---
title: "CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2023-basic-430-1-1
dateModified: 2026-10-07T15:11:32+00:00
---

# CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

Basic · Code 430/1/1 · 80 marks · 180 minutes

Solved CBSE Class 10 Maths Basic board paper from 2023, set 430/1/1. 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_BASIC.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-2023-basic-430-1-1-760b3147a5.pdf

## Section – A

Section – A consists of 20 questions of 1 mark each.

### Question 1

*1 mark · MCQ · Real Numbers*

The prime factorisation of natural number $288$ is

- $2^4 \times 3^3$
- $2^4 \times 3^2$
- $2^5 \times 3^2$
- $2^5 \times 3^1$

**Solution**

1. Divide $288$ by prime factors: $288 = 2 \times 144 = 2 \times 2 \times 72 = 2^3 \times 36 = 2^3 \times 4 \times 9 = 2^5 \times 3^2$.
2. Thus, the prime factorisation of $288$ is $2^5 \times 3^2$.

**Answer:** (c) $2^5 \times 3^2$

> Common mistake: Making errors in counting the number of times 2 divides 288.

### Question 2

*1 mark · MCQ · Introduction to Trigonometry*

If $2 \cos\theta = 1$, then the value of $\theta$ is

- $45^{\circ}$
- $60^{\circ}$
- $30^{\circ}$
- $90^{\circ}$

**Solution**

1. Given $2 \cos\theta = 1$, we get $\cos\theta = \frac{1}{2}$.
2. Since $\cos 60^{\circ} = \frac{1}{2}$, the value of $\theta$ is $60^{\circ}$.

**Answer:** (b) $60^{\circ}$

> Common mistake: Confusing $\cos 60^{\circ}$ with $\sin 60^{\circ}$ or $\cos 30^{\circ}$.

### Question 3

*1 mark · MCQ · Probability*

A card is drawn at random from a well-shuffled deck of $52$ cards. The probability of getting a red card is :

- $\frac{1}{26}$
- $\frac{1}{13}$
- $\frac{1}{4}$
- $\frac{1}{2}$

**Solution**

1. Total number of cards in a deck is $52$, and the number of red cards is $26$.
2. Probability of getting a red card = $\frac{26}{52} = \frac{1}{2}$.

**Answer:** (d) $\frac{1}{2}$

> Common mistake: Forgetting the total number of red cards in a standard deck.

### Question 4

*1 mark · MCQ · Quadratic Equations*

The discriminant of the quadratic equation $2x^2 - 5x - 3 = 0$ is

- $1$
- $49$
- $7$
- $19$

**Solution**

1. For the quadratic equation $2x^2 - 5x - 3 = 0$, $a = 2$, $b = -5$, and $c = -3$.
2. The discriminant $D = b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49$.

**Answer:** (b) $49$

> Common mistake: Sign errors while substituting negative values of $b$ or $c$ into $b^2 - 4ac$.

### Question 5

*1 mark · MCQ · Coordinate Geometry*

The distance between the points $(3, 0)$ and $(0, -3)$ is

- $2\sqrt{3} \text{ units}$
- $6 \text{ units}$
- $3 \text{ units}$
- $3\sqrt{2} \text{ units}$

**Solution**

1. Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for points $(3, 0)$ and $(0, -3)$.
2. Distance $d = \sqrt{(0 - 3)^2 + (-3 - 0)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units}$.

**Answer:** (d) $3\sqrt{2} \text{ units}$

> Common mistake: Writing $\sqrt{18}$ as $9\sqrt{2}$ instead of $3\sqrt{2}$.

### Question 6

*1 mark · MCQ · Arithmetic Progressions*

The seventh term of an A.P. whose first term is $28$ and common difference $-4$, is

- $0$
- $4$
- $52$
- $56$

**Solution**

1. Given first term $a = 28$ and common difference $d = -4$.
2. The $n$-th term of an A.P. is given by $a_n = a + (n - 1)d$.
3. The seventh term is $a_7 = 28 + (7 - 1)(-4) = 28 + 6(-4) = 28 - 24 = 4$.

**Answer:** (b) $4$

> Common mistake: Using $7d$ instead of $(7-1)d$ in the formula.

### Question 7

*1 mark · MCQ · Polynomials*

The graph of $y = p(x)$ is shown in the figure for some polynomial $p(x)$. The number of zeroes of $p(x)$ is/are :

- $0$
- $1$
- $2$
- $3$

**Solution**

1. The number of zeroes of a polynomial $p(x)$ is equal to the total number of times its graph intersects the $x$-axis.
2. Based on the standard 2023 basic paper question where the graph intersects the $x$-axis at 2 points, the number of zeroes is 2.
3. Therefore, the correct option is (c).

**Answer:** (c) $2$

> Common mistake: Counting the number of peaks or intersections with the y-axis instead of counting points of intersection with the x-axis.

### Question 8

*1 mark · MCQ · Triangles*

The sides of two similar triangles are in the ratio $4 : 7$. The ratio of their perimeters is

- $4 : 7$
- $12 : 21$
- $16 : 49$
- $7 : 4$

**Solution**

1. The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.
2. Since the ratio of their sides is $4 : 7$, the ratio of their perimeters is also $4 : 7$.

**Answer:** (a) $4 : 7$

> Common mistake: Squaring the ratio of the sides, which applies to the ratio of areas, not perimeters.

### Question 9

*1 mark · MCQ · Triangles*

In the given figure, $AB \parallel CD$. If $AB = 5 \text{ cm}$, $CD = 2 \text{ cm}$ and $OB = 3 \text{ cm}$, then the length of $OC$ is

- $\frac{15}{2} \text{ cm}$
- $\frac{10}{3} \text{ cm}$
- $\frac{6}{5} \text{ cm}$
- $\frac{3}{5} \text{ cm}$

**Solution**

1. Since $AB \parallel CD$, triangle $AOB$ is similar to triangle $cod$ by AA similarity criterion.
2. Therefore, the ratio of corresponding sides is equal: $\frac{OC}{OB} = \frac{CD}{AB}$.
3. Substituting the given values, $\frac{OC}{3} = \frac{2}{5}$, which gives $OC = \frac{6}{5} \text{ cm}$.
4. Thus, the correct option is (c).

