---
title: "CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2026-standard-30-1-1
dateModified: 2026-10-07T15:33:09+00:00
---

# CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

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

Solved CBSE Class 10 Maths Standard board paper from 2026, 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/2026/X/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-2026-standard-30-1-1-ef83aa7196.pdf

## SECTION - A

Question numbers 1 to 20 are multiple choice questions of 1 mark each.

### Question 1

*1 mark · MCQ · Real Numbers*

The HCF of 960 and 432 is :

- 48
- 54
- 72
- 36

**Solution**

1. Find the prime factorisation of 960 and 432: $960 = 2^6 \times 3 \times 5$ and $432 = 2^4 \times 3^3$.
2. The HCF is the product of the smallest power of each common prime factor in the numbers: $\text{HCF}(960, 432) = 2^4 \times 3^1 = 16 \times 3 = 48$.

**Answer:** (A) 48

> Common mistake: Taking the highest powers instead of the lowest powers for HCF.

### Question 2

*1 mark · MCQ · Real Numbers*

The natural number $2$ is :

- a prime number
- a composite number
- prime as well as composite
- neither prime nor composite

**Solution**

1. A natural number greater than 1 is called a prime number if it has exactly two distinct factors, 1 and itself.
2. The number 2 has only two factors, 1 and 2, and hence it is a prime number.

**Answer:** (a) a prime number

> Common mistake: Students sometimes think 2 is a composite number or neither.

### Question 3

*1 mark · MCQ · Real Numbers*

For any natural number $n$, $6^n$ ends with the digit :

- 0
- 6
- 3
- 2

**Solution**

1. If any number $6^n$ were to end with the digit 0, it would be divisible by 5, meaning its prime factorisation must contain the prime 5.
2. The prime factorisation of $6^n$ is $(2 \times 3)^n = 2^n \times 3^n$, which shows that 2 and 3 are the only prime factors, so by the uniqueness of the Fundamental Theorem of Arithmetic, 5 is not present.

**Answer:** (B) 6

> Common mistake: Assuming $6^n$ ends in 0 for some large $n$ because $6 \times 6 = 36$ ends in 6.

### Question 4

*1 mark · MCQ · Polynomials*

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

- 0
- 1
- 2
- 4

**Solution**

1. The zeroes of a polynomial $f(x)$ are the x-coordinates of the points where the graph of $y = f(x)$ intersects the x-axis.
2. From the given graph in the textbook (Fig. 2.10), the curve intersects the x-axis at 3 points, so the number of zeroes is 3.

**Answer:** (c) 3

> Common mistake: Counting the intersections with the y-axis instead of the x-axis.

### Question 5

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

If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has :

- a unique solution
- two solutions
- no solution
- an infinite number of solutions

**Solution**

1. For a pair of linear equations represented by coincident lines, every point on the line is a common solution.
2. Therefore, coincident lines have infinitely many solutions.

**Answer:** (d) an infinite number of solutions

> Common mistake: Confusing coincident lines with parallel lines which have no solution.

### Question 6

*1 mark · MCQ · Arithmetic Progressions*

The common difference of the AP : $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$ is :

- \sqrt{2}
- 1
- 2\sqrt{2}
- -\sqrt{2}

**Solution**

1. The given AP is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$.
2. The common difference $d$ is obtained by subtracting the first term from the second term: $d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$.

**Answer:** (a) $\sqrt{2}$

> Common mistake: Subtracting terms incorrectly or getting confused with surds.

### Question 7

*1 mark · MCQ · Triangles*

If $\Delta \text{ABC} \sim \Delta \text{DEF}$ are similar such that $2\text{AB} = \text{DE}$ and $\text{BC} = 8\text{ cm}$, then $\text{EF}$ is equal to :

- 4\text{ cm}
- 8\text{ cm}
- 12\text{ cm}
- 16\text{ cm}

**Solution**

1. Given $\Delta \text{ABC} \sim \Delta \text{DEF}$, the ratio of their corresponding sides is equal: $\frac{\text{AB}}{\text{DE}} = \frac{\text{BC}}{\text{EF}}$.
2. From $2\text{AB} = \text{DE}$, we get $\frac{\text{AB}}{\text{DE}} = \frac{1}{2}$. Substituting $\text{BC} = 8\text{ cm}$, $\frac{1}{2} = \frac{8}{\text{EF}}$, which gives $\text{EF} = 16\text{ cm}$.

**Answer:** (d) $16\text{ cm}$

> Common mistake: Taking the ratio of sides incorrectly as $\frac{\text{AB}}{\text{DE}} = 2$ instead of $\frac{1}{2}$.

### Question 8

*1 mark · MCQ · Coordinate Geometry*

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

- x\text{-axis}
- y\text{-axis}
- origin
- neither } x\text{-axis nor } y\text{-axis

**Solution**

1. The mid-point of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).$
2. Substituting the given points, the mid-point is $\left(\frac{5 + 6}{2}, \frac{-4 + 4}{2}\right) = \left(\frac{11}{2}, 0\right).$
3. Since the y-coordinate is 0, the mid-point lies on the x-axis.

**Answer:** (a) x\text{-axis}

> Common mistake: Selecting y-axis by looking at the y-values before finding the coordinate.

### Question 9

*1 mark · MCQ · Introduction to Trigonometry*

Given that $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to :

- \frac{b}{\sqrt{b^2 - a^2}}
- \frac{b}{a}
- \frac{\sqrt{b^2 - a^2}}{b}
- \frac{a}{\sqrt{b^2 - a^2}}

**Solution**

1. Given $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{b}$.
2. By using identity $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}$.

**Answer:** (c) \frac{\sqrt{b^2 - a^2}}{b}

> Common mistake: Inverting the numerator and denominator.

