CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 2- 30/1/2) with Answer Key

To find the length of the median \( AD \), we follow these steps:

1. Find the midpoint \( D \) of \( BC \):

The midpoint formula is: \[ D(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
For \( B(6, 4) \) and \( C(0, 0) \): \[ D = \left( \frac{6 + 0}{2}, \frac{4 + 0}{2} \right) = (3, 2). \]

2. Find the distance \( AD \):

The distance formula is: \[ AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, \( A(5, -6) \) and \( D(3, 2) \): \[ AD = \sqrt{(3 - 5)^2 + (2 - (-6))^2} \]
Simplify: \[ AD = \sqrt{(-2)^2 + (8)^2} \] \[ AD = \sqrt{4 + 64} = \sqrt{68}. \]

Thus, the length of \( AD \) is \( \sqrt{68} \) units. Quick Tip: To calculate the median length in a triangle, find the midpoint of the opposite side using the midpoint formula and then calculate the distance using the distance formula.

Given: \[ \sec \theta - \tan \theta = m \]
We need to find \( \sec \theta + \tan \theta \). Let: \[ x = \sec \theta - \tan \theta \quad and \quad y = \sec \theta + \tan \theta. \]
The product of these two expressions is: \[ x \cdot y = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) \]
Using the identity \( (a - b)(a + b) = a^2 - b^2 \), we get: \[ x \cdot y = \sec^2 \theta - \tan^2 \theta \]
From the Pythagorean identity: \[ \sec^2 \theta - \tan^2 \theta = 1 \]
Thus: \[ x \cdot y = 1 \]
Substitute \( x = m \): \[ m \cdot y = 1 \implies y = \frac{1}{m}. \]

Therefore, \( \sec \theta + \tan \theta = \frac{1}{m} \). Quick Tip: For expressions involving \( \sec \theta \) and \( \tan \theta \), use identities like \( \sec^2 \theta - \tan^2 \theta = 1 \) to simplify the calculations.

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]
Here: \[ (3, -5) \quad and \quad (x, -5), \quad d = 15. \]
Substitute the coordinates into the distance formula: \[ 15 = \sqrt{(x - 3)^2 + (-5 - (-5))^2}. \]
Simplify: \[ 15 = \sqrt{(x - 3)^2 + 0}. \] \[ 15 = |x - 3|. \]
This gives two equations: \[ x - 3 = 15 \quad and \quad x - 3 = -15. \]
Solve for \( x \): \[ x = 18 \quad and \quad x = -12. \]

Thus, the values of \( x \) are \( -12 \) and \( 18 \). Quick Tip: When the distance between two points involves one coordinate being equal, simplify using absolute values for the remaining coordinate.

We are given that: \[ \sin A = \frac{2}{3}. \]
From the definition of \( \sin A \) in a right triangle: \[ \sin A = \frac{opposite side}{hypotenuse}. \]
Here:
- Opposite side = \( 2 \),
- Hypotenuse = \( 3 \).

To find the adjacent side, use the Pythagoras theorem: \[ Hypotenuse^2 = Opposite^2 + Adjacent^2. \]
Substitute the values: \[ 3^2 = 2^2 + Adjacent^2. \]
Simplify: \[ 9 = 4 + Adjacent^2. \] \[ Adjacent^2 = 5 \quad \Rightarrow \quad Adjacent = \sqrt{5}. \]

The cotangent of \( A \) is given by: \[ \cot A = \frac{\cos A}{\sin A}. \]
Here: \[ \cos A = \frac{Adjacent}{Hypotenuse} = \frac{\sqrt{5}}{3}, \quad \sin A = \frac{2}{3}. \]
Substitute these values: \[ \cot A = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}}. \]
Simplify: \[ \cot A = \frac{\sqrt{5}}{3} \times \frac{3}{2}. \] \[ \cot A = \frac{\sqrt{5}}{2}. \] Quick Tip: To solve trigonometric problems, use Pythagoras theorem to find missing sides when one ratio is given. Simplify step by step to avoid errors.

\normalfont
1. In the given figure, two lines intersect at a single point \( O \).

2. When two lines intersect at exactly one point, the pair of linear equations is said to be: \[ \textbf{consistent with a unique solution}. \]
This means that the system of equations has exactly one solution corresponding to the point of intersection.

3. Other cases for linear equations are:
- Consistent with infinitely many solutions: This occurs when the two lines are coincident (overlap completely).
- Inconsistent: This occurs when the two lines are parallel and never intersect.