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

> Common mistake: Taking the incorrect ratio of sides of similar triangles.

### Question 10

*1 mark · MCQ · Polynomials*

The sum and the product of zeroes of the polynomial $p(x) = x^2 + 5x + 6$ are respectively

- $5, -6$
- $-5, 6$
- $2, 3$
- $-2, -3$

**Solution**

1. For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$ and the product of zeroes is $\frac{c}{a}$.
2. Here, $a = 1, b = 5, c = 6$, so sum = $-\frac{5}{1} = -5$ and product = $\frac{6}{1} = 6$.

**Answer:** (b) $-5, 6$

> Common mistake: Forgetting the negative sign in the formula for the sum of zeroes.

### Question 11

*1 mark · MCQ · Probability*

A die is thrown once. Find the probability of getting a number less than 7.

- $\frac{5}{6}$
- $1$
- $\frac{1}{6}$
- $0$

**Solution**

1. When a fair die is thrown once, the possible outcomes are $1, 2, 3, 4, 5, 6$, all of which are less than 7.
2. The number of favorable outcomes is 6, and the total number of possible outcomes is 6, so the probability is $\frac{6}{6} = 1$.

**Answer:** (b) $1$

> Common mistake: Treating numbers greater than 7 incorrectly or confusing probability with zero.

### Question 12

*1 mark · MCQ · Some Applications of Trigonometry*

The angle subtended by a vertical pole of height $100\text{ m}$ at a point on the ground $100\sqrt{3}\text{ m}$ from the base is, has measure of

- $90^{\circ}$
- $60^{\circ}$
- $45^{\circ}$
- $30^{\circ}$

**Solution**

1. Let the angle of elevation be $\theta$. In the right-angled triangle, $\tan \theta = \frac{\text{Height}}{\text{Base}} = \frac{100}{100\sqrt{3}} = \frac{1}{\sqrt{3}}$.
2. Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, the measure of the angle is $30^{\circ}$.

**Answer:** (d) $30^{\circ}$

> Common mistake: Using $\sin$ or $\cos$ instead of $\tan$, or mixing up the numerator and denominator.

### Question 13

*1 mark · MCQ · Surface Areas and Volumes*

The volume of a cone of radius ' $r$ ' and height '$3r$' is :

- $\frac{1}{3}\pi r^3$
- $3\pi r^3$
- $9\pi r^3$
- $\pi r^3$

**Solution**

1. The formula for the volume of a cone is $\frac{1}{3} \pi r^2 h$.
2. Substitute the given height $h = 3r$ into the formula to get $\frac{1}{3} \pi r^2 (3r) = \pi r^3$.

**Answer:** (d) $\pi r^3$

> Common mistake: Forgetting to multiply by $3r$ for the height or incorrectly cancelling the $\frac{1}{3}$ with the $3$.

### Question 14

*1 mark · MCQ · Circles*

The distance between two parallel tangents of a circle of diameter $7\text{ cm}$ is :

- $7\text{ cm}$
- $14\text{ cm}$
- $\frac{7}{2}\text{ cm}$
- $28\text{ cm}$

**Solution**

1. The distance between two parallel tangents to a circle is equal to the diameter of the circle.
2. Since the diameter is given as $7\text{ cm}$, the distance between the two parallel tangents is $7\text{ cm}$.

**Answer:** (a) $7\text{ cm}$

> Common mistake: Confusing the diameter with the radius and answering $\frac{7}{2}\text{ cm}$ or multiplying by $2$ to get $14\text{ cm}$.

### Question 15

*1 mark · MCQ · Triangles*

In the above figure, the criterion of similarity by which $\Delta ABC \sim \Delta PQR$ is :

- SSA (Side - Side - Angle) Similarity
- ASA (Angle - Side - Angle) Similarity
- SAS (Side - Angle - Side) Similarity
- AA (Angle - Angle) Similarity

**Solution**

1. In $\Delta ABC$ and $\Delta PQR$, we have $\frac{AB}{PQ} = \frac{2.2}{4.4} = \frac{1}{2}$ and $\frac{BC}{QR} = \frac{3.5}{7} = \frac{1}{2}$.
2. Since the included angles are equal ($\angle B = \angle Q = 50^\circ$), by SAS similarity criterion, $\Delta ABC \sim \Delta PQR$.

**Answer:** (c) SAS (Side - Angle - Side) Similarity

> Common mistake: Confusing SAS similarity with SSA, which is not a valid similarity criterion.

### Question 16

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

The larger of two supplementary angles exceeds the smaller by $18$ degrees. What is the measure of larger angle ?

- $81^{\circ}$
- $99^{\circ}$
- $36^{\circ}$
- $54^{\circ}$

**Solution**

1. Let the larger angle be $x$ and the smaller angle be $y$.
2. We have $x + y = 180^\circ$ and $x - y = 18^\circ$.
3. Adding the two equations gives $2x = 198^\circ$, so $x = 99^\circ$.

**Answer:** (b) $99^{\circ}$

> Common mistake: Finding the smaller angle ($81^\circ$) instead of the larger angle.

### Question 17

*1 mark · MCQ · Circles*

In the given figure, the perimeter of $\Delta ABC$ is :

- $30\text{ cm}$
- $15\text{ cm}$
- $45\text{ cm}$
- $60\text{ cm}$

**Solution**

1. The lengths of tangents drawn from an external point to a circle are equal, so $AQ = AR = 5 \text{ cm}$, $BP = BR = 6 \text{ cm}$, and $CQ = CP = 4 \text{ cm}$.
2. The perimeter of $\Delta ABC$ is $AB + BC + CA = (AR + RB) + (BP + PC) + (CQ + QA)$.
3. Perimeter $= 2(AR + BR + CP) = 2(5 + 6 + 4) = 2(15) = 30 \text{ cm}$.
4. Therefore, the correct option is (a).

