---
title: "CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2024-standard-30-1-1
dateModified: 2026-10-07T15:20:37+00:00
---

# CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

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

Solved CBSE Class 10 Maths Standard board paper from 2024, 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/2024/X/MATHEMATICS_STANDARD.zip. Solutions written by SwaVid.

Free PDF (23 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2024-standard-30-1-1-1fcc3bcf51.pdf

## SECTION - A

This section consists of 20 questions of 1 mark each.

### Question 1

*1 mark · MCQ · Polynomials*

If the sum of zeroes of the polynomial $p(x) = 2x^2 - k\sqrt{2}x + 1$ is $\sqrt{2}$, then value of $k$ is :

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

**Solution**

1. For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is given by $-\frac{b}{a}$.
2. Here, $a = 2$, $b = -k\sqrt{2}$, and the sum of zeroes is $\sqrt{2}$.
3. $-\frac{-k\sqrt{2}}{2} = \sqrt{2} \implies \frac{k\sqrt{2}}{2} = \sqrt{2} \implies k = 2$.

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

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

### Question 2

*1 mark · MCQ · Probability*

If the probability of a player winning a game is $0.79$, then the probability of his losing the same game is :

- $1.79$
- $0.31$
- $0.21\%$
- $0.21$

**Solution**

1. Let $E$ be the event of winning the game, so $P(E) = 0.79$.
2. The probability of losing the game is $P(\bar{E}) = 1 - P(E) = 1 - 0.79 = 0.21$.

**Answer:** (d) $0.21$

> Common mistake: Subtracting incorrectly or confusing percentage with decimal.

### Question 3

*1 mark · MCQ · Quadratic Equations*

If the roots of equation $ax^2 + bx + c = 0, a \neq 0$ are real and equal, then which of the following relation is true ?

- $a = \frac{b^2}{c}$
- $b^2 = ac$
- $ac = \frac{b^2}{4}$
- $c = \frac{b^2}{a}$

**Solution**

1. For a quadratic equation $ax^2 + bx + c = 0$ with real and equal roots, the discriminant must be zero, so $b^2 - 4ac = 0$.
2. Rearranging this gives $b^2 = 4ac$, which can be written as $ac = \frac{b^2}{4}$.

**Answer:** (c) $ac = \frac{b^2}{4}$

> Common mistake: Students often forget the factor of 4 and mistakenly write $b^2 = ac$ or $ac = b^2$.

### Question 4

*1 mark · MCQ · Arithmetic Progressions*

In an A.P., if the first term $a = 7$, $n$th term $a_n = 84$ and the sum of first $n$ terms $s_n = \frac{2093}{2}$, then $n$ is equal to :

- $22$
- $24$
- $23$
- $26$

**Solution**

1. The sum of $n$ terms of an A.P. is given by $S_n = \frac{n}{2}(a + a_n)$.
2. Substituting the given values, $\frac{2093}{2} = \frac{n}{2}(7 + 84)$, which gives $2093 = n(91)$ and $n = \frac{2093}{91} = 23$.

**Answer:** (c) $23$

> Common mistake: Students may try to use the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ and get stuck due to the unknown common difference $d$.

### Question 5

*1 mark · MCQ · Real Numbers*

If two positive integers $p$ and $q$ can be expressed as $p = 18a^2b^4$ and $q = 20a^3b^2$, where $a$ and $b$ are prime numbers, then $\text{LCM}(p, q)$ is :

- $2a^2b^2$
- $180a^2b^2$
- $12a^3b^2$
- $180a^3b^4$

**Solution**

1. Given $p = 18a^2b^4 = 2 \times 3^2 \times a^2 \times b^4$ and $q = 20a^3b^2 = 2^2 \times 5 \times a^3 \times b^2$, where $a$ and $b$ are prime numbers.
2. The LCM of two numbers is the product of the greatest power of each prime factor involved in the numbers: $\text{LCM}(p, q) = 2^2 \times 3^2 \times 5 \times a^3 \times b^4 = 180a^3b^4$.

**Answer:** (d) $180a^3b^4$

> Common mistake: Students often find HCF instead of LCM by taking the lowest powers of the prime factors.

### Question 6

*1 mark · MCQ · Coordinate Geometry*

AD is a median of $\Delta ABC$ with vertices $A(5, -6), B(6, 4)$ and $C(0, 0)$. Length AD is equal to :

- $\sqrt{68}$ units
- $2\sqrt{15}$ units
- $\sqrt{101}$ units
- $10$ units

**Solution**

1. Since $AD$ is a median, $D$ is the midpoint of $BC$. Using the midpoint formula, coordinates of $D$ are $\left(\frac{6 + 0}{2}, \frac{4 + 0}{2}\right) = (3, 2)$.
2. Using the distance formula between $A(5, -6)$ and $D(3, 2)$, $AD = \sqrt{(3 - 5)^2 + (2 - (-6))^2} = \sqrt{(-2)^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68}$ units.

**Answer:** (a) $\sqrt{68}$ units

> Common mistake: Students sometimes mistakenly calculate the midpoint of $AB$ or $AC$ instead of $BC$ for median $AD$.

### Question 7

*1 mark · MCQ · Introduction to Trigonometry*

If $\sec \theta - \tan \theta = m$, then the value of $\sec \theta + \tan \theta$ is :

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

**Solution**

1. We know the identity $\sec^2 \theta - \tan^2 \theta = 1$, which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
2. Substituting the given value $m(\sec \theta + \tan \theta) = 1$, we get $\sec \theta + \tan \theta = \frac{1}{m}$.

**Answer:** (c) $\frac{1}{m}$

> Common mistake: Students often forget the fundamental identity $\sec^2 \theta - \tan^2 \theta = 1$ and try to solve using triangle ratios.

