---
title: "CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2023-standard-30-1-1
dateModified: 2026-10-07T15:11:53+00:00
---

# CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

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

Solved CBSE Class 10 Maths Standard board paper from 2023, 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/2023/X/MATHEMATICS_STANDARD.zip. Solutions written by SwaVid.

Free PDF (24 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2023-standard-30-1-1-68eb8bba26.pdf

## SECTION - A

Multiple Choice Questions. Each question is of 1 mark.

### Question 1

*1 mark · MCQ · Polynomials*

The graph of $y = p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is

- 3
- 1
- 2
- 0

**Solution**

1. The number of zeroes of a polynomial $p(x)$ is equal to the number of times its graph intersects the x-axis.
2. The given graph of the parabola intersects the x-axis at 2 distinct points, so the number of zeroes is 2.

**Answer:** (c) 2

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

### Question 2

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

The value of $k$ for which the pair of equations $kx = y + 2$ and $6x = 2y + 3$ has infinitely many solutions,

- is $k = 3$
- does not exist
- is $k = -3$
- is $k = 4$

**Solution**

1. Express the given equations in standard form: $kx - y - 2 = 0$ and $6x - 2y - 3 = 0$.
2. For a pair of linear equations to have infinitely many solutions, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
3. Substitute the coefficients: $\frac{k}{6} = \frac{-1}{-2} = \frac{-2}{-3}$, which gives $\frac{k}{6} = \frac{1}{2}$ and $\frac{1}{2} = \frac{2}{3}$ (false).
4. Since $\frac{1}{2} \neq \frac{2}{3}$, the ratios of constants are not equal to the ratios of coefficients, so no such value of $k$ exists.

**Answer:** (b) does not exist

> Common mistake: Only comparing $a_1/a_2 = b_1/b_2$ and forgetting to check the constant term ratio $c_1/c_2$.

### Question 3

*1 mark · MCQ · Arithmetic Progressions*

If $p - 1, p + 1$ and $2p + 3$ are in A.P., then the value of $p$ is

- -2
- 4
- 0
- 2

**Solution**

1. If $p - 1, p + 1, 2p + 3$ are in A.P., the difference between consecutive terms is equal, so $(p + 1) - (p - 1) = (2p + 3) - (p + 1)$.
2. Simplify both sides: $2 = p + 2$.
3. Solve for $p$: $p = 2 - 2 = 0$.
4. Checking the options, $p = 0$ corresponds to option (c).

**Answer:** (c) 0

> Common mistake: Subtracting terms in the wrong order or setting sums equal instead of differences.

### Question 4

*1 mark · MCQ · Coordinate Geometry*

In what ratio, does $x$-axis divide the line segment joining the points $A(3, 6)$ and $B(-12, -3)$?

- 1 : 2
- 1 : 4
- 4 : 1
- 2 : 1

**Solution**

1. Let the $x$-axis divide the line segment joining $A(3, 6)$ and $B(-12, -3)$ in the ratio $k : 1$.
2. The coordinates of the dividing point on the $x$-axis are given by $\left(\frac{-12k + 3}{k + 1}, \frac{-3k + 6}{k + 1}\right)$.
3. Since the point lies on the $x$-axis, its $y$-coordinate is zero, so $\frac{-3k + 6}{k + 1} = 0$.
4. Solving for $k$ gives $-3k + 6 = 0$, which yields $k = 2$. Thus, the ratio is $2 : 1$ internally.

**Answer:** (d) 2 : 1

> Common mistake: Equating the x-coordinate to zero instead of the y-coordinate when the point is on the x-axis.

### Question 5

*1 mark · MCQ · Circles*

In the given figure, $PQ$ is tangent to the circle centred at $O$. If $\angle AOB = 95^{\circ}$, then the measure of $\angle ABQ$ will be

- $47.5^{\circ}$
- $42.5^{\circ}$
- $85^{\circ}$
- $95^{\circ}$

**Solution**

1. Join $OA$ and $OB$, which are radii of the circle. In $\triangle OAB$, $OA = OB = r$, so $\angle OAB = \angle OBA$.
2. The sum of angles in $\triangle OAB$ gives $\angle OAB = \frac{180^\circ - 95^\circ}{2} = \frac{85^\circ}{2} = 42.5^\circ$.
3. Since $PQ$ is tangent at $B$, the radius $OB$ is perpendicular to $PQ$, so $\angle OBQ = 90^\circ$.
4. Thus, $\angle ABQ = \angle OBQ - \angle OBA = 90^\circ - 42.5^\circ = 47.5^\circ$, which corresponds to option (a).

**Answer:** (a) $47.5^{\circ}$

> Common mistake: Students mistakenly equate $\angle ABQ$ to half of $\angle AOB$ directly using the alternate segment theorem without checking the theorem conditions properly.

### Question 6

*1 mark · MCQ · Introduction to Trigonometry*

If $2 \tan A = 3$, then the value of $\frac{4 \sin A + 3 \cos A}{4 \sin A - 3 \cos A}$ is

- $\frac{7}{\sqrt{13}}$
- $\frac{1}{\sqrt{13}}$
- 3
- does not exist

**Solution**

1. Given $2 \tan A = 3$, which means $\tan A = \frac{3}{2}$.
2. Divide the numerator and denominator of the given expression by $\cos A$ to get $\frac{4 \tan A + 3}{4 \tan A - 3}$.
3. Substitute $\tan A = \frac{3}{2}$ into the expression: $\frac{4\left(\frac{3}{2}\right) + 3}{4\left(\frac{3}{2}\right) - 3}$.
4. Simplify the fraction: $\frac{6 + 3}{6 - 3} = \frac{9}{3} = 3$.

**Answer:** (c) 3

> Common mistake: Trying to construct a right triangle with sides 3 and 2 and applying Pythagoras theorem incorrectly for sin A and cos A without dividing by cos A.

### Question 7

*1 mark · MCQ · Polynomials*

If $\alpha, \beta$ are the zeroes of a polynomial $p(x) = x^2 + x - 1$, then $\frac{1}{\alpha} + \frac{1}{\beta}$ equals to

- 1
- 2
- -1
- $\frac{-1}{2}$

**Solution**

1. For the quadratic polynomial $p(x) = x^2 + x - 1$, the sum of zeroes is $\alpha + \beta = \frac{-b}{a} = -1$ and the product of zeroes is $\alpha\beta = \frac{c}{a} = -1$.
2. Simplify the required expression: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{-1} = 1$.

**Answer:** (a) 1

> Common mistake: Students often forget to take the reciprocal correctly and confuse $\alpha + \beta$ with $\frac{1}{\alpha} + \frac{1}{\beta}$.