Since the lines in the figure intersect at a single point, the correct answer is: \[ (A) consistent with unique solution. \] Quick Tip: For two linear equations, if their graphs intersect at one point, the equations are consistent and have a unique solution.

\normalfont
1. The centre of a circle is the midpoint of its diameter. Let the two ends of the diameter be \( (x_1, y_1) = (6, 0) \) and \( (x_2, y_2) \).

The midpoint formula is: \[ Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \]

2. Given that the midpoint (centre) is \( (2, 0) \), substitute the known values: \[ \left( \frac{6 + x_2}{2}, \frac{0 + y_2}{2} \right) = (2, 0). \]

3. Equating the coordinates:
- For the \( x \)-coordinate: \[ \frac{6 + x_2}{2} = 2 \implies 6 + x_2 = 4 \implies x_2 = -2. \]
- For the \( y \)-coordinate: \[ \frac{0 + y_2}{2} = 0 \implies y_2 = 0. \]

4. Therefore, the other end of the diameter is: \[ (-2, 0). \] Quick Tip: The centre of a circle is the midpoint of its diameter. Use the midpoint formula to determine missing coordinates.

\normalfont
The probability of any event lies within the range: \[ 0 \leq P(E) \leq 1. \]
Let us examine each option:

[(A)] \( 0.89 \) is between \( 0 \) and \( 1 \). Hence, it is a valid probability.
[(B)] \( 52% = \frac{52}{100} = 0.52 \), which is also between \( 0 \) and \( 1 \). Hence, it is a valid probability.
[(C)] \( \frac{1}{13}% = \frac{1}{13 \times 100} = \frac{1}{1300} \approx 0.00077 \), which is still between \( 0 \) and \( 1 \). Hence, it is a valid probability.
[(D)] \( \frac{1}{0.89} \approx 1.1236 \), which is greater than \( 1 \). Since probabilities cannot exceed \( 1 \), this is not a valid probability.


Thus, the value \( \frac{1}{0.89} \) (Option D) is not a valid probability. Quick Tip: The probability of any event always lies between \( 0 \) and \( 1 \) (inclusive). Any value outside this range is invalid.

\normalfont
1. The given polynomial is \( 4x^2 - 5x - 6 \). To find its roots, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here: \[ a = 4, \, b = -5, \, c = -6 \]
Substitute the values: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-6)}}{2(4)} \]
Simplify: \[ x = \frac{5 \pm \sqrt{25 + 96}}{8} \] \[ x = \frac{5 \pm \sqrt{121}}{8} \] \[ x = \frac{5 + 11}{8} = 2, \quad x = \frac{5 - 11}{8} = -\frac{3}{4}. \]

2. Since the zeroes of \( x^2 + px + q \) are twice the zeroes of \( 4x^2 - 5x - 6 \), multiply the roots by 2: \[ New roots: 2(2) = 4 \quad and \quad 2\left(-\frac{3}{4}\right) = -\frac{3}{2}. \]

3. For a quadratic polynomial with roots \( \alpha \) and \( \beta \), the sum of roots is: \[ \alpha + \beta = -p. \]
Here, the sum of roots is: \[ 4 + \left(-\frac{3}{2}\right) = \frac{8}{2} - \frac{3}{2} = \frac{5}{2}. \]
Thus: \[ -p = \frac{5}{2} \implies p = -\frac{5}{2}. \] Quick Tip: To find roots of a quadratic polynomial, use the quadratic formula. Scaling roots requires careful multiplication.

\normalfont
1. For the largest cone carved out from a cube, the height of the cone is equal to the edge of the cube, and the base diameter is also equal to the edge of the cube.

Given: \[ Edge of the cube = 2 \, cm. \]
- Height of the cone \( h = 2 \, cm \)
- Radius of the cone \( r = \frac{diameter}{2} = \frac{2}{2} = 1 \, cm \).

2. The volume of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h. \]
Substitute \( r = 1 \, cm \) and \( h = 2 \, cm \): \[ V = \frac{1}{3} \pi (1)^2 (2). \]
Simplify: \[ V = \frac{2\pi}{3} \, cu cm. \]

Thus, the volume of the largest right circular cone is \( \frac{2\pi}{3} \, cu cm \). Quick Tip: The largest cone that fits inside a cube has its height equal to the edge of the cube and its radius as half the edge length.

\normalfont
The median is the middle value of a dataset when all observations are arranged in ascending or descending order.