### Question 8

*1 mark · MCQ · Probability*

From the data $1, 4, 7, 9, 16, 21, 25$, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :

- $\frac{2}{5}$
- $\frac{1}{5}$
- $\frac{1}{7}$
- $\frac{2}{7}$

**Solution**

1. The given data is $1, 4, 7, 9, 16, 21, 25$. Removing all even numbers ($4, 16$), the remaining numbers are $1, 7, 9, 21, 25$.
2. The total number of remaining outcomes is $5$. Among these, the prime number is only $7$, so the number of favorable outcomes is $1$.
3. The probability is $\frac{1}{5}$.

**Answer:** (b) $\frac{1}{5}$

> Common mistake: Students often consider $1$ as a prime number.

### Question 9

*1 mark · MCQ · Statistics*

For some data $x_1, x_2, ......, x_n$ with respective frequencies $f_1, f_2, ......, f_n$, the value of $\sum_{i=1}^{n} f_i (x_i - \bar{x})$ is equal to :

- $n\bar{x}$
- $1$
- $\sum f_i$
- $0$

**Solution**

1. Expand the given expression: $\sum_{i=1}^{n} f_i (x_i - \bar{x}) = \sum_{i=1}^{n} f_i x_i - \bar{x} \sum_{i=1}^{n} f_i$.
2. Since $\sum f_i x_i = N\bar{x}$ and $\sum f_i = N$, the expression becomes $N\bar{x} - \bar{x}(N) = 0$.

**Answer:** (d) $0$

> Common mistake: Students confuse $\sum f_i x_i$ with $\bar{x}$.

### Question 10

*1 mark · MCQ · Polynomials*

The zeroes of a polynomial $x^2 + px + q$ are twice the zeroes of the polynomial $4x^2 - 5x - 6$. The value of $p$ is :

- $-\frac{5}{2}$
- $\frac{5}{2}$
- $-5$
- $10$

**Solution**

1. Find the zeroes of $4x^2 - 5x - 6 = 0$ by splitting the middle term: $4x^2 - 8x + 3x - 6 = 0$, giving zeroes as $2$ and $-\frac{3}{4}$.
2. The zeroes of $x^2 + px + q$ are twice these zeroes, so they are $2(2) = 4$ and $2\left(-\frac{3}{4}\right) = -\frac{3}{2}$.
3. The sum of zeroes is $4 + \left(-\frac{3}{2}\right) = \frac{5}{2}$, which equals $-p$, so $p = -\frac{5}{2}$.

**Answer:** (a) $-\frac{5}{2}$

> Common mistake: Forgetting the negative sign while relating the sum of zeroes to the coefficient $p$.

### Question 11

*1 mark · MCQ · Coordinate Geometry*

If the distance between the points $(3, -5)$ and $(x, -5)$ is $15$ units, then the values of $x$ are :

- $12, -18$
- $-12, 18$
- $18, 5$
- $-9, -12$

**Solution**

1. Use the distance formula between $(3, -5)$ and $(x, -5)$: $\sqrt{(x - 3)^2 + (-5 - (-5))^2} = 15$.
2. This simplifies to $|x - 3| = 15$, which gives $x - 3 = 15$ or $x - 3 = -15$.
3. Solving these gives $x = 18$ or $x = -12$.

**Answer:** (b) $-12, 18$

> Common mistake: Students often miss the negative value when removing the square root.

### Question 12

*1 mark · MCQ · Introduction to Trigonometry*

If $\cos(\alpha + \beta) = 0$, then value of $\cos\left(\frac{\alpha + \beta}{2}\right)$ is equal to :

- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2}$
- $0$
- $\sqrt{2}$

**Solution**

1. Given $\cos(\alpha + \beta) = 0$. Since $\cos 90^\circ = 0$, we have $\alpha + \beta = 90^\circ$.
2. We need to find $\cos\left(\frac{\alpha + \beta}{2}\right) = \cos\left(\frac{90^\circ}{2}\right) = \cos 45^\circ$.
3. The value of $\cos 45^\circ$ is $\frac{1}{\sqrt{2}}$.

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

> Common mistake: Students incorrectly substitute $\alpha + \beta = 0$ instead of $90^\circ$.

### Question 13

*1 mark · MCQ · Surface Areas and Volumes*

A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together, is :

- $1 : 1$
- $1 : 4$
- $2 : 3$
- $3 : 2$

**Solution**

1. The surface area of a sphere of radius $r$ is $4\pi r^2$.
2. When cut into two hemispheres, the total surface area of two hemispheres is $2 \times 3\pi r^2 = 6\pi r^2$.
3. The ratio of the surface area of the sphere to that of two hemispheres is $\frac{4\pi r^2}{6\pi r^2} = \frac{4}{6} = \frac{2}{3}$.

**Answer:** (c) $2 : 3$

> Common mistake: Taking curved surface area of hemispheres instead of total surface area.

### Question 14

*1 mark · MCQ · Statistics*

The middle most observation of every data arranged in order is called :

- mode
- median
- mean
- deviation

**Solution**

1. By definition in statistics, the median is the middle-most observation of a data arranged in ascending or descending order.

**Answer:** (b) median

> Common mistake: Confusing median with mean or mode.

### Question 15

*1 mark · MCQ · Surface Areas and Volumes*

The volume of the largest right circular cone that can be carved out from a solid cube of edge $2\text{ cm}$ is :

- $\frac{4\pi}{3}\text{ cu cm}$
- $\frac{5\pi}{3}\text{ cu cm}$
- $\frac{8\pi}{3}\text{ cu cm}$
- $\frac{2\pi}{3}\text{ cu cm}$

**Solution**

1. The largest right circular cone that can be carved out of a cube of edge $2\text{ cm}$ will have a diameter equal to the edge of the cube ($2\text{ cm}$) and height equal to the edge of the cube ($2\text{ cm}$).
2. The radius of the cone is $r = 1\text{ cm}$ and height is $h = 2\text{ cm}$.
3. Volume of the cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1)^2 (2) = \frac{2\pi}{3}\text{ cu cm}$.

**Answer:** (d) $\frac{2\pi}{3}\text{ cu cm}$

> Common mistake: Taking diameter equal to the diagonal of the cube's face.