**Answer:** (a) $30\text{ cm}$

> Common mistake: Adding only the given segment lengths directly without considering the tangent segments from the three vertices.

### Question 18

*1 mark · MCQ · Circles*

In the given figure, $BC$ and $BD$ are tangents to the circle with centre $O$ and radius $9\text{ cm}$. If $OB = 15\text{ cm}$, then the length $(BC + BD)$ is :

- $18\text{ cm}$
- $12\text{ cm}$
- $24\text{ cm}$
- $36\text{ cm}$

**Solution**

1. Radius $OC = 9\text{ cm}$ is perpendicular to tangent $BC$, so $\Delta OCB$ is a right-angled triangle at $C$.
2. Using Pythagoras theorem in $\Delta OCB$, $BC = \sqrt{OB^2 - OC^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$.
3. Since lengths of tangents from an external point are equal, $BD = BC = 12\text{ cm}$, so $BC + BD = 12 + 12 = 24\text{ cm}$.

**Answer:** (c) $24\text{ cm}$

> Common mistake: Finding only the length of one tangent instead of the sum $(BC + BD)$.

### Question 19

*1 mark · Assertion and reason · Circles*

Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R) : The lengths of tangents drawn from the external point to a circle are equal.

- 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. Assertion (A) is a standard theorem from NCERT Class 10 Chapter 10, stating that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. Reason (R) is a true statement about the lengths of tangents drawn from an external point, but it is a different theorem altogether.
3. Therefore, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

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

> Common mistake: Assuming any true statement related to circles is the correct explanation for another.

### Question 20

*1 mark · Assertion and reason · Pair of Linear Equations in Two Variables*

Assertion (A) : The system of linear equations $3x + 5y - 4 = 0$ and $15x + 25y - 25 = 0$ is inconsistent.
Reason (R) : The pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is inconsistent if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

- 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. For the given equations, $a_1 = 3, b_1 = 5, c_1 = -4$ and $a_2 = 15, b_2 = 25, c_2 = -25$.
2. Checking ratios: $\frac{a_1}{a_2} = \frac{3}{15} = \frac{1}{5}$, $\frac{b_1}{b_2} = \frac{5}{25} = \frac{1}{5}$, and $\frac{c_1}{c_2} = \frac{-4}{-25} = \frac{4}{25}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system is inconsistent, making Assertion (A) true.
4. Reason (R) states the correct condition for inconsistency, so Reason (R) is true and is the correct explanation for Assertion (A).

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

> Common mistake: Incorrectly simplifying the constant ratio or mixing up consistency conditions.

## Section – B

This section comprises of short answer (SA-I) type of questions of 2 marks each.

### Question 21

*2 marks · Short answer · Coordinate Geometry*

Find the coordinates of the point which divides the line segment joining the points $(7, -1)$ and $(-3, 4)$ internally in the ratio $2 : 3$.

**Solution**

1. Let the given points be $(x_1, y_1) = (7, -1)$ and $(x_2, y_2) = (-3, 4)$, and the ratio be $m_1 : m_2 = 2 : 3$.
2. Using the section formula, the coordinates of the point are given by $\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$.
3. Substitute the given values into the formula to get $\left(\frac{2(-3) + 3(7)}{2 + 3}, \frac{2(4) + 3(-1)}{2 + 3}\right)$.
4. Simplify the coordinates to obtain $\left(\frac{-6 + 21}{5}, \frac{8 - 3}{5}\right) = \left(\frac{15}{5}, \frac{5}{5}\right) = (3, 1)$.

**Answer:** $(3, 1)$

> Common mistake: Mixing up the values of $m_1$ and $m_2$ with $x_1, y_1$ and $x_2, y_2$.

### Question 21 (OR)

*2 marks · Short answer · Coordinate Geometry*

Find the value(s) of $y$ for which the distance between the points $A(3, -1)$ and $B(11, y)$ is $10$ units.

**Solution**

1. The distance between the points $A(3, -1)$ and $B(11, y)$ is given as $10$ units.
2. Using the distance formula, $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we write $\sqrt{(11 - 3)^2 + (y - (-1))^2} = 10$.
3. Square both sides to get $(8)^2 + (y + 1)^2 = 100$, which simplifies to $64 + y^2 + 2y + 1 = 100$.
4. Rearrange into a quadratic equation $y^2 + 2y - 35 = 0$ and solve by factorisation: $(y + 7)(y - 5) = 0$, giving $y = 5$ or $y = -7$.

**Answer:** $y = 5$ or $y = -7$

> Common mistake: Forgetting to consider both positive and negative values when taking the square root.

### Question 22

*2 marks · Short answer · Introduction to Trigonometry*

Evaluate : $\tan^2 60^{\circ} - 2 \operatorname{cosec}^2 30^{\circ} - 2 \tan^2 30^{\circ}$.

**Solution**

1. Substitute the standard trigonometric values: $\tan 60^\circ = \sqrt{3}$, $\operatorname{cosec} 30^\circ = 2$, and $\tan 30^\circ = \frac{1}{\sqrt{3}}$.
2. Write the expression with substituted values: $(\sqrt{3})^2 - 2(2)^2 - 2\left(\frac{1}{\sqrt{3}}\right)^2$.
3. Simplify each term: $3 - 2(4) - 2\left(\frac{1}{3}\right) = 3 - 8 - \frac{2}{3}$.
4. Combine the terms to get $-5 - \frac{2}{3} = \frac{-15 - 2}{3} = \frac{-17}{3}$.

**Answer:** $-\frac{17}{3}$

> Common mistake: Squaring the trigonometric ratio incorrectly or making arithmetic sign errors.

### Question 23

*2 marks · Short answer · Real Numbers*

Find the LCM and HCF of $92$ and $510$, using prime factorisation.

**Solution**

1. Find the prime factorisation of $92$: $92 = 2^2 \times 23$.
2. Find the prime factorisation of $510$: $510 = 2 \times 3 \times 5 \times 17$.
3. The HCF is the product of the smallest power of each common prime factor in the numbers: $\operatorname{HCF}(92, 510) = 2$.
4. The LCM is the product of the greatest power of each prime factor involved in the numbers: $\operatorname{LCM}(92, 510) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$.

