency} & 17 & f_1 & 32 & f_2 & 19 \\ \hline \end{array}$$ [5]
SECTION E β CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) Find the number of seats in the 15th row. [1]
(b) Determine the difference in the number of seats between the 20th row and the 10th row. [1]
(c) If there are 30 rows in total in the auditorium, calculate its total seating capacity. [2]
An auditorium has a tiered seating layout designed for optimal acoustics. The first row contains 20 seats, the second row contains 24 seats, the third row contains 28 seats, and this regular pattern continues for all successive rows.
Based on the above information, answer the following questions:
(a) Find the number of seats in the 15th row. [1]
(b) Determine the difference in the number of seats between the 20th row and the 10th row. [1]
(c) If there are 30 rows in total in the auditorium, calculate its total seating capacity. [2]
OR
If the management decides to construct 25 rows in total, what will be the total seating capacity? [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Calculate the distance between landing pad $A$ and landing pad $B$. [1]
(c) Find the coordinates of a refueling waypoint $P(x, 0)$ on the horizontal transit line (x-axis) that is equidistant from pad $A(3, 4)$ and pad $B(7, 7)$. [2]
To automate courier dispatches, an aerial delivery drone operates relative to a main logistics hub located at the origin $O(0, 0)$. Two designated landing pads are situated at coordinates $A(3, 4)$ and $B(7, 7)$, where 1 unit on the grid corresponds to 1 kilometre.
(a) Find the direct flight distance from the Hub $O(0, 0)$ to landing pad $A$. [1]
(b) Calculate the distance between landing pad $A$ and landing pad $B$. [1]
(c) Find the coordinates of a refueling waypoint $P(x, 0)$ on the horizontal transit line (x-axis) that is equidistant from pad $A(3, 4)$ and pad $B(7, 7)$. [2]
OR
Check whether the Hub $O(0, 0)$, pad $A(3, 4)$, and pad $B(7, 7)$ lie in a straight collinear line. [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Find the distances of the observation point $P$ from the base of each tower. [1]
(c) Calculate the height of the transmission towers. (Take $\sqrt{3} \approx 1.732$). [2]
Two transmission towers of equal heights stand directly opposite each other on either side of an $80\text{ m}$ wide straight highway. From an observation point $P$ located on the road between the two towers, the angles of elevation of the tops of the towers are observed to be $60^\circ$ and $30^\circ$.
(a) Draw a neat labelled mathematical diagram representing this physical arrangement. [1]
(b) Find the distances of the observation point $P$ from the base of each tower. [1]
(c) Calculate the height of the transmission towers. (Take $\sqrt{3} \approx 1.732$). [2]
OR
Find the line-of-sight distance from the observation point $P$ to the top of the tower which has an elevation angle of $60^\circ$. [2]
CBSE CLASS X MATHEMATICS (STANDARD) β SOLUTIONS & MARKING SCHEME (SET β 3)
SECTION A SOLUTIONS
Q1. (c) 4 [1 Mark]
Explanation: $144 = 16 \times 9 = 2^4 \times 3^2$. The exponent of 2 is 4.
Explanation: $144 = 16 \times 9 = 2^4 \times 3^2$. The exponent of 2 is 4.
Q2. (b) $-\frac{67}{6}$ [1 Mark]
Explanation: $\alpha + \beta = 5$, $\alpha\beta = 6$.
$\frac{1}{\alpha} + \frac{1}{\beta} – 2\alpha\beta = \frac{\alpha+\beta}{\alpha\beta} – 2\alpha\beta = \frac{5}{6} – 2(6) = \frac{5}{6} – 12 = \frac{5 – 72}{6} = -\frac{67}{6}$.
Explanation: $\alpha + \beta = 5$, $\alpha\beta = 6$.
$\frac{1}{\alpha} + \frac{1}{\beta} – 2\alpha\beta = \frac{\alpha+\beta}{\alpha\beta} – 2\alpha\beta = \frac{5}{6} – 2(6) = \frac{5}{6} – 12 = \frac{5 – 72}{6} = -\frac{67}{6}$.
Q3. (a) $am \neq bl$ [1 Mark]
Explanation: For a unique solution, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \frac{a}{l} \neq \frac{b}{m} \implies am \neq bl$.
Explanation: For a unique solution, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \frac{a}{l} \neq \frac{b}{m} \implies am \neq bl$.