### Question 16

*1 mark · MCQ · Probability*

Two dice are rolled together. The probability of getting sum of numbers on the two dice as $2, 3$ or $5$, is :

- $\frac{7}{36}$
- $\frac{11}{36}$
- $\frac{5}{36}$
- $\frac{4}{9}$

**Solution**

1. Total number of possible outcomes when two dice are rolled is $36$.
2. Outcomes with sum $2$: $(1, 1)$ -> $1$ outcome.
3. Outcomes with sum $3$: $(1, 2), (2, 1)$ -> $2$ outcomes.
4. Outcomes with sum $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ -> $4$ outcomes.
5. Total favourable outcomes = $1 + 2 + 4 = 7$, so the probability is $\frac{7}{36}$.

**Answer:** (a) $\frac{7}{36}$

> Common mistake: Forgetting to include some pairs or counting pairs twice.

### Question 17

*1 mark · MCQ · Coordinate Geometry*

The centre of a circle is at $(2, 0)$. If one end of a diameter is at $(6, 0)$, then the other end is at :

- $(0, 0)$
- $(4, 0)$
- $(-2, 0)$
- $(-6, 0)$

**Solution**

1. The centre of a circle is the midpoint of any of its diameters.
2. Let the other end of the diameter be $(x, y)$. Using the midpoint formula, $\frac{6 + x}{2} = 2$ and $\frac{0 + y}{2} = 0$.
3. Solving for $x$ gives $6 + x = 4 \implies x = -2$, and $y = 0$. Thus, the other end is $(-2, 0)$.

**Answer:** (c) $(-2, 0)$

> Common mistake: Multiplying the centre coordinates by 2 instead of using the midpoint relation.

### Question 18

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

In the given figure, graphs of two linear equations are shown. The pair of these linear equations is :

- consistent with unique solution.
- consistent with infinitely many solutions.
- inconsistent.
- inconsistent but can be made consistent by extending these lines.

**Solution**

1. The figure shows two lines intersecting at a single point on the Cartesian plane.
2. Intersecting lines represent a pair of linear equations that have a unique solution, hence they are consistent.

**Answer:** (a) consistent with unique solution.

> Common mistake: Confusing intersecting lines with parallel lines (inconsistent).

### Question 19

*1 mark · Assertion and reason · Circles*

Assertion (A) : The tangents drawn at the end points of a diameter of a circle, are parallel.
Reason (R) : Diameter of a circle is the longest chord.

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

**Solution**

1. The tangents drawn at the end points of a diameter of a circle are perpendicular to the radius at the points of contact, making alternate interior angles equal.
2. Thus, the tangents are parallel, making Assertion (A) true.
3. The reason that the diameter is the longest chord is a true statement about chords, but it does not explain why the tangents at its ends are parallel.

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

> Common mistake: Assuming that a true statement about a circle is always the correct explanation for another true property.

### Question 20

*1 mark · Assertion and reason · Polynomials*

Assertion (A) : If the graph of a polynomial touches $x$-axis at only one point, then the polynomial cannot be a quadratic polynomial.
Reason (R) : A polynomial of degree $n (n > 1)$ can have at most $n$ zeroes.

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

**Solution**

1. A quadratic polynomial can touch the $x$-axis at exactly one point if its discriminant is zero, meaning it has two coincident real roots.
2. Thus, Assertion (A) is false because a quadratic polynomial can indeed touch the $x$-axis at only one point.
3. Reason (R) is a standard true theorem stating that a polynomial of degree $n$ can have at most $n$ zeroes.

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

> Common mistake: Confusing intersecting the $x$-axis at two distinct points with touching the $x$-axis at one point for a quadratic graph.

## SECTION - B

This section consists of 5 questions of 2 marks each.

### Question 21

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

Solve the following system of linear equations $7x - 2y = 5$ and $8x + 7y = 15$ and verify your answer.

**Solution**

1. Multiply the first equation by $7$ and the second equation by $2$ to equate coefficients of $y$: $49x - 14y = 35$ and $16x + 14y = 30$.
2. Add the two new equations: $(49x + 16x) = 35 + 30$, which gives $65x = 65$, so $x = 1$.
3. Substitute $x = 1$ in $7x - 2y = 5$: $7(1) - 2y = 5$, giving $2y = 2$, so $y = 1$.
4. Verification: For $7(1) - 2(1) = 5$ and $8(1) + 7(1) = 15$, both equations are satisfied.

**Answer:** $x = 1$, $y = 1$

> Common mistake: Arithmetic errors while multiplying equations for elimination method.

### Question 22

*2 marks · Very short answer · Probability*

In a pack of 52 playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.

**Solution**

1. Total number of cards in a standard deck is $52$.
2. Given that one lost card is a black card, the remaining deck still contains all $4$ cards of hearts, including the queen of hearts.
3. The total number of remaining cards is $52 - 1 = 51$.
4. The probability of drawing a queen of heart from the remaining cards is $\frac{1}{51}$.

**Answer:** $$\frac{1}{51}$$

> Common mistake: Taking the total remaining cards as $52$ or assuming the lost card could be the queen of hearts.

### Question 23

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

Evaluate : $2\sqrt{2} \cos 45^\circ \sin 30^\circ + 2\sqrt{3} \cos 30^\circ$

**Solution**

1. Substitute standard trigonometric values: $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 30^\circ = \frac{1}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
2. Expression becomes: $2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2\sqrt{3} \times \frac{\sqrt{3}}{2}$.
3. Simplify the terms: $1 + 3 = 4$.

**Answer:** $4$

> Common mistake: Incorrect substitution of trigonometric table values.

### Question 23 (OR)

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

If $A = 60^\circ$ and $B = 30^\circ$, verify that : $\sin(A + B) = \sin A \cos B + \cos A \sin B$

**Solution**

1. LHS = $\sin(A + B) = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.
2. RHS = $\sin A \cos B + \cos A \sin B = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$.
3. Substitute values: $\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1$.
4. Since LHS = RHS, the identity is verified.

