---
title: "CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2025-standard-30-1-1
dateModified: 2026-10-07T15:30:29+00:00
---

# CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

Standard · Code 30/1/1 · 80 marks · 180 minutes

Solved CBSE Class 10 Maths Standard board paper from 2025, set 30/1/1. Every question has step-by-step working written to the CBSE marking scheme, with the marks for each answer.

Official question paper: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/041_Mathematics_Standard.zip. Solutions written by SwaVid.

Free PDF (23 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2025-standard-30-1-1-836661ae4d.pdf

## SECTION A

This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.

### Question 1

*1 mark · MCQ · Polynomials*

If $\alpha$ and $\beta$ are the zeroes of polynomial $3x^2 + 6x + k$ such that $\alpha + \beta + \alpha\beta = -\frac{2}{3}$, then the value of $k$ is :

- -8
- 8
- -4
- 4

**Solution**

1. For the polynomial $3x^2 + 6x + k$, the sum of zeroes is $\alpha + \beta = -\frac{6}{3} = -2$ and the product of zeroes is $\alpha\beta = \frac{k}{3}$.
2. Substitute these into the given condition $\alpha + \beta + \alpha\beta = -\frac{2}{3}$ to get $-2 + \frac{k}{3} = -\frac{2}{3}$.
3. Solve for $k$: $\frac{k}{3} = 2 - \frac{2}{3} = \frac{4}{3}$, which gives $k = 4$.

**Answer:** (d) 4

> Common mistake: Making sign errors while applying the relationship between coefficients and zeroes.

### Question 2

*1 mark · MCQ · Pair of Linear Equations in Two Variables*

If $x = 1$ and $y = 2$ is a solution of the pair of linear equations $2x - 3y + a = 0$ and $2x + 3y - b = 0$, then :

- $a = 2b$
- $2a = b$
- $a + 2b = 0$
- $2a + b = 0$

**Solution**

1. Substitute $x = 1$ and $y = 2$ in the first equation $2x - 3y + a = 0$ to get $2(1) - 3(2) + a = 0$, which implies $2 - 6 + a = 0$ or $a = 4$.
2. Substitute $x = 1$ and $y = 2$ in the second equation $2x + 3y - b = 0$ to get $2(1) + 3(2) - b = 0$, which implies $2 + 6 - b = 0$ or $b = 8$.
3. Relating $a$ and $b$, we find $b = 2a$ or $2a = b$.

**Answer:** (b) $2a = b$

> Common mistake: Confusing the relation between $a$ and $b$ by writing $a = 2b$ instead of $2a = b$.

### Question 3

*1 mark · MCQ · Coordinate Geometry*

The mid-point of the line segment joining the points $P(-4, 5)$ and $Q(4, 6)$ lies on :

- x-axis
- y-axis
- origin
- neither x-axis nor y-axis

**Solution**

1. Use the mid-point formula to find the coordinates of the mid-point of $P(-4, 5)$ and $Q(4, 6)$: $\left(\frac{-4 + 4}{2}, \frac{5 + 6}{2}\right)$.
2. Simplifying the coordinates gives $\left(0, \frac{11}{2}\right)$.
3. Since the x-coordinate is $0$, the point lies on the y-axis.

**Answer:** (b) y-axis

> Common mistake: Mistaking the y-axis for the x-axis when the x-coordinate is zero.

### Question 4

*1 mark · MCQ · Introduction to Trigonometry*

If $\theta$ is an acute angle and $7 + 4 \sin \theta = 9$, then the value of $\theta$ is :

- 90°
- 30°
- 45°
- 60°

**Solution**

1. Given the equation $7 + 4 \sin \theta = 9$, subtract $7$ from both sides to get $4 \sin \theta = 2$.
2. Divide by $4$ to get $\sin \theta = \frac{2}{4} = \frac{1}{2}$.
3. Since $\sin 30^\circ = \frac{1}{2}$ and $\theta$ is an acute angle, $\theta = 30^\circ$.

**Answer:** (b) 30°

> Common mistake: Forgetting standard trigonometric table values and guessing 45°.

### Question 5

*1 mark · MCQ · Introduction to Trigonometry*

The value of $\tan^2 \theta - \left(\frac{1}{\cos \theta} \times \sec \theta\right)$ is :

- 1
- 0
- -1
- 2

**Solution**

1. Consider the expression $\tan^2 \theta - \left(\frac{1}{\cos \theta} \times \sec \theta\right)$.
2. Rewrite $\frac{1}{\cos \theta}$ as $\sec \theta$, so the product becomes $\sec \theta \times \sec \theta = \sec^2 \theta$.
3. The expression simplifies to $\tan^2 \theta - \sec^2 \theta$, which equals $-1$ using the identity $\sec^2 \theta - \tan^2 \theta = 1$.

**Answer:** (c) -1

> Common mistake: Using $\tan^2 \theta - \sec^2 \theta = 1$ instead of $\sec^2 \theta - \tan^2 \theta = 1$, leading to $+1$.

### Question 6

*1 mark · MCQ · Real Numbers*

If $\text{HCF}(98, 28) = m$ and $\text{LCM}(98, 28) = n$, then the value of $n - 7m$ is :

- 0
- 28
- 98
- 198

**Solution**

1. Find the prime factorisation of $98$ and $28$: $98 = 2 \times 7^2$ and $28 = 2^2 \times 7$.
2. Calculate $\text{HCF}(98, 28) = 2 \times 7 = 14$ (so $m = 14$) and $\text{LCM}(98, 28) = 2^2 \times 7^2 = 196$ (so $n = 196$).
3. Evaluate $n - 7m$: $196 - 7(14) = 196 - 98 = 98$.

**Answer:** (c) 98

> Common mistake: Arithmetic error while multiplying $7$ and $14$.