Q4. (a) $\pm 0.2$ [1 Mark]
Explanation: $x^2 = 0.04 \implies x = \pm \sqrt{0.04} = \pm 0.2$.
Explanation: $x^2 = 0.04 \implies x = \pm \sqrt{0.04} = \pm 0.2$.
Q5. (a) 8 [1 Mark]
Explanation: $a_{18} – a_{14} = (a + 17d) – (a + 13d) = 4d = 32 \implies d = 8$.
Explanation: $a_{18} – a_{14} = (a + 17d) – (a + 13d) = 4d = 32 \implies d = 8$.
Q6. (b) $\sqrt{a^2 + b^2}$ [1 Mark]
Explanation: $\text{Distance} = \sqrt{(a\cos\theta + b\sin\theta – 0)^2 + (0 – (a\sin\theta – b\cos\theta))^2}$
$= \sqrt{a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta + a^2\sin^2\theta + b^2\cos^2\theta – 2ab\sin\theta\cos\theta} = \sqrt{a^2 + b^2}$.
Explanation: $\text{Distance} = \sqrt{(a\cos\theta + b\sin\theta – 0)^2 + (0 – (a\sin\theta – b\cos\theta))^2}$
$= \sqrt{a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta + a^2\sin^2\theta + b^2\cos^2\theta – 2ab\sin\theta\cos\theta} = \sqrt{a^2 + b^2}$.
Q7. (c) 9 cm [1 Mark]
Explanation: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2}{3} = \frac{6}{EC} \implies EC = \frac{6 \times 3}{2} = 9\text{ cm}$.
Explanation: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2}{3} = \frac{6}{EC} \implies EC = \frac{6 \times 3}{2} = 9\text{ cm}$.
Q8. (b) 30Β° [1 Mark]
Explanation: $\sqrt{3}\tan 2\theta = 3 \implies \tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3} \implies 2\theta = 60^\circ \implies \theta = 30^\circ$.
Explanation: $\sqrt{3}\tan 2\theta = 3 \implies \tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3} \implies 2\theta = 60^\circ \implies \theta = 30^\circ$.
Q9. (a) $\cos 90^\circ$ [1 Mark]
Explanation: $\frac{1 – (1)^2}{1 + (1)^2} = \frac{0}{2} = 0 = \cos 90^\circ$.
Explanation: $\frac{1 – (1)^2}{1 + (1)^2} = \frac{0}{2} = 0 = \cos 90^\circ$.
Q10. (c) 45Β° [1 Mark]
Explanation: $\tan\theta = \frac{\text{Height}}{\text{Base}} = \frac{30}{30} = 1 \implies \theta = 45^\circ$.
Explanation: $\tan\theta = \frac{\text{Height}}{\text{Base}} = \frac{30}{30} = 1 \implies \theta = 45^\circ$.
Q11. (a) 50Β° [1 Mark]
Explanation: In quadrilateral $OAPB$, $\angle AOB = 180^\circ – 80^\circ = 100^\circ$. $\triangle POA \cong \triangle POB \implies \angle POA = \frac{1}{2}\angle AOB = 50^\circ$.
Explanation: In quadrilateral $OAPB$, $\angle AOB = 180^\circ – 80^\circ = 100^\circ$. $\triangle POA \cong \triangle POB \implies \angle POA = \frac{1}{2}\angle AOB = 50^\circ$.
Q12. (b) 125% [1 Mark]
Explanation: Initial Area $A_1 = \pi r^2$. New radius $r’ = 1.5r \implies A_2 = \pi (1.5r)^2 = 2.25\pi r^2$. Increase $= \frac{2.25 – 1}{1} \times 100\% = 125\%$.
Explanation: Initial Area $A_1 = \pi r^2$. New radius $r’ = 1.5r \implies A_2 = \pi (1.5r)^2 = 2.25\pi r^2$. Increase $= \frac{2.25 – 1}{1} \times 100\% = 125\%$.
Q13. (c) $3\pi r^2$ [1 Mark]
Explanation: Total surface area of a solid hemisphere $= \text{Curved surface area} + \text{Base area} = 2\pi r^2 + \pi r^2 = 3\pi r^2$.
Explanation: Total surface area of a solid hemisphere $= \text{Curved surface area} + \text{Base area} = 2\pi r^2 + \pi r^2 = 3\pi r^2$.
Q14. (b) 33 [1 Mark]
Explanation: $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1320}{40} = 33$.
Explanation: $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1320}{40} = 33$.