**Answer:** LHS = RHS = 1, hence verified.

> Common mistake: Errors in substituting trigonometric values for specific angles.

### Question 24

*2 marks · Proof · Triangles*

In the given figure, ABCD is a quadrilateral. Diagonal BD bisects $\angle B$ and $\angle D$ both. Prove that : (i) $\Delta ABD \sim \Delta ACBD$ (ii) $AB = BC$

**Part (i)**

1. Given that diagonal BD bisects $\angle B$ and $\angle D$ of quadrilateral ABCD.
2. In $\triangle ABD$ and $\triangle CBD$, $\angle ABD = \label{CBD}$ since BD bisects $\angle B$.
3. Also, $\angle ADB = \angle CDB$ since BD bisects $\angle D$.
4. By AA similarity criterion, $\triangle ABD \sim \triangle CBD$.

Answer (i): \triangle ABD \sim \triangle CBD

**Part (ii)**

1. From the similarity of $\triangle ABD$ and $\triangle CBD$, the corresponding sides are proportional.
2. Therefore, $\frac{AB}{CB} = \frac{AD}{CD}$.

Answer (ii): AB = BC

**Answer:** Hence proved.

> Common mistake: Confusing the order of vertices in similar triangles.

### Question 25

*2 marks · Proof · Real Numbers*

Prove that $5 - 2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number.

**Solution**

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

**Answer:** $5 - 2\sqrt{3}$ is irrational.

> Common mistake: Not explicitly stating the contradiction with the given fact that root 3 is irrational.

### Question 25 (OR)

*2 marks · Proof · Real Numbers*

Show that the number $5 \times 11 \times 17 + 3 \times 11$ is a composite number.

**Solution**

1. Consider the given number: $5 \times 11 \times 17 + 3 \times 11$.
2. Take out the common factor $11$ from both terms: $11 \times (5 \times 17 + 3)$.
3. Simplify the expression inside the bracket: $11 \times (85 + 3) = 11 \times 88$.
4. Express $88$ as product of primes: $11 \times (11 \times 8) = 11^2 \times 2^3$.
5. Since the number can be expressed as a product of prime factors greater than 1, it is a composite number.

**Answer:** Hence proved that the given number is a composite number.

> Common mistake: Multiplying out the entire large number instead of taking out common factors to show prime factorization.

## SECTION - C

This section consists of 6 questions of 3 marks each.

### Question 26

*3 marks · Short answer · Coordinate Geometry*

Find the ratio in which the point $\left(\frac{8}{5}, y\right)$ divides the line segment joining the points $(1, 2)$ and $(2, 3)$. Also, find the value of $y$.

**Solution**

1. Let the point $\left(\frac{8}{5}, y\right)$ divide the line segment joining $(1, 2)$ and $(2, 3)$ in the ratio $k : 1$.
2. Using the section formula for the x-coordinate, we get $\frac{8}{5} = \frac{2k + 1}{k + 1}$.
3. Solving for $k$, we get $8(k + 1) = 5(2k + 1)$, which gives $8k + 8 = 10k + 5$, so $2k = 3$, giving $k = \frac{3}{2}$.
4. Thus, the required ratio is $3 : 2$.
5. Using the section formula for the y-coordinate with $k = \frac{3}{2}$, we get $y = \frac{\frac{3}{2}(3) + 1(2)}{\frac{3}{2} + 1} = \frac{\frac{9}{2} + 2}{\frac{5}{2}} = \frac{13}{5}$.

**Answer:** Ratio is $3 : 2$ and $y = \frac{13}{5}$

> Common mistake: Taking the ratio as $m:n$ and getting stuck with two variables instead of using $k:1$.

### Question 26 (OR)

*3 marks · Proof · Coordinate Geometry*

ABCD is a rectangle formed by the points $A(-1, -1), B(-1, 6), C(3, 6)$ and $D(3, -1)$. P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.

**Solution**

1. Given vertices of rectangle ABCD are $A(-1, -1), B(-1, 6), C(3, 6)$ and $D(3, -1)$.
2. P is the mid-point of AB, so its coordinates are $\left(\frac{-1-1}{2}, \frac{-1+6}{2}\right) = \left(-1, \frac{5}{2}\right)$.
3. Q is the mid-point of BC, so its coordinates are $\left(\frac{-1+3}{2}, \frac{6+6}{2}\right) = (1, 6)$.
4. R is the mid-point of CD, so its coordinates are $\left(\frac{3+3}{2}, \frac{6-1}{2}\right) = \left(3, \frac{5}{2}\right)$.
5. S is the mid-point of DA, so its coordinates are $\left(\frac{3-1}{2}, \frac{-1-1}{2}\right) = (1, -1)$.
6. The coordinates of the mid-point of diagonal PR are $\left(\frac{-1+3}{2}, \frac{\frac{5}{2}+\frac{5}{2}}{2}\right) = (1, \frac{5}{2})$.
7. The coordinates of the mid-point of diagonal QS are $\left(\frac{1+1}{2}, \frac{6+(-1)}{2}\right) = (1, \frac{5}{2})$.
8. Since the mid-points of both diagonals PR and QS are the same $(1, \frac{5}{2})$, the diagonals bisect each other. Hence proved.

**Answer:** Diagonals bisect each other as their mid-points are identical.

> Common mistake: Making calculation errors while finding coordinates of mid-points.

### Question 27

*3 marks · Short answer · Real Numbers*

In a teachers' workshop, the number of teachers teaching French, Hindi and English are $48, 80$ and $144$ respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject.

**Solution**

1. Given number of teachers teaching French, Hindi and English are $48, 80$ and $144$ respectively.
2. To find the minimum number of rooms, the number of teachers in each room must be the maximum possible, which is the HCF of $48, 80$ and $144$.
3. Prime factorisation of $48 = 2^4 \times 3$
4. Prime factorisation of $80 = 2^4 \times 5$
5. Prime factorisation of $144 = 2^4 \times 3^2$
6. HCF$(48, 80, 144) = 2^4 = 16$.
7. Maximum number of teachers in each room = $16$.
8. Total number of teachers = $48 + 80 + 144 = 272$.
9. Minimum number of rooms required = $\frac{272}{16} = 17$.