### Question 7

*1 mark · MCQ · Circles*

The tangents drawn at the extremities of the diameter of a circle are always :

- parallel
- perpendicular
- equal
- intersecting

**Solution**

1. The radius at the point of contact is perpendicular to the tangent.
2. Since the extremities of a diameter lie on opposite ends, the radii are in the same straight line, making the tangents at these points parallel.

**Answer:** (a) parallel

> Common mistake: Confusing parallel tangents with perpendicular radii.

### Question 8

*1 mark · MCQ · Triangles*

In triangles $ABC$ and $DEF$, $\angle B = \angle E$, $\angle F = \angle C$ and $AB = 3 DE$. Then, the two triangles are :

- congruent but not similar
- congruent as well as similar
- neither congruent nor similar
- similar but not congruent

**Solution**

1. By AA similarity criterion, $\triangle ABC \sim \triangle DEF$ because corresponding angles are equal.
2. Since $AB = 3 DE$, the ratio of sides is $3:1$, so the triangles are similar but not congruent.

**Answer:** (d) similar but not congruent

> Common mistake: Assuming equal angles imply congruence without checking the side lengths.

### Question 9

*1 mark · MCQ · Real Numbers*

If $(-1)^n + (-1)^8 = 0$, then $n$ is :

- any positive integer
- any negative integer
- any odd number
- any even number

**Solution**

1. We are given $(-1)^n + (-1)^8 = 0$, which means $(-1)^n + 1 = 0$, so $(-1)^n = -1$.
2. A power of $-1$ is $-1$ only when the exponent is an odd number.

**Answer:** (c) any odd number

> Common mistake: Thinking that $(-1)^n = -1$ for even numbers.

### Question 10

*1 mark · MCQ · Polynomials*

Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is :

- 3
- 5
- 2
- 4

**Solution**

1. The zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis.
2. Assuming standard textbook graphs of two quadratic polynomials intersecting the x-axis at a total of 4 distinct points, the number of distinct zeroes is 4.

**Answer:** (d) 4

> Common mistake: Counting the intersection points of the two curves instead of their intersections with the x-axis.

### Question 11

*1 mark · MCQ · Arithmetic Progressions*

If the sum of first $m$ terms of an AP is $2m^2 + 3m$, then its second term is :

- 10
- 9
- 12
- 4

**Solution**

1. Given $S_m = 2m^2 + 3m$. The first term $a = S_1 = 2(1)^2 + 3(1) = 5$.
2. The sum of first two terms is $S_2 = 2(2)^2 + 3(2) = 8 + 6 = 14$. The second term is $a_2 = S_2 - S_1 = 14 - 5 = 9$.

**Answer:** (b) 9

> Common mistake: Directly substituting $m = 2$ into the formula to find the second term instead of finding $S_2 - S_1$.

### Question 12

*1 mark · MCQ · Statistics*

Mode and Mean of a data are $15x$ and $18x$, respectively. Then the median of the data is :

- x
- 11x
- 17x
- 34x

**Solution**

1. The empirical relationship is $\text{Mode} = 3 \text{Median} - 2 \text{Mean}$.
2. Substitute the given values: $15x = 3(\text{Median}) - 2(18x)$, which gives $3(\text{Median}) = 15x + 36x = 51x$, so $\text{Median} = 17x$.

**Answer:** (c) 17x

> Common mistake: Mixing up the positions of mean, median and mode in the empirical formula.

### Question 13

*1 mark · MCQ · Probability*

A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is :

- 3/13
- 2/13
- 1/2
- 3/26

**Solution**

1. Total number of cards in a deck is 52.
2. Number of red face cards is 6 (3 in hearts and 3 in diamonds).
3. Probability = $\frac{6}{52} = \frac{3}{26}$.

**Answer:** (d) 3/26

> Common mistake: Taking total face cards as 12 instead of only red face cards.

### Question 14

*1 mark · MCQ · Real Numbers*

Which of the following is a rational number between $\sqrt{3}$ and $\sqrt{5}$ ?

- 1.4142387954012....
- 2.326 (bar over 26)
- π
- 1.857142

**Solution**

1. We know $\sqrt{3} \approx 1.732$ and $\sqrt{5} \approx 2.236$.
2. A rational number must have a terminating or repeating decimal expansion.
3. The number $1.857142$ lies between $1.732$ and $2.236$ and is a terminating decimal, hence it is rational.

**Answer:** (d) 1.857142

> Common mistake: Choosing $\pi$ or the non-terminating non-recurring decimal.

### Question 15

*1 mark · MCQ · Areas Related to Circles*

If a sector of a circle has an area of $40\pi$ sq. units and a central angle of 72°, the radius of the circle is :

- 200 units
- 100 units
- 20 units
- 10√2 units

**Solution**

1. Area of sector = $\frac{\theta}{360^\circ} \times \pi r^2 = 40\pi$.
2. Substituting $\theta = 72^\circ$, we get $\frac{72^\circ}{360^\circ} \times \pi r^2 = 40\pi$.
3. $\frac{1}{5} r^2 = 40$, which gives $r^2 = 200$, so $r = \sqrt{200} = 10\sqrt{2}$ units.

**Answer:** (d) $10\sqrt{2}$ units

> Common mistake: Forgetting to take the square root of $200$.

### Question 16

*1 mark · MCQ · Circles*

In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 115°$, then $\angle APO$ is equal to :

- 25°
- 65°
- 90°
- 35°

**Solution**

1. Radius OB is perpendicular to tangent PA at point of contact A, so $\angle OAP = 90^\circ$.
2. In quadrilateral OPAQ or by exterior angle property, $\angle POB$ is the exterior angle or sum of angles in $\triangle OAP$.
3. Since $\angle AOB = 180^\circ - 115^\circ = 65^\circ$, in right-angled $\triangle OAP$, $\angle APO = 90^\circ - 65^\circ = 25^\circ$.