### Question 10

*1 mark · MCQ · Introduction to Trigonometry*

If $\cos A = \frac{1}{2}$, then the value of $\sin^2 A + 2\cos^2 A$ is :

- \frac{3}{2}
- \frac{5}{4}
- -1
- \frac{1}{2}

**Solution**

1. Given $\cos A = \frac{1}{2}$, which implies $A = 60^\circ$.
2. Therefore, $\sin A = \frac{\sqrt{3}}{2}$, so $\sin^2 A = \frac{3}{4}$ and $\cos^2 A = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
3. Substitute these values into the expression: $\sin^2 A + 2\cos^2 A = \frac{3}{4} + 2\left(\frac{1}{4}\right) = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$.

**Answer:** (b) \frac{5}{4}

> Common mistake: Forgetting to square the trigonometric ratios or substituting incorrect values.

### Question 11

*1 mark · MCQ · Some Applications of Trigonometry*

A car is moving away from the base of a $30\text{ m}$ high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3}\text{ m}$ away from the base of the tower, is :

- 30^\circ
- 45^\circ
- 90^\circ
- 60^\circ

**Solution**

1. Let the height of the tower be $AB = 30\text{ m}$ and the distance of the car from the base be $BC = 10\sqrt{3}\text{ m}$.
2. Let the angle of elevation be $\theta$. In right triangle $ABC$, $\tan \theta = \frac{AB}{BC} = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
3. Since $\tan 60^\circ = \sqrt{3}$, we get $\theta = 60^\circ$.

**Answer:** (d) 60^\circ

> Common mistake: Using sine or cosine instead of tangent ratio.

### Question 12

*1 mark · MCQ · Circles*

If $\text{TP}$ and $\text{TQ}$ are two tangents to a circle with centre $\text{O}$ from an external point $\text{T}$ so that $\angle \text{POQ} = 120^\circ$, then $\angle \text{PTQ}$ is equal to :

- 60^\circ
- 70^\circ
- 80^\circ
- 90^\circ

**Solution**

1. The sum of angles opposite to each other in quadrilateral PAOR is $180^\circ$, where $\angle OPA = \angle OQA = 90^\circ$.
2. Therefore, $\angle \text{PTQ} = 180^\circ - \angle \text{POQ} = 180^\circ - 120^\circ = 60^\circ$.

**Answer:** (a) $60^\circ$

> Common mistake: Assuming $\angle \text{PTQ}$ is equal to $\angle \text{POQ}$ or half of it without geometric proof.

### Question 13

*1 mark · MCQ · Circles*

In the given figure, $\text{PA}$ is a tangent from an external point $\text{P}$ to a circle with centre $\text{O}$. If $\angle \text{POB} = 125^\circ$, then $\angle \text{APO}$ is equal to :

- 25^\circ
- 65^\circ
- 90^\circ
- 35^\circ

**Solution**

1. Radius $OA$ is perpendicular to tangent $PA$, so $\angle OAP = 90^\circ$.
2. The angle $\angle AOB = 180^\circ - \angle POB = 180^\circ - 125^\circ = 55^\circ$.
3. In triangle $OAP$, the sum of angles is $180^\circ$, so $\angle APO = 180^\circ - 90^\circ - 55^\circ = 35^\circ$.

**Answer:** (d) 35^\circ

> Common mistake: Assuming triangle OAP is isosceles without checking side lengths.

### Question 14

*1 mark · MCQ · Areas Related to Circles*

The length of the arc of the sector of a circle with radius $21\text{ cm}$ and of central angle $60^\circ$, is :

- 22\text{ cm}
- 44\text{ cm}
- 88\text{ cm}
- 11\text{ cm}

**Solution**

1. The formula for the length of an arc of a sector is $\text{Length} = \frac{\theta}{360^\circ} \times 2\pi r$.
2. Substitute $\theta = 60^\circ$, $r = 21\text{ cm}$, and $\pi = \frac{22}{7}$ into the formula.
3. Length $= \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22\text{ cm}$.

**Answer:** (a) $22\text{ cm}$

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

### Question 15

*1 mark · MCQ · Areas Related to Circles*

The hour hand of a clock is $7\text{ cm}$ long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is :

- \left(\frac{35}{4}\right)^\circ
- \left(\frac{35}{2}\right)^\circ
- 35^\circ
- 70^\circ

**Solution**

1. The hour hand completes $360^\circ$ in 12 hours, meaning it sweeps $\frac{360^\circ}{12} = 30^\circ$ per hour or $0.5^\circ$ per minute.
2. The total time elapsed from 7:00 a.m. to 8:10 a.m. is 1 hour and 10 minutes, which is 70 minutes.
3. The angle swept is $70 \times 0.5^\circ = 35^\circ$.

**Answer:** (c) 35^\circ

> Common mistake: Calculating the angle swept by the minute hand instead of the hour hand.

### Question 16

*1 mark · MCQ · Surface Areas and Volumes*

The total surface area of a solid hemisphere of diameter '$2d$' is :

- 3\pi d^2
- 2\pi d^2
- \frac{1}{2}\pi d^2
- \frac{3}{4}\pi d^2

**Solution**

1. The diameter of the solid hemisphere is given as $2d$, so its radius $r$ is $d$.
2. The total surface area of a solid hemisphere is given by the formula $3\pi r^2$.
3. Substituting $r = d$ gives $3\pi d^2$.

**Answer:** (a) 3\pi d^2

> Common mistake: Using $2\pi r^2$ which is only the curved surface area of a hemisphere.

### Question 17

*1 mark · MCQ · Statistics*

If the mean and mode of a data are 12 and 21 respectively, then its median is :

- 6
- 13.5
- 15
- 14

**Solution**

1. The empirical relationship between mean, median, and mode is: $\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}$.
2. Substitute the given values: $21 = 3 \text{ Median} - 2(12)$.
3. Solve for median: $21 = 3 \text{ Median} - 24$, so $3 \text{ Median} = 45$, giving $\text{Median} = 15$.