**Answer:** 17 rooms

> Common mistake: Dividing individual counts incorrectly or finding LCM instead of HCF.

### Question 28

*3 marks · Proof · Introduction to Trigonometry*

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

**Solution**

1. Consider the LHS: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$.
2. Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$.
3. Substitute these into the expression: $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$.
4. Simplify the denominators: $\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$.
5. Rewrite as: $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}$.
6. Take the LCM as $\sin \theta \cos \theta (\sin \theta - \cos \theta)$ to get $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
7. Use the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ in the numerator: $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
8. Cancel $(\sin \theta - \cos \theta)$ and substitute $\sin^2 \theta + \cos^2 \theta = 1$ to get $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$.
9. Split the fraction: $\frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} = \sec \theta \csc \theta + 1 = 1 + \sec \theta \csc \theta = \text{RHS}$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Sign errors while changing $\cos \theta - \sin \theta$ to $\sin \theta - \cos \theta$.

### Question 29

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

Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now ?

**Solution**

1. Let the present age of Rashmi be $x$ years and the present age of Nazma be $y$ years.
2. Three years ago, Rashmi's age was $(x - 3)$ and Nazma's age was $(y - 3)$.
3. According to the first condition, $x - 3 = 3(y - 3)$, which simplifies to $x - 3 = 3y - 9$, or $x - 3y = -6$ (Equation 1).
4. Ten years later, Rashmi's age will be $(x + 10)$ and Nazma's age will be $(y + 10)$.
5. According to the second condition, $x + 10 = 2(y + 10)$, which simplifies to $x + 10 = 2y + 20$, or $x - 2y = 10$ (Equation 2).
6. Subtracting Equation 1 from Equation 2: $(x - 2y) - (x - 3y) = 10 - (-6)$, which gives $y = 16$.
7. Substitute $y = 16$ in Equation 2: $x - 2(16) = 10 \implies x - 32 = 10 \implies x = 42$.

**Answer:** Rashmi's age is 42 years and Nazma's age is 16 years.

> Common mistake: Forgetting to add or subtract years for both persons in age word problems.

### Question 30

*3 marks · Proof · Circles*

In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $\angle POQ = 90^\circ$.

**Solution**

1. Given AB is a diameter of the circle with centre O, and AQ, BP, PQ are tangents.
2. Let C be the point of contact of tangent PQ with the circle.
3. The tangents drawn from an external point to a circle are equal in length, so QA = QC and QB = QC.
4. In $\triangle AOQ$ and $\triangle COQ$, $OA = OC$ (radii), $OQ = OQ$ (common), and $QA = QC$ (tangents).
5. Therefore, $\triangle AOQ \cong \triangle COQ$ by SSS congruency, which gives $\angle AOQ = \angle COQ$.
6. Similarly, $\triangle BOP \cong \triangle COP$, which gives $\angle BOP = \angle COP$.
7. Since AB is a diameter, it is a straight line, so $\angle AOQ + \angle COQ + \angle COP + \angle BOP = 180^\circ$.
8. Thus, $2\angle COQ + 2\angle COP = 180^\circ$, meaning $\angle COQ + \angle COP = 90^\circ$.
9. Since $\angle POQ = \angle COQ + \angle COP$, we get $\angle POQ = 90^\circ$. Hence proved.

**Answer:** $\angle POQ = 90^\circ$

> Common mistake: Not stating the reason for tangents from an external point or straight line angle sum.

### Question 30 (OR)

*3 marks · Short answer · Circles*

A circle with centre O and radius $8\text{ cm}$ is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, $\text{BC} = 30\text{ cm}$ and $\text{BS} = 24\text{ cm}$, then find the length DC.

**Solution**

1. We are given that $\text{AD} \perp \text{DC}$ and the circle has radius $8\text{ cm}$ with centre O, making quadrilateral OPDS a square of side $8\text{ cm}$, so $\text{DS} = 8\text{ cm}$.
2. Tangents drawn from an external point to a circle are equal in length, so $\text{BQ} = \text{BS} = 24\text{ cm}$, $\text{CQ} = \text{CR}$, and $\text{DR} = \text{DS} = 8\text{ cm}$.
3. Given $\text{BC} = 30\text{ cm}$, we find $\text{CQ} = \text{BC} - \text{BQ} = 30 - 24 = 6\text{ cm}$, which means $\text{CR} = 6\text{ cm}$.
4. The length of side $\text{DC}$ is $\text{DR} + \text{CR} = 8\text{ cm} + 6\text{ cm} = 14\text{ cm}$.

**Answer:** $14\text{ cm}$

> Common mistake: Confusing the lengths of tangents from vertices or miscalculating the segments of DC.

### Question 31

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

The difference between the outer and inner radii of a hollow right circular cylinder of length $14\text{ cm}$ is $1\text{ cm}$. If the volume of the metal used in making the cylinder is $176\text{ cm}^3$, find the outer and inner radii of the cylinder.

**Solution**

1. Let the outer radius be $R$ and the inner radius be $r$. Given $R - r = 1$, so $R = r + 1$, and length $h = 14\text{ cm}$.
2. The volume of the metal in the hollow cylinder is given by $\pi (R^2 - r^2) h = 176\text{ cm}^3$.
3. Substitute the values: $\frac{22}{7} \times (R - r)(R + r) \times 14 = 176$, which gives $44 \times 1 \times (R + r) = 176$.
4. Solve for $R + r$: $R + r = \frac{176}{44} = 4$.
5. Solve the simultaneous equations $R - r = 1$ and $R + r = 4$ to get $2R = 5$, so $R = 2.5\text{ cm}$ and $r = 1.5\text{ cm}$.

**Answer:** Outer radius $= 2.5\text{ cm}$, inner radius $= 1.5\text{ cm}$

> Common mistake: Using total surface area formula instead of the volume formula for the hollow cylinder.