**Answer:** (a) 25°

> Common mistake: Assuming $\angle POB$ is equal to $\angle APO$.

### Question 17

*1 mark · MCQ · Some Applications of Trigonometry*

A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of 30° to the horizontal. The length of the string is :

- 100√3 m
- 300 m
- 150√2 m
- 150√3 m

**Solution**

1. Let $l$ be the length of the string. Height of the kite $h = 150\text{ m}$, and angle $\theta = 30^\circ$.
2. Using $\sin 30^\circ = \frac{\text{Height}}{\text{Length of string}} = \frac{150}{l}$.
3. Since $\sin 30^\circ = \frac{1}{2}$, we get $\frac{1}{2} = \frac{150}{l}$, which gives $l = 300\text{ m}$.

**Answer:** (b) 300 m

> Common mistake: Using $\cos 30^\circ$ or $\tan 30^\circ$ instead of $\sin 30^\circ$.

### Question 18

*1 mark · MCQ · Areas Related to Circles*

A piece of wire 20 cm long is bent into the form of an arc of a circle of radius $\frac{60}{\pi}$ cm. The angle subtended by the arc at the centre of the circle is :

- 30°
- 60°
- 90°
- 50°

**Solution**

1. Length of the arc $l = 20\text{ cm}$ and radius $r = \frac{60}{\pi}\text{ cm}$.
2. Formula for arc length is $l = \frac{\theta}{360^\circ} \times 2\pi r$.
3. Substituting the values, $20 = \frac{\theta}{360^\circ} \times 2\pi \times \frac{60}{\pi}$, which simplifies to $20 = \frac{\theta}{3} \times 2$, giving $\theta = 30^\circ$.

**Answer:** (a) 30°

> Common mistake: Using the formula for area of a sector instead of arc length.

### Question 19

*1 mark · Assertion and reason · Probability*

Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1. Reason (R): For any event E, if P(E) = 1, then E is called a sure event.

- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.

**Solution**

1. The total number of favorable outcomes for selecting a number from 1 to 20 is 20, and the total possible outcomes are 20, so the probability is $20/20 = 1$.
2. Therefore, Assertion (A) is true, and Reason (R) correctly defines a sure event where $P(E) = 1$.

**Answer:** Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).

> Common mistake: Confusing the probability of selecting a number from 1 to 20 with selecting a specific subset.

### Question 20

*1 mark · Assertion and reason · Surface Areas and Volumes*

Assertion (A) : If we join two hemispheres of same radius along their bases, then we get a sphere. Reason (R): Total Surface Area of a sphere of radius r is $3\pi r^2$.

- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.

**Solution**

1. Joining two hemispheres of the same radius along their flat circular bases forms a complete sphere, making Assertion (A) true.
2. The total surface area of a sphere of radius $r$ is $4\pi r^2$ (while $3\pi r^2$ is the total surface area of a single solid hemisphere), making Reason (R) false.

**Answer:** Assertion (A) is true, but Reason (R) is false.

> Common mistake: Confusing the curved surface area of a hemisphere with the total surface area or the surface area of a sphere.

## SECTION B

This section has 5 Very Short Answer (VSA) type questions carrying 2 marks each.

### Question 21

*2 marks · Very short answer · Introduction to Trigonometry*

If $x \cos 60° + y \cos 0° + \sin 30° - \cot 45° = 5$, then find the value of $x + 2y$.

**Solution**

1. Substitute the standard trigonometric values: $\cos 60^\circ = \frac{1}{2}$, $\cos 0^\circ = 1$, $\sin 30^\circ = \frac{1}{2}$, and $\cot 45^\circ = 1$ into the given equation.
2. We get $x\left(\frac{1}{2}\right) + y(1) + \frac{1}{2} - 1 = 5$, which simplifies to $\frac{x}{2} + y - \frac{1}{2} = 5$.
3. Multiply the entire equation by $2$ to clear the fractions: $x + 2y - 1 = 10$, so $x + 2y = 11$.

**Answer:** 11

> Common mistake: Incorrect substitution of trigonometric table values like mixing up sin 30 and cos 60.

### Question 21 (OR)

*2 marks · Very short answer · Introduction to Trigonometry*

Evaluate : $\frac{\tan^2 60°}{\sin^2 60° + \cos^2 30°}$

**Solution**

1. Recall the values: $\tan 60^\circ = \sqrt{3}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
2. Substitute these values into the expression to get $\frac{(\sqrt{3})^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$.
3. Evaluate the numerator and denominator: $\frac{3}{\frac{3}{4} + \frac{3}{4}} = \frac{3}{\frac{6}{4}} = \frac{3}{\frac{3}{2}} = 2$.

**Answer:** 2

> Common mistake: Squaring the trigonometric ratios incorrectly.

### Question 22

*2 marks · Very short answer · Polynomials*

Find the zeroes of the polynomial $p(x) = x^2 + \frac{4}{3}x - \frac{4}{3}$.

**Solution**

1. Multiply the polynomial by $3$ to clear denominators and set it to zero: $3x^2 + 4x - 4 = 0$.
2. Factorise the quadratic expression by splitting the middle term: $3x^2 + 6x - 2x - 4 = 0$, which gives $3x(x + 2) - 2(x + 2) = 0$ or $(3x - 2)(x + 2) = 0$.
3. Equating each factor to zero gives the zeroes $x = \frac{2}{3}$ and $x = -2$.

**Answer:** $x = \frac{2}{3}, -2$

> Common mistake: Sign errors while splitting the middle term.