Q15. (a) 1/7 [1 Mark]
Explanation: A non-leap year has 365 days $= 52\text{ weeks} + 1\text{ extra day}$. The single extra day can be any of the 7 days of the week $\implies P(53\text{ Sundays}) = \frac{1}{7}$.
Explanation: A non-leap year has 365 days $= 52\text{ weeks} + 1\text{ extra day}$. The single extra day can be any of the 7 days of the week $\implies P(53\text{ Sundays}) = \frac{1}{7}$.
Q16. (a) $(-4, 2)$ [1 Mark]
Explanation: $\text{Midpoint} = \left(\frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{-8}{2}, \frac{4}{2}\right) = (-4, 2)$.
Explanation: $\text{Midpoint} = \left(\frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2}\right) = \left(\frac{-8}{2}, \frac{4}{2}\right) = (-4, 2)$.
Q17. (d) 0.95 [1 Mark]
Explanation: $P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95$.
Explanation: $P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95$.
Q18. (a) 20 [1 Mark]
Explanation: Smallest two-digit composite number $= 10$, smallest composite number $= 4$. $\text{LCM}(10, 4) = 20$.
Explanation: Smallest two-digit composite number $= 10$, smallest composite number $= 4$. $\text{LCM}(10, 4) = 20$.
Q19. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: 2 is prime, so $\sqrt{2}$ is irrational by theorem.
Explanation: 2 is prime, so $\sqrt{2}$ is irrational by theorem.
Q20. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: Any point on the y-axis has $x = 0$, so $(0, 4)$ lies on the y-axis.
Explanation: Any point on the y-axis has $x = 0$, so $(0, 4)$ lies on the y-axis.
SECTION B SOLUTIONS
Q21. Prime Factorisation of 72 and 120: [2 Marks]
$72 = 2^3 \times 3^2$ [0.5 Mark]
$120 = 2^3 \times 3 \times 5$ [0.5 Mark]
$\text{HCF}(72, 120) = 2^3 \times 3 = \mathbf{24}$ [0.5 Mark]
$\text{LCM}(72, 120) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = \mathbf{360}$. [0.5 Mark]
$72 = 2^3 \times 3^2$ [0.5 Mark]
$120 = 2^3 \times 3 \times 5$ [0.5 Mark]
$\text{HCF}(72, 120) = 2^3 \times 3 = \mathbf{24}$ [0.5 Mark]
$\text{LCM}(72, 120) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = \mathbf{360}$. [0.5 Mark]
Q22. Ratio of Sides: [2 Marks]
$XY \parallel BC \implies \triangle AXY \sim \triangle ABC$ by AA similarity. [0.5 Mark]
$\frac{\text{Area}(\triangle AXY)}{\text{Area}(\triangle ABC)} = \frac{1}{2} \implies \left(\frac{AX}{AB}\right)^2 = \frac{1}{2}$ [1 Mark]
$\frac{AX}{AB} = \frac{1}{\sqrt{2}} = \mathbf{\frac{\sqrt{2}}{2}}$. [0.5 Mark]
OR
In $\triangle AOB$ and $\triangle COD$: $\angle AOB = \angle COD$ (Vertically opposite), $\angle OAB = \angle OCD$ (Alt. interior as $AB \parallel DC$). [1 Mark]
$\triangle AOB \sim \triangle COD$ (by AA) $\implies \frac{AO}{CO} = \frac{BO}{DO} \implies \mathbf{\frac{AO}{BO} = \frac{CO}{DO}}$. [1 Mark]
$XY \parallel BC \implies \triangle AXY \sim \triangle ABC$ by AA similarity. [0.5 Mark]
$\frac{\text{Area}(\triangle AXY)}{\text{Area}(\triangle ABC)} = \frac{1}{2} \implies \left(\frac{AX}{AB}\right)^2 = \frac{1}{2}$ [1 Mark]
$\frac{AX}{AB} = \frac{1}{\sqrt{2}} = \mathbf{\frac{\sqrt{2}}{2}}$. [0.5 Mark]
OR
In $\triangle AOB$ and $\triangle COD$: $\angle AOB = \angle COD$ (Vertically opposite), $\angle OAB = \angle OCD$ (Alt. interior as $AB \parallel DC$). [1 Mark]
$\triangle AOB \sim \triangle COD$ (by AA) $\implies \frac{AO}{CO} = \frac{BO}{DO} \implies \mathbf{\frac{AO}{BO} = \frac{CO}{DO}}$. [1 Mark]
Q23. Equidistant Point Relation: [2 Marks]
$PA^2 = PB^2 \implies (x – 3)^2 + (y – 6)^2 = (x + 3)^2 + (y – 4)^2$ [0.5 Mark]
$x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$ [0.5 Mark]
$-6x – 12y + 45 = 6x – 8y + 25 \implies 12x + 4y = 20$ [0.5 Mark]
Dividing by 4: $\mathbf{3x + y = 5}$. [0.5 Mark]
$PA^2 = PB^2 \implies (x – 3)^2 + (y – 6)^2 = (x + 3)^2 + (y – 4)^2$ [0.5 Mark]
$x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$ [0.5 Mark]
$-6x – 12y + 45 = 6x – 8y + 25 \implies 12x + 4y = 20$ [0.5 Mark]
Dividing by 4: $\mathbf{3x + y = 5}$. [0.5 Mark]
Q24. Trigonometric Angles: [2 Marks]
$\sin(A + B) = 1 \implies A + B = 90^\circ$ …(1)
$\cos(A – B) = \frac{\sqrt{3}}{2} \implies A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 120^\circ \implies \mathbf{A = 60^\circ}$.