## SECTION - D

This section consists of 4 questions of 5 marks each.

### Question 32

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

An arc of a circle of radius $21\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find : (i) the length of the arc. (ii) the area of the minor segment of the circle made by the corresponding chord.

**Part (i)**

1. Given radius $r = 21\text{ cm}$ and central angle $\theta = 60^\circ$.
2. Length of the arc = $\frac{\theta}{360^\circ} \times 2 \pi r$
3. Length of the arc = $\frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$
4. Length of the arc = $\frac{1}{6} \times 2 \times 22 \times 3 = 22\text{ cm}$.

Answer (i): $22\text{ cm}$

**Part (ii)**

1. Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$
2. Area of the minor sector = $\frac{1}{6} \times 22 \times 3 \times 21 = 231\text{ cm}^2$
3. Area of the minor triangle formed by two radii and chord = $\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times (21)^2 \times \sin 60^\circ = \frac{441 \sqrt{3}}{4}\text{ cm}^2$
4. Area of the minor segment = Area of minor sector - Area of triangle = $\left(231 - \frac{441 \sqrt{3}}{4}\right)\text{ cm}^2$.

Answer (ii): $\left(231 - \frac{441 \sqrt{3}}{4}\right)\text{ cm}^2$

**Answer:** (i) $22\text{ cm}$, (ii) $21\left(\frac{44}{7} - \frac{\sqrt{3}}{4}\right)\text{ cm}^2$

> Common mistake: Using incorrect formula for area of triangle or using radians instead of degrees.

### Question 33

*5 marks · Long answer · Arithmetic Progressions*

The sum of first and eighth terms of an A.P. is $32$ and their product is $60$. Find the first term and common difference of the A.P. Hence, also find the sum of its first $20$ terms.

**Solution**

1. Given $a_1 + a_8 = 32$, which implies $a + (a + 7d) = 32$, so $2a + 7d = 32$.
2. Given $a_1 \times a_8 = 60$, which implies $a(a + 7d) = 60$.
3. Substitute $2a + 7d = 32$ or express terms: since $a_1 + a_8 = 32$ and product is $60$, let the two terms be $x$ and $y$ where $x+y=32$ and $xy=60$.
4. Solving $z^2 - 32z + 60 = 0$, we get $(z - 30)(z - 2) = 0$, so the first and eighth terms are $2$ and $30$ (or vice versa).
5. Case 1: $a = 2$ and $a_8 = 30$. Since $a + 7d = 30$, we get $2 + 7d = 30$, so $7d = 28$, giving $d = 4$.
6. Case 2: $a = 30$ and $a_8 = 2$. Since $a + 7d = 2$, we get $30 + 7d = 2$, so $7d = -28$, giving $d = -4$.
7. For $a = 2, d = 4$, the sum of first $20$ terms is $S_{20} = \frac{20}{2} [2(2) + (20-1)(4)] = 10 [4 + 76] = 800$.
8. For $a = 30, d = -4$, the sum of first $20$ terms is $S_{20} = \frac{20}{2} [2(30) + (20-1)(-4)] = 10 [60 - 76] = -160$.

**Answer:** First term $2$, common difference $4$, sum of $20$ terms $800$ (or first term $30$, common difference $-4$, sum $-160$).

> Common mistake: Failing to consider both possible cases for first term and common difference.

### Question 33 (OR)

*5 marks · Long answer · Arithmetic Progressions*

In an A.P. of $40$ terms, the sum of first $9$ terms is $153$ and the sum of last $6$ terms is $687$. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

**Solution**

1. Given an A.P. of $n = 40$ terms. Sum of first $9$ terms is $S_9 = 153$, so $\frac{9}{2}[2a + 8d] = 153$, which gives $2a + 8d = 153 \times \frac{2}{9} = 34$, or $a + 4d = 17$ (Equation 1).
2. The sum of the last $6$ terms is the sum of all $40$ terms minus the sum of the first $34$ terms: $S_{40} - S_{34} = 687$.
3. Alternatively, the last $6$ terms start from the $35^{\text{th}}$ term: $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$.
4. Using $a_n = a + (n-1)d$, this gives $(a + 34d) + (a + 35d) + (a + 36d) + (a + 37d) + (a + 38d) + (a + 39d) = 687$, which simplifies to $6a + 219d = 687$, or $2a + 73d = 229$ (Equation 2).
5. Multiply Equation 1 by 2: $2a + 8d = 34$. Subtract from Equation 2: $(2a + 73d) - (2a + 8d) = 229 - 34$, so $65d = 195$, giving $d = 3$.
6. Substitute $d = 3$ into Equation 1: $a + 4(3) = 17$, so $a = 5$.
7. Find the sum of all $40$ terms: $S_{40} = \frac{40}{2} [2(5) + (40-1)(3)] = 20 [10 + 39 \times 3] = 20 [10 + 117] = 20 \times 127 = 2540$.

**Answer:** First term $5$, common difference $3$, sum of all terms $2540$.

> Common mistake: Errors in writing the sum of the last few terms.

### Question 34

*5 marks · Proof · Triangles*

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then 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 AB at D and AC at E.
2. To prove: $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$
3. Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
4. Proof: Area of triangle ADE is $\text{ar}(\text{ADE}) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AD} \times EN$.
5. Similarly, $\text{ar}(\text{BDE}) = \frac{1}{2} \times \text{DB} \times EN$.
6. Dividing the two areas: $\frac{\text{ar}(\text{ADE})}{\text{ar}(\text{BDE})} = \frac{\frac{1}{2} \times \text{AD} \times EN}{\frac{1}{2} \times \text{DB} \times EN} = \frac{\text{AD}}{\text{DB}}$ (Equation 1).
7. Similarly, considering base AE on AC with height DM: $\frac{\text{ar}(\text{ADE})}{\text{ar}(\text{DEC})} = \frac{\frac{1}{2} \times \text{AE} \times DM}{\frac{1}{2} \times \text{EC} \times DM} = \frac{\text{AE}}{\text{EC}}$ (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: $\text{ar}(\text{BDE}) = \text{ar}(\text{DEC})$ (Equation 3).
9. From Equations 1, 2, and 3, the left-hand sides are equal, so the right-hand sides are equal: $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Omitting reasons for equal areas of triangles on the same base and between same parallels.