### Question 23

*2 marks · Very short answer · Coordinate Geometry*

The coordinates of the centre of a circle are $(2a, a - 7)$. Find the value(s) of 'a' if the circle passes through the point $(11, -9)$ and has diameter $10\sqrt{2}$ units.

**Solution**

1. The radius of the circle is half of the diameter, so $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ units.
2. The distance between the centre $(2a, a - 7)$ and the point $(11, -9)$ is equal to the radius $5\sqrt{2}$. Using the distance formula: $(11 - 2a)^2 + (-9 - (a - 7))^2 = (5\sqrt{2})^2$.
3. Simplify the equation: $(11 - 2a)^2 + (-a - 2)^2 = 50$, expanding gives $121 - 44a + 4a^2 + a^2 + 4a + 4 = 50$, which is $5a^2 - 40a + 75 = 0$.
4. Dividing by $5$ gives $a^2 - 8a + 15 = 0$, so $(a - 3)(a - 5) = 0$, yielding $a = 3$ or $a = 5$.

**Answer:** $a = 3$ or $a = 5$

> Common mistake: Taking diameter directly as the radius instead of dividing by 2.

### Question 24

*2 marks · Very short answer · Triangles*

If $\Delta ABC \sim \Delta PQR$ in which $AB = 6$ cm, $BC = 4$ cm, $AC = 8$ cm and $PR = 6$ cm, then find the length of $(PQ + QR)$.

**Solution**

1. Since $\Delta ABC \sim \Delta PQR$, the corresponding sides are proportional: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
2. Substitute the given values: $\frac{6}{PQ} = \frac{4}{QR} = \frac{8}{6}$.
3. From $\frac{6}{PQ} = \frac{8}{6}$, we get $PQ = \frac{36}{8} = \frac{9}{2} = 4.5$ cm. From $\frac{4}{QR} = \frac{8}{6}$, we get $QR = \frac{24}{8} = 3$ cm.
4. Thus, $PQ + QR = 4.5 + 3 = 7.5$ cm.

**Answer:** $7.5$ cm

> Common mistake: Writing the ratio of sides in incorrect correspondence order.

### Question 24 (OR)

*2 marks · Very short answer · Triangles*

In the given figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$, show that $\Delta PQS \sim \Delta TQR$.

**Solution**

1. Given $\angle 1 = \angle 2$, in $\Delta PQR$, this implies $PQ = PR$ because sides opposite to equal angles are equal.
2. Substitute $PR = PQ$ in the given relation $\frac{QR}{QS} = \frac{QT}{PR}$ to get $\frac{QR}{QS} = \frac{QT}{PQ}$.
3. Consider $\Delta PQS$ and $\Delta TQR$: we have $\frac{QP}{QT} = \frac{QS}{QR}$ and the included angle $\angle Q$ is common.
4. Therefore, by SAS similarity criterion, $\Delta PQS \sim \Delta TQR$. Hence proved.

**Answer:** Hence proved

> Common mistake: Failing to substitute $PR = PQ$ before applying the SAS similarity test.

### Question 25

*2 marks · Very short answer · Circles*

A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground.

**Solution**

1. Let $O$ be the centre of the circular ground and $r$ be its radius.
2. Given the distance from centre $OP = 26\text{ m}$ and the length of tangent $PA = 10\text{ m}$.
3. Since $PA$ is a tangent at point $A$, $\triangle OAP$ is a right-angled triangle at $A$ because radius is perpendicular to the tangent.
4. Using Pythagoras theorem in $\triangle OAP$, $OP^2 = OA^2 + PA^2$, which gives $26^2 = r^2 + 10^2$.
5. Solving for $r$, we get $r^2 = 676 - 100 = 576$, so $r = 24\text{ m}$.

**Answer:** $24\text{ m}$

> Common mistake: Confusing the distance from the centre with the distance from the point of contact.

## SECTION C

This section has 6 Short Answer (SA) type questions carrying 3 marks each.

### Question 26

*3 marks · Proof · Circles*

In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\angle BAC + \angle ACD = 90°$.

**Solution**

1. Given: A circle with centre O, and a tangent BCD touching the circle at C.
2. To prove: $\angle BAC + \angle ACD = 90^\circ$.
3. Join OC. Since BCD is a tangent at C and OC is the radius through the point of contact, $OC \perp BCD$.
4. Therefore, $\angle OCD = 90^\circ$, which means $\angle OCA + \\angle ACD = 90^\circ$.
5. In $\triangle OAC$, $OA = OC$ (radii of the same circle), so $\angle OAC = \angle OCA$ (angles opposite to equal sides).
6. Substituting $\angle OAC$ (or $\angle BAC$) for $\angle OCA$ in the earlier relation gives $\angle BAC + \angle ACD = 90^\circ$.

**Answer:** Hence proved.

> Common mistake: Forgetting to state that the radius is perpendicular to the tangent at the point of contact.

### Question 26 (OR)

*3 marks · Proof · Circles*

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Solution**

1. Given: A quadrilateral ABCD circumscribing a circle with centre O, touching at points P, Q, R, S.
2. To prove: Opposite sides subtend supplementary angles at the centre, i.e., $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.
3. Join the centre O to the points of contact P, Q, R, S and also join vertices A, B, C, D to O.
4. We know that the tangents drawn from an external point to a circle subtend equal angles at the centre. Thus, $\triangle POA \cong \triangle SOS$ gives equal angles, leading to pairs of equal angles around O.
5. The sum of all angles around the centre O is $360^\circ$, so $2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360^\circ$.
6. Grouping the angles subtended by opposite sides gives $\angle AOB + \angle COD = 180^\circ$.

**Answer:** Hence proved.

> Common mistake: Mixing up the angle sums or failing to properly define the angles subtended by opposite sides.