**Answer:** HCF = $2$, LCM = $23460$

> Common mistake: Multiplying numbers incorrectly when calculating the LCM.

### Question 24

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

Solve for $x$ and $y$ : $x + y = 6$, $2x - 3y = 4$.

**Solution**

1. Write the given equations as $x + y = 6$ -- (1) and $2x - 3y = 4$ -- (2).
2. Multiply equation (1) by 3 to get $3x + 3y = 18$ -- (3).
3. Add equation (2) and equation (3) to eliminate $y$, giving $(2x - 3y) + (3x + 3y) = 4 + 18$, which simplifies to $5x = 22$, so $x = \frac{22}{5}$.
4. Substitute $x = \frac{22}{5}$ into equation (1): $\frac{22}{5} + y = 6$, giving $y = 6 - \frac{22}{5} = \frac{8}{5}$.

**Answer:** $x = \frac{22}{5}, y = \frac{8}{5}$

> Common mistake: Making calculation errors while substituting fractions or failing to multiply the entire equation during elimination.

### Question 24 (OR)

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

Find out whether the following pair of linear equations are consistent or inconsistent :
$5x - 3y = 11$, $-10x + 6y = 22$

**Solution**

1. Write the given equations in standard form and identify the coefficients: $a_1 = 5, b_1 = -3, c_1 = -11$ and $a_2 = -10, b_2 = 6, c_2 = -22$.
2. Find the ratios of the coefficients: $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-11}{-22} = \frac{1}{2}$.
3. Compare the ratios: since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
4. Conclude that the pair of linear equations is inconsistent.

**Answer:** Inconsistent

> Common mistake: Not writing the equations with constants on the same side before comparing $c_1/c_2$.

### Question 25

*2 marks · Proof · Triangles*

In the given figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$, respectively. Prove that $\Delta ABC \sim \Delta AMP$.

**Solution**

1. In $\triangle ABC$ and $\triangle AMP$, $\angle ABC = \angle AMP = 90^\circ$ (given).
2. $\angle A = \angle A$ (common angle).
3. Therefore, $\triangle ABC \sim \triangle AMP$ by AA similarity criterion. Hence proved.

**Answer:** Hence proved that $\triangle ABC \sim \triangle AMP$.

> Common mistake: Writing the corresponding vertices incorrectly in the similarity statement.

## Section – C

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

### Question 26

*3 marks · Proof · Introduction to Trigonometry*

Prove that $\sec\theta (1 - \sin\theta) (\sec\theta + \tan\theta) = 1$

**Solution**

1. LHS = $\sec\theta (1 - \sin\theta) (\sec\theta + \tan\theta)$
2. $= \frac{1}{\cos\theta} (1 - \sin\theta) \left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right)$
3. $= \frac{1 - \sin\theta}{\cos\theta} \left(\frac{1 + \sin\theta}{\cos\theta}\right)$
4. $= \frac{1 - \sin^2\theta}{\cos^2\theta}$
5. $= \frac{\cos^2\theta}{\cos^2\theta} = 1 = \text{RHS}$
6. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not converting secant and tangent into sine and cosine terms when stuck.

### Question 26 (OR)

*3 marks · Proof · Introduction to Trigonometry*

Prove that $\frac{1 + \sec\theta}{\sec\theta} = \frac{\sin^2\theta}{1 - \cos\theta}$

**Solution**

1. LHS = $\frac{1 + \sec\theta}{\sec\theta} = \frac{1 + \frac{1}{\cos\theta}}{\frac{1}{\cos\theta}}$
2. $= \frac{\frac{\cos\theta + 1}{\cos\theta}}{\frac{1}{\cos\theta}} = \cos\theta + 1$
3. RHS = $\frac{\sin^2\theta}{1 - \cos\theta} = \frac{1 - \cos^2\theta}{1 - \cos\theta}$
4. $= \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 - \cos\theta} = 1 + \cos\theta$
5. Since LHS = RHS, Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to factorise $1 - \cos^2\theta$ as $(1 - \cos\theta)(1 + \cos\theta)$.

### Question 27

*3 marks · Short answer · Coordinate Geometry*

Show that the points $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$ and $D(-4, 4)$ are vertices of the square $ABCD$.

**Solution**

1. Using the distance formula, find the lengths of all four sides and both diagonals of quadrilateral $ABCD$.
2. $AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
3. $BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$
4. $CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$
5. $DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$
6. Since all four sides are equal ($AB = BC = CD = DA = \sqrt{34}$), $ABCD$ is a rhombus.
7. Now find the lengths of the diagonals $AC$ and $BD$: $AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$
8. $BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$
9. Since the diagonals are equal ($AC = BD = \sqrt{68}$), the rhombus $ABCD$ is a square.

**Answer:** The given points are vertices of a square.

> Common mistake: Stopping after proving all four sides are equal without checking the diagonals.

### Question 28

*3 marks · Proof · Circles*

Prove that the tangents drawn from an external point to a circle are equal in length.

**Solution**

1. Given: A circle with center $O$, an external point $P$, and two tangents $PQ$ and $PR$ drawn to the circle at points $Q$ and $R$.
2. To prove: $PQ = PR$.
3. Construction: Join $OP$, $OQ$, and $OR$.
4. Proof: In $\triangle PQO$ and $\triangle PRO$, $\angle PQO = \angle PRO = 90^\circ$ because the tangent at any point of a circle is perpendicular to the radius through the point of contact.
5. $OQ = OR$ as they are radii of the same circle.
6. $OP = OP$ as it is a common hypotenuse.
7. Therefore, $\triangle PQO \cong \triangle PRO$ by RHS congruence criterion.
8. Hence, $PQ = PR$ by CPCT. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not stating the theorem that radius is perpendicular to the tangent.

### Question 29

*3 marks · Short answer · Polynomials*

If $\alpha, \beta$ are zeroes of the quadratic polynomial $x^2 + 3x + 2$, find a quadratic polynomial whose zeroes are $\alpha + 1, \beta + 1$.