From (1): $B = 90^\circ – 60^\circ \implies \mathbf{B = 30^\circ}$. [1 Mark]
OR
$\text{Expression} = \frac{1/2 + 1 – 2/\sqrt{3}}{2/\sqrt{3} + 1/2 + 1} = \frac{3/2 – 2/\sqrt{3}}{3/2 + 2/\sqrt{3}} = \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4}$ [1 Mark]
Rationalising: $\frac{(3\sqrt{3} – 4)^2}{(3\sqrt{3})^2 – 4^2} = \frac{27 + 16 – 24\sqrt{3}}{27 – 16} = \mathbf{\frac{43 – 24\sqrt{3}}{11}}$. [1 Mark]
$\sin(A + B) = 1 \implies A + B = 90^\circ$ …(1)
$\cos(A – B) = \frac{\sqrt{3}}{2} \implies A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 120^\circ \implies \mathbf{A = 60^\circ}$.
From (1): $B = 90^\circ – 60^\circ \implies \mathbf{B = 30^\circ}$. [1 Mark]
OR
$\text{Expression} = \frac{1/2 + 1 – 2/\sqrt{3}}{2/\sqrt{3} + 1/2 + 1} = \frac{3/2 – 2/\sqrt{3}}{3/2 + 2/\sqrt{3}} = \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4}$ [1 Mark]
Rationalising: $\frac{(3\sqrt{3} – 4)^2}{(3\sqrt{3})^2 – 4^2} = \frac{27 + 16 – 24\sqrt{3}}{27 – 16} = \mathbf{\frac{43 – 24\sqrt{3}}{11}}$. [1 Mark]
Q25. Concentric Circles Chord Length: [2 Marks]
Let $AB$ be the chord of the larger circle touching the smaller circle at $P$. Centre is $O$.
$OP \perp AB$ and $OP$ bisects $AB$. Radius of smaller circle $OP = 5\text{ cm}$, larger circle $OA = 13\text{ cm}$. [0.5 Mark]
In right $\triangle OPA$: $AP = \sqrt{OA^2 – OP^2} = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12\text{ cm}$. [1 Mark]
Length of chord $AB = 2 \times AP = 2 \times 12 = \mathbf{24\text{ cm}}$. [0.5 Mark]
Let $AB$ be the chord of the larger circle touching the smaller circle at $P$. Centre is $O$.
$OP \perp AB$ and $OP$ bisects $AB$. Radius of smaller circle $OP = 5\text{ cm}$, larger circle $OA = 13\text{ cm}$. [0.5 Mark]
In right $\triangle OPA$: $AP = \sqrt{OA^2 – OP^2} = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12\text{ cm}$. [1 Mark]
Length of chord $AB = 2 \times AP = 2 \times 12 = \mathbf{24\text{ cm}}$. [0.5 Mark]
SECTION C SOLUTIONS
Q26. Ratio of Zeroes: [3 Marks]
Let the zeroes be $2k$ and $3k$.