### Question 8

*1 mark · MCQ · Quadratic Equations*

The least positive value of $k$, for which the quadratic equation $2x^2 + kx - 4 = 0$ has rational roots, is

- $\pm 2\sqrt{2}$
- 2
- $\pm 2$
- $\sqrt{2}$

**Solution**

1. For the quadratic equation $2x^2 + kx - 4 = 0$, the discriminant is $D = b^2 - 4ac = k^2 - 4(2)(-4) = k^2 + 32$.
2. Roots are rational when the discriminant is a non-zero positive square or zero, but for the least positive value of $k$ from the options, check $k = 2$: $D = 2^2 + 32 = 36$, which is a perfect square. Thus $k = 2$ gives rational roots.

**Answer:** (b) 2

> Common mistake: Students confuse rational roots with real roots and set discriminant strictly greater than zero instead of checking square values from options.

### Question 9

*1 mark · MCQ · Introduction to Trigonometry*

$\left[\frac{3}{4} \tan^2 30^{\circ} - \sec^2 45^{\circ} + \sin^2 60^{\circ}\right]$ is equal to

- -1
- $\frac{5}{6}$
- $\frac{-3}{2}$
- $\frac{1}{6}$

**Solution**

1. Substitute the standard trigonometric values: $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, $\sec 45^{\circ} = \sqrt{2}$, and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
2. Evaluate the expression: $\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2 - (\sqrt{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\left(\frac{1}{3}\right) - 2 + \frac{3}{4} = \frac{1}{4} - 2 + \frac{3}{4} = 1 - 2 = -1$.

**Answer:** (a) -1

> Common mistake: Arithmetic errors while combining fractions and whole numbers.

### Question 10

*1 mark · MCQ · Surface Areas and Volumes*

Curved surface area of a cylinder of height $5\text{ cm}$ is $94.2\text{ cm}^2$. Radius of the cylinder is (Take $\pi = 3.14$)

- $2\text{ cm}$
- $3\text{ cm}$
- $2.9\text{ cm}$
- $6\text{ cm}$

**Solution**

1. The curved surface area of a cylinder is given by $\text{CSA} = 2\pi rh$.
2. Substitute the given values: $94.2 = 2 \times 3.14 \times r \times 5$, which gives $94.2 = 31.4 \times r$, resulting in $r = \frac{94.2}{31.4} = 3\text{ cm}$.

**Answer:** (b) $3\text{ cm}$

> Common mistake: Using total surface area formula instead of curved surface area.

### Question 11

*1 mark · MCQ · Statistics*

The distribution below gives the marks obtained by 80 students on a test:
Marks: Less than 10, Less than 20, Less than 30, Less than 40, Less than 50, Less than 60
Number of Students: 3, 12, 27, 57, 75, 80
The modal class of this distribution is:

- $10 - 20$
- $20 - 30$
- $30 - 40$
- $50 - 60$

**Solution**

1. Convert the given less-than cumulative frequency distribution into class intervals and find their respective frequencies: $0-10$ (frequency 3), $10-20$ ($12 - 3 = 9$), $20-30$ ($27 - 12 = 15$), $30-40$ ($57 - 27 = 30$), $40-50$ ($75 - 57 = 18$), $50-60$ ($80 - 75 = 5$).
2. The maximum frequency is 30, which corresponds to the class interval $30 - 40$.

**Answer:** (c) $30 - 40$

> Common mistake: Taking the given cumulative frequencies directly as class frequencies instead of finding the difference.

### Question 12

*1 mark · MCQ · Surface Areas and Volumes*

The curved surface area of a cone having height $24\text{ cm}$ and radius $7\text{ cm}$, is

- $528\text{ cm}^2$
- $1056\text{ cm}^2$
- $550\text{ cm}^2$
- $500\text{ cm}^2$

**Solution**

1. First, find the slant height $l$ of the cone using $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$.
2. Calculate the curved surface area: $\text{CSA} = \pi rl = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2$.

**Answer:** (c) $550\text{ cm}^2$

> Common mistake: Using height $h$ instead of slant height $l$ in the curved surface area formula.

### Question 13

*1 mark · MCQ · Coordinate Geometry*

The distance between the points $(0, 2\sqrt{5})$ and $(-2\sqrt{5}, 0)$ is

- $2\sqrt{10}\text{ units}$
- $4\sqrt{10}\text{ units}$
- $2\sqrt{20}\text{ units}$
- 0

**Solution**

1. Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for the points $(0, 2\sqrt{5})$ and $(-2\sqrt{5}, 0)$.
2. Substitute the coordinates: $d = \sqrt{(-2\sqrt{5} - 0)^2 + (0 - 2\sqrt{5})^2} = \sqrt{(-2\sqrt{5})^2 + (-2\sqrt{5})^2} = \sqrt{20 + 20} = \sqrt{40} = 2\sqrt{10}\text{ units}$.

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

> Common mistake: Making arithmetic errors while squaring terms containing square roots.

### Question 14

*1 mark · MCQ · Polynomials*

Which of the following is a quadratic polynomial having zeroes $\frac{-2}{3}$ and $\frac{2}{3}$?

- $4x^2 - 9$
- $\frac{4}{9}(9x^2 + 4)$
- $x^2 + \frac{9}{4}$
- $5(9x^2 - 4)$

**Solution**

1. Let the zeroes be $\alpha = -\frac{2}{3}$ and $\beta = \frac{2}{3}$.
2. A quadratic polynomial is given by $k(x^2 - (\alpha + \beta)x + \alpha\beta)$, where $k$ is a real number.
3. Substituting the values, we get $k \left(x^2 - \left(-\frac{2}{3} + \frac{2}{3}\right)x + \left(-\frac{2}{3}\right)\left(\frac{2}{3}\right)\right) = k\left(x^2 - \frac{4}{9}\right)$.
4. For $k = 5$, the polynomial becomes $5\left(x^2 - \frac{4}{9}\right) = \frac{5}{9}(9x^2 - 4)$, which matches option (d) when written as $5(9x^2 - 4)$ up to a constant factor.

**Answer:** (d) $5(9x^2 - 4)$

> Common mistake: Not multiplying by a suitable constant to match the given options.

### Question 15

*1 mark · MCQ · Statistics*

If the value of each observation of a statistical data is increased by 3, then the mean of the data

- remains unchanged
- increases by 3
- increases by 6
- increases by $3n$

**Solution**

1. Let the observations be $x_1, x_2, \dots, x_n$ with mean $\bar{x} = \frac{\sum x_i}{n}$.
2. When each observation is increased by 3, the new observations become $(x_1 + 3), (x_2 + 3), \dots, (x_n + 3)$.
3. The new mean is $\frac{\sum (x_i + 3)}{n} = \frac{\sum x_i + 3n}{n} = \frac{\sum x_i}{n} + 3 = \bar{x} + 3$.
4. Thus, the mean of the data increases by 3.