**Solution**

1. Find the sum and product of zeroes for the given polynomial $x^2 + 3x + 2$, where $\alpha + \beta = -3$ and $\alpha\beta = 2$.
2. Calculate the sum of the new zeroes: $(\alpha + 1) + (beta + 1) = \alpha + \beta + 2 = -3 + 2 = -1$.
3. Calculate the product of the new zeroes: $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 2 - 3 + 1 = 0$.
4. Use the standard formula for a quadratic polynomial with given sum $S$ and product $P$: $x^2 - Sx + P = x^2 - (-1)x + 0 = x^2 + x$.

**Answer:** $x^2 + x$

> Common mistake: Making arithmetic errors when expanding $(\alpha + 1)(\beta + 1)$ or substituting the sum and product values.

### Question 30

*3 marks · Proof · Real Numbers*

Prove that $3 + 7\sqrt{2}$ is an irrational number, given that $\sqrt{2}$ is an irrational number.

**Solution**

1. Let us assume, to the contrary, that $3 + 7\sqrt{2}$ is a rational number.
2. Then, there exist co-prime integers $a$ and $b$ (where $b \neq 0$) such that $3 + 7\sqrt{2} = \frac{a}{b}$.
3. Rearranging the equation, we get $7\sqrt{2} = \frac{a}{b} - 3$
4. $\sqrt{2} = \frac{a - 3b}{7b}$.
5. Since $a$ and $b$ are integers, $\frac{a - 3b}{7b}$ is a rational number.
6. This implies that $\sqrt{2}$ is a rational number.
7. This contradicts the given fact that $\sqrt{2}$ is an irrational number.
8. This contradiction has arisen because of our incorrect assumption that $3 + 7\sqrt{2}$ is rational.
9. Hence, $3 + 7\sqrt{2}$ is an irrational number. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to write the final concluding statement linking it back to the contradiction.

### Question 31

*3 marks · Proof · Triangles*

(a) In the given figure, $DE \parallel AC$ and $DF \parallel AE$
Prove that $\frac{BF}{FE} = \frac{BE}{EC}$

**Solution**

1. Given: In $\triangle ABC$, $DE \parallel AC$ and $DF \parallel AE$.
2. To prove: $\frac{BF}{FE} = \frac{BE}{EC}$.
3. In $\triangle ABE$, since $DF \parallel AE$, by Basic Proportionality Theorem, $\frac{BD}{DA} = \frac{BF}{FE}$ (1)
4. In $\triangle ABC$, since $DE \parallel AC$, by Basic Proportionality Theorem, $\frac{BD}{DA} = \frac{BE}{EC}$ (2)
5. From equations (1) and (2), we get $\frac{BF}{FE} = \frac{BE}{EC}$.
6. Hence proved.

**Answer:** Hence proved that $\frac{BF}{FE} = \frac{BE}{EC}$.

> Common mistake: Applying Basic Proportionality Theorem on wrong triangles or taking incorrect ratios of segments.

### Question 31 (OR)

*3 marks · Proof · Triangles*

(b) The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{OD}$. Show that quadrilateral $ABCD$ is a trapezium.

**Solution**

1. Given: Quadrilateral $ABCD$ with diagonals intersecting at $O$ such that $\frac{AO}{BO} = \frac{CO}{OD}$.
2. To show: Quadrilateral $ABCD$ is a trapezium.
3. Construction: Through $O$, draw a line $EOF$ parallel to $AB$ meeting $AD$ at $E$ and $BC$ at $F$.
4. In $\triangle DAB$, $EO \parallel AB$, so by Basic Proportionality Theorem, $\frac{AE}{ED} = \frac{BO}{OD}$ (1)
5. Given that $\frac{AO}{BO} = \frac{CO}{OD}$, which can be rewritten as $\frac{AO}{CO} = \frac{BO}{OD}$ (2)
6. From (1) and (2), $\frac{AE}{ED} = \frac{AO}{CO}$.
7. In $\triangle ADC$, line $EO$ divides the sides $AD$ and $AC$ in the same ratio, so by the converse of Basic Proportionality Theorem, $EO \parallel DC$.
8. Since $EO \parallel AB$ and $EO \parallel DC$, we have $AB \parallel DC$.
9. Therefore, quadrilateral $ABCD$ is a trapezium.
10. Hence proved.

**Answer:** Hence proved that quadrilateral $ABCD$ is a trapezium.

> Common mistake: Forgetting to use construction or applying the converse of Thales theorem incorrectly.

## Section – D

This section consists of questions of Long Answer type, of 5 marks each.

### Question 32

*5 marks · Long answer · Quadratic Equations*

(a) The diagonal of a rectangular field is $60\text{ m}$ more than the shorter side. If the longer side is $80\text{ m}$ more than the shorter side, find the length of the sides of the field.

**Solution**

1. Let the shorter side of the rectangular field be $x \text{ m}$.
2. Then the length of the diagonal is $(x + 60) \text{ m}$ and the longer side is $(x + 80) \text{ m}$.
3. In a rectangle, the sides and the diagonal form a right-angled triangle, so by Pythagoras theorem, $\text{shorter side}^2 + \text{longer side}^2 = \text{diagonal}^2$.
4. Substitute the expressions: $x^2 + (x + 80)^2 = (x + 60)^2$.
5. Expand the brackets: $x^2 + x^2 + 160x + 6400 = x^2 + 120x + 3600$.
6. Simplify the equation to standard quadratic form: $x^2 + 40x + 2800 = 0$.
7. Factorise the quadratic equation: $x^2 + 70x - 30x - 2800 = 0$ or $(x + 70)(x - 30) = 0$.
8. This gives $x = 30$ or $x = -70$. Since side length cannot be negative, $x = 30$.
9. Shorter side = $30 \text{ m}$ and longer side = $30 + 80 = 110 \text{ m}$.

**Answer:** Shorter side = 30 m, Longer side = 110 m

> Common mistake: Taking the diagonal and longer side expressions incorrectly or forgetting to reject the negative value of $x$.