Sum of zeroes: $2k + 3k = p \implies 5k = p \implies k = \frac{p}{5}$ …(1) [1 Mark]
Product of zeroes: $(2k)(3k) = q \implies 6k^2 = q$ …(2) [1 Mark]
Substituting (1) into (2): $6\left(\frac{p}{5}\right)^2 = q \implies 6\left(\frac{p^2}{25}\right) = q \implies \mathbf{6p^2 = 25q}$. Hence Proved. [1 Mark]
Let the zeroes be $2k$ and $3k$.
Sum of zeroes: $2k + 3k = p \implies 5k = p \implies k = \frac{p}{5}$ …(1) [1 Mark]
Product of zeroes: $(2k)(3k) = q \implies 6k^2 = q$ …(2) [1 Mark]
Substituting (1) into (2): $6\left(\frac{p}{5}\right)^2 = q \implies 6\left(\frac{p^2}{25}\right) = q \implies \mathbf{6p^2 = 25q}$. Hence Proved. [1 Mark]
Q27. Age Word Problem: [3 Marks]
Let Jacob’s present age be $x$ years and son’s present age be $y$ years.
Five years hence: $(x + 5) = 3(y + 5) \implies x – 3y = 10$ …(1) [1 Mark]
Five years ago: $(x – 5) = 7(y – 5) \implies x – 7y = -30$ …(2) [1 Mark]
Subtracting (2) from (1): $4y = 40 \implies \mathbf{y = 10\text{ years}}$.
From (1): $x – 3(10) = 10 \implies \mathbf{x = 40\text{ years}}$.
Jacob’s present age is 40 years, son’s present age is 10 years. [1 Mark]
OR
$152x – 378y = -74$ …(1); $-378x + 152y = -604$ …(2)
Adding (1) and (2): $-226x – 226y = -678 \implies x + y = 3$ …(3) [1.5 Marks]
Subtracting (2) from (1): $530x – 530y = 530 \implies x – y = 1$ …(4) [1 Mark]
Adding (3) and (4): $2x = 4 \implies \mathbf{x = 2}$; from (3): $\mathbf{y = 1}$. [0.5 Mark]
Let Jacob’s present age be $x$ years and son’s present age be $y$ years.
Five years hence: $(x + 5) = 3(y + 5) \implies x – 3y = 10$ …(1) [1 Mark]
Five years ago: $(x – 5) = 7(y – 5) \implies x – 7y = -30$ …(2) [1 Mark]
Subtracting (2) from (1): $4y = 40 \implies \mathbf{y = 10\text{ years}}$.
From (1): $x – 3(10) = 10 \implies \mathbf{x = 40\text{ years}}$.
Jacob’s present age is 40 years, son’s present age is 10 years. [1 Mark]
OR
$152x – 378y = -74$ …(1); $-378x + 152y = -604$ …(2)
Adding (1) and (2): $-226x – 226y = -678 \implies x + y = 3$ …(3) [1.5 Marks]
Subtracting (2) from (1): $530x – 530y = 530 \implies x – y = 1$ …(4) [1 Mark]
Adding (3) and (4): $2x = 4 \implies \mathbf{x = 2}$; from (3): $\mathbf{y = 1}$. [0.5 Mark]
Q28. AP Term Relation: [3 Marks]
$a_8 = a + 7d = 0 \implies a = -7d$ …(1) [1 Mark]
38th term: $a_{38} = a + 37d = -7d + 37d = 30d$ …(2) [1 Mark]
18th term: $a_{18} = a + 17d = -7d + 17d = 10d$ …(3) [0.5 Mark]
From (2) and (3): $a_{38} = 3(10d) = \mathbf{3 \times a_{18}}$. Hence Proved. [0.5 Mark]
$a_8 = a + 7d = 0 \implies a = -7d$ …(1) [1 Mark]
38th term: $a_{38} = a + 37d = -7d + 37d = 30d$ …(2) [1 Mark]
18th term: $a_{18} = a + 17d = -7d + 17d = 10d$ …(3) [0.5 Mark]
From (2) and (3): $a_{38} = 3(10d) = \mathbf{3 \times a_{18}}$. Hence Proved. [0.5 Mark]
Q29. Trigonometric Identity: [3 Marks]
$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 – \sin A)(1 + \sin A)}}$ [1 Mark]
$= \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$ [1 Mark]
$= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \mathbf{\sec A + \tan A} = \text{RHS}$. Hence Proved. [1 Mark]
$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 – \sin A)(1 + \sin A)}}$ [1 Mark]
$= \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$ [1 Mark]
$= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \mathbf{\sec A + \tan A} = \text{RHS}$. Hence Proved. [1 Mark]
Q30. Circumscribed Parallelogram is Rhombus: [3 Marks]
$ABCD$ is a parallelogram $\implies AB = CD$ and $AD = BC$. [0.5 Mark]
Tangents from external points: $AP = AS, BP = BQ, CR = CQ, DR = DS$.