### Question 34 (OR)

*5 marks · Proof · Triangles*

In the given figure PA, QB and RC are each perpendicular to AC. If $\text{AP} = x, \text{BQ} = y$ and $\text{CR} = z$, then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$

**Solution**

1. Given: PA, QB and RC are perpendicular to AC. $\text{AP} = x$, $\text{BQ} = y$, $\text{CR} = z$. Let $\text{AQ} = a$ and $\text{QC} = b$.
2. To prove: $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$
3. Proof: Since $PA \perp AC$ and $QB \perp AC$, $PA \parallel QB$.
4. In triangle PAC, since $BQ \parallel AP$, triangle CBQ is similar to triangle CAP by AA similarity.
5. Therefore, $\frac{BQ}{AP} = \frac{CQ}{CA}$, which gives $\frac{y}{x} = \frac{b}{a+b}$ (Equation 1).
6. Similarly, since $QB \perp AC$ and $RC \perp AC$, $QB \parallel RC$.
7. In triangle RAC, since $BQ \parallel RC$, triangle ABQ is similar to triangle ACR by AA similarity.
8. Therefore, $\frac{BQ}{RC} = \frac{AQ}{AC}$, which gives $\frac{y}{z} = \frac{a}{a+b}$ (Equation 2).
9. Adding Equation 1 and Equation 2: $\frac{y}{x} + \frac{y}{z} = \frac{b}{a+b} + \frac{a}{a+b} = \frac{a+b}{a+b} = 1$.
10. Dividing both sides by $y$, we get $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
11. Hence proved.

**Answer:** Hence proved.

> Common mistake: Incorrectly setting up the ratios of corresponding sides of similar triangles.

### Question 35

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

A pole $6\text{ m}$ high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $60^\circ$ and the angle of depression of the point P from the top of the tower is $45^\circ$. Find the height of the tower and the distance of point P from the foot of the tower. (Use $\sqrt{3} = 1.73$)

**Solution**

1. Let the height of the tower be $h$ and the distance of point P from the foot of the tower be $d$.
2. Let the tower be AB and the pole on top be BC, so $\text{BC} = 6\text{ m}$ and $\text{AC} = h + 6$.
3. From point P on the ground, the angle of depression of P from the top of the tower (point A) is $45^\circ$, so the angle of elevation of A from P is $45^\circ$.
4. In right-angled triangle PAB: $\tan 45^\circ = \frac{\text{AB}}{\text{PB}} = \frac{h}{d}$.
5. Since $\tan 45^\circ = 1$, we get $\frac{h}{d} = 1$, which implies $h = d$ (Equation 1).
6. The angle of elevation of the top of the pole (point C) from P is given as $60^\circ$.
7. In right-angled triangle PAC: $\tan 60^\circ = \frac{\text{AC}}{\text{PB}} = \frac{h + 6}{d}$.
8. Since $\tan 60^\circ = \sqrt{3}$ and $h = d$, substitute $d = h$: $\sqrt{3} = \frac{h + 6}{h}$.
9. Multiply both sides by $h$: $h\sqrt{3} = h + 6$, so $h(\sqrt{3} - 1) = 6$.
10. Solve for $h$: $h = \frac{6}{\sqrt{3} - 1} = \frac{6(\sqrt{3} + 1)}{3 - 1} = \frac{6(\sqrt{3} + 1)}{2} = 3(\sqrt{3} + 1)$.
11. Substitute $\sqrt{3} = 1.73$: $h = 3(1.73 + 1) = 3(2.73) = 8.19\text{ m}$.
12. Since $d = h$, the distance of point P from the foot of the tower is $8.19\text{ m}$.

**Answer:** Height of the tower is $8.19\text{ m}$ and the distance of point P from the foot of the tower is $8.19\text{ m}$.

> Common mistake: Rationalizing incorrectly or substituting $\sqrt{3} = 1.73$ too early before rationalization.

## SECTION - E

This section consists of 3 Case-Study Based Questions of 4 marks each.

### Question 36

*4 marks · Case-based · Quadratic Equations*

A rectangular floor area can be completely tiled with $200$ square tiles. If the side length of each tile is increased by $1$ unit, it would take only $128$ tiles to cover the floor.
(i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) (a) Find the value of $x$, the length of side of a tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.

**Part (i)**

1. Let the original side length of each square tile be $x$ units.
2. The area of one tile is $x^2$ square units, so the total floor area is $200x^2$.
3. When the side length is increased by $1$ unit, the new side length is $(x + 1)$ units and the area of one tile is $(x + 1)^2$ square units.
4. Since $128$ such tiles cover the same floor, the total floor area is also $128(x + 1)^2$.
5. Equating the two expressions for the floor area gives $200x^2 = 128(x + 1)^2$.

Answer (i): $200x^2 = 128(x + 1)^2$

**Part (ii)**

1. Start with the equation $200x^2 = 128(x + 1)^2$.
2. Divide both sides by $8$ to simplify: $25x^2 = 16(x^2 + 2x + 1)$.
3. Expand the right side: $25x^2 = 16x^2 + 32x + 16$.
4. Transpose all terms to the left-hand side: $25x^2 - 16x^2 - 32x - 16 = 0$.
5. Combine like terms to get the standard form: $9x^2 - 32x - 16 = 0$.

Answer (ii): $9x^2 - 32x - 16 = 0$

**Part (iii)(a)**

1. Consider the quadratic equation $9x^2 - 32x - 16 = 0$.
2. Split the middle term: $9x^2 - 36x + 4x - 16 = 0$.
3. Factor by grouping: $9x(x - 4) + 4(x - 4) = 0$.
4. $(9x + 4)(x - 4) = 0$.
5. Since $x$ is a side length, $x = 4$ or $x = -\frac{4}{9}$ (rejected). Thus $x = 4$.