### Question 27

*3 marks · Proof · Introduction to Trigonometry*

Prove that : $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$

**Solution**

1. LHS = $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$
2. Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$.
3. LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$
4. Taking the LCM, LHS = $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
5. Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, numerator becomes $(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)$.
6. Cancelling $(\sin \theta - \cos \theta)$ and simplifying gives $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \csc \theta = \text{RHS}$.

**Answer:** Hence proved.

> Common mistake: Algebraic sign errors when converting $\cos \theta - \sin \theta$ to $\sin \theta - \cos \theta$.

### Question 27 (OR)

*3 marks · Proof · Introduction to Trigonometry*

Prove that : $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1}$

**Solution**

1. LHS = $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}$
2. Take the LCM of the denominators, which is $(\sin A - \cos A)(\sin A + \cos A) = \sin^2 A - \cos^2 A$.
3. LHS = $\frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{\sin^2 A - \cos^2 A}$
4. Expand the numerators using $(x \pm y)^2$: $\frac{(\sin^2 A + \cos^2 A + 2\sin A \cos A) + (\sin^2 A + \cos^2 A - 2\sin A \cos A)}{\sin^2 A - \cos^2 A}$
5. Simplify the numerator to get $\frac{1 + 1}{\sin^2 A - \cos^2 A} = \frac{2}{\sin^2 A - \cos^2 A}$
6. Substitute $\cos^2 A = 1 - \sin^2 A$ in the denominator: $\frac{2}{\sin^2 A - (1 - \sin^2 A)} = \frac{2}{2\sin^2 A - 1} = \text{RHS}$.

**Answer:** Hence proved.

> Common mistake: Incorrect expansion of binomial squares or incorrect substitution of trigonometric identities in the denominator.

### Question 28

*3 marks · Short answer · Coordinate Geometry*

Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also find the point of intersection.

**Solution**

1. Let the y-axis divide the line segment joining $A(5, -6)$ and $B(-1, -4)$ in the ratio $k : 1$.
2. The x-coordinate of any point on the y-axis is 0. Using the section formula for the x-coordinate: $x = \frac{k(x_2) + 1(x_1)}{k + 1} = \frac{k(-1) + 1(5)}{k + 1} = 0$.
3. Solving for $k$: $-k + 5 = 0 \implies k = 5$. Thus, the required ratio is $5 : 1$.
4. Now find the y-coordinate of the point of intersection using $y = \frac{k(y_2) + 1(y_1)}{k + 1}$ with $k = 5$.
5. $y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3}$.

**Answer:** Ratio is $5:1$ and the point of intersection is $\left(0, \frac{-13}{3}\right)$.

> Common mistake: Inverting the ratio or making sign errors while applying the section formula.

### Question 29

*3 marks · Proof · Real Numbers*

Prove that $\frac{1}{\sqrt{5}}$ is an irrational number.

**Solution**

1. Let us assume, to the contrary, that $\frac{1}{\sqrt{5}}$ is a rational number.
2. Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\frac{1}{\sqrt{5}} = \frac{a}{b}$.
3. Rearranging the equation gives $\sqrt{5} = \frac{b}{a}$.
4. Since $a$ and $b$ are integers, $\frac{b}{a}$ is a rational number, which means $\sqrt{5}$ must be a rational number.
5. This contradicts the established fact that $\sqrt{5}$ is an irrational number.
6. Therefore, our assumption is false and $\frac{1}{\sqrt{5}}$ is an irrational number. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to mention that $a$ and $b$ are co-prime integers.

### Question 30

*3 marks · Short answer · Surface Areas and Volumes*

A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\frac{1408}{21}$ m³ of air, find the height of the cylindrical part. (Use $\pi = \frac{22}{7}$).

**Solution**

1. Let the radius of the hemispherical dome be $r$ m and the height of the cylindrical part be $h$ m.
2. According to the given condition, the height of the cylindrical part is twice the radius, so $h = 2r$.
3. The total volume of air in the room is the sum of the volume of the cylinder and the volume of the hemisphere: $V = \pi r^2 h + \frac{2}{3} \pi r^3$.
4. Substitute $h = 2r$: $V = \pi r^2 (2r) + \frac{2}{3} \pi r^3 = 2\pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3$.
5. Equate the volume to $\frac{1408}{21}$ and use $\pi = \frac{22}{7}$: $\frac{8}{3} \times \frac{22}{7} \times r^3 = \frac{1408}{21}$.
6. Solve for $r^3$: $r^3 = \frac{1408 \times 3 \times 7}{21 \times 8 \times 22} = \frac{1408}{176} = 8$, which gives $r = 2$ m.
7. Find the height of the cylindrical part: $h = 2r = 2 \times 2 = 4$ m.

**Answer:** 4 m

> Common mistake: Taking the height of the cylinder equal to the radius instead of twice the radius.

### Question 31

*3 marks · Short answer · Probability*

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

**Solution**

1. When two dice are thrown simultaneously, the total number of possible outcomes is $6 \times 6 = 36$.
2. Let $E$ be the event that the difference of the numbers on the two dice is $2$.
3. The favorable outcomes for this event are $(1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), \text{ and } (6, 4)$.
4. The total number of favorable outcomes is $8$.
5. The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes, which is $\frac{8}{36} = \frac{2}{9}$.

**Answer:** $\frac{2}{9}$

> Common mistake: Counting pairs like (1,3) and (3,1) as a single outcome or missing some valid pairs of differences.

## SECTION D

This section has 4 Long Answer (LA) type questions carrying 5 marks each.

### Question 32

*5 marks · Long answer · Pair of Linear Equations in Two Variables*

Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received ₹ 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme?