**Answer:** (c) 15

> Common mistake: Confusing the empirical formula coefficients for mean and median.

### Question 18

*1 mark · MCQ · Probability*

A die is thrown once. Probability of getting a number other than 3 is :

- \frac{1}{6}
- \frac{3}{6}
- \frac{5}{6}
- 1

**Solution**

1. When a die is thrown once, the total number of possible outcomes is 6.
2. The outcomes other than 3 are 1, 2, 4, 5, and 6, which are 5 outcomes in total.
3. The required probability is $\frac{5}{6}$.

**Answer:** (c) \frac{5}{6}

> Common mistake: Writing the probability of getting the number 3 instead of a number other than 3.

### Question 19

*1 mark · Assertion and reason · Probability*

Assertion (A) : The probability that a leap year has 53 Mondays is $\frac{2}{7}$.
Reason (R) : The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$.

- 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. A leap year has 366 days, which consists of 52 complete weeks and 2 extra days.
2. The 2 extra days can be {Sun, Mon}, {Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, or {Sat, Sun}.
3. Out of these 7 possible outcomes, 2 outcomes contain Monday, so the probability of 53 Mondays in a leap year is $\frac{2}{7}$, making Assertion (A) true.
4. A non-leap year has 365 days, which consists of 52 complete weeks and 1 extra day.
5. The 1 extra day can be any of the 7 days of the week, so the probability of 53 Mondays in a non-leap year is $\frac{1}{7}$, making Reason (R) false.

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

> Common mistake: Confusing the number of extra days in a non-leap year and incorrectly calculating the probability as 5/7.

### Question 20

*1 mark · Assertion and reason · Polynomials*

Assertion (A) : The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes.
Reason (R) : A quadratic polynomial can have at most two zeroes.

- 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 given polynomial is $p(y) = y^2 + 4y + 3$, which can be factored as $(y + 1)(y + 3)$.
2. The zeroes of $p(y)$ are $-1$ and $-3$, so the polynomial has two distinct zeroes, making Assertion (A) true.
3. A quadratic polynomial is of the degree 2 and can have at most two zeroes, which is a standard theorem and makes Reason (R) true.
4. Reason (R) correctly explains why a quadratic polynomial like $p(y)$ can have up to two zeroes.

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

> Common mistake: Thinking that a quadratic polynomial always has two distinct zeroes without checking for coincidence or non-real roots.

## SECTION - B

Question numbers 21 to 25 are Very Short Answer (VSA) type questions, carrying 2 marks each.

### Question 21

*2 marks · Very short answer · Polynomials*

If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 3x - 1$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

**Solution**

1. For the quadratic polynomial $p(x) = x^2 - 3x - 1$, the sum of zeroes $\alpha + \beta = -\frac{-3}{1} = 3$ and the product of zeroes $\alpha\beta = \frac{-1}{1} = -1$.
2. Consider the expression $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.
3. Substitute the values of $\alpha + \beta$ and $\alpha\beta$ to get $\frac{3}{-1} = -3$.

**Answer:** -3

> Common mistake: Making a sign error while substituting the coefficients into the sum and product formulas.

### Question 22

*2 marks · Very short answer · Triangles*

(A) In $\Delta \text{ABC}$, $\text{DE} \parallel \text{BC}$. If $\text{AD} = x$, $\text{DB} = x - 2$, $\text{AE} = x + 2$ and $\text{EC} = x - 1$, then find the value of $x$.

**Solution**

1. Since $DE \parallel BC$ in $\triangle ABC$, by the Basic Proportionality Theorem, $\frac{AD}{DB} = \frac{AE}{EC}$.
2. Substitute the given lengths into the relation: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$.
3. Cross-multiply to get $x(x - 1) = (x + 2)(x - 2)$, which simplifies to $x^2 - x = x^2 - 4$.
4. Solving for $x$ gives $x = 4$.

**Answer:** $4$

> Common mistake: Incorrect cross-multiplication or algebraic expansion of terms.

### Question 22 (OR)

*2 marks · Very short answer · Triangles*

(B) In the figure given above, $\Delta \text{ABC} \sim \Delta \text{XYZ}$, then find the values of $x$ and $y$.

**Solution**

1. Since $\triangle ABC \sim \triangle XYZ$, the corresponding sides are proportional: $\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
2. Substitute the given values $\frac{4}{x} = \frac{y}{6} = \frac{6}{7.2}$.
3. From $\frac{4}{x} = \frac{6}{7.2}$, we get $x = \frac{4 \times 7.2}{6} = 4.8$.
4. From $\frac{y}{6} = \frac{6}{7.2}$, we get $y = \frac{6 \times 6}{7.2} = 5$.

**Answer:** $x = 4.8$, $y = 5$

> Common mistake: Matching the incorrect corresponding sides of the similar triangles.

### Question 23

*2 marks · Very short answer · Coordinate Geometry*

The coordinates of the centre of a circle are $(x - 7, 2x)$. Find the value(s) of '$x$', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.

**Solution**

1. Let the centre of the circle be $C(x - 7, 2x)$ and the given point on the circle be $P(-9, 11)$.
2. The radius of the circle is the distance between the centre and point $P$, which is given as $5\sqrt{2}$.
3. Using the distance formula, $((x - 7) - (-9))^2 + (2x - 11)^2 = (5\sqrt{2})^2$.
4. Simplify to $(x + 2)^2 + (2x - 11)^2 = 50$, which expands to $x^2 + 4x + 4 + 4x^2 - 44x + 121 = 50$.
5. Simplify the quadratic equation to $5x^2 - 40x + 75 = 0$, or $x^2 - 8x + 15 = 0$.
6. Factorising gives $(x - 3)(x - 5) = 0$, so $x = 3$ or $x = 5$.