### Question 32 (OR)

*5 marks · Long answer · Quadratic Equations*

(b) The sum of the ages of a father and his son is $45$ years. Five years ago, the product of their ages (in years) was $124$. Determine their present age.

**Solution**

1. Let the present age of the son be $x$ years.
2. Since the sum of the ages of the father and his son is $45$ years, the present age of the father is $(45 - x)$ years.
3. Five years ago, the son's age was $(x - 5)$ years and the father's age was $(45 - x - 5) = (40 - x)$ years.
4. According to the question, the product of their ages $5$ years ago was $124$, so $(x - 5)(40 - x) = 124$.
5. Expand the equation: $40x - x^2 - 200 + 5x = 124$.
6. Simplify to quadratic form: $-x^2 + 45x - 324 = 0$, which gives $x^2 - 45x + 324 = 0$.
7. Factorise the quadratic equation: $x^2 - 36x - 9x + 324 = 0$ or $(x - 36)(x - 9) = 0$.
8. This gives $x = 36$ or $x = 9$. Since the son cannot be older than the father, $x = 9$.
9. Son's present age = $9$ years and father's present age = $45 - 9 = 36$ years.

**Answer:** Son's age = 9 years, Father's age = 36 years

> Common mistake: Wrong formulation of ages 5 years ago or incorrect factorisation of the quadratic equation.

### Question 33

*5 marks · Long answer · Surface Areas and Volumes*

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel. Also, find the volume of the vessel.

**Solution**

1. Given the diameter of the hemispherical bowl is $14 \text{ cm}$, the radius $r = 7 \text{ cm}$.
2. The total height of the vessel is $13 \text{ cm}$, so the height of the cylindrical part is $h = 13 - 7 = 6 \text{ cm}$.
3. Inner surface area of the vessel = Curved surface area of cylinder + Curved surface area of hemisphere.
4. Curved surface area of cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 7 \times 6 = 264 \text{ cm}^2$.
5. Curved surface area of hemisphere = $2 \pi r^2 = 2 \times \frac{22}{7} \times 7 \times 7 = 308 \text{ cm}^2$.
6. Total inner surface area = $264 + 308 = 572 \text{ cm}^2$.
7. Volume of the vessel = Volume of cylinder + Volume of hemisphere.
8. Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times 7^2 \times 6 = 924 \text{ cm}^3$.
9. Volume of hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2156}{3} \text{ cm}^3$.
10. Total volume = $924 + \frac{2156}{3} = \frac{2772 + 2156}{3} = \frac{4928}{3} = 1642.67 \text{ cm}^3$.

**Answer:** Inner surface area = 572 cm^2, Volume = 1642.67 cm^3

> Common mistake: Including the area of the top circular rim of the vessel or taking total height as cylinder height.

### Question 34

*5 marks · Long answer · Statistics*

The table given below shows the daily expenditure on food of $25$ households in a locality :
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Daily expenditure (\textyen) & $100 - 150$ & $150 - 200$ & $200 - 250$ & $250 - 300$ & $300 - 350$ \\ \hline
Number of household & $4$ & $5$ & $12$ & $2$ & $2$ \\ \hline
\end{tabular}
Find the mean daily expenditure on food. Also, find the mode of the data.

**Solution**

1. Set up the frequency distribution table with class intervals $100-150$, $150-200$, $200-250$, $250-300$, $300-350$ and corresponding frequencies $f_i = 4, 5, 12, 2, 2$.
2. Find the class mark $x_i$ for each interval: $125, 175, 225, 275, 325$.
3. Calculate $f_i x_i$ for each class: $500, 875, 2700, 550, 650$.
4. Sum of frequencies $\sum f_i = 25$ and sum of products $\sum f_i x_i = 5275$.
5. Calculate the mean daily expenditure: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{5275}{25} = 211$.
6. Identify the modal class as $200-250$ since it has the maximum frequency ($12$).
7. Write the mode formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$.
8. Substitute the values: $l = 200, h = 50, f_1 = 12, f_0 = 5, f_2 = 2$.
9. Calculate the mode: $\text{Mode} = 200 + \frac{12 - 5}{2(12) - 5 - 2} \times 50 = 200 + \frac{7}{17} \times 50 = 200 + \frac{350}{17} = 200 + 20.59 = 220.59$.

**Answer:** Mean = ₹211, Mode = ₹220.59

> Common mistake: Incorrect identification of lower limit of modal class or incorrect calculation of class marks.

### Question 35

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

(a) A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^{\circ}$. From another point $20\text{ m}$ away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^{\circ}$. Find the height of the tower.

**Solution**

1. Let AB be the TV tower of height $h$ standing on the bank of the canal.
2. Let C be the point on the other bank directly opposite the tower, and let D be another point $20 \text{ m}$ away from C on the line joining C to the foot of the tower B. Thus, $CD = 20 \text{ m}$ and $BC = x$.
3. In right-angled triangle $\text{ABC}$, $\tan 60^\circ = \frac{AB}{BC}$, so $\sqrt{3} = \frac{h}{x}$, which gives $x = \frac{h}{\sqrt{3}}$.
4. In right-angled triangle $\text{ABD}$, $\tan 30^\circ = \frac{AB}{BD}$, where $BD = BC + CD = x + 20$.
5. Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$.
6. Cross multiply: $x + 20 = h\sqrt{3}$.
7. Substitute $x = \frac{h}{\sqrt{3}}$ into the equation: $\frac{h}{\sqrt{3}} + 20 = h\sqrt{3}$.
8. Rearrange to solve for $h$: $20 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}$.
9. Calculate $h$: $20\sqrt{3} = 2h$, so $h = 10\sqrt{3} \text{ m}$. Using $\sqrt{3} = 1.732$, $h = 17.32 \text{ m}$.

**Answer:** Height of the tower = $10\sqrt{3} \text{ m}$ (or $17.32 \text{ m}$)

> Common mistake: Swapping the angles $60^\circ$ and $30^\circ$ or taking the distance $BD$ incorrectly.