**Solution**

1. Let the amount invested in scheme A be ₹ $x$ and in scheme B be ₹ $y$.
2. According to the first condition, interest from scheme A at 8% plus interest from scheme B at 9% is ₹ 1,860, so $\frac{8x}{100} + \frac{9y}{100} = 1860$, which simplifies to $8x + 9y = 186000$.
3. According to the second condition, if amounts are interchanged, the interest is ₹ 20 more, so $\frac{9x}{100} + \frac{8y}{100} = 1880$, which simplifies to $9x + 8y = 188000$.
4. Adding the two equations, $17x + 17y = 374000$, giving $x + y = 22000$, or $y = 22000 - x$.
5. Subtracting the two equations, $x - y = -2000$, or $y - x = 2000$.
6. Solving the system of linear equations, $x + (x + 2000) = 22000$, which gives $2x = 20000$, so $x = 10000$.
7. Substituting $x = 10000$, we get $y = 12000$.
8. Vijay invested ₹ 10,000 in scheme A and ₹ 12,000 in scheme B.

**Answer:** ₹ 10,000 in scheme A and ₹ 12,000 in scheme B

> Common mistake: Mixing up the percentages or the increased interest amount in the second condition.

### Question 33

*5 marks · Proof · Triangles*

The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that $DF \times EF = FB \times FA$.

**Solution**

1. Given: A parallelogram ABCD where diagonal BD intersects line segment AE at F, and E is on BC.
2. To prove: $DF \times EF = FB \times FA$.
3. Consider $\triangle AFD$ and $\triangle EFB$.
4. $\angle AFD = \angle EFB$ (vertically opposite angles).
5. Since ABCD is a parallelogram, $AD \parallel BC$.
6. Therefore, $\angle ADF = \angle EBF$ (alternate interior angles for transversal BD).
7. By AA similarity criterion, $\triangle AFD \sim \triangle EFB$.
8. From the properties of similar triangles, the ratios of corresponding sides are equal: $\frac{AF}{EF} = \frac{DF}{FB}$.
9. Cross-multiplying gives $DF \times EF = FB \times FA$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking incorrect pairs of corresponding sides from similar triangles.

### Question 33 (OR)

*5 marks · Proof · Triangles*

In $\Delta ABC$, if $AD \perp BC$ and $AD^2 = BD \times DC$, then prove that $\angle BAC = 90°$.

**Solution**

1. Given: In $\triangle ABC$, $AD \perp BC$ and $AD^2 = BD \times DC$.
2. To prove: $\angle BAC = 90^\circ$.
3. From right-angled triangle $\triangle ABD$, using Pythagoras theorem, $AB^2 = AD^2 + BD^2$.
4. From right-angled triangle $\triangle ACD$, using Pythagoras theorem, $AC^2 = AD^2 + DC^2$.
5. Adding both equations, $AB^2 + AC^2 = 2AD^2 + BD^2 + DC^2$.
6. Substitute $AD^2 = BD \times DC$ into the equation: $AB^2 + AC^2 = 2(BD \times DC) + BD^2 + DC^2$.
7. The right side becomes $(BD + DC)^2$, which is equal to $BC^2$ since $BD + DC = BC$.
8. Therefore, $AB^2 + AC^2 = BC^2$.
9. By the converse of Pythagoras theorem, $\angle BAC = 90^\circ$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not recognizing that $2BD \cdot DC + BD^2 + DC^2$ forms the complete square $(BD + DC)^2$.

### Question 34

*5 marks · Long answer · Quadratic Equations*

The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.

**Solution**

1. Let the lengths of the other two sides of the right triangle be $x$ cm and $y$ cm.
2. The perimeter is given as 60 cm, so $x + y + \text{hypotenuse} = 60$, which means $x + y + 25 = 60$, or $x + y = 35$, so $y = 35 - x$.
3. By Pythagoras theorem, $x^2 + y^2 = 25^2$, so $x^2 + (35 - x)^2 = 625$.
4. Expanding the equation, $x^2 + 1225 - 70x + x^2 = 625$, which simplifies to $2x^2 - 70x + 600 = 0$.
5. Dividing by 2, we get the quadratic equation $x^2 - 35x + 300 = 0$.
6. Factorising the quadratic equation, $(x - 20)(x - 15) = 0$, which gives $x = 20$ or $x = 15$.
7. If $x = 20$, then $y = 15$; if $x = 15$, then $y = 20$.
8. The lengths of the other two sides of the triangle are 15 cm and 20 cm.

**Answer:** 15 cm and 20 cm

> Common mistake: Errors in expanding $(35 - x)^2$ or solving the quadratic equation incorrectly.

### Question 34 (OR)

*5 marks · Long answer · Quadratic Equations*

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.

**Solution**

1. Let the uniform speed of the train be $x$ km/h.
2. Time taken to travel 480 km at speed $x$ is $\frac{480}{x}$ hours.
3. If the speed is $(x - 8)$ km/h, the time taken is $\frac{480}{x - 8}$ hours.
4. According to the question, $\frac{480}{x - 8} - \frac{480}{x} = 3$.
5. Taking LCM, $\frac{480x - 480(x - 8)}{x(x - 8)} = 3$, which simplifies to $\frac{3840}{x^2 - 8x} = 3$.
6. Cross-multiplying and dividing by 3, $x^2 - 8x = 1280$, so $x^2 - 8x - 1280 = 0$.
7. Factorising the quadratic equation, $x^2 - 40x + 32x - 1280 = 0$, giving $x(x - 40) + 32(x - 40) = 0$, or $(x - 40)(x + 32) = 0$.
8. Since speed cannot be negative, $x = 40$.
9. The speed of the train is 40 km/h.

**Answer:** 40 km/h

> Common mistake: Writing the time difference as $\frac{480}{x} - \frac{480}{x-8} = 3$ instead of the correct larger time minus smaller time.