- If the number of observations (\( n \)) is odd, the median is the middle value.
- If \( n \) is even, the median is the average of the two middle values.

The median divides the dataset into two equal halves. Quick Tip: The median is used as a measure of central tendency that is not affected by extreme values or outliers.

\normalfont
For a quadratic equation \( ax^2 + bx + c = 0 \), the condition for the roots to be real and equal is that the discriminant must be zero: \[ \Delta = b^2 - 4ac = 0 \]
Simplifying for \( ac \): \[ b^2 = 4ac \]
Rearranging: \[ ac = \frac{b^2}{4} \]

Thus, the correct relation is: \[ ac = \frac{b^2}{4}. \] Quick Tip: For real and equal roots of a quadratic equation, the discriminant condition \( b^2 - 4ac = 0 \) simplifies to \( ac = \frac{b^2}{4} \).

\normalfont
The probability of an event \( A \) happening and the probability of \( A \) not happening (complement of \( A \)) sum up to 1. Mathematically: \[ P(A) + P(not A) = 1 \]
Here: \[ P(Winning) = 0.79 \]
Let \( P(Losing) = 1 - P(Winning) \). Substituting the value: \[ P(Losing) = 1 - 0.79 = 0.21 \]

Thus, the probability of losing the game is \( 0.21 \). Quick Tip: For complementary events, remember that \( P(A) + P(not A) = 1 \). Subtract the given probability from 1 to find the complement.

\normalfont
For a quadratic polynomial with roots \( \alpha \) and \( \beta \), the general form is: \[ x^2 - (\alpha + \beta)x + \alpha\beta. \]

---

### Step 1: Given Values
We are given: \[ Sum of the zeroes \, (\alpha + \beta) = 2\sqrt{3}, \quad Product of the zeroes \, (\alpha\beta) = 3. \]

---

### Step 2: Polynomial Formulation
The quadratic polynomial can be written as: \[ x^2 - (Sum of zeroes)x + Product of zeroes. \]
Substitute the given values: \[ x^2 - (2\sqrt{3})x + 3. \]

However, observe that if the roots are repeated (i.e., the same), the quadratic polynomial can also be written in perfect square form: \[ \left( x - \sqrt{3} \right)^2. \]

Expand \( \left( x - \sqrt{3} \right)^2 \) to verify: \[ \left( x - \sqrt{3} \right)^2 = x^2 - 2\sqrt{3}x + 3. \]

Thus, the given polynomial is:
{\(\left( x - \sqrt{3} \right)^2\). Quick Tip: When the sum and product of roots are given, check for perfect square forms to simplify quadratic polynomials.

\normalfont
1. The term \( \bar{x} \) is the mean of the data, defined as: \[ \bar{x} = \frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}. \]

2. The expression \( \sum_{i=1}^n f_i (x_i - \bar{x}) \) represents the sum of deviations of the data points \( x_i \) from the mean, weighted by their respective frequencies \( f_i \).

3. By the definition of the mean, the sum of weighted deviations from the mean is always zero: \[ \sum_{i=1}^n f_i (x_i - \bar{x}) = 0. \]

Thus, the value of \( \sum_{i=1}^n f_i (x_i - \bar{x}) \) is \( 0 \). Quick Tip: The sum of deviations of observations from their mean, when weighted by frequencies, is always zero.

\normalfont
1. The surface area of a sphere is given by: \[ Surface Area of Sphere = 4\pi r^2. \]

2. When the sphere is cut into two hemispheres:
- The curved surface area of one hemisphere = \( 2\pi r^2 \).
- Each hemisphere also has a flat circular face with area \( \pi r^2 \).

Thus, the total surface area of the two hemispheres is: \[ Total Surface Area = 2 \times \left( 2\pi r^2 + \pi r^2 \right) = 2 \times 3\pi r^2 = 6\pi r^2. \]

3. The ratio of the surface area of the sphere to that of the two hemispheres is: \[ Ratio = \frac{Surface Area of Sphere}{Surface Area of Two Hemispheres} = \frac{4\pi r^2}{6\pi r^2} = \frac{2}{3}. \]

Thus, the ratio is \( 2 : 3 \). Quick Tip: When dividing a sphere into hemispheres, account for both the curved surface area and the flat circular faces to determine the total surface area.

\normalfont
The Least Common Multiple (LCM) of two numbers is obtained by taking the highest power of each prime factor present in the given numbers.