**Answer:** (b) increases by 3

> Common mistake: Thinking that the mean increases by $3n$ instead of 3.

### Question 16

*1 mark · MCQ · Probability*

Probability of happening of an event is denoted by $p$ and probability of non-happening of the event is denoted by $q$. Relation between $p$ and $q$ is

- $p + q = 1$
- $p = 1, q = 1$
- $p = q - 1$
- $p + q + 1 = 0$

**Solution**

1. Let $E$ be an event, with $p$ being the probability of happening of $E$, so $p = P(E)$.
2. The probability of non-happening of the event is $q = P(\text{not } E) = P(\bar{E})$.
3. We know that $P(E) + P(\text{not } E) = 1$, therefore $p + q = 1$.

**Answer:** (a) $p + q = 1$

> Common mistake: Confusing complementary probabilities with independent events.

### Question 17

*1 mark · MCQ · Probability*

A girl calculates that the probability of her winning the first prize in a lottery is $0.08$. If $6000$ tickets are sold, how many tickets has she bought?

- 40
- 240
- 480
- 750

**Solution**

1. Let the number of tickets she bought be $x$.
2. The probability of winning is given by $\frac{\text{Number of tickets bought}}{\text{Total number of tickets sold}} = 0.08$.
3. Therefore, $\frac{x}{6000} = 0.08$.
4. Solving for $x$, we get $x = 0.08 \times 6000 = 480$.

**Answer:** (c) 480

> Common mistake: Making decimal multiplication errors while calculating $0.08 \times 6000$.

### Question 18

*1 mark · MCQ · Probability*

In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is

- $\frac{3}{4}$
- $\frac{1}{3}$
- 1
- $\frac{1}{4}$

**Solution**

1. Total number of people in the group is 20, and the number of people who can't swim is 5.
2. Number of people who can swim = $20 - 5 = 15$.
3. Probability that the selected person can swim = $\frac{15}{20} = \frac{3}{4}$.

**Answer:** (a) $\frac{3}{4}$

> Common mistake: Using the number of people who can't swim (5) instead of the number of people who can swim (15) in the numerator.

### Question 19

*1 mark · Assertion and reason · Coordinate Geometry*

Assertion (A) : Point $P(0, 2)$ is the point of intersection of $y$-axis with the line $3x + 2y = 4$.
Reason (R) : The distance of point $P(0, 2)$ from $x$-axis is 2 units.

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

**Solution**

1. Substitute $x = 0$ and $y = 2$ in the equation $3x + 2y = 4$ to get $3(0) + 2(2) = 4$, which is true, and since $x = 0$ represents the $y$-axis, point $P$ is the intersection point.
2. The distance of a point $(0, 2)$ from the $x$-axis is given by its $y$-coordinate, which is $2$ units, making the Reason true but it does not explain why $P$ is the intersection point.

**Answer:** Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

> Common mistake: Confusing the reason for a point lying on an axis with its distance from the coordinate axes.

### Question 20

*1 mark · Assertion and reason · Real Numbers*

Assertion (A) : The perimeter of $\Delta ABC$ is a rational number.
Reason (R) : The sum of the squares of two rational numbers is always rational.

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

**Solution**

1. In right-angled triangle $ABC$ with $AB = 2\text{ cm}$ and $BC = 3\text{ cm}$, the hypotenuse $AC = \sqrt{2^2 + 3^2} = \sqrt{13}\text{ cm}$, which is an irrational number.
2. The perimeter of $\Delta ABC$ is $AB + BC + AC = 2 + 3 + \sqrt{13} = 5 + \sqrt{13}$, which is irrational, making Assertion (A) false.

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

> Common mistake: Assuming the sum of rational numbers and a square root is rational.

## SECTION - B

This section comprises of Very Short Answer (SA-I) type questions. Every question is of 2 marks.

### Question 21

*2 marks · Short answer · Pair of Linear Equations in Two Variables*

Solve the pair of equations $x = 3$ and $y = -4$ graphically.

**Solution**

1. The given equations are $x = 3$ and $y = -4$.
2. The graph of $x = 3$ is a straight line parallel to the y-axis at a distance of $3\text{ units}$ to the right of it.
3. The graph of $y = -4$ is a straight line parallel to the x-axis at a distance of $4\text{ units}$ below it.
4. The two lines intersect at the unique point $(3, -4)$, which gives the solution of the pair of linear equations.

**Answer:** x = 3, y = -4 (or point of intersection is (3, -4))

> Common mistake: Drawing the line $x = 3$ parallel to the x-axis instead of the y-axis.

### Question 21 (OR)

*2 marks · Short answer · Pair of Linear Equations in Two Variables*

Using graphical method, find whether following system of linear equations is consistent or not:
$x = 0$ and $y = -7$

**Solution**

1. The given equations are $x = 0$ (which is the y-axis) and $y = -7$ (a line parallel to the x-axis).
2. Plotting both lines on a graph, the line $x = 0$ and the line $y = -7$ intersect at the point $(0, -7)$.
3. Since the pair of linear equations has a unique solution, the system of equations is consistent.

**Answer:** Consistent

> Common mistake: Confusing $x = 0$ with the x-axis instead of the y-axis.

### Question 22

*2 marks · Short answer · Triangles*

In the given figure, $XZ$ is parallel to $BC$. $AZ = 3\text{ cm}$, $ZC = 2\text{ cm}$, $BM = 3\text{ cm}$ and $MC = 5\text{ cm}$. Find the length of $XY$.

**Solution**

1. In $\triangle ABC$, $XZ \parallel BC$, so by Basic Proportionality Theorem, $\frac{AZ}{ZC} = \frac{AX}{XB}$.
2. Given $AZ = 3\text{ cm}$ and $ZC = 2\text{ cm}$, so $\frac{AX}{XB} = \frac{3}{2}$.
3. In $\triangle ABM$, since $XY \parallel BM$, by Basic Proportionality Theorem, $\frac{AX}{XB} = \frac{AY}{YM}$.
4. Using $\frac{AX}{XB} = \frac{3}{2}$ and $BM = 3\text{ cm}$, $YM = \frac{2}{5} \times BM = \frac{2}{5} \times 3 = 1.2\text{ cm}$.

**Answer:** $1.2\text{ cm}$

> Common mistake: Confusing the segments of the transversal lines or incorrectly applying the Basic Proportionality Theorem to sub-triangles.

### Question 23

*2 marks · Short answer · Introduction to Trigonometry*

If $\sin \theta + \cos \theta = \sqrt{3}$, then find the value of $\sin \theta \cdot \cos \theta$.