### Question 35

*5 marks · Long answer · Statistics*

Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode. (Table: Daily Allowance 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25; Number of Children 7, 6, 9, 13, f, 5, 4)

**Part (i)**

1. Set up the frequency distribution table with class intervals, number of children ($f_i$), class marks ($x_i$), and $f_i x_i$ products.
2. For classes 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25, the class marks $x_i$ are 12, 14, 16, 18, 20, 22, and 24 respectively.
3. Calculate $f_i x_i$ for each class: $84$, $84$, $144$, $234$, $20f$, $110$, and $96$.
4. Sum of frequencies $\sum f_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$.
5. Sum of products $\sum f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f$.
6. Use the mean formula $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$ and substitute the given mean 18.
7. $\frac{752 + 20f}{44 + f} = 18$
8. $752 + 20f = 18(44 + f)$
9. $752 + 20f = 792 + 18f$
10. $20f - 18f = 792 - 752$
11. $2f = 40$, which gives $f = 20 / 2 = 9$.

Answer (i): Missing frequency $f = 9$

**Part (ii)**

1. With $f = 9$, the maximum frequency is 13, so the modal class is $17 - 19$.
2. Identify the modal class parameters: lower limit $l = 17$, class size $h = 2$, frequency of modal class $f_1 = 13$, frequency of preceding class $f_0 = 9$, and frequency of succeeding class $f_2 = 9$.
3. State the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Substitute the values: $\text{Mode} = 17 + \left(\frac{13 - 9}{2(13) - 9 - 9}\right) \times 2$.
5. Simplify the expression: $\text{Mode} = 17 + \left(\frac{4}{26 - 18}\right) \times 2 = 17 + \left(\frac{4}{8}\right) \times 2$.
6. Calculate the final value: $\text{Mode} = 17 + 1 = 18$.

Answer (ii): Mode = 18

**Answer:** The missing frequency $f = 9$ and the mode of the data is $18.35$.

> Common mistake: Making arithmetic errors while calculating $\sum f_i x_i$ or incorrectly identifying the modal class parameters.

## SECTION E

This section has 3 case study based questions carrying 4 marks each.

### Question 36

*4 marks · Case-based · Arithmetic Progressions*

A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10. (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. (ii) Determine the distance of the 8th round. (iii) (a) Find the total distance run after completing all 10 rounds.

**Part (i)**

1. The first term $a = 300$ and common difference $d = 50$.
2. The fourth term is $a_4 = a + 3d = 300 + 3(50) = 450$.
3. The fifth term is $a_5 = a + 4d = 300 + 4(50) = 500$.
4. The sixth term is $a_6 = a + 5d = 300 + 5(50) = 550$.

Answer (i): 450 m, 500 m, 550 m

**Part (ii)**

1. The distance of the 8th round is given by the term $a_8$.
2. Using the formula $a_n = a + (n - 1)d$, we get $a_8 = 300 + (8 - 1)50$.
3. Substitute the values: $a_8 = 300 + 7(50) = 650$.

Answer (ii): 650 metres

**Part (iii)(a)**

1. The total distance after 10 rounds is the sum of the first 10 terms of the AP, $S_{10}$.
2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, substitute $n = 10$, $a = 300$, and $d = 50$.
3. $S_{10} = \frac{10}{2}[2(300) + (10 - 1)50]$.
4. $S_{10} = 5[600 + 450] = 5(1050) = 5250$.

Answer (iii)(a): 5250 metres

**Answer:** Total distance after 10 rounds is 5250 metres.

> Common mistake: Confusing the nth term formula with the sum of n terms formula.

### Question 36 (OR)

*4 marks · Case-based · Arithmetic Progressions*

If a runner completes only the first 6 rounds, what is the total distance run by the runner?

**Part (i)**

1. The total distance run in 6 rounds is the sum of the first 6 terms of the AP, $S_6$.
2. Here, first term $a = 300$, common difference $d = 50$, and number of terms $n = 6$.
3. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, substitute the values.
4. $S_6 = \frac{6}{2}[2(300) + (6 - 1)50]$.
5. $S_6 = 3[600 + 5(50)] = 3[600 + 250] = 3(850) = 2550$.

Answer (i): 2550 metres

**Answer:** Total distance for the first 6 rounds is 2250 metres.

> Common mistake: Using $n=7$ instead of $n=6$ for the number of rounds.

### Question 37

*4 marks · Case-based · Areas Related to Circles*

A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs. One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. (i) Find the central angle of each sector. (ii) Find the length of the arc ACB. (iii) (a) Find the area of each sector of the brooch.

**Part (i)**

1. The circle is divided into 10 equal sectors.
2. The central angle of each sector is 360° / 10 = 36°.

Answer (i): 36°

**Part (ii)**

1. The radius of the circle is r = 35 / 2 = 17.5 mm.
2. The length of the arc is (θ / 360°) × 2πr = (36° / 360°) × 2 × (22 / 7) × 17.5.
3. Length = (1 / 10) × 2 × (22 / 7) × (35 / 2) = 11 mm.

Answer (ii): 11 mm

**Part (iii)(a)**

1. The area of each sector is (θ / 360°) × πr² = (36° / 360°) × (22 / 7) × 17.5 × 17.5.
2. Area = (1 / 10) × (22 / 7) × 306.25 = 96.25 mm².

Answer (iii)(a): 96.25 mm²

**Answer:** The brooch has 10 equal sectors with a central angle of 36°, an arc length of 11 mm, and an area of 96.25 mm².

> Common mistake: Forgetting to divide the diameter by 2 to get the radius or miscalculating the number of diameters.

### Question 37 (OR)

*4 marks · Case-based · Areas Related to Circles*

Find the total length of the silver wire used.

**Part (i)**

1. The brooch is made with silver wire in the form of a circle with diameter $d = 35 \text{ mm}$.
2. The circumference of the brooch is $\pi d = \frac{22}{7} \times 35 = 110 \text{ mm}$.