Answer (iii)(a): $x = 4$

**Answer:** The side length of each tile is $4$ units.

> Common mistake: Taking negative value of side length.

### Question 37

*4 marks · Case-based · Statistics*

BINGO is game of chance. The host has $75$ balls numbered $1$ through $75$. Each player has a BINGO card with some numbers written on it.
The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.
The table given below, shows the data of one such game where $48$ balls were used before Tara said 'BINGO'.
\begin{tabular}{|c|c|}
\hline
Numbers announced & Number of times \\ \hline
$0-15$ & $8$ \\ \hline
$15-30$ & $9$ \\ \hline
$30-45$ & $10$ \\ \hline
$45-60$ & $12$ \\ \hline
$60-75$ & $9$ \\ \hline
\end{tabular}
Based on the above information, answer the following :
(i) Write the median class.
(ii) When first ball was picked up, what was the probability of calling out an even number ?
(iii) (a) Find median of the given data.
OR
(b) Find mode of the given data.

**Part (i)**

1. Total number of balls used is $N = 48$, so $\frac{N}{2} = 24$.
2. Write the cumulative frequencies for the intervals: $0-15$ ($8$), $15-30$ ($17$), $30-45$ ($27$), $45-60$ ($39$), $60-75$ ($48$).
3. The cumulative frequency just greater than or equal to $24$ is $27$, which corresponds to the interval $30-45$.

Answer (i): $30-45$

**Part (ii)**

1. Total number of balls is $75$, numbered from $1$ to $75$.
2. The even numbers from $1$ to $75$ are $2, 4, 6, \dots, 74$, which are $37$ in total.
3. The probability of calling out an even number is $\frac{37}{75}$.

Answer (ii): \frac{37}{75}

**Part (iii)(a)**

1. The median class is $30-45$, so lower limit $l = 30$, class size $h = 15$, frequency $f = 10$, and cumulative frequency of preceding class $cf = 17$.
2. Apply the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
3. Substitute the values: $\text{Median} = 30 + \left(\frac{24 - 17}{10}\right) \times 15$.
4. Simplify: $\text{Median} = 30 + \frac{7}{10} \times 15 = 30 + 10.5 = 40.5$.

Answer (iii)(a): 40.5

**Answer:** Median class is $30-45$, probability is $\frac{37}{75}$, and median is $42$.

> Common mistake: Taking wrong cumulative frequency for the median formula.

### Question 38

*4 marks · Case-based · Circles*

A backyard is in the shape of a triangle ABC with right angle at B. $\text{AB} = 7\text{ m}$ and $\text{BC} = 15\text{ m}$. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that $\text{AP} = x\text{ m}$.
Based on the above information, answer the following questions :
(i) Find the length of AR in terms of $x$.
(ii) Write the type of quadrilateral BQOR.
(iii) (a) Find the length PC in terms of $x$ and hence find the value of $x$.
OR
(b) Find $x$ and hence find the radius $r$ of circle.

**Part (i)**

1. We know that the lengths of tangents drawn from an external point to a circle are equal.
2. Since the tangents from point A to the circle are AP and AR, we have $\text{AR} = \text{AP}$.
3. Given that $\text{AP} = x\text{ m}$, therefore $\text{AR} = x\text{ m}$.

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

**Part (ii)**

1. OQ and OR are radii drawn to the points of contact Q and R on the sides BC and AB respectively.
2. Since $\angle\text{B} = 90^\circ$ and the radii are perpendicular to the tangents, $\angle\text{ORB} = \angle\text{OQB} = 90^\circ$.
3. Also, adjacent sides OR and OQ are equal to the radius $r$.
4. Therefore, quadrilateral BQOR is a square.

Answer (ii): Square

**Part (iii)(a)**

1. From the external point C, the tangents are CQ and CP, so $\text{CQ} = \text{CP}$.
2. Since BQOR is a square with side equal to the radius $r$, we have $\text{BQ} = \text{BR} = r$.
3. From $\text{AB} = 7$, we get $\text{BR} = 7 - x$, so $r = 7 - x$.
4. Then $\text{BC} = \text{BQ} + \text{QC} \implies 15 = (7 - x) + \text{PC}$, which gives $\text{PC} = 15 - (7 - x) = 8 + x$.
5. Using $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$, we have $\text{AC}^2 = 7^2 + 15^2 = 49 + 225 = 274$, so $\text{AC} = \sqrt{274}$.
6. Also $\text{AC} = \text{AP} + \text{PC} = x + (8 + x) = 2x + 8$.
7. Equating the two expressions for AC gives $2x + 8 = \sqrt{274}$, leading to $x$ or using the tangent length property $\text{AC}^2 = (x+7)^2 + (x+8)^2$ is not needed since $x$ can be found using $\text{AC} = \text{AP} + \text{PC} = x + 8 + x = 2x + 8$.

Answer (iii)(a): $x = 3$

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

> Common mistake: Confusing the equality of tangent lengths from an external point or miscalculating the segments of sides.

## Frequently asked questions

### What is the paper pattern and structure of the CBSE Class 10 Maths Standard Question Paper 2024 Set 30/1/1?

The question paper carries 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 with internal choices in some sections.

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

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

### Which chapters carry the highest weightage in the CBSE Class 10 Maths Standard 2024 Set 30/1/1 paper?

Circles carries the highest weightage with 8 marks, followed by Introduction to Trigonometry and Triangles with 7 marks each. Arithmetic Progressions, Real Numbers, and Coordinate Geometry each carry 6 marks in this paper.

### How should students write their answers to score full marks in the board exam?

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

### Is the solutions PDF for this question paper available for free on SwaVid?

Yes, the complete solutions PDF for the CBSE Class 10 Maths Standard Question Paper 2024 Set 30/1/1 is available for free download on SwaVid. Students can use these detailed solutions to verify their answers and understand the correct marking scheme.

## 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