Adding gives: $AB + CD = AD + BC \implies AB + AB = AD + AD \implies 2AB = 2AD \implies AB = AD$. [2 Marks]
Since adjacent sides of the parallelogram are equal, $\mathbf{ABCD\text{ is a rhombus}}$. [0.5 Mark]
OR
In $\triangle PAB$, $PA = PB \implies \angle PAB = \angle PBA = \frac{180^\circ – \angle APB}{2} = 90^\circ – \frac{1}{2}\angle APB$. [1.5 Marks]
Since $OA \perp PA$, $\angle OAP = 90^\circ$.
$\angle OAB = \angle OAP – \angle PAB = 90^\circ – \left(90^\circ – \frac{1}{2}\angle APB\right) = \frac{1}{2}\angle APB \implies \mathbf{\angle APB = 2\angle OAB}$. [1.5 Marks]
$ABCD$ is a parallelogram $\implies AB = CD$ and $AD = BC$. [0.5 Mark]
Tangents from external points: $AP = AS, BP = BQ, CR = CQ, DR = DS$.
Adding gives: $AB + CD = AD + BC \implies AB + AB = AD + AD \implies 2AB = 2AD \implies AB = AD$. [2 Marks]
Since adjacent sides of the parallelogram are equal, $\mathbf{ABCD\text{ is a rhombus}}$. [0.5 Mark]
OR
In $\triangle PAB$, $PA = PB \implies \angle PAB = \angle PBA = \frac{180^\circ – \angle APB}{2} = 90^\circ – \frac{1}{2}\angle APB$. [1.5 Marks]
Since $OA \perp PA$, $\angle OAP = 90^\circ$.
$\angle OAB = \angle OAP – \angle PAB = 90^\circ – \left(90^\circ – \frac{1}{2}\angle APB\right) = \frac{1}{2}\angle APB \implies \mathbf{\angle APB = 2\angle OAB}$. [1.5 Marks]
Q31. Marbles Probability: [3 Marks]
Total marbles $= 24$. Let number of green marbles be $x$.
$P(\text{Green}) = \frac{x}{24} = \frac{2}{3} \implies x = \frac{2 \times 24}{3} = 16\text{ green marbles}$. [1.5 Marks]
Number of blue marbles $= \text{Total} – \text{Green} = 24 – 16 = \mathbf{8\text{ blue marbles}}$. [1.5 Marks]
Total marbles $= 24$. Let number of green marbles be $x$.
$P(\text{Green}) = \frac{x}{24} = \frac{2}{3} \implies x = \frac{2 \times 24}{3} = 16\text{ green marbles}$. [1.5 Marks]
Number of blue marbles $= \text{Total} – \text{Green} = 24 – 16 = \mathbf{8\text{ blue marbles}}$. [1.5 Marks]
SECTION D SOLUTIONS
Q32. Aeroplane Speed Problem: [5 Marks]
Let usual speed of the plane be $x\text{ km/h}$. Increased speed $= (x + 250)\text{ km/h}$.
Difference in time $= 30\text{ minutes} = \frac{1}{2}\text{ hour}$.
$\frac{1500}{x} – \frac{1500}{x + 250} = \frac{1}{2}$ [1.5 Marks]
$1500\left[\frac{x + 250 – x}{x(x + 250)}\right] = \frac{1}{2} \implies \frac{375000}{x^2 + 250x} = \frac{1}{2} \implies x^2 + 250x – 750000 = 0$ [1.5 Marks]
$(x + 1000)(x – 750) = 0 \implies x = 750\text{ or } x = -1000$ (Speed cannot be negative).
The usual speed of the aeroplane is $\mathbf{750\text{ km/h}}$. [2 Marks]
OR
$\frac{(x – 7) – (x + 4)}{(x + 4)(x – 7)} = \frac{11}{30} \implies \frac{-11}{x^2 – 3x – 28} = \frac{11}{30}$ [2 Marks]
$\frac{-1}{x^2 – 3x – 28} = \frac{1}{30} \implies x^2 – 3x – 28 = -30 \implies x^2 – 3x + 2 = 0$ [1.5 Marks]
$(x – 1)(x – 2) = 0 \implies \mathbf{x = 1\text{ or } x = 2}$. [1.5 Marks]
Let usual speed of the plane be $x\text{ km/h}$. Increased speed $= (x + 250)\text{ km/h}$.