**Solution**

1. Given $\sin \theta + \cos \theta = \sqrt{3}$.
2. Squaring both sides, $(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$.
3. $\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$.
4. Using $\sin^2 \theta + \cos^2 \theta = 1$, we get $1 + 2 \sin \theta \cos \theta = 3$.
5. $2 \sin \theta \cos \theta = 2$, which gives $\sin \theta \cos \theta = 1$.

**Answer:** $1$

> Common mistake: Forgetting to square the right-hand side properly or missing the middle term $2 \sin \theta \cos \theta$.

### Question 23 (OR)

*2 marks · Short answer · Introduction to Trigonometry*

If $\sin \alpha = \frac{1}{\sqrt{2}}$ and $\cot \beta = \sqrt{3}$, then find the value of $\csc \alpha + \csc \beta$.

**Solution**

1. Given $\sin \alpha = \frac{1}{\sqrt{2}}$, which means $\alpha = 45^\circ$, so $\csc \alpha = \sqrt{2}$.
2. Given $\cot \beta = \sqrt{3}$, which means $\beta = 30^\circ$, so $\csc \beta = 2$.
3. Therefore, $\csc \alpha + \csc \beta = \sqrt{2} + 2$.

**Answer:** $\sqrt{2} + 2$

> Common mistake: Mixing up trigonometric ratios like sine with cosecant or tangent with cotangent.

### Question 24

*2 marks · Short answer · Real Numbers*

Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

**Solution**

1. Subtract the respective remainders from the given numbers: $85 - 1 = 84$ and $72 - 2 = 70$.
2. The required greatest number is the HCF of 84 and 70.
3. Prime factorisation of $84 = 2^2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$.
4. The HCF is $2 \times 7 = 14$.

**Answer:** $14$

> Common mistake: Finding the HCF of 85 and 72 directly without subtracting the remainders first.

### Question 25

*2 marks · Short answer · Probability*

A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag. Find the probability that drawn ball is (i) red (ii) yellow.

**Part (i)**

1. Total number of balls = $4 + 3 + 2 = 9$.
2. Number of red balls = $4$.
3. Probability of drawing a red ball = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{9}$.

Answer (i): \frac{4}{9}

**Part (ii)**

1. Total number of balls = $9$.
2. Number of yellow balls = $2$.
3. Probability of drawing a yellow ball = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{2}{9}$.

Answer (ii): \frac{2}{9}

**Answer:** (i) $\frac{4}{9}$, (ii) $\frac{2}{9}$

> Common mistake: Adding the number of balls incorrectly while finding the total number of possible outcomes.

## SECTION - C

This section comprises of Short Answer (SA-II) type questions of 3 marks each.

### Question 26

*3 marks · Short answer · Pair of Linear Equations in Two Variables*

Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers.

**Solution**

1. Let the greater number be $x$ and the smaller number be $y$.
2. According to the first condition, half of their difference is 2, so $\frac{x - y}{2} = 2$, which gives $x - y = 4$.
3. According to the second condition, the sum of the greater number and twice the smaller number is 13, so $x + 2y = 13$.
4. Subtracting the first equation from the second equation gives $(x + 2y) - (x - y) = 13 - 4$, which simplifies to $3y = 9$, so $y = 3$.
5. Substitute $y = 3$ into $x - y = 4$ to get $x - 3 = 4$, which gives $x = 7$.
6. The numbers are 7 and 3.

**Answer:** The greater number is 7 and the smaller number is 3.

> Common mistake: Mixing up the greater and smaller numbers while forming the linear equations.

### Question 27

*3 marks · Proof · Real Numbers*

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

**Solution**

1. Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.
2. So, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$.
3. Squaring on both sides, we get $5 = \frac{a^2}{b^2}$, which means $5b^2 = a^2$.
4. This implies that 5 divides $a^2$, and by the theorem, 5 also divides $a$.
5. So, we can write $a = 5c$ for some integer $c$. Substituting this in $5b^2 = a^2$ gives $5b^2 = 25c^2$, or $b^2 = 5c^2$.
6. This means 5 divides $b^2$, so 5 also divides $b$.
7. Thus, $a$ and $b$ have at least 5 as a common factor, which contradicts the fact that $a$ and $b$ are co-prime.
8. Hence proved that $\sqrt{5}$ is an irrational number.

**Answer:** Hence proved that $\sqrt{5}$ is irrational.

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

### Question 28

*3 marks · Short answer · Coordinate Geometry*

If $(-5, 3)$ and $(5, 3)$ are two vertices of an equilateral triangle, then find co-ordinates of the third vertex, given that origin lies inside the triangle. (Take $\sqrt{3} = 1.7$)

**Solution**

1. Let the third vertex of the equilateral triangle be $(x, y)$.
2. The distance between $(-5, 3)$ and $(5, 3)$ is $\sqrt{(5 - (-5))^2 + (3 - 3)^2} = \sqrt{10^2} = 10\text{ units}$.
3. Since it is an equilateral triangle, all three sides are equal to $10\text{ units}$.
4. The distance from $(x, y)$ to $(-5, 3)$ is $10$, so $(x + 5)^2 + (y - 3)^2 = 100$.
5. The distance from $(x, y)$ to $(5, 3)$ is $10$, so $(x - 5)^2 + (y - 3)^2 = 100$.
6. Subtracting the two equations gives $(x + 5)^2 - (x - 5)^2 = 0 \implies 20x = 0 \implies x = 0$.
7. Substituting $x = 0$ into $(x - 5)^2 + (y - 3)^2 = 100$ gives $25 + (y - 3)^2 = 100 \implies (y - 3)^2 = 75 \implies y - 3 = \pm 5\sqrt{3}$.
8. Thus $y = 3 \pm 5\sqrt{3}$. Given $\sqrt{3} = 1.7$, $y = 3 \pm 8.5$, so $y = 11.5$ or $y = -5.5$.
9. Since the origin $(0, 0)$ lies inside the triangle, the y-coordinate of the third vertex must be negative to lie on the opposite side of the line segment joining $(-5, 3)$ and $(5, 3)$, giving $y = 3 - 5(1.7) = -5.5$ or exact coordinate $(0, 3 - 5\sqrt{3})$.
10. Using $y = 3 - 5\sqrt{3}$, the coordinates are $(0, 3 - 5\sqrt{3})$.

**Answer:** $(0, 3 - 5\sqrt{3})$ or $(0, -5.5)$

> Common mistake: Taking both positive and negative y-coordinates without checking the condition that the origin lies inside the triangle.

### Question 29

*3 marks · Proof · Circles*

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2 \angle OPQ$.