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

> Common mistake: Forgetting to square the radius when equating it to the square of the distance formula.

### Question 24

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

(A) If $\tan \theta = \frac{24}{7}$, then find the value of $\sin \theta + \cos \theta$.

**Solution**

1. Given $\tan \theta = \frac{24}{7} = \frac{\text{Perpendicular}}{\text{Base}}$.
2. Let perpendicular be $24k$ and base be $7k$. By Pythagoras theorem, hypotenuse $=\sqrt{(24k)^2 + (7k)^2} = 25k$.
3. Therefore, $\sin \theta = \frac{24}{25}$ and $\cos \theta = \frac{7}{25}$.
4. Add the values: $\sin \theta + \cos \theta = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}$.

**Answer:** $\frac{31}{25}$

> Common mistake: Mixing up the perpendicular and base sides for the given trigonometric ratio.

### Question 24 (OR)

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

(B) If $\cot \theta = \frac{7}{8}$, then find the value of $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$.

**Solution**

1. Given expression is $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$.
2. Using trigonometric identities, $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = (\cot \theta)^2$.
3. Substitute $\cot \theta = \frac{7}{8}$ into the expression to get $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

**Answer:** $\frac{49}{64}$

> Common mistake: Squaring the numerator and denominator incorrectly or not simplifying the expression using identities first.

### Question 25

*2 marks · Very short answer · Circles*

Two concentric circles are of radii $5\text{ cm}$ and $4\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let the concentric circles be with center $O$. Let $AB$ be the chord of the larger circle of radius $5\text{ cm}$ which touches the smaller circle of radius $4\text{ cm}$ at point $P$.
2. Since $AB$ is a tangent to the smaller circle at $P$, $OP \perp AB$.
3. In right-angled triangle $\triangle OPA$, $\angle OPA = 90^\circ$, $OA = 5\text{ cm}$, and $OP = 4\text{ cm}$.
4. By Pythagoras theorem, $AP = \sqrt{OA^2 - OP^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$.
5. Since the perpendicular from the center to a chord bisects the chord, $AB = 2AP = 2 \times 3\text{ cm} = 6\text{ cm}$.

**Answer:** $6\text{ cm}$

> Common mistake: Students often forget to multiply the length of the segment $AP$ by 2 to get the total length of the chord $AB$.

## SECTION - C

Question numbers 26 to 31 are Short Answer (SA) type questions, carrying 3 marks each.

### Question 26

*3 marks · Proof · Real Numbers*

Prove that $\sqrt{3}$ is an irrational number.

**Solution**

1. Let us assume, to the contrary, that $\sqrt{3}$ is a rational number.
2. So, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \frac{a}{b}$.
3. Squaring both sides, we get $3 = \frac{a^2}{b^2}$, which means $3b^2 = a^2$.
4. Therefore, $3$ divides $a^2$, and consequently $3$ divides $a$ by Theorem 1.3.
5. So, we can write $a = 3c$ for some integer $c$.
6. Substituting $a = 3c$ in $3b^2 = a^2$, we get $3b^2 = 9c^2$, which gives $b^2 = 3c^2$.
7. This means $3$ divides $b^2$, and so $3$ divides $b$.
8. Thus, $a$ and $b$ have at least $3$ as a common factor, which contradicts the fact that $a$ and $b$ are co-prime.
9. This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.
10. Hence, $\sqrt{3}$ is an irrational number.

**Answer:** Hence proved.

> Common mistake: Stating that $a$ and $b$ are integers instead of co-prime integers.

### Question 27

*3 marks · Short answer · Coordinate Geometry*

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

**Solution**

1. Let the $x$-axis divide the line segment joining $(-6, 5)$ and $(-4, -1)$ in the ratio $k : 1$ at the point $(x, 0).$
2. Using the section formula for the $y$-coordinate, we have $0 = \frac{k(-1) + 1(5)}{k + 1}$.
3. Solving for $k$, we get $-k + 5 = 0$, which gives $k = 5$.
4. Thus, the required ratio is $5 : 1$ internally.
5. Now, find the $x$-coordinate of the point of intersection: $x = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3}$.
6. Therefore, the point of intersection is $\left(-\frac{13}{3}, 0\right).$

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

> Common mistake: Forgetting to state that the $y$-coordinate on the $x$-axis is zero.

### Question 28

*3 marks · Short answer · Introduction to Trigonometry*

(A) If $x = h + a \cos \theta, y = k + b \sin \theta$, then prove that : $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1$

**Solution**

1. Given equations are $x = h + a \cos \theta$ and $y = k + b \n \theta$.
2. Rearrange the equations to find $\cos \theta$ and $\sin \theta$: $\frac{x - h}{a} = \cos \theta$ and $\frac{y - k}{b} = \sin \theta$.
3. Square both equations: $\left(\frac{x - h}{a}\right)^2 = \cos^2 \theta$ and $\left(\frac{y - k}{b}\right)^2 = \sin^2 \theta$.
4. Add the two squared equations: $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = \cos^2 \theta + \sin^2 \theta$.
5. Since $\cos^2 \theta + \sin^2 \theta = 1$, we get $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1$.
6. Hence proved.

**Answer:** Hence proved.

> Common mistake: Algebraic errors while squaring the fractions.