Difference in time $= 30\text{ minutes} = \frac{1}{2}\text{ hour}$.
$\frac{1500}{x} – \frac{1500}{x + 250} = \frac{1}{2}$ [1.5 Marks]
$1500\left[\frac{x + 250 – x}{x(x + 250)}\right] = \frac{1}{2} \implies \frac{375000}{x^2 + 250x} = \frac{1}{2} \implies x^2 + 250x – 750000 = 0$ [1.5 Marks]
$(x + 1000)(x – 750) = 0 \implies x = 750\text{ or } x = -1000$ (Speed cannot be negative).
The usual speed of the aeroplane is $\mathbf{750\text{ km/h}}$. [2 Marks]
OR
$\frac{(x – 7) – (x + 4)}{(x + 4)(x – 7)} = \frac{11}{30} \implies \frac{-11}{x^2 – 3x – 28} = \frac{11}{30}$ [2 Marks]
$\frac{-1}{x^2 – 3x – 28} = \frac{1}{30} \implies x^2 – 3x – 28 = -30 \implies x^2 – 3x + 2 = 0$ [1.5 Marks]
$(x – 1)(x – 2) = 0 \implies \mathbf{x = 1\text{ or } x = 2}$. [1.5 Marks]
Q33. BPT Theorem and Application: [5 Marks]
- BPT Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. [1 Mark]
- Application:
In $\triangle ABC$, $DE \parallel BC \implies \frac{AD}{AB} = \frac{AE}{AC}$ …(1) (by BPT corollary) [1.5 Marks]
In $\triangle ADC$, $FE \parallel DC \implies \frac{AF}{AD} = \frac{AE}{AC}$ …(2) (by BPT corollary) [1.5 Marks]
From (1) and (2): $\frac{AD}{AB} = \frac{AF}{AD} \implies \mathbf{AD^2 = AB \times AF}$. Hence Proved. [1 Mark]
Q34. Ice Cream Cones: [5 Marks]
Cylinder: $R = 6\text{ cm}, H = 15\text{ cm} \implies V_{\text{cyl}} = \pi R^2 H = \pi (6)^2 (15) = 540\pi\text{ cm}^3$. [1.5 Marks]
One Ice Cream Cone (Cone + Hemisphere): $r = 3\text{ cm}, h = 12\text{ cm}$.
$V_{\text{ice cream}} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi (3)^2 (12) + \frac{2}{3}\pi (3)^3 = 36\pi + 18\pi = 54\pi\text{ cm}^3$. [2 Marks]
$\text{Number of cones} = \frac{\text{Volume of Cylinder}}{\text{Volume of one ice cream cone}} = \frac{540\pi}{54\pi} = \mathbf{10\text{ cones}}$. [1.5 Marks]
OR
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, Cone height $h = 15.5 – 3.5 = 12\text{ cm}$.
Slant height $l = \sqrt{h^2 + r^2} = \sqrt{12^2 + (3.5)^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5\text{ cm}$. [1 Mark]
$\text{Total Surface Area} = \pi r l + 2\pi r^2 = \frac{22}{7} \times \frac{7}{2} \times [12.5 + 2(3.5)] = 11 \times [12.5 + 7] = 11 \times 19.5 = \mathbf{214.5\text{ cm}^2}$. [2 Marks]
$\text{Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 [h + 2r] = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times [12 + 7] = \frac{77}{6} \times 19 = \mathbf{\frac{1463}{6}\text{ cm}^3} \approx 243.83\text{ cm}^3$. [2 Marks]
Cylinder: $R = 6\text{ cm}, H = 15\text{ cm} \implies V_{\text{cyl}} = \pi R^2 H = \pi (6)^2 (15) = 540\pi\text{ cm}^3$. [1.5 Marks]
One Ice Cream Cone (Cone + Hemisphere): $r = 3\text{ cm}, h = 12\text{ cm}$.
$V_{\text{ice cream}} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi (3)^2 (12) + \frac{2}{3}\pi (3)^3 = 36\pi + 18\pi = 54\pi\text{ cm}^3$. [2 Marks]
$\text{Number of cones} = \frac{\text{Volume of Cylinder}}{\text{Volume of one ice cream cone}} = \frac{540\pi}{54\pi} = \mathbf{10\text{ cones}}$. [1.5 Marks]
OR
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$, Cone height $h = 15.5 – 3.5 = 12\text{ cm}$.
