📌 General Instructions This question paper contains 38 questions. All questions are compulsory.
Section A: Q1-18 MCQs + Q19-20 Assertion-Reason (1 mark each)
Section B: Q21-25 VSA (2 marks each)
Section C: Q26-31 SA (3 marks each)
Section D: Q32-35 LA (5 marks each)
Section E: Q36-38 Case-based (4 marks each)
Internal choices in Sections B, C, D, E. Draw neat diagrams. Use $\pi = \frac{22}{7}$ if not stated. No calculators.

📝 SECTION A

20 × 1 = 20 marks

Q1. Real Numbers

The LCM of 960 and 240 is:

(A) 960

(B) 240

(D) 15

🔍 VIEW ANSWER

✅ ANSWER: A (960)

📐 EXPLANATION:

LCM Calculation: Prime Factorization

First, find the prime factors of each number:
\[ 240 = 2^4 \times 3 \times 5 \]
\[ 960 = 2^6 \times 3 \times 5 \]
To find the LCM, take the highest power of each prime factor:

Highest power of $ 2 $: $ 2^6 $
Highest power of $ 3 $: $ 3^1 $
Highest power of $ 5 $: $ 5^1 $

Multiply them together: \[ \text{LCM} = 2^6 \times 3 \times 5 \]
\[ \text{LCM} = 64 \times 15 = 960 \]

Q2. Number System

The natural number 1 is :

(A) a prime number.

(B) a composite number.

(D) neither prime nor composite.

🔍 VIEW ANSWER

✅ ANSWER: D (neither prime nor composite)

📐 EXPLANATION: A prime number has exactly two factors: 1 and itself. A composite number has more than two factors. The number 1 has only one factor (itself), so it is neither prime nor composite.

Q3. Number Theory

For any natural number $n$, $5^{n}$ ends with the digit :

(A) 0

(B) 5

(D) 2

🔍 VIEW ANSWER

✅ ANSWER: B (5)

📐 EXPLANATION: Any power of 5 ends with digit 5.
Check: $5^1=5$,
$5^2=25$,
$5^3=125$, etc.
All end with 5.

Q4. 📈 Polynomials

The graph of $y = f(x)$ is given. The number of distinct zeroes of $y = f(x)$ is :

polynomial

(A) 0

(B) 1

(D) 3

🔍 VIEW ANSWER

✅ ANSWER: C (2)

📐 EXPLANATION: The zeroes of a function are the x-coordinates where the graph intersects the x-axis. From the given graph (parabola), it crosses the x-axis at two distinct points, so there are 2 distinct zeroes.

Q5. 🔢 Polynomials

If $\alpha$ and $\beta$ are two zeroes of a polynomial $f(x) = px^2 – 2x + 3p$ and $\alpha + \beta = \alpha \beta$, then value of $p$ is :

(A) $-\frac{2}{3}$

(B) $\frac{2}{3}$

(D) $-\frac{1}{3}$

🔍 VIEW ANSWER

✅ ANSWER: B ($\frac{2}{3}$)

📐 EXPLANATION: For quadratic $px^2 – 2x + 3p$:
Sum of zeroes $\alpha + \beta = \frac{-(-2)}{p} = \frac{2}{p}$
Product of zeroes $\alpha \beta = \frac{3p}{p} = 3$
Given $\alpha + \beta = \alpha \beta$ ⇒ $\frac{2}{p} = 3$ ⇒ $p = \frac{2}{3}$

Q6. 📐 Linear Equations

If the pair of linear equations : $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ is consistent and dependent, then

(A) $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$

(B) $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

(D) $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

🔍 VIEW ANSWER

✅ ANSWER: D ($\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$)

📐 EXPLANATION: For a pair of linear equations to be consistent and dependent (coincident lines), the ratios of coefficients must be equal: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Q7.📊 Arithmetic Progression

Which of the following sequence is not an A.P.?

(A) $2,\frac{5}{2},3,\frac{7}{2},\dots$

(B) $-1.2, -3.2, -5.2, -7.2,\dots$

(D) $1^{2},3^{2},5^{2},7^{2},\dots$

🔍 VIEW ANSWER

✅ ANSWER: D ($1^{2},3^{2},5^{2},7^{2},\dots$)

📐 EXPLANATION: Check common difference:
(A) $d = \frac{5}{2}-2 = \frac{1}{2}$ constant → AP
(B) $d = -3.2-(-1.2) = -2$ constant → AP
(C) $\sqrt{8}-\sqrt{2}$$ = 2\sqrt{2}-\sqrt{2} $$= \sqrt{2}$, $\sqrt{18}-\sqrt{8} $$= 3\sqrt{2}-2\sqrt{2}$$ = \sqrt{2}$ constant → AP
(D) $1,9,25,49$ differences: 8,16,24 → not constant → not AP

Q8.🔺 Triangles

In triangles ABC and PQR, $\angle A = \angle Q$ and $\angle B = \angle R$, then AB : AC is equal to:

(A) PQ:PR

(B) PQ:QR

(D) PR:QR

🔍 VIEW ANSWER

✅ ANSWER: ANSWER: C (QR:QP)