### Question 28 (OR)

*3 marks · Short answer · Introduction to Trigonometry*

(B) Prove that : $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A$

**Solution**

1. Consider the LHS: $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A}$.
2. Take $\tan A$ common from both terms: $\tan A \left(\frac{1}{1 + \sec A} - \frac{1}{1 - \sec A}\right)$.
3. Take the LCM inside the bracket: $\tan A \left[\frac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A)(1 - \sec A)}\right]$.
4. Simplify the numerator and denominator: $\tan A \left[\frac{-2 \sec A}{1 - \sec^2 A}\right]$.
5. Use the identity $1 - \sec^2 A = -\tan^2 A$: $\tan A \left[\frac{-2 \sec A}{-\tan^2 A}\right]$.
6. Cancel $-\tan A$ from numerator and denominator: $\frac{2 \sec A}{\tan A}$.
7. Convert into sine and cosine: $\frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Sign errors while simplifying the numerator.

### Question 29

*3 marks · Short answer · Circles*

(A) In the given figure, $\Delta \text{ABC}$ is a right triangle in which $\angle \text{B} = 90^\circ$, $\text{AB} = 4\text{ cm}$ and $\text{BC} = 3\text{ cm}$. Find the radius of the circle inscribed in the triangle $\text{ABC}$.

**Solution**

1. Given $\text{AB} = 4\text{ cm}$ and $\text{BC} = 3\text{ cm}$, find the hypotenuse $\text{AC}$ using Pythagoras theorem: $\text{AC} = \sqrt{4^2 + 3^2} = 5\text{ cm}$.
2. Let $r$ be the radius of the inscribed circle. The circle touches $\text{AB}$ and $\text{BC}$ at points making squares with the vertex $\text{B}$, so the tangents from $\text{B}$ have length $r$.
3. The lengths of tangents from vertex $\text{A}$ are $4 - r$ and from vertex $\text{C}$ are $3 - r$.
4. The sum of the lengths of tangents from $\text{C}$ and $\text{A}$ equals $\text{AC}$: $(4 - r) + (3 - r) = 5$.
5. Simplify the equation: $7 - 2r = 5$, which gives $2r = 2$, so $r = 1\text{ cm}$.
6. Thus, the radius of the inscribed circle is $1\text{ cm}$.

**Answer:** $1\text{ cm}$

> Common mistake: Confusing the radius with the side lengths of the formed square at vertex B.

### Question 29 (OR)

*3 marks · Proof · Circles*

(B) In the given figure, if a circle touches the side $\text{QR}$ of $\Delta \text{PQR}$ at $\text{S}$ and extended sides $\text{PQ}$ and $\text{PR}$ at $\text{M}$ and $\text{N}$ respectively, then prove that : $\text{PM} = \frac{1}{2}(\text{PQ} + \text{QR} + \text{PR})$

**Solution**

1. We know that the lengths of tangents drawn from an external point to a circle are equal.
2. Therefore, from point $\text{P}$, tangents are $\text{PM} = \text{PN}$.
3. From point $\text{Q}$, tangents are $\text{QM} = \text{QS}$.
4. From point $\text{R}$, tangents are $\text{RN} = \text{RS}$.
5. Consider the RHS: $\text{PQ} + \text{QR} + \text{PR} = (\text{PM} - \text{QM}) + (\text{QS} + \text{RS}) + (\text{PN} - \text{RN})$.
6. Substitute $\text{QM} = \text{QS}$ and $\text{RN} = \text{RS}$: $\text{PQ} + \text{QR} + \text{PR} = (\text{PM} - \text{QS}) + (\text{QS} + \text{RS}) + (\text{PN} - \text{RS})$.
7. Cancel out the terms to get: $\text{PQ} + \text{QR} + \text{PR} = \text{PM} + \text{PN}$.
8. Since $\text{PM} = \text{PN}$, we can write $\text{PQ} + \text{QR} + \text{PR} = 2\text{PM}$.
9. Hence, $\text{PM} = \frac{1}{2}(\text{PQ} + \text{QR} + \text{PR})$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Wrong substitution of extended segments in terms of tangents.

### Question 30

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

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $20\text{ cm}$ and the diameter of the cylinder is $7\text{ cm}$. Find the total volume of the solid. $\left(\text{Use } \pi = \frac{22}{7}\right)$

**Solution**

1. Diameter of the cylinder = $7\text{ cm}$, so the radius of the cylinder and hemispherical ends is $r = \frac{7}{2}\text{ cm}$.
2. Height of the cylindrical part $h = \text{Total height} - 2 \times \text{radius} = 20 - 2 \times \frac{7}{2} = 20 - 7 = 13\text{ cm}$.
3. Volume of the solid = Volume of the cylinder + 2 $\times$ Volume of a hemisphere = $\pi r^2 h + 2 \times \left(\frac{2}{3} \pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)$.
4. Substitute the values: $\frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left(13 + \frac{4}{3} \times \frac{7}{2}\right) = \frac{22}{7} \times \frac{49}{4} \times \left(13 + \frac{14}{3}\right) = \frac{77}{2} \times \frac{53}{3}$.
5. Calculate the total volume: $\frac{4081}{6} = 680.17\text{ cm}^3$ (or $680\frac{1}{6}\text{ cm}^3$).

**Answer:** $680.17\text{ cm}^3$

> Common mistake: Forgetting to subtract the radii of the two hemispherical ends from the total height to find the height of the cylinder.

### Question 31

*3 marks · Short answer · Probability*

Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :
(i) same number appears on both the dice ?
(ii) different number appears on both the dice ?

**Part (i)**

1. Total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
2. Outcomes where the same number appears on both dice are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$, which are 6 in number.
3. Probability = $\frac{6}{36} = \frac{1}{6}$.

Answer (i): \frac{1}{6}

**Part (ii)**

1. Outcomes where different numbers appear on both dice are the remaining outcomes, or $36 - 6 = 30$.
2. Probability = $\frac{30}{36} = \frac{5}{6}$.

Answer (ii): \frac{5}{6}

**Answer:** Total possible outcomes are 36.

> Common mistake: Listing outcomes incorrectly or miscalculating the number of favorable outcomes for different numbers.