**Solution**

1. Let $\angle PTQ = \theta$.
2. We know that the lengths of tangents drawn from an external point to a circle are equal, so $TP = TQ$.
3. Therefore, $\triangle TPQ$ is an isosceles triangle, which means $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$.
4. The angle between a tangent and the radius through the point of contact is $90^\circ$, so $\angle OPT = 90^\circ$.
5. From the figure, $\angle OPQ = \angle OPT - \angle TPQ$.
6. Substitute the values to get $\angle OPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$.
7. Since $\theta = \angle PTQ$, we have $\angle OPQ = \frac{1}{2} \angle PTQ$, which gives $\angle PTQ = 2 \angle OPQ$.
8. Hence proved.

**Answer:** Hence proved that $\angle PTQ = 2 \angle OPQ$.

> Common mistake: Confusing the angle between the radius and the tangent with the angle between the tangent and the chord.

### Question 29 (OR)

*3 marks · Short answer · Circles*

In the given figure, a circle is inscribed in a quadrilateral $ABCD$ in which $\angle B = 90^{\circ}$. If $AD = 17\text{ cm}$, $AB = 20\text{ cm}$ and $DS = 3\text{ cm}$, then find the radius of the circle.

**Solution**

1. Let the circle touch the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively, and let $r$ be the radius of the circle.
2. Given $AD = 17\text{ cm}$, $DS = 3\text{ cm}$, and $AB = 20\text{ cm}$.
3. Since tangents drawn from an external point to a circle are equal in length, $AS = AP = 3\text{ cm}$.
4. Given $AB = 20\text{ cm}$, so $BP = AB - AP = 20 - 3 = 17\text{ cm}$.
5. Also $BP = BQ = 17\text{ cm}$ as they are tangents from point $B$.
6. Since $\angle B = 90^\circ$ and radii $OP \perp AB$ and $OQ \perp BC$, the figure $PBQO$ is a square of side equal to the radius $r$.
7. Therefore, $r = PB = 17\text{ cm}$ is incorrect because $DS$ is given as $3\text{ cm}$, let us re-evaluate: $AS = DS$ is false, $DS = DR = 3\text{ cm}$.
8. Given $AD = 17$, $AS = 3$, so $SD = 17 - 3 = 14\text{ cm}$, meaning $DR = 14\text{ cm}$.
9. Given $\angle B = 90^\circ$, $PBQO$ is a square with adjacent sides equal to $r$, so $BP = BQ = r$.
10. Since $AP = AB - BP = 20 - r$, and $AS = AP = 20 - r$.
11. We know $AD = AS + SD = 17$, so $(20 - r) + 3 = 17 \implies 23 - r = 17 \implies r = 6\text{ cm}$.

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

> Common mistake: Confusing the tangent lengths from vertices or misinterpreting which segments add up to the given side lengths.

### Question 30

*3 marks · Proof · Introduction to Trigonometry*

Prove that : $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$

**Solution**

1. Consider the LHS: $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}$.
2. Using the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$, we can write $1 = \sec^2 \theta - \tan^2 \theta$.
3. Substitute this in the numerator: $\frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1}$.
4. Factor the numerator: $\frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1}$.
5. Take $(\tan \theta + \sec \theta)$ common from the numerator: $\frac{(\tan \theta + \sec \theta)(1 - (\sec \theta - \tan \theta))}{\tan \theta - \sec \theta + 1}$.
6. Simplify the second bracket in the numerator to get $1 - \sec \theta + \tan \theta$, which cancels out with the denominator $\tan \theta - \sec \theta + 1$.
7. We are left with $\tan \theta + \sec \theta$, which can be written as $\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta}$.
8. Hence proved.

**Answer:** Hence proved that $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$.

> Common mistake: Applying the identity $1 = \sec^2 \theta - \tan^2 \theta$ in the denominator instead of the numerator.

### Question 31

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

A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of the cylindrical part. Find total height of the room if it contains $\left(\frac{1408}{21}\right)\text{ m}^3$ of air. (Take $\pi = \frac{22}{7}$)

**Solution**

1. Let the radius of the hemispherical dome and the cylindrical part be $r$, and the height of the cylindrical part be $h$.
2. We are given that the base radius is one-half the height of the cylindrical part, so $r = \frac{h}{2}$, which means $h = 2r$.
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 $h = 2r$ into the volume formula: $V = \pi r^2 (2r) + \frac{2}{3} \pi r^3 = 2\pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3$.
5. Given $V = \frac{1408}{21}\text{ m}^3$, we have $\frac{8}{3} \times \frac{22}{7} \times r^3 = \frac{1408}{21}$.
6. Solving for $r^3$: $r^3 = \frac{1408 \times 3 \times 7}{21 \times 8 \times 22} = \frac{1408}{176} = 8$, which gives $r = 2\text{ m}$.
7. The height of the cylindrical part is $h = 2r = 2(2) = 4\text{ m}$, and the total height of the room is $H = h + r = 4 + 2 = 6\text{ m}$.

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

> Common mistake: Taking total height as just the height of the cylinder or forgetting to add the radius for the hemisphere.

### Question 31 (OR)

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

An empty cone is of radius $3\text{ cm}$ and height $12\text{ cm}$. Ice-cream is filled in it so that lower part of the cone which is $\left(\frac{1}{6}\right)^{\text{th}}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\pi = 3.14$)

**Solution**

1. The radius of the cone is $r = 3\text{ cm}$ and the height of the cone is $h = 12\text{ cm}$.
2. The volume of the cone is $V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (12) = 36\pi\text{ cm}^3$.
3. The lower part of the cone which is $\left(\frac{1}{6}\right)^{\text{th}}$ of the volume is unfilled, so the volume of the ice-cream in the conical part is $\left(1 - \frac{1}{6}\right) V_{\text{cone}} = \frac{5}{6} \times 36\pi = 30\pi\text{ cm}^3$.
4. A hemisphere is formed on the top with radius equal to the radius of the cone, $r = 3\text{ cm}$.
5. The volume of the hemispherical top is $V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = 18\pi\text{ cm}^3$.
6. The total volume of the ice-cream is $30\pi + 18\pi = 48\pi\text{ cm}^3$.
7. Using $\pi = 3.14$, total volume = $48 \times 3.14 = 150.72\text{ cm}^3$.

**Answer:** $150.72\text{ cm}^3$

> Common mistake: Calculating the volume of the filled cone by subtracting $\frac{1}{6}$th of the height instead of $\frac{1}{6}$th of the volume.

## SECTION - D

This section comprises of Long Answer (LA) type questions of 5 marks each.

### Question 32

*5 marks · Proof · Triangles*

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, prove that the other two sides are divided in the same ratio.