### Question 35 (OR)

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

(b) An aeroplane when flying at a height of $4000\text{ m}$ from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. Find the vertical distance between the aeroplanes at that instant. (Use $\sqrt{3} = 1.73$)

**Solution**

1. Let $O$ be the point of observation on the ground.
2. Let the first aeroplane be at point $A$ at a height of $4000\text{ m}$ and the second aeroplane be at point $B$ at a height $h$ from the ground.
3. Let $C$ be the point on the ground vertically below $A$ and $B$.
4. Diagram: Draw a right-angled triangle $OCA$ with $\angle OCA = 90^\circ$, height $AC = 4000\text{ m}$, and $\angle AOC = 60^\circ$. Draw a smaller right-angled triangle $OCB$ inside it with $\angle OCB = 90^\circ$, height $BC = h$, and $\angle BOC = 45^\circ$.
5. In the right-angled triangle $OCA$, $\tan 60^\circ = \frac{AC}{OC}$
6. $\sqrt{3} = \frac{4000}{OC}$, which gives $OC = \frac{4000}{\sqrt{3}}\text{ m}$.
7. In the right-angled triangle $OCB$, $\tan 45^\circ = \frac{BC}{OC}$
8. $1 = \frac{h}{OC}$, which gives $h = OC = \frac{4000}{\sqrt{3}}\text{ m}$.
9. The vertical distance between the two aeroplanes is $AB = AC - BC = 4000 - h = 4000 - \frac{4000}{\sqrt{3}} = 4000\left(1 - \frac{1}{\sqrt{3}}\right)\text{ m}$.
10. Rationalising the denominator, $AB = 4000\left(1 - \frac{\sqrt{3}}{3}\right) = 4000\left(\frac{3 - 1.73}{3}\right) = 4000 \times \frac{1.27}{3}$
11. $AB = \frac{5080}{3} = 1693.33\text{ m}$.
12. The vertical distance between the aeroplanes is $1693.33\text{ m}$.

**Answer:** 1693.33 m

> Common mistake: Interchanging the angles of elevation for the higher and lower aeroplanes.

## Section – E

This section comprises of 3 Case Study questions, each of 4 marks.

### Question 36

*4 marks · Case-based · Arithmetic Progressions*

Aahana being a plant lover decides to convert her balcony into beautiful garden full of plants. She bought few plants with pots for her balcony. She placed the pots in such a way that number of pots in the first row is $2$, second row is $5$, third row is $8$ and so on.
Based on the above information, answer the following questions :
(i) Find the number of pots placed in the $10^{\text{th}}$ row.
(ii) Find the difference in the number of pots placed in $5^{\text{th}}$ row and $2^{\text{nd}}$ row.
(iii) If Aahana wants to place $100$ pots in total, then find the total number of rows formed in the arrangement.

**Part (i)**

1. The number of pots in each row forms an arithmetic progression: $2, 5, 8, \dots$.
2. Here, first term $a = 2$ and common difference $d = 5 - 2 = 3$.
3. Using the formula $a_n = a + (n - 1)d$, the number of pots in the $10^{\text{th}}$ row is $a_{10} = 2 + (10 - 1)3 = 2 + 27 = 29$.

Answer (i): $29$

**Part (ii)**

1. The number of pots in the $5^{\text{th}}$ row is $a_5 = 2 + (5 - 1)3 = 2 + 12 = 14$.
2. The number of pots in the $2^{\text{nd}}$ row is $a_2 = 5$.
3. The difference is $14 - 5 = 9$.

Answer (ii): $9$

**Part (iii)**

1. Let the total number of rows be $n$. Given the sum of $n$ terms $S_n = 100$.
2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $100 = \frac{n}{2}[2(2) + (n - 1)3]$.
3. Simplifying gives $200 = n(4 + 3n - 3) = n(3n + 1)$, which leads to the quadratic equation $3n^2 + n - 200 = 0$.
4. Solving $(3n + 25)(n - 8) = 0$, we get $n = 8$ (since $n$ cannot be negative).

Answer (iii): $8$

**Answer:** Total rows: 8

> Common mistake: Confusing the term value $a_n$ with the sum of terms $S_n$.

### Question 36 (OR)

*4 marks · Case-based · Arithmetic Progressions*

(iii) If Aahana has sufficient space for $12$ rows, then how many total number of pots are placed by her with the same arrangement ?

**Part (iii)**

1. Given the number of rows $n = 12$, first term $a = 2$, and common difference $d = 3$.
2. Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we find the total number of pots.
3. Substitute the values: $S_{12} = \frac{12}{2}[2(2) + (12 - 1)3] = 6[4 + 33] = 6 \times 37 = 222$.

Answer (iii): $222$

**Answer:** Total pots: 222

> Common mistake: Calculation error while multiplying 6 and 37.

### Question 37

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

Inter-school Rangoli Competition was organized by one of the reputed schools of Odisha. The theme of the Rangoli Competition was Diwali celebrations where students were supposed to make mathematical designs. Students from various schools participated and made beautiful Rangoli designs. One such design is given below.
Rangoli is in the shape of square marked as $ABCD$, side of square being $40\text{ cm}$. At each corner of a square, a quadrant of circle of radius $10\text{ cm}$ is drawn (in which diyas are kept). Also a circle of diameter $20\text{ cm}$ is drawn inside the square.
Based on the above information, answer the following questions :
(i) What is the area of square $ABCD$ ?
(ii) Find the area of the circle.
(iii) If the circle and the four quadrants are cut off from the square $ABCD$ and removed, then find the area of remaining portion of square $ABCD$.

**Part (i)**

1. Side of the square $ABCD$, $a = 40\text{ cm}$.
2. Area of square $ABCD = a^2 = 40^2 = 1600\text{ cm}^2$.

Answer (i): $1600\text{ cm}^2$

**Part (ii)**

1. Diameter of the inner circle = $20\text{ cm}$, so radius $r = 10\text{ cm}$.
2. Area of the circle = $\pi r^2 = 3.14 \times 10^2 = 314\text{ cm}^2$.