## SECTION - D

Question numbers 32 to 35 are Long Answer (LA) type questions, carrying 5 marks each.

### Question 32

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

Determine graphically, the coordinates of vertices of a triangle whose equations are $2x - 3y + 6 = 0$; $2x + 3y - 18 = 0$ and $x = 0$. Also, find the area of this triangle.

**Solution**

1. For the equation $2x - 3y + 6 = 0$, when $x = 0$, $y = 2$; when $x = -3$, $y = 0$; when $x = 3$, $y = 4$.
2. For the equation $2x + 3y - 18 = 0$, when $x = 0$, $y = 6$; when $x = 3$, $y = 4$; when $x = 9$, $y = 0$.
3. For the equation $x = 0$, it represents the y-axis.
4. Plot the lines on a graph paper and find the intersection points which form the vertices of the triangle.
5. The three lines intersect at $(0, 2)$, $(3, 4)$, and $(0, 6)$, which are the vertices of the triangle.
6. The base of the triangle along the y-axis is from $y = 2$ to $y = 6$, so base = $6 - 2 = 4\text{ units}$.
7. The corresponding height is the x-coordinate of the vertex $(3, 4)$, which is height = $3\text{ units}$.
8. Area of the triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6\text{ square units}$.

**Answer:** Vertices are $(0, 2)$, $(3, 4)$, and $(0, 6)$; Area = $6\text{ sq. units}$.

> Common mistake: Wrong calculation of base and height from the plotted graph.

### Question 33

*5 marks · Long answer · Quadratic Equations*

(A) A faster train takes one hour less than a slower train for a journey of $200\text{ km}$. If the speed of the slower train is $10\text{ km/hr}$ less than that of the faster train, find the speeds of the two trains.

**Solution**

1. Let the speed of the slower train be $x\text{ km/hr}$.
2. Then the speed of the faster train is $(x + 10)\text{ km/hr}$.
3. Distance to be covered = $200\text{ km}$.
4. Time taken by the slower train = $\frac{200}{x}\text{ hours}$ and time taken by the faster train = $\frac{200}{x + 10}\text{ hours}$.
5. According to the given condition, $\frac{200}{x} - \frac{200}{x + 10} = 1$.
6. Simplifying the equation gives $200(x + 10) - 200x = x(x + 10)$, which leads to $x^2 + 10x - 2000 = 0$.
7. Solving by splitting the middle term: $x^2 + 50x - 40x - 2000 = 0$, so $(x + 50)(x - 40) = 0$.
8. Since speed cannot be negative, $x = 40$.
9. Thus, the speed of the slower train is $40\text{ km/hr}$ and the speed of the faster train is $40 + 10 = 50\text{ km/hr}$.

**Answer:** Speed of slower train = $40\text{ km/hr}$, Speed of faster train = $50\text{ km/hr}$.

> Common mistake: Taking speed of slower train as $x - 10$ and faster as $x$ and getting negative time differences.

### Question 33 (OR)

*5 marks · Long answer · Quadratic Equations*

(B) The sum of the areas of two squares is $640\text{ m}^2$. If the difference in their perimeters is $64\text{ m}$, find the sides of the two squares.

**Solution**

1. Let the side of the first square be $x\text{ m}$ and the side of the second square be $y\text{ m}$, where $x > y$.
2. Perimeter of the first square is $4x$ and of the second is $4y$.
3. Given the difference in perimeters is $64\text{ m}$, so $4x - 4y = 64$, which gives $x - y = 16$ or $x = y + 16$.
4. Given the sum of their areas is $640\text{ m}^2$, so $x^2 + y^2 = 640$.
5. Substitute $x = y + 16$ into the area equation: $(y + 16)^2 + y^2 = 640$.
6. Expanding and simplifying: $y^2 + 32y + 256 + y^2 = 640$, which gives $2y^2 + 32y - 384 = 0$.
7. Divide by 2: $y^2 + 16y - 192 = 0$.
8. Factorising the quadratic equation: $y^2 + 24y - 8y - 192 = 0$, so $(y + 24)(y - 8) = 0$.
9. Since side length cannot be negative, $y = 8$.
10. Then $x = 8 + 16 = 24$.
11. The sides of the two squares are $24\text{ m}$ and $8\text{ m}$.

**Answer:** Sides of the two squares are $24\text{ m}$ and $8\text{ m}$.

> Common mistake: Forgetting to divide the perimeter equation by 4 or expanding $(y+16)^2$ incorrectly.

### Question 34

*5 marks · Proof · Triangles*

(A) State and prove Basic Proportionality Theorem.

**Solution**

1. Given: A triangle $\Delta ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively.
2. To prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
3. Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
4. Proof: Area of $\Delta ADE$ = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$.
5. Area of $\Delta BDE$ = $\frac{1}{2} \times DB \times EN$.
6. Dividing the two areas: $\frac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ (Equation 1).
7. Similarly, $\frac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ (Equation 2).
8. Since $\Delta BDE$ and $\Delta DEC$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$, their areas are equal: $\text{ar}(\Delta BDE) = \text{ar}(\Delta DEC)$.
9. From Equations 1, 2, and the equal areas, we get $\frac{AD}{DB} = \frac{AE}{EC}$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not mentioning the reason why $\text{ar}(\Delta BDE) = \text{ar}(\Delta DEC)$.