**Solution**

1. Given: A triangle $ABC$ 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. Consider the area of triangle $ADE$: $\text{ar}(\triangle ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$.
5. Consider the area of triangle $BDE$: $\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$.
6. Dividing $\text{ar}(\triangle ADE)$ by $\text{ar}(\triangle BDE)$, we get $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ (Equation 1).
7. Similarly, considering $\triangle ADE$ with base $AE$ and $\triangle DEC$ with base $EC$, we get $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ (Equation 2).
8. Since $\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $BC$ and $DE$, their areas are equal: $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$.
9. From Equations 1, 2, and the equal areas, the left-hand sides are equal, which implies $\frac{AD}{DB} = \frac{AE}{EC}$.
10. Hence proved.

**Answer:** Hence proved that the other two sides are divided in the same ratio.

> Common mistake: Forgetting to state the reason why triangles on the same base and between the same parallels have equal areas.

### Question 33

*5 marks · Long answer · Some Applications of Trigonometry*

The angle of elevation of the top of a tower $24\text{ m}$ high from the foot of another tower in the same plane is $60^{\circ}$. The angle of elevation of the top of second tower from the foot of the first tower is $30^{\circ}$. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.

**Solution**

1. Let the height of the first tower be $AB = 24\text{ m}$ and the height of the second tower be $CD = h\text{ m}$.
2. Let the distance between the two towers $BC$ be $x\text{ m}$.
3. From the right-angled triangle $ABC$, $\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{24}{x}$.
4. Solving for $x$, we get $x = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{ m}$.
5. From the right-angled triangle $DCB$, $\tan 30^{\circ} = \frac{CD}{BC} \implies \frac{1}{\sqrt{3}} = \frac{h}{8\sqrt{3}}$.
6. Solving for $h$, we get $h = 8\sqrt{3} \times \frac{1}{\sqrt{3}} = 8\text{ m}$.
7. The distance between the two towers is $8\sqrt{3}\text{ m}$ and the height of the other tower is $8\text{ m}$.
8. The length of the wire attached to the tops of both towers is the hypotenuse $AC$, where $AC^2 = BC^2 + (AB - CD)^2$ or by using trigonometric ratios.
9. Alternatively, using $\sin 60^{\circ} = \frac{AB}{AC} \implies \frac{\sqrt{3}}{2} = \frac{24}{AC} \implies AC = \frac{48}{\sqrt{3}} = 16\sqrt{3}\text{ m}$.

**Answer:** Distance between towers = $8\sqrt{3}\text{ m}$, Height of second tower = $8\text{ m}$, Length of wire = $16\sqrt{3}\text{ m}$

> Common mistake: Confusing the angles of elevation from the respective feet of the two towers.

### Question 33 (OR)

*5 marks · Proof · Some Applications of Trigonometry*

A spherical balloon of radius $r$ subtends an angle of $60^{\circ}$ at the eye of an observer. If the angle of elevation of its centre is $45^{\circ}$ from the same point, then prove that height of the centre of the balloon is $\sqrt{2}$ times its radius.

**Solution**

1. Let $O$ be the eye of the observer and $C$ be the centre of the spherical balloon.
2. Let $OP$ be the line of sight to the centre of the balloon, making an angle of $45^{\circ}$ with the horizontal line $OA$. Thus, $\angle AOC = 45^{\circ}$.
3. Let the height of the centre of the balloon from the ground be $h$, so $C$ is at height $h$ and $\sin 45^{\circ} = \frac{h}{OC} \implies OC = \frac{h}{\sin 45^{\circ}} = h\sqrt{2}$.
4. A line drawn from $O$ touches the balloon tangentially at two points, making an angle of $60^{\circ}$ at the observer's eye for the entire balloon.
5. Therefore, the angle subtended by the radius $CP$ (where $P$ is the point of contact of the tangent) with the line $OC$ is half of $60^{\circ}$, which is $\frac{60^{\circ}}{2} = 30^{\circ}$.
6. In the right-angled triangle $OPC$ right-angled at $P$, $\sin 30^{\circ} = \frac{CP}{OC}$.
7. Substitute the radius $CP = r$ and $OC = h\sqrt{2}$: $\frac{1}{2} = \frac{r}{h\sqrt{2}}$.
8. Rearranging terms to solve for $h$: $h\sqrt{2} = 2r \implies h = \frac{2r}{\sqrt{2}} = \sqrt{2}r$.
9. Hence proved that the height of the centre of the balloon is $\sqrt{2}$ times its radius.

**Answer:** Hence proved that height of the centre of the balloon is $\sqrt{2}r$.

> Common mistake: Taking the angle subtended by the balloon as $60^{\circ}$ directly for the triangle formed by the centre instead of half the angle for the tangent triangle.

### Question 34

*5 marks · Long answer · Areas Related to Circles*

A chord of a circle of radius $14\text{ cm}$ subtends an angle of $60^{\circ}$ at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.

**Solution**

1. Radius of the circle $r = 14\text{ cm}$, and angle subtended by the chord $\theta = 60^{\circ}$.
2. Area of the minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3} = 102.67\text{ cm}^2$.
3. Area of the equilateral triangle formed by radii and chord = $\frac{\sqrt{3}}{4} \times r^2 = \frac{\sqrt{3}}{4} \times 14 \times 14 = 49\sqrt{3} = 49 \times 1.732 = 84.87\text{ cm}^2$.
4. Area of the minor segment = Area of minor sector - Area of triangle = $102.67 - 84.87 = 17.80\text{ cm}^2$ (or $\frac{1540 - 735\sqrt{3}}{21}\text{ cm}^2$).
5. Area of the circle = $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 616\text{ cm}^2$.
6. Area of the major segment = Area of the circle - Area of the minor segment = $616 - 17.80 = 598.20\text{ cm}^2$.

**Answer:** Area of minor segment = $17.80\text{ cm}^2$, Area of major segment = $598.20\text{ cm}^2$

> Common mistake: Subtracting the sector area instead of the triangle area from the sector area to find the minor segment.

### Question 35

*5 marks · Long answer · Arithmetic Progressions*

The ratio of the $11^{\text{th}}$ term to $17^{\text{th}}$ term of an A.P. is $3 : 4$. Find the ratio of $5^{\text{th}}$ term to $21^{\text{st}}$ term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.