📐 EXPLANATION:

Triangle Similarity Problem

Step 1: Establish Similarity

In $\triangle ABC$ and $\triangle QRP$, we are given:

$\angle A = \angle Q$
$\angle B = \angle R$

According to the AA (Angle-Angle) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. Therefore:

$$\triangle ABC \sim \triangle QRP$$

Step 2: Determine Corresponding Sides

When two triangles are similar, the ratios of their corresponding sides are equal. Based on the similarity $\triangle ABC \sim \triangle QRP$, the corresponding sides are:

$AB$ corresponds to $QR$
$BC$ corresponds to $RP$
$AC$ corresponds to $QP$

Setting up the ratio for the sides mentioned in the question:

$$\frac{AB}{QR} = \frac{AC}{QP}$$

Step 3: Solve for the Required Ratio

To find what $AB : AC$ is equal to, we rearrange the proportion:

$$\frac{AB}{AC} = \frac{QR}{QP}$$

In ratio form, this is $AB : AC = QR : QP$.

Answer:

✅ CORRECT ANSWER: C (QR:QP)

Q9.📐 Coordinate Geometry

The distance of the point A(4a, 3a) from x-axis is:

(A) 3a

(B) -3a

(D) -4a

🔍 VIEW ANSWER

✅ ANSWER: A (3a)

📐 EXPLANATION: Distance from x-axis = |y-coordinate| = |3a| = 3a (distance is always positive).

Q10. 📐 Trigonometry

If cos A = $\frac{4}{5}$, then the value of tan A is :

(A) $\frac{3}{5}$

(B) $\frac{3}{4}$

(D) $\frac{5}{3}$

🔍 VIEW ANSWER

✅ ANSWER: B ($\frac{3}{4}$)

📐 EXPLANATION: Using identity $\sin^2 A + \cos^2 A = 1$: $\sin^2 A = 1 – \left(\frac{4}{5}\right)^2$$ = 1 – \frac{16}{25}$$ = \frac{9}{25}$ $\sin A = \frac{3}{5}$ (positive, assuming acute angle) $\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4}$

Q11. 🔢 Trigonometry

If 2 sin A = 1, then the value of tan A + cot A is :

(A) $\sqrt{3}$

(B) $\frac{4}{\sqrt{3}}$

(D) 1

🔍 VIEW ANSWER

✅ ANSWER: B ($\frac{4}{\sqrt{3}}$)

📐 EXPLANATION:
2 sin A = 1 ⇒ sin A = $\frac{1}{2}$ ⇒ A = 30° (or 150°, but take acute) cos 30° = $\frac{\sqrt{3}}{2}$
tan 30° = $\frac{1}{\sqrt{3}}$,
cot 30° = $\sqrt{3}$
tan A + cot A
= $\frac{1}{\sqrt{3}} + \sqrt{3}
= \frac{1 + 3}{\sqrt{3}}
= \frac{4}{\sqrt{3}}$

Q12. 📈 Height & Distance

From a point on the ground, which is 60 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 45°. The height (in metres) of the tower is :

(A) $10\sqrt{3}$

(B) $30\sqrt{3}$

(D) 30

🔍 VIEW ANSWER

✅ ANSWER: C (60)

📐 EXPLANATION:

Application-of-Trignometry

Height of a Vertical Tower

Step 1: Identify Given Information

The problem describes a right-angled triangle formed by:

The vertical tower (the opposite side, $$h$$).
The distance from the foot of the tower (the adjacent side), which is $$60 \text{ m}$$.
The angle of elevation to the top of the tower, which is $$45^\circ$$.

Step 2: Use Trigonometric Ratios

We use the tangent ratio, which relates the opposite side to the adjacent side:

$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$$

Substitute the given values:

$$\tan(45^\circ) = \frac{h}{60}$$

Step 3: Solve for Height

Since $$\tan(45^\circ) = 1$$, the equation becomes:

$$1 = \frac{h}{60}$$
$$h = 60 \text{ m}$$

Answer:

The height of the tower is $$60 \text{ m}$$.

Q13. ⭕ Circles

In the given figure, PA and PB are tangents to a circle centred at O. If $\angle OAB = 15^\circ$, then $\angle APB$ equals :

(A) $30^\circ$

(B) $15^\circ$

(D) $10^\circ$

🔍 VIEW ANSWER

✅ ANSWER: A (30°)

📐 EXPLANATION:
Area-related-to-circle
In triangle OAB,
OA = OB (radii),
so it’s isosceles. ∠OAB = 15°
⇒ ∠OBA = 15°
∠AOB = 180° – 15° – 15° = 150°
In quadrilateral OAPB, ∠OAP = ∠OBP = 90° (radius ⊥ tangent)
∠APB = 360° – 90° – 90° – 150° = 30°

Q14. ⭕ Circles

In the given figure, PA and PB are tangents to a circle centred at O. If $\angle AOB = 130^\circ$, then $\angle APB$ is equal to :

(A) 130°

(B) 50°

(D) 90°

🔍 VIEW ANSWER

✅ ANSWER: B (50°)