### Question 34 (OR)

*5 marks · Proof · Triangles*

(B) In the given figure, $\text{CM}$ and $\text{RN}$ are respectively the medians of $\Delta \text{ABC}$ and $\Delta \text{PQR}$. If $\Delta \text{ABC} \sim \Delta \text{PQR}$, then prove that :
(i) $\Delta \text{AMC} \sim \Delta \text{PNR}$
(ii) $\Delta \text{CMB} \sim \Delta \text{RNQ}$

**Part (i)**

1. Given $\Delta ABC \sim \Delta PQR$, therefore $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
2. Since $CM$ and $RN$ are medians, $BM = \frac{1}{2}BC$ and $QN = \frac{1}{2}QR$.
3. Thus, $\frac{AB}{PQ} = \frac{2BM}{2QN} = \frac{BM}{QN}$, which can also be written as $\frac{AB}{PQ} = \frac{AM}{PN}$ since $AM$ and $PN$ are corresponding halves of proportional sides.
4. In $\Delta AMC$ and $\Delta PNR$, $\frac{AC}{PR} = \frac{AM}{PN}$ and $\angle A = \angle P$.
5. Therefore, by SAS similarity criterion, $\Delta AMC \sim \Delta PNR$.

Answer (i): Hence proved.

**Part (ii)**

1. From $\Delta ABC \sim \Delta PQR$, we have $\angle B = \angle Q$.
2. Also, we know that $\frac{BC}{QR} = \frac{BM}{QN}$.
3. In $\Delta CMB$ and $\Delta RNQ$, $\frac{BC}{QR} = \frac{BM}{QN}$ translates to proportional sides including the median segments, and $\angle B = \angle Q$.
4. Therefore, by SAS similarity criterion, $\Delta CMB \sim \Delta RNQ$.

Answer (ii): Hence proved.

**Answer:** Hence proved.

> Common mistake: Using incorrect corresponding sides when applying the SAS similarity criterion.

### Question 35

*5 marks · Long answer · Statistics*

The mean of the following frequency distribution is 35. Find the values of $x$ and $y$, if the sum of frequencies is 25 :
Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: 1, x, 5, 7, y, 3, 1

**Solution**

1. Construct the frequency distribution table with columns: Class, Frequency ($f_i$), Class mark ($x_i$), and $f_i x_i$.
2. For class 0-10, $f_1 = 1$, $x_1 = 5$, $f_1 x_1 = 5$.
3. For class 10-20, $f_2 = x$, $x_2 = 15$, $f_2 x_2 = 15x$.
4. For class 20-30, $f_3 = 5$, $x_3 = 25$, $f_3 x_3 = 125$.
5. For class 30-40, $f_4 = 7$, $x_4 = 35$, $f_4 x_4 = 245$.
6. For class 40-50, $f_5 = y$, $x_5 = 45$, $f_5 x_5 = 45y$.
7. For class 50-60, $f_6 = 3$, $x_6 = 55$, $f_6 x_6 = 165$.
8. For class 60-70, $f_7 = 1$, $x_7 = 65$, $f_7 x_7 = 65$.
9. Given the sum of frequencies is 25: $1 + x + 5 + 7 + y + 3 + 1 = 25 \implies x + y + 17 = 25 \implies x + y = 8$ (Equation 1).
10. Sum of $f_i x_i = 5 + 15x + 125 + 245 + 45y + 165 + 65 = 15x + 45y + 605$.
11. Given Mean = 35, so $\frac{\sum f_i x_i}{\sum f_i} = 35 \implies \frac{15x + 45y + 605}{25} = 35$.
12. Multiply by 25: $15x + 45y + 605 = 875 \implies 15x + 45y = 270 \implies x + 3y = 18$ (Equation 2).
13. Subtract Equation 1 from Equation 2: $(x + 3y) - (x + y) = 18 - 8 \implies 2y = 10 \implies y = 5$.
14. Substitute $y = 5$ into Equation 1: $x + 5 = 8 \implies x = 3$.

**Answer:** $x = 3$ and $y = 5$.

> Common mistake: Arithmetic errors while summing $f_i x_i$ terms.

## SECTION - E

Question numbers 36 to 38 are Case Study Based questions, carrying 4 marks each.

### Question 36

*4 marks · Case-based · Arithmetic Progressions*

In a potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato. The other potatoes are arranged $3\text{ m}$ apart in a straight line, with a total of 10 potatoes, as shown in the figure :
A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
(i) What is the distance covered to pick up the first potato and drop it in bucket ? [1]
(ii) What is the distance covered to pick up the second potato and drop it in bucket ? [1]
(iii) (a) What is the total distance the competitor has to run ? [2]
OR
(iii) (b) If average speed of competitor is $5\text{ m/s}$, then find the average time taken by competitor to put all the potatoes in the bucket. [2]

**Part (i)**

1. The distance of the first potato from the bucket is $5\text{ m}$.
2. The competitor runs to the potato and back to the bucket, so the total distance is $2 \times 5 = 10\text{ m}$.

Answer (i): $10\text{ m}$

**Part (ii)**

1. The distance of the second potato from the bucket is $5 + 3 = 8\text{ m}$.
2. The total distance covered to pick up the second potato and drop it in the bucket is $2 \times 8 = 16\text{ m}$.

Answer (ii): $16\text{ m}$

**Part (iii) (a)**

1. The distances form an arithmetic progression: $10, 16, 22, \dots$ with first term $a = 10$ and common difference $d = 6$.
2. The total distance for 10 potatoes is given by the sum of 10 terms of the AP: $S_{10} = \frac{10}{2} [2(10) + (10 - 1)6]$.
3. $S_{10} = 5 [20 + 54] = 5 \times 74 = 370\text{ m}$.

Answer (iii) (a): $370\text{ m}$

**Part (iii) (b)**

1. The total distance covered is $370\text{ m}$ and the average speed is $5\text{ m/s}$.
2. Time taken = $\frac{\text{Total Distance}}{\text{Speed}} = \frac{370}{5} = 74\text{ s}$.