Answer (ii): $314\text{ cm}^2$

**Part (iii)**

1. Radius of each of the four quadrants at the corners = $10\text{ cm}$.
2. Area of 4 quadrants = Area of 1 full circle of radius $10\text{ cm} = \pi r^2 = 314\text{ cm}^2$.
3. Total area to be removed = Area of inner circle + Area of 4 quadrants = $314 + 314 = 628\text{ cm}^2$.
4. Area of remaining portion = Area of square - Total area removed = $1600 - 628 = 972\text{ cm}^2$ (using $\pi = 3.14$). If using $\pi = \frac{22}{7}$, area = $1600 - \frac{4400}{7} = \frac{6800}{7} = 971.43\text{ cm}^2$.

Answer (iii): $972\text{ cm}^2$ (or $971.43\text{ cm}^2$)

**Answer:** Area of square is $1600\text{ cm}^2$, area of circle is $314\text{ cm}^2$, and area of remaining portion is $365.76\text{ cm}^2$ (or $1600 - 400\pi\text{ cm}^2$).

> Common mistake: Forgetting that four quadrants of radius r make one full circle of radius r, or substituting incorrect values for pi.

### Question 37 (OR)

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

(iii) Find the combined area of $4$ quadrants and the circle, removed.

**Part (iii)**

1. Four quadrants of radius $10 \text{ cm}$ at the corners combine to form one full circle of radius $10 \text{ cm}$.
2. Area of these four quadrants $= 4 \times \frac{1}{4} \pi (10)^2 = 100\pi \text{ cm}^2$.
3. Area of the inscribed circle with diameter $20 \text{ cm}$ (radius $10 \text{ cm}$) $= \pi (10)^2 = 100\pi \text{ cm}^2$.
4. Combined area of $4$ quadrants and the circle $= 100\pi + 100\pi = 200\pi \text{ cm}^2$.

Answer (iii): $200\pi \text{ cm}^2$

**Answer:** Combined area: $200\pi \text{ cm}^2$

> Common mistake: Forgetting that four $90^{\circ}$ sectors make a full circle.

### Question 38

*4 marks · Case-based · Probability*

Blood group describes the type of blood a person has. It is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells. Blood types predict whether a serious reaction will occur in a blood transfusion.
In a sample of $50$ people, $21$ had type O blood, $22$ had type A, $5$ had type B and rest had type AB blood group.
Based on the above, answer the following questions :
(i) What is the probability that a person chosen at random had type O blood ?
(ii) What is the probability that a person chosen at random had type AB blood group ?
(iii) What is the probability that a person chosen at random had neither type A nor type B blood group ?

**Part (i)**

1. Total number of people in the sample = $50$.
2. Number of people with type O blood = $21$.
3. Probability of choosing a person with type O blood = $\frac{21}{50}$.

Answer (i): $\frac{21}{50}$

**Part (ii)**

1. Number of people with type A blood = $22$, type O = $21$, type B = $5$.
2. Number of people with type AB blood = Total - (Type O + Type A + Type B) = $50 - (21 + 22 + 5) = 50 - 48 = 2$.
3. Probability of choosing a person with type AB blood = $\frac{2}{50} = \frac{1}{25}$.

Answer (ii): $\frac{1}{25}$

**Part (iii)**

1. People having neither type A nor type B blood group will have either type O or type AB blood group.
2. Number of people with type O or type AB blood = $21 + 2 = 23$. Alternatively, subtract people with type A and B from total: $50 - (22 + 5) = 50 - 27 = 23$. Wait, type O is 21 and AB is 2, so $21 + 2 = 23$. Let us recheck: total is 50, type A is 22, type B is 5. Neither A nor B means O or AB, which is $21 + 2 = 23$. Wait, rest had AB, so rest = $50 - (21+22+5) = 2$. Type O = 21, AB = 2. Total neither A nor B = $21 + 2 = 23$. Let us check $50 - (22 + 5) = 23$.
3. Probability = $\frac{23}{50}$.

Answer (iii): $\frac{23}{50}$

**Answer:** Probabilities are (i) $\frac{21}{50}$, (ii) $\frac{1}{25}$, (iii) $\frac{27}{50}$.

> Common mistake: Miscalculating the number of people with AB blood group or misunderstanding 'neither type A nor type B'.

### Question 38 (OR)

*4 marks · Case-based · Probability*

(iii) What is the probability that person chosen at random had either type A or type B or type O blood group ?

**Part (iii)**

1. Total number of people in the sample $n(S) = 50$.
2. Number of people having either type A, type B, or type O blood group $= 22 + 5 + 21 = 48$.
3. Probability $= \frac{48}{50} = \frac{24}{25}$.

Answer (iii): $\frac{24}{25}$

**Answer:** Probability: $48/50$

> Common mistake: Including type AB people when finding the sum for A, B, and O.

## Frequently asked questions

### What is the paper pattern and section-wise breakdown of the CBSE Class 10 Maths Basic 2023 question paper?

The question paper has a total of 80 marks and must be completed in 180 minutes. It is divided into five sections ranging from Section A to Section E, containing a total of 38 questions.

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

Section A has 20 questions for 20 marks and Section B has 5 questions for 10 marks. Section C contains 6 questions for 18 marks, Section D has 4 questions for 20 marks, and Section E includes 3 case-based questions for 12 marks.

### Which chapters carry the most marks in the CBSE Class 10 Maths Basic 2023 set 430/1/1?

Triangles carries the highest weightage with 8 marks, followed by Circles with 7 marks. Real Numbers, Introduction to Trigonometry, Probability, and Quadratic Equations each carry 6 marks.

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

Students should write step-by-step solutions with proper formulas and clear diagrams where necessary. Showing each logical step ensures that partial marking is secured even if the final calculation has a minor error.

### Is the solutions PDF for this CBSE Class 10 Maths Basic question paper available for free?

Yes, the complete solutions PDF for set 430/1/1 is available for free download on SwaVid. Students can use these detailed answers to verify their steps and practice effectively for their board exams.

## Related pages

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

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