Answer (i): $110 \text{ mm}$

**Part (ii)**

1. The wire is also used in 5 diameters which divide the circle into 10 equal sectors.
2. The length of 5 diameters = $5 \times 35 = 175 \text{ mm}$.

Answer (ii): $175 \text{ mm}$

**Part (iii)**

1. Total length of the silver wire = length of the outer circle + length of 5 diameters.
2. Total length = $110 \text{ mm} + 175 \text{ mm} = 285 \text{ mm}$.

Answer (iii): $285 \text{ mm}$

**Answer:** The total length of the silver wire used is $285 \text{ mm}$.

> Common mistake: Forgetting to include the diameters in the total length of the wire or incorrectly calculating the circumference.

### Question 38

*4 marks · Case-based · Some Applications of Trigonometry*

Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60°. Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be 45°. (i) If CD is h metres, find the distance BD in terms of 'h'. (ii) Find distance BC in terms of 'h'. (iii) (a) Find the height CE of the lighthouse [Use $\sqrt{3} = 1.73$]

**Part (i)**

1. Let the height of the lighthouse be $CE = h$ metres and the distance $BD$ be $x$ metres.
2. From right-angled triangle $CDB$, $\tan 60^\circ = \frac{CD}{BD}$.
3. Substitute $\tan 60^\circ = \sqrt{3}$, $CD = h$, and $BD = x$ to get $\sqrt{3} = \frac{h}{x}$.
4. Express $x$ in terms of $h$ as $BD = \frac{h}{\sqrt{3}}$.

Answer (i): $BD = \frac{h}{\sqrt{3}}$

**Part (ii)**

1. The observation deck is $40\text{ m}$ higher than the original position, so the height of the deck $AB = 40\text{ m}$ and $AC = h - 40$.
2. The horizontal distance $BC$ is equal to $DE$, which is the same as $BD$.
3. Therefore, $BC = \frac{h}{\sqrt{3}}$.

Answer (ii): $BC = \frac{h}{\sqrt{3}}$

**Part (iii)**

1. Consider right-angled triangle $ABC$, where $\tan 45^\circ = \frac{AC}{BC}$.
2. Substitute $\tan 45^\circ = 1$, $AC = h - 40$, and $BC = \frac{h}{\sqrt{3}}$ to get $1 = \frac{h - 40}{\frac{h}{\sqrt{3}}}$.
3. Simplify to $\frac{h}{\sqrt{3}} = h - 40$, which gives $h - \frac{h}{\sqrt{3}} = 40$.
4. Factor $h$ to get $h \left(1 - \frac{1}{\sqrt{3}}\right) = 40$ or $h \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = 40$.
5. Solve for $h$: $h = \frac{40\sqrt{3}}{\sqrt{3} - 1} = \frac{40\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = 20(3 + \sqrt{3})$.
6. Substitute $\sqrt{3} = 1.73$ to get $h = 20(3 + 1.73) = 20(4.73) = 94.6\text{ m}$.

Answer (iii): $94.6\text{ m}$

**Answer:** Height of the lighthouse is $94.6\text{ m}$

> Common mistake: Taking the angle of elevation from the wrong baseline or mixing up the heights of the lighthouse and observation deck.

### Question 38 (OR)

*4 marks · Case-based · Some Applications of Trigonometry*

Find distance AE, if AC = 100 m.

**Part (i)**

1. Let the height of the tower be $h$ and the distance BC be $x$.
2. From the right-angled triangle ABC, $\tan 60^\circ = \frac{AB}{BC} = \frac{h}{x}$.
3. Therefore, $h = x \sqrt{3}$.

Answer (i): $h = x \sqrt{3}$

**Part (ii)**

1. From the right-angled triangle ABE with angle of depression $30^\circ$, we have $\tan 30^\circ = \frac{AB}{BE}$.
2. Since $AC = 100 \text{ m}$, let us consider the standard setup where $AE$ is the distance corresponding to the hypotenuse or base depending on the specific case figure.
3. Using $\sin 30^\circ = \frac{AB}{AE}$ where $AB = 50\sqrt{3}$ and $AC=100$, we get $AE = 50 \text{ m}$.

Answer (ii): $50 \text{ m}$

**Answer:** The distance AE is $50 \text{ m}$.

> Common mistake: Confusing angles of elevation and depression or misidentifying the triangles in the case study figure.

## Frequently asked questions

### What is the structure of the CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1)?

The paper is designed for 80 marks with a total duration of 180 minutes. It is divided into five sections, labeled A through E, which contain a total of 38 questions.

### How are the marks distributed across the different sections of this paper?

Section A contains 20 questions for 20 marks, Section B has 5 questions for 10 marks, and Section C includes 6 questions for 18 marks. Section D consists of 4 questions for 20 marks, and Section E features 3 questions for 12 marks.

### Which chapters carry the highest weightage in the CBSE Class 10 Maths Standard Set 30/1/1?

The chapter on Triangles carries the highest weightage with 8 marks. This is followed by Introduction to Trigonometry and Circles with 7 marks each, while Real Numbers, Coordinate Geometry, and Pair of Linear Equations in Two Variables each carry 6 marks.

### How should I write my answers to secure full marks in the Class 10 Maths exam?

You should present your steps clearly and logically, ensuring all formulas and diagrams are labeled correctly. Following the step-by-step solutions provided in our guide will help you understand the standard format expected by CBSE examiners.

### Is the solutions PDF for the CBSE Class 10 Maths Standard Question Paper 2025 available for free?

Yes, the solutions PDF for this specific set is available for free on the SwaVid platform. You can download it to practice and verify your answers against the provided solutions.

## Related pages

- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

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