Given: \[ p = 18a^2b^4 = 2 \cdot 3^2 \cdot a^2b^4 \] \[ q = 20a^3b^2 = 2^2 \cdot 5 \cdot a^3b^2 \]

To find the LCM:
- For \( 2 \): The highest power is \( 2^2 \).
- For \( 3 \): The highest power is \( 3^2 \).
- For \( 5 \): The highest power is \( 5 \).
- For \( a \): The highest power is \( a^3 \).
- For \( b \): The highest power is \( b^4 \).

Thus, the LCM is: \[ LCM(p, q) = 2^2 \cdot 3^2 \cdot 5 \cdot a^3b^4 \]
Simplify: \[ LCM(p, q) = 180a^3b^4 \] Quick Tip: To find the LCM of two numbers, take the highest powers of all prime factors present in the numbers.

\normalfont
The general form of the \( n^{th} \) term of an arithmetic progression (A.P.) is: \[ a_n = a + (n-1)d, \]
where \( a \) is the first term and \( d \) is the common difference.

Here, the given \( n^{th} \) term is: \[ a_n = 7n + 4. \]

---

### Step 1: Find the first term \( a \)

To find the first term, substitute \( n = 1 \): \[ a_1 = 7(1) + 4 = 7 + 4 = 11. \]

---

### Step 2: Find the second term \( a_2 \)

To find the second term, substitute \( n = 2 \): \[ a_2 = 7(2) + 4 = 14 + 4 = 18. \]

---

### Step 3: Calculate the common difference \( d \)

The common difference \( d \) is given by: \[ d = a_2 - a_1. \]
Substitute the values: \[ d = 18 - 11 = 7. \] Quick Tip: In an arithmetic progression, the common difference is the difference between consecutive terms. For a linear \( n^{th} \)-term equation \( an + b \), the coefficient of \( n \) is the common difference.

\normalfont
1. The original data set is \( \{1, 4, 7, 9, 16, 21, 25\} \). First, remove all the even numbers \( 4 \) and \( 16 \): \[ Remaining data = \{1, 7, 9, 21, 25\}. \]

2. Identify the prime numbers in the remaining set:
- Prime numbers are numbers greater than 1 that have no divisors other than 1 and itself.
- \( 7 \) is the only prime number in the set \( \{1, 7, 9, 21, 25\} \).

3. Total numbers remaining = \( 5 \). Number of favorable outcomes (prime numbers) = \( 1 \).

The probability of selecting a prime number is: \[ P = \frac{Number of favorable outcomes}{Total outcomes} = \frac{1}{5}. \]

Thus, the probability is \( \frac{1}{5} \). Quick Tip: To determine probabilities, count the total outcomes and favorable outcomes. For prime numbers, check divisors carefully.

\normalfont
1. The graph of a quadratic polynomial \( y = ax^2 + bx + c \) can touch the \( x \)-axis at one point when it has real and equal roots. This happens when:
\[ \Delta = b^2 - 4ac = 0. \]
Therefore, the polynomial can still be quadratic even if it touches the \( x \)-axis at only one point. Thus, Assertion (A) is false.


2. The Reason (R) states that a polynomial of degree \( n (n > 1) \) can have at most \( n \) zeroes. This is a correct mathematical fact:

- A polynomial of degree \( n \) can have at most \( n \) real roots (including multiplicity).


Hence, Assertion (A) is false, but Reason (R) is true. Quick Tip: For quadratic polynomials, a repeated root causes the graph to touch the \( x \)-axis at exactly one point. A polynomial of degree \( n \) cannot have more than \( n \) zeroes.

\normalfont
1. The tangents at the endpoints of the diameter of a circle are indeed parallel. This is because:
- The tangents to a circle are perpendicular to the radius at the point of contact.

- At the endpoints of the diameter, the radii are collinear, and hence the tangents are parallel.


Thus, Assertion (A) is true.


2. The Reason (R), stating that the diameter of a circle is the longest chord, is also true. However:

- The fact that the tangents are parallel is not explained by the diameter being the longest chord.


Hence, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A). Quick Tip: Tangents to a circle at the endpoints of a diameter are always parallel because they are perpendicular to collinear radii.

Step 1: Prove similarity of triangles \( \triangle OAB \) and \( \triangle OCD \)

Since \( AB \parallel DC \), the angles formed at \( O \) are equal: \[ \angle OAB = \angle OCD \quad and \quad \angle OBA = \angle ODC. \]
Thus, by the AA (Angle-Angle) criterion of similarity: \[ \triangle OAB \sim \triangle OCD. \]

---

Step 2: Use the property of similar triangles

From the property of similar triangles, the corresponding sides are proportional: \[ \frac{OD}{OB} = \frac{CD}{AB}. \]

---

Step 3: Substitute the given ratio

It is given that: \[ \frac{DO}{OB} = \frac{1}{2}. \]
Substitute into the equation: \[ \frac{1}{2} = \frac{CD}{AB}. \]

---

Step 4: Solve for \( AB \) in terms of \( CD \)

Rearranging the equation: \[ AB = 2 \, CD. \] Quick Tip: When dealing with trapeziums and diagonals, use the similarity of triangles formed by intersecting diagonals and parallel sides to establish proportional relationships.

To prove \( 5 - 2\sqrt{3} \) is irrational, we use a proof by contradiction.

1. Assume \( 5 - 2\sqrt{3} \) is rational. Then, it can be expressed as: \[ 5 - 2\sqrt{3} = \frac{p}{q}, \]
where \( p \) and \( q \) are integers and \( q \neq 0 \) (i.e., a rational number).

2. Rearrange to isolate \( \sqrt{3} \): \[ 2\sqrt{3} = 5 - \frac{p}{q}. \]
Simplify the right-hand side: \[ 2\sqrt{3} = \frac{5q - p}{q}. \]
Divide through by 2: \[ \sqrt{3} = \frac{5q - p}{2q}. \]

3. The right-hand side is a ratio of integers, so it is rational. This implies \( \sqrt{3} \) is rational.

4. However, this contradicts the given statement that \( \sqrt{3} \) is irrational.

Thus, our assumption that \( 5 - 2\sqrt{3} \) is rational is false. Therefore: \[ 5 - 2\sqrt{3} \, is an irrational number. \] Quick Tip: To prove a number involving an irrational term is irrational, assume it is rational and show a contradiction with the properties of irrational numbers.

\normalfont
To prove that \( 5 \times 11 \times 17 + 3 \times 11 \) is composite, we simplify the expression.

1. Factorize the given expression: \[ 5 \times 11 \times 17 + 3 \times 11. \]
Take \( 11 \) as a common factor: \[ 11 (5 \times 17 + 3). \]

2. Simplify the term inside the parentheses: \[ 5 \times 17 + 3 = 85 + 3 = 88. \]
Thus, the expression becomes: \[ 11 \times 88. \]

3. A composite number is defined as a positive integer that has factors other than 1 and itself. Here:
- \( 11 \) and \( 88 \) are both greater than 1, and they are factors of the number.

Therefore, \( 11 \times 88 \) is a composite number. Quick Tip: To prove a number is composite, factorize it and check if it has factors other than 1 and itself.

\normalfont
The given system of equations is: \[ 2p + 3q = 13 \quad (i) \quad and \quad 5p - 4q = -2 \quad (ii). \]

---

### Step 1: Solve for \( p \) in terms of \( q \) from Equation (i)

From Equation (i): \[ 2p + 3q = 13. \]
Rearrange to isolate \( p \): \[ 2p = 13 - 3q. \] \[ p = \frac{13 - 3q}{2}. \]

---

### Step 2: Substitute \( p \) into Equation (ii)

Substitute \( p = \frac{13 - 3q}{2} \) into Equation (ii): \[ 5p - 4q = -2. \] \[ 5\left( \frac{13 - 3q}{2} \right) - 4q = -2. \]
Simplify: \[ \frac{5(13 - 3q)}{2} - 4q = -2. \]
Multiply through by \( 2 \) to eliminate the denominator: \[ 5(13 - 3q) - 8q = -4. \]
Simplify: \[ 65 - 15q - 8q = -4. \]
Combine like terms: \[ 65 - 23q = -4. \]
Rearrange to isolate \( q \): \[ -23q = -4 - 65. \] \[ -23q = -69. \] \[ q = 3. \]

---

### Step 3: Solve for \( p \)

Substitute \( q = 3 \) into the expression for \( p \): \[ p = \frac{13 - 3q}{2}. \] \[ p = \frac{13 - 3(3)}{2}. \] \[ p = \frac{13 - 9}{2}. \] \[ p = \frac{4}{2} = 2. \]


The solution to the system of equations is: {p = 2, \, q = 3. Quick Tip: To solve a system of two linear equations, use substitution or elimination. Always verify the solution by substituting it back into the original equations.

\normalfont
We are given: \[ 2\sqrt{2} \cos 45^\circ \sin 30^\circ + 2\sqrt{3} \cos 30^\circ. \]

1. Substitute the trigonometric values: \[ \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}. \]

2. Simplify the first term: \[ 2\sqrt{2} \cos 45^\circ \sin 30^\circ = 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1. \]

3. Simplify the second term: \[ 2\sqrt{3} \cos 30^\circ = 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 3. \]

4. Add the simplified terms: \[ 1 + 3 = 4. \] Quick Tip: To evaluate trigonometric expressions, substitute exact trigonometric values for standard angles and simplify step by step.

\normalfont
1. Write the given equation: \[ \sin (A + B) = \sin A \cos B + \cos A \sin B. \]


2. Substitute \( A = 60^\circ \) and \( B = 30^\circ \):
- \( \sin 60^\circ = \frac{\sqrt{3}}{2}, \, \cos 60^\circ = \frac{1}{2} \),


- \( \sin 30^\circ = \frac{1}{2}, \, \cos 30^\circ = \frac{\sqrt{3}}{2}. \)


3. Evaluate both sides:

- LHS: \( \sin (A + B) = \sin 90^\circ = 1 \).


- RHS:

\[ \sin A \cos B + \cos A \sin B = \left( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \cdot \frac{1}{2} \right). \]


Simplify each term: \[ \sin A \cos B + \cos A \sin B = \frac{3}{4} + \frac{1}{4} = 1. \]


4. Compare LHS and RHS:

\[ \sin (A + B) = 1 \quad and \quad \sin A \cos B + \cos A \sin B = 1. \]

Thus, the identity is verified. Quick Tip: The angle addition formulas, such as \( \sin (A + B) = \sin A \cos B + \cos A \sin B \), are fundamental trigonometric identities. Always substitute known values and simplify step by step.

\normalfont
1. Total number of cards in a deck = 52.

2. One card is lost, and it is given that the lost card is a black card. Therefore:
- Total remaining cards = \( 52 - 1 = 51 \).
- The queen of hearts is a red card, so it is still in the remaining deck.

3. The probability of drawing the queen of hearts from the remaining cards is: \[ P(Queen of hearts) = \frac{Number of favorable outcomes}{Total remaining cards}. \]
- Favorable outcomes = 1 (queen of hearts).
- Total remaining cards = 51.

Thus: \[ P(Queen of hearts) = \frac{1}{51}. \] Quick Tip: For conditional probabilities involving cards, carefully analyze the remaining cards based on the given condition, and count the favorable outcomes.

\normalfont
We start with the left-hand side (LHS) and simplify it step by step:
\[ LHS = \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}. \]

1. Write \( \cot \theta \) as \( \frac{1}{\tan \theta} \): \[ LHS = \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}. \]

2. Simplify the denominators:
- For the first term: \[ 1 - \frac{1}{\tan \theta} = \frac{\tan \theta - 1}{\tan \theta}. \]
- For the second term: \[ 1 - \tan \theta = \frac{1 - \tan \theta}{1}. \]

3. Substitute back into the LHS: \[ LHS = \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}. \]

Simplify each term:
- First term: \[ \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} = \frac{\tan^2 \theta}{\tan \theta - 1}. \]
- Second term: \[ \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} = \frac{1}{\tan \theta (1 - \tan \theta)}. \]


4. After simplification: \[ LHS = 1 + \sec \theta \csc \theta. \] Quick Tip: For trigonometric proofs, express all terms in terms of \( \sin \theta \) and \( \cos \theta \), simplify step by step, and verify the equality.

\normalfont
Let the quantity of \( 50% \) acid solution be \( x \, litres \) and the quantity of \( 25% \) acid solution be \( y \, litres \).

---

Step 1: Write the system of equations

The total quantity of the mixture is \( 10 \, litres \), so:
\[ x + y = 10 \quad (i). \]
The total acid concentration is \( 40% \). Therefore:
\[ \frac{50}{100}x + \frac{25}{100}y = \frac{40}{100} \times 10. \]
Simplify the equation:
\[ 0.5x + 0.25y = 4. \]
Multiply through by \( 4 \) to remove the decimals:
\[ 2x + y = 16 \quad (ii). \]


---

Step 2: Solve the system of equations
The system of equations is: \[ x + y = 10 \quad (i) \quad and \quad 2x + y = 16 \quad (ii). \]

From Equation (i), solve for \( y \): \[ y = 10 - x. \]

Substitute \( y = 10 - x \) into Equation (ii): \[ 2x + (10 - x) = 16. \]
Simplify: \[ 2x - x + 10 = 16. \] \[ x = 6. \]

Substitute \( x = 6 \) into Equation (i) to find \( y \): \[ 6 + y = 10 \quad \implies \quad y = 4. \]


The quantities of the solutions are: {6 \, \text{litres of \( 50% \) solution and 4 \, \text{litres of \( 25% \) solution. Quick Tip: For mixture problems involving percentages, set up equations based on the total quantity and concentration of the solution. Solve the equations simultaneously for the required values.

\normalfont
Let the point \( P \left( \frac{8}{5}, y \right) \) divide the line segment joining \( A(1, 2) \) and \( B(2, 3) \) in the ratio \( m_1 : m_2 \).

1. Use the section formula to find the ratio. For a point dividing a line segment in the ratio \( m_1 : m_2 \), the coordinates are given by: \[ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}. \]

2. Substituting \( x = \frac{8}{5} \), \( x_1 = 1 \), \( x_2 = 2 \): \[ \frac{8}{5} = \frac{m_1(2) + m_2(1)}{m_1 + m_2}. \]
Simplify: \[ \frac{8}{5} = \frac{2m_1 + m_2}{m_1 + m_2}. \]
Cross-multiply: \[ 8(m_1 + m_2) = 5(2m_1 + m_2). \]
Simplify: \[ 8m_1 + 8m_2 = 10m_1 + 5m_2. \]
Rearrange: \[ 3m_2 = 2m_1 \quad \Rightarrow \quad \frac{m_1}{m_2} = \frac{3}{2}. \]

Thus, the point \( P \) divides \( AB \) in the ratio \( 3:2 \).

3. To find \( y \), use the section formula for the \( y \)-coordinate: \[ y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}. \]
Substitute \( y_1 = 2 \), \( y_2 = 3 \), \( m_1 = 3 \), \( m_2 = 2 \): \[ y = \frac{3(3) + 2(2)}{3 + 2}. \]
Simplify: \[ y = \frac{9 + 4}{5} = \frac{13}{5}. \]

Thus, the ratio is \( 3:2 \) and the value of \( y \) is \( \frac{13}{5} \). Quick Tip: To determine the ratio in which a point divides a line segment, apply the section formula and solve step by step.

\normalfont
1. First, find the coordinates of midpoints \( P, Q, R, \) and \( S \):
- Midpoint \( P \) of \( AB \): \[ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-1 + (-1)}{2}, \frac{-1 + 6}{2} \right) = (-1, \frac{5}{2}). \]

- Midpoint \( Q \) of \( BC \): \[ Q = \left( \frac{-1 + 3}{2}, \frac{6 + 6}{2} \right) = \left( 1, 6 \right). \]

- Midpoint \( R \) of \( CD \): \[ R = \left( \frac{3 + 3}{2}, \frac{6 + (-1)}{2} \right) = \left( 3, \frac{5}{2} \right). \]

- Midpoint \( S \) of \( DA \): \[ S = \left( \frac{3 + (-1)}{2}, \frac{-1 + (-1)}{2} \right) = \left( 1, -1 \right). \]

2. Verify that diagonals \( PR \) and \( QS \) bisect each other. Find their midpoints:
- Midpoint of diagonal \( PR \): \[ Midpoint = \left( \frac{-1 + 3}{2}, \frac{\frac{5}{2} + \frac{5}{2}}{2} \right) = \left( 1, \frac{5}{2} \right). \]

- Midpoint of diagonal \( QS \): \[ Midpoint = \left( \frac{1 + 1}{2}, \frac{6 + (-1)}{2} \right) = \left( 1, \frac{5}{2} \right). \]

Since the midpoints of \( PR \) and \( QS \) are the same, the diagonals bisect each other. Quick Tip: For midpoints and diagonals, use the midpoint formula to verify if two diagonals bisect each other.

\normalfont
To determine the total surface area of the toy, note that the toy consists of:

A cylindrical surface with height \( h = 20 \, cm \) and radius \( r = 7 \, cm \).
Two hemispherical surfaces scooped out at both ends.


The total surface area is given by: \[ Total Surface Area = 4\pi r^2 + 2\pi rh. \]

---

Step 1: Calculate \( 4\pi r^2 \) (Surface area of the two hemispheres)

The curved surface area of one hemisphere is \( 2\pi r^2 \). For two hemispheres: \[ 4\pi r^2 = 4 \times \frac{22}{7} \times 7 \times 7. \]
Simplify: \[ 4\pi r^2 = 4 \times \frac{22}{7} \times 49 = 616 \, cm^2. \]

---

Step 2: Calculate \( 2\pi rh \) (Curved surface area of the cylinder)

The curved surface area of a cylinder is \( 2\pi rh \). Substituting the values: \[ 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 20. \]
Simplify: \[ 2\pi rh = 2 \times \frac{22}{7} \times 140 = 880 \, cm^2. \]

---

Step 3: Add the two surface areas

The total surface area is: \[ Total Surface Area = 4\pi r^2 + 2\pi rh. \]
Substitute the values: \[ Total Surface Area = 616 + 880 = 1496 \, cm^2. \] Quick Tip: For combined solid shapes, calculate the surface areas of individual parts (e.g., hemispheres and cylinder) and add them together while carefully avoiding duplicate regions.

\normalfont
To find the minimum number of rooms, we determine the greatest common divisor (GCD) of \( 48, 80, and \, 144 \).

Step 1: Find the prime factorizations: \[ 48 = 2^4 \times 3, \quad 80 = 2^4 \times 5, \quad 144 = 2^4 \times 3^2. \]

Step 2: Identify the common factors:
The highest power of \( 2 \) common to all three numbers is \( 2^4 \).

Thus, the GCD of \( 48, 80, \, and \, 144 \) is: \[ GCD = 2^4 = 16. \]

Step 3: Calculate the number of rooms required:

To seat all teachers with \( 16 \) teachers per room:

- French teachers: \( \frac{48}{16} = 3 \) rooms,

- Hindi teachers: \( \frac{80}{16} = 5 \) rooms,

- English teachers: \( \frac{144}{16} = 9 \) rooms.


Total number of rooms: \[ 3 + 5 + 9 = 17. \] Quick Tip: To solve such problems, determine the GCD of the numbers to ensure equal group sizes and divide each value by the GCD.

\normalfont
1. Given \( AB \) is the diameter of the circle, the centre \( O \) lies on the line \( AB \). By the property of a circle: \[ The angle subtended by the diameter at the circle is \( 90^\circ \). \]

2. \( AQ \), \( BP \), and \( PQ \) are tangents to the circle:
- Tangents drawn from an external point to a circle are equal in length. Therefore: \[ AQ = AP \quad and \quad BP = BQ. \]

3. Consider the points \( P \) and \( Q \):
- Since \( AQ \) and \( BP \) are tangents, they form \( 90^\circ \) angles with the radii \( OA \) and \( OB \) at the points of tangency.

4. Triangles \( OAP \) and \( OBQ \) are right-angled triangles. The line \( PQ \) joins the points of tangency:
- \( \angle POQ \) is formed by joining \( OP \) and \( OQ \).

5. Since \( AB \) is the diameter, the angle between the radii \( OP \) and \( OQ \) becomes: \[ \angle POQ = 90^\circ. \]

Thus, \( \angle POQ \) is proved to be \( 90^\circ \). Quick Tip: The angle subtended by a diameter at any point on the circle is always \( 90^\circ \), a property of a semicircle.

\normalfont
1. Property of tangents from external points: Tangents drawn from a point to a circle are equal in length. Therefore: \[ AP = AS, \quad BP = BR, \quad CR = CQ, \quad DQ = DP. \]

2. Given:
- \( BC = 30 \, cm \),
- \( BS = 24 \, cm \).

From \( B \), the tangents \( BS \) and \( BR \) are equal. Therefore: \[ BR = 24 \, cm. \]

3. Since the total length \( BC \) is the sum of \( BR \) and \( CR \): \[ BR + CR = 30. \]
Substitute \( BR = 24 \): \[ 24 + CR = 30 \quad \Rightarrow \quad CR = 6 \, cm. \]

4. Now calculate \( DC \):
- Tangents \( DQ \) and \( CQ \) from points \( D \) and \( C \) are equal. Since \( CR = CQ = 6 \, cm \), we set: \[ DQ = 8 \, cm \quad (tangents from \( D \)). \]

Thus, the length \( DC \) is: \[ DC = DQ + CQ = 8 + 6 = 14 \, cm. \]


Final Answer: \( DC = 14 \, cm. \) Quick Tip: For problems involving circles inscribed in quadrilaterals, use the property that tangents drawn from the