Answer (iii) (b): $74\text{ seconds}$

**Answer:** Refer to individual parts for solutions.

> Common mistake: Forgetting to multiply the distance to the potato by 2 for the round trip.

### Question 37

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

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure).
Distance between the base of the tower and point 'O' is $6\text{ m}$. From point 'O', the angle of elevation of the top of the section 'B' is $30^\circ$ and the angle of elevation of the top of section 'A' is $60^\circ$.
Based on the above information, answer the following questions :
(i) Find the length of the wire from the point 'O' to the top of section 'B'. [1]
(ii) Find the length of the wire from the point 'O' to the top of section 'A'. [1]
(iii) (a) Find the distance $\text{AB}$. [2]
OR
(iii) (b) Find the area of $\Delta \text{OPB}$. [2]

**Part (i)**

1. Let the base of the tower be $B$ and point $O$ be on the ground at a distance of $6\text{ m}$ from the base.
2. In the right triangle formed with the top of section B, the angle of elevation is $30^\circ$.
3. Using $\cos 30^\circ = \frac{\text{Base}}{\text{Hypotenuse}}$, let length of wire be $L_B$.
4. $\cos 30^\circ = \frac{6}{L_B} \implies \frac{\sqrt{3}}{2} = \frac{6}{L_B} \implies L_B = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ m}$.

Answer (i): $4\sqrt{3}\text{ m}$

**Part (ii)**

1. In the right triangle formed with the top of section A, the angle of elevation is $60^\circ$.
2. Using $\cos 60^\circ = \frac{\text{Base}}{\text{Hypotenuse}}$, let length of wire be $L_A$.
3. $\cos 60^\circ = \frac{6}{L_A} \implies \frac{1}{2} = \frac{6}{L_A} \implies L_A = 12\text{ m}$.

Answer (ii): $12\text{ m}$

**Part (iii) (a)**

1. Height of top of section B ($h_B$) is given by $\tan 30^\circ = \frac{h_B}{6} \implies h_B = \frac{6}{\sqrt{3}} = 2\sqrt{3}\text{ m}$.
2. Height of top of section A ($h_A$) is given by $\tan 60^\circ = \frac{h_A}{6} \implies h_A = 6\sqrt{3}\text{ m}$.
3. Distance $AB = h_A - h_B = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}\text{ m}$.

Answer (iii) (a): $4\sqrt{3}\text{ m}$

**Part (iii) (b)**

1. The height of triangle OPB where P is the top of section B is $h_B = 2\sqrt{3}\text{ m}$ and the base on the ground is $6\text{ m}$.
2. Area of $\Delta OPB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 2\sqrt{3} = 6\sqrt{3}\text{ m}^2$.

Answer (iii) (b): $6\sqrt{3}\text{ m}^2$

**Answer:** Refer to individual parts for solutions.

> Common mistake: Using sine or tangent instead of cosine when finding the length of the wire (hypotenuse).

### Question 38

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

A brooch is crafted from silver wire in the shape of a circle with a diameter of $35\text{ cm}$. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure.
Based on the above information, answer the following questions :
(i) What is the radius of circle ? [1]
(ii) What is the circumference of the brooch ? [1]
(iii) (a) What is the total length of silver wire required ? [2]
OR
(iii) (b) What is the area of each sector of the brooch ? [2]

**Part (i)**

1. The diameter of the brooch is given as $d = 35\text{ cm}$.
2. Radius $r = \frac{d}{2} = \frac{35}{2} = 17.5\text{ cm}$.

Answer (i): $17.5\text{ cm}$ (or $\frac{35}{2}\text{ cm}$)

**Part (ii)**

1. Circumference of the brooch (outer silver wire) = $\pi d$.
2. Circumference = $\frac{22}{7} \times 35 = 110\text{ cm}$.

Answer (ii): $110\text{ cm}$

**Part (iii) (a)**

1. The total length of silver wire includes the circumference and the 5 diameters.
2. Circumference = $110\text{ cm}$ and length of 5 diameters = $5 \times 35 = 175\text{ cm}$.
3. Total length of wire = $110 + 175 = 285\text{ cm}$.

Answer (iii) (a): $285\text{ cm}$

**Part (iii) (b)**

1. The circle is divided into 10 equal sectors, so the angle of each sector is $\theta = \frac{360^\circ}{10} = 36^\circ$.
2. Area of one sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{36^\circ}{360^\circ} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}$.
3. Area = $\frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25\text{ cm}^2$.

Answer (iii) (b): $96.25\text{ cm}^2$

**Answer:** Refer to individual parts for solutions.

> Common mistake: Forgetting to include the lengths of the 5 diameters when calculating total wire required in part (iii)(a).

## Frequently asked questions

### What is the paper pattern and section breakdown for the CBSE Class 10 Maths Standard Question Paper 2026 Set 30/1/1?

This 80-mark question paper is for a duration of 180 minutes and is divided into five sections from A to E. It contains a total of 38 questions with internal choices provided in some of the questions.

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

Section A contains 20 questions carrying 1 mark each. Section B has 5 questions of 2 marks each, Section C has 6 questions of 3 marks each, Section D includes 4 questions of 5 marks each, and Section E consists of 3 case-based questions of 4 marks each.

### Which chapters carry the most marks in the CBSE Class 10 Maths Standard 2026 Set 30/1/1 paper?

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

### How should students write their answers to score full marks in this Mathematics examination?

Students should write step-by-step solutions with proper mathematical reasoning and formulas. Drawing neat diagrams wherever necessary and stating the final answer clearly with appropriate units helps secure maximum marks.

### Is the solutions PDF for this specific set available for free on SwaVid?

Yes, the complete and detailed solutions PDF for this question paper can be accessed and downloaded for free on SwaVid. It helps students verify their answers and understand the correct method of solving each problem.

## Related pages

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

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