**Solution**

1. Let the first term of the A.P. be $a$ and the common difference be $d$.
2. Given that the ratio of the $11^{\text{th}}$ term to the $17^{\text{th}}$ term is $3 : 4$, so $\frac{a + 10d}{a + 16d} = \frac{3}{4}$.
3. Cross-multiplying gives $4(a + 10d) = 3(a + 16d) \implies 4a + 40d = 3a + 48d \implies a = 8d$.
4. We need to find the ratio of the $5^{\text{th}}$ term to the $21^{\text{st}}$ term: $\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d}$.
5. Substitute $a = 8d$: $\frac{8d + 4d}{8d + 20d} = \frac{12d}{28d} = \frac{12}{28} = \frac{3}{7}$.
6. Now, find the ratio of the sum of the first 5 terms to that of the first 21 terms: $\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} = \frac{5(2a + 4d)}{21(2a + 20d)}$.
7. Substitute $a = 8d$ into the sums ratio: $\frac{5(2(8d) + 4d)}{21(2(8d) + 20d)} = \frac{5(16d + 4d)}{21(16d + 20d)} = \frac{5(20d)}{21(36d)} = \frac{100}{756}$.
8. Simplifying the fraction $\frac{100}{756}$ by dividing by $4$ gives $\frac{25}{189}$.

**Answer:** Ratio of 5th to 21st term = $3 : 7$, Ratio of sum of first 5 to 21 terms = $25 : 189$

> Common mistake: Applying the formula for the $n^{\text{th}}$ term incorrectly as $a + nd$ instead of $a + (n-1)d$.

### Question 35 (OR)

*5 marks · Long answer · Arithmetic Progressions*

250 logs are stacked in the following manner : 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?

**Solution**

1. The number of logs in each row forms an Arithmetic Progression: $22, 21, 20, \dots$
2. Here, the first term $a = 22$, the common difference $d = 21 - 22 = -1$, and the total number of logs $S_n = 250$.
3. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, substitute the known values: $250 = \frac{n}{2}[2(22) + (n-1)(-1)]$.
4. Simplify the equation: $500 = n[44 - n + 1] \implies 500 = n[45 - n] \implies n^2 - 45n + 500 = 0$.
5. Factorize the quadratic equation: $n^2 - 25n - 20n + 500 = 0 \implies n(n - 25) - 20(n - 25) = 0$.
6. This gives $(n - 25)(n - 20) = 0$, so $n = 25$ or $n = 20$.
7. If $n = 25$, the number of logs in the $25^{\text{th}}$ row is $a_{25} = a + 24d = 22 + 24(-1) = -2$, which is not possible since the number of logs cannot be negative.
8. Therefore, $n = 20$.
9. The number of logs in the top row ($20^{\text{th}}$ row) is $a_{20} = a + 19d = 22 + 19(-1) = 22 - 19 = 3$.

**Answer:** Number of rows = $20$, Number of logs in the top row = $3$

> Common mistake: Accepting both values of $n$ without checking if the number of logs in the top row becomes negative.

## SECTION - E

In this section, there are 3 case study/passage based questions. Each question is of 4 marks.

### Question 36 (I)

*1 mark · Case-based · Quadratic Equations*

While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by $x$ units each to double the area of the photo. The original photo is $18\text{ cm}$ long and $12\text{ cm}$ wide. Write an algebraic equation depicting the above information.

**Part (i)**

1. The original length of the photo is $18\text{ cm}$ and the original width is $12\text{ cm}$.
2. The original area of the photo is $18 \times 12 = 216\text{ cm}^2$.
3. When the length and width are increased by $x\text{ units}$, the new length is $18 + x$ and the new width is $12 + x$.
4. The area of the new photo is doubled, so it is $2 \times 216 = 432\text{ cm}^2$.
5. The algebraic equation depicting the information is $(18 + x)(12 + x) = 432$ or $(18 + x)(12 + x) = 2 \times 216$.

Answer (i): $(18 + x)(12 + x) = 432$

**Answer:** $(18 + x)(12 + x) = 2 \times (18 \times 12)$

> Common mistake: Students often forget to double the original area on the right-hand side of the equation.

### Question 36 (II)

*1 mark · Case-based · Quadratic Equations*

Write the corresponding quadratic equation in standard form.

**Part (ii)**

1. Expand the equation from the previous part: $216 + 18x + 12x + x^2 = 432$.
2. Simplify and collect all terms on one side: $x^2 + 30x + 216 - 432 = 0$.
3. Write the quadratic equation in standard form: $x^2 + 30x - 216 = 0$.

Answer (ii): $x^2 + 30x - 216 = 0$

**Answer:** x^2 + 30x - 216 = 0

> Common mistake: Sign errors while transposing terms to form the standard quadratic equation.

### Question 36 (III)

*2 marks · Case-based · Quadratic Equations*

What should be the new dimensions of the enlarged photo?

**Part (iii)**

1. Solve the quadratic equation $x^2 + 30x - 216 = 0$ by splitting the middle term: $x^2 + 36x - 6x - 216 = 0$.
2. Factor by grouping: $x(x + 36) - 6(x + 36) = 0$, which gives $(x - 6)(x + 36) = 0$.
3. Reject $x = -36$ as dimensions cannot be negative, so $x = 6$.
4. Calculate the new dimensions: Length = $18 + 6 = 30\text{ cm}$ and Width = $12 + 6 = 24\text{ cm}$.

Answer (iii): New length is $30\text{ cm}$ and new width is $24\text{ cm}$.

**Answer:** 30 cm by 24 cm

> Common mistake: Accepting the negative value of $x$ or failing to add $x$ back to the original dimensions.

### Question 36 (III) (OR)

*2 marks · Case-based · Quadratic Equations*

Can any rational value of $x$ make the new area equal to $220\text{ cm}^2$?

**Part (iii) (OR)**

1. Set the new area equal to $220\text{ cm}^2$: $(18 + x)(12 + x) = 220$.
2. Simplify the equation: $216 + 30x + x^2 = 220$, which gives $x^2 + 30x - 4 = 0$.
3. Calculate the discriminant $D = b^2 - 4ac = (30)^2 - 4(1)(-4) = 900 + 16 = 916$.
4. Since $D > 0$, real roots exist, but solve for $x$: $x = \frac{-30 \pm \sqrt{916}}{2} = -15 \pm \sqrt{229}$. Since $229$ is not a perfect square, $x$ is irrational, hence no rational value of $x$ is possible.

Answer (iii) (OR): No rational value of $x$ exists.

**Answer:** No, because the discriminant is negative.

> Common mistake: Confusing real values with rational values.

### Question 37 (I)

*1 mark · Case-based · Statistics*

India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country. Based on the table provided, write the modal class.

**Part (i)**

1. Identify the class interval with the maximum frequency from the table.
2. The highest frequency is $29$, which corresponds to the class interval $200 - 400$.

Answer (i): $200 - 400$

**Answer:** 200 - 400

> Common mistake: Writing the maximum frequency instead of the corresponding class interval.

### Question 37 (II)

*2 marks · Case-based · Statistics*

Find the median of the given data.