Slant height $l = \sqrt{h^2 + r^2} = \sqrt{12^2 + (3.5)^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5\text{ cm}$. [1 Mark]
$\text{Total Surface Area} = \pi r l + 2\pi r^2 = \frac{22}{7} \times \frac{7}{2} \times [12.5 + 2(3.5)] = 11 \times [12.5 + 7] = 11 \times 19.5 = \mathbf{214.5\text{ cm}^2}$. [2 Marks]
$\text{Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 [h + 2r] = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times [12 + 7] = \frac{77}{6} \times 19 = \mathbf{\frac{1463}{6}\text{ cm}^3} \approx 243.83\text{ cm}^3$. [2 Marks]
Q35. Missing Frequencies: [5 Marks]
$\sum f_i = 120 \implies 17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52$ …(1) [1.5 Marks]
$\text{Class marks } (x_i): 10, 30, 50, 70, 90$.
$\sum f_i x_i = 17(10) + 30f_1 + 32(50) + 70f_2 + 19(90) = 170 + 30f_1 + 1600 + 70f_2 + 1710 = 3480 + 30f_1 + 70f_2$. [1.5 Marks]
$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \implies 50 = \frac{3480 + 30f_1 + 70f_2}{120} \implies 6000 = 3480 + 30f_1 + 70f_2$
$30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252$ …(2) [1 Mark]
From (1), $3f_1 + 3f_2 = 156$. Subtracting from (2): $4f_2 = 96 \implies \mathbf{f_2 = 24}$.
$f_1 = 52 – 24 \implies \mathbf{f_1 = 28}$. [1 Mark]
$\sum f_i = 120 \implies 17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52$ …(1) [1.5 Marks]
$\text{Class marks } (x_i): 10, 30, 50, 70, 90$.
$\sum f_i x_i = 17(10) + 30f_1 + 32(50) + 70f_2 + 19(90) = 170 + 30f_1 + 1600 + 70f_2 + 1710 = 3480 + 30f_1 + 70f_2$. [1.5 Marks]
$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \implies 50 = \frac{3480 + 30f_1 + 70f_2}{120} \implies 6000 = 3480 + 30f_1 + 70f_2$
$30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252$ …(2) [1 Mark]
From (1), $3f_1 + 3f_2 = 156$. Subtracting from (2): $4f_2 = 96 \implies \mathbf{f_2 = 24}$.
$f_1 = 52 – 24 \implies \mathbf{f_1 = 28}$. [1 Mark]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) $a = 20, d = 4 \implies a_{15} = 20 + 14(4) = 20 + 56 = \mathbf{76\text{ seats}}$. [1 Mark]
- (b) $a_{20} – a_{10} = (a + 19d) – (a + 9d) = 10d = 10(4) = \mathbf{40\text{ seats}}$. [1 Mark]
- (c) $S_{30} = \frac{30}{2}[2(20) + (30 – 1)4] = 15[40 + 116] = 15(156) = \mathbf{2340\text{ seats}}$. [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) $OA = \sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5\text{ km}}$ (5 units). [1 Mark]
- (b) $AB = \sqrt{(7 – 3)^2 + (7 – 4)^2} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5\text{ km}}$ (5 units). [1 Mark]
- (c) $P(x, 0)$ is equidistant $\implies PA^2 = PB^2 \implies (x – 3)^2 + 16 = (x – 7)^2 + 49$
$x^2 – 6x + 25 = x^2 – 14x + 98 \implies 8x = 73 \implies x = \frac{73}{8} \implies \mathbf{P\left(\frac{73}{8}, 0\right)}$. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Neat diagram showing road width 80 m, two poles of equal height $h$, and point $P$ with angles $60^\circ$ and $30^\circ$. [1 Mark]
- (b) Let distance of $P$ from first pole be $x \implies$ from second pole is $(80 – x)$.
$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3}$; $\tan 30^\circ = \frac{h}{80 – x} \implies \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{80 – x} \implies 3x = 80 – x \implies 4x = 80 \implies \mathbf{x = 20\text{ m}}$.
Distances are 20 m and 60 m. [1 Mark] - (c) $h = x\sqrt{3} = 20\sqrt{3} = 20(1.732) = \mathbf{34.64\text{ m}}$. [2 Marks]