📐 EXPLANATION: In quadrilateral OAPB, ∠OAP = ∠OBP = 90° (radius ⊥ tangent) ∠AOB = 130° ∠APB = 360° – 90° – 90° – 130° = 50°

Q15.📐 Area related to circles

Area of a segment of a circle of radius ‘r’ and central angle 60° is :

(A) $\frac{\pi r^2}{2} – \frac{1}{2}r^2$

(B) $\frac{2\pi r}{4} – \frac{\sqrt{3}}{4}r^2$

(D) $\frac{2\pi r}{4} – r^2 \sin 60^\circ$

🔍 VIEW ANSWER

✅ ANSWER: C ($\frac{\pi r^2}{6} – \frac{\sqrt{3}}{4}r^2$)

📐 EXPLANATION: Area of sector with angle θ = $\frac{\theta}{360°} \times \pi r^2$ For θ = 60°, sector area = $\frac{60}{360} \pi r^2 = \frac{\pi r^2}{6}$ Area of equilateral triangle formed by radii and chord = $\frac{\sqrt{3}}{4}r^2$ Area of segment = sector area – triangle area = $\frac{\pi r^2}{6} – \frac{\sqrt{3}}{4}r^2$

Q16.🔮 Surface Area & Volume

A hemispherical bowl is made of steel of thickness 1 cm. The outer radius of the bowl is 6 cm. The volume of steel used (in cm³) is :

(A) 182 π

(B) $\frac{182}{3} \pi$

(D) $\frac{364}{3} \pi$

🔍 VIEW ANSWER

✅ ANSWER: D ($\frac{364}{3} \pi$)

📐 EXPLANATION: Outer radius R = 6 cm, inner radius r = 6 – 1 = 5 cm (thickness 1 cm) Volume of hemisphere = $\frac{2}{3}\pi r^3$ Volume of steel = outer volume – inner volume $ = \frac{2}{3}\pi (6^3) – \frac{2}{3}\pi (5^3)$ $= \frac{2}{3}\pi (216 – 125)$ $= \frac{2}{3}\pi \times 91 = \frac{182}{3}\pi$ cm³ Wait, 2/3 × 91 = 182/3, so answer is $\frac{182}{3}\pi$ → option B.

✅ CORRECT ANSWER: B ($\frac{182}{3} \pi$)

Q17.📊 Statistics

The mean and median of a frequency distribution are 43 and 43.4 respectively. The mode of the distribution is:

(A) 43.4

(B) 42.4

(D) 49.3

🔍 VIEW ANSWER

✅ ANSWER: C (44.2)

📐 EXPLANATION: Using empirical formula: Mode = 3(Median) – 2(Mean) Mode = 3(43.4) – 2(43) = 130.2 – 86 = 44.2

Q18. 🎲 Probability

The probability for a randomly selected number out of 1, 2, 3, 4, …, 25 to be a composite number is:

(A) $\frac{15}{25}$

(B) $\frac{10}{25}$

(D) $\frac{9}{25}$

🔍 VIEW ANSWER

✅ ANSWER: A ($\frac{15}{25}$)

📐 EXPLANATION: Numbers from 1 to 25: total 25 numbers. Composite numbers (not prime and not 1): 4,6,8,9,10,12,14,15,16,18,20,21,22,24,25 → 15 numbers. Probability = $\frac{15}{25}$

Q19. ⚖️ Assertion-Reason

Assertion (A): The surface area of the cuboid formed by joining two cubes of sides 4 cm each, end-to-end, is 160 cm². Reason (R): The surface area of a cuboid of dimensions $l \times b \times h$ is $(lb + bh + hl)$.

Choices: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

🔍 VIEW ANSWER

✅ ANSWER: (d) A is false but R is true.

📐 EXPLANATION: Reason (R) is the correct formula for surface area of cuboid: $2(lb + bh + hl)$ (missing factor 2 in given statement). The given R says $(lb + bh + hl)$ without the factor 2, which is incorrect. So R is false. Assertion (A): Two cubes of side 4 cm joined end-to-end form a cuboid of dimensions 8 cm × 4 cm × 4 cm. Surface area = $2(8×4 + 4×4 + 4×8) $$= 2(32 + 16 + 32) $$= 2×80 = 160$ cm². So A is true. But since R is false, correct option is (c) A true, R false.

✅ CORRECT ANSWER: (c) A is true but R is false.

Q20. ⚖️ Assertion-Reason

Assertion (A): The mean of first ‘n’ natural numbers is $\frac{n – 1}{2}$. Reason (R): The sum of first ‘n’ natural numbers is $\frac{n(n + 1)}{2}$.

🔍 VIEW ANSWER

✅ ANSWER: (d) A is false but R is true.

📐 EXPLANATION: Reason (R) is correct: sum of first n natural numbers = $\frac{n(n+1)}{2}$. Assertion (A): Mean = sum/n = $\frac{n(n+1)}{2n} = \frac{n+1}{2}$, not $\frac{n-1}{2}$. So A is false. Thus, A false, R true → option (d).

📌 SECTION A COMPLETE · 20/20 SOLVED

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