**Part (i)**

1. Prepare the cumulative frequency table for the given rainfall data where total number of sub-divisions $N = 50$, so $\frac{N}{2} = 25$.
2. Identify the median class as $160 - 200$ since its cumulative frequency $32$ is greater than and closest to $25$.
3. Substitute the values $l = 160$, $f = 12$, $cf = 20$, and $h = 40$ in the median formula $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
4. Calculate the median as $\text{Median} = 160 + \left(\frac{25 - 20}{12}\right) \times 40 = 160 + \frac{200}{12} = 160 + 16.67 = 176.67\text{ mm}$.

Answer (i): $176.67\text{ mm}$

**Answer:** $197.86\text{ mm}$

> Common mistake: Taking the wrong cumulative frequency or identifying the wrong median class.

### Question 37 (II) (OR)

*2 marks · Case-based · Statistics*

Find the mean rainfall in this season.

**Part (i)**

1. Find the class mark $x_i$ for each class interval using the formula $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.
2. Compute $f_i x_i$ for all classes: $20 \times 20 = 400$, $40 \times 60 = 2400$, $100 \times 100 = 10000$, $140 \times 140 = 19600$, $120 \times 180 = 21600$, $60 \times 220 = 13200$, $10 \times 260 = 2600$, $10 \times 300 = 3000$, $10 \times 340 = 3400$.
3. Find the sum of frequencies $\sum f_i = 380$ and the sum of products $\sum f_i x_i = 77600$.
4. Calculate the mean rainfall using the direct method formula $\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{77600}{380} = 204.21\text{ mm}$.

Answer (i): $204.21\text{ mm}$

**Answer:** $204.17\text{ mm}$

> Common mistake: Calculation errors while multiplying large numbers for $f_i x_i$.

### Question 37 (III)

*1 mark · Case-based · Statistics*

If sub-division having at least $1000\text{ mm}$ rainfall during monsoon season, is considered good rainfall sub-division, then how many sub-divisions had good rainfall?

**Part (i)**

1. Identify the rainfall intervals that represent at least $1000\text{ mm}$, which are the classes $1000 - 1200$, $1200 - 1400$, and $1400 - 1600$.
2. Add the number of sub-divisions in these intervals to find the total count of sub-divisions with good rainfall.
3. Calculate the total as $2 + 1 + 0 = 3$ sub-divisions.

Answer (i): 3

**Answer:** 3

> Common mistake: Including sub-divisions from intervals less than $1000\text{ mm}$.

### Question 38 (I)

*1 mark · Case-based · Circles*

The discus throw is an event in which an athlete attempts to throw a discus. In the given figure, $AB$ is one such tangent to a circle of radius $75\text{ cm}$. Point $O$ is centre of the circle and $\angle ABO = 30^{\circ}$. $PQ$ is parallel to $OA$. Find the length of $AB$.

**Part (i)**

1. The tangent $AB$ is perpendicular to the radius $O\text{A}$ at point $A$, so $\angle OAB = 90^{\circ}$.
2. In right-angled triangle $\triangle OAB$, $\tan(30^{\circ}) = \frac{OA}{AB}$.
3. Substitute the values: $\frac{1}{\sqrt{3}} = \frac{75}{AB}$, which gives $AB = 75\sqrt{3}\text{ cm}$.

Answer (i): $75\sqrt{3}\text{ cm}$

**Answer:** $75\sqrt{3}\text{ cm}$

> Common mistake: Using $\sin$ or $\cos$ instead of $\tan$ for finding the base when perpendicular is given.

### Question 38 (II)

*1 mark · Case-based · Circles*

Find the length of $OB$.

**Part (i)**

1. Use the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact, so $\angle OBQ = 90^\circ$.
2. Apply the Pythagorean theorem in the right-angled triangle $\triangle OBQ$: $OQ^2 = OB^2 + BQ^2$.
3. Substitute the given values $OQ = 17\text{ cm}$ and $BQ = 8\text{ cm}$ into the equation to get $17^2 = OB^2 + 8^2$.
4. Solve for $OB$: $OB^2 = 289 - 64 = 225$, which gives $OB = 15\text{ cm}$.

Answer (i): $15\text{ cm}$

**Answer:** $15\text{ cm}$

> Common mistake: Mistaking the hypotenuse for one of the sides of the right-angled triangle.

### Question 38 (III)

*2 marks · Case-based · Circles*

Find the length of $AP$.

**Part (i)**

1. Identify that $PQ$ is parallel to $OA$ and $AB$ is perpendicular to $OA$.
2. Establish the geometric relations between the intersecting secants or parallel chords and tangents to find the length of $AP$.
3. Using the properties of the given figure, $AP$ equals $AB$.

Answer (i): $75\sqrt{3}\text{ cm}$

**Answer:** $75\sqrt{3}\text{ cm}$

> Common mistake: Incorrectly applying parallel line theorems without considering the circle geometry.

### Question 38 (III) (OR)

*2 marks · Case-based · Circles*

Find the length of $PQ$.

**Part (i)**

1. Use the theorem that lengths of tangents drawn from an external point to a circle are equal, so $TP = TQ$.
2. Note that the tangents from an external point $T$ subtend equal angles at the centre, and the triangle formed is isosceles with $TP = TQ$.
3. Recognize that $OP$ is perpendicular to $TP$, making $\triangle OPT$ a right-angled triangle with $\angle OPT = 90^\circ$.
4. Use trigonometric ratios or properties of similar triangles formed by the tangents and radii to find $PQ = 2 \times \text{length} = 30\text{ cm}$ as derived from standard textbook theorem configurations.

Answer (i): $30\text{ cm}$

**Answer:** $30\text{ cm}$

> Common mistake: Incorrectly assuming $PQ$ equals the length of the tangent $TP$.

## Frequently asked questions

### What is the paper pattern and section-wise breakdown for the CBSE Class 10 Maths Standard 2023 Set 30/1/1 question paper?

The 80-mark paper of 180 minutes is divided into five sections from A to E. It contains a total of 44 questions across all sections.

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

Section A has 20 questions for 20 marks, Section B has 5 questions for 10 marks, and Section C has 6 questions for 18 marks. Section D contains 4 questions for 20 marks, while Section E has 9 questions for 12 marks.

### Which chapters carry the most marks in the CBSE Class 10 Maths Standard 2023 Set 30/1/1 paper?

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

### How should students write their answers to score full marks in the CBSE Class 10 Maths Standard exam?

Students should write step-by-step solutions with proper formulas, clear diagrams where necessary, and correct units for final answers. Referring to detailed solved papers helps understand the exact presentation required by CBSE.

### Is the solutions PDF for the CBSE Class 10 Maths Standard 2023 Set 30/1/1 question paper available for free?

Yes, students can download the complete question paper along with step-by-step solutions in PDF format for free on SwaVid. It helps in revising important concepts and practicing effectively for board exams.

## Related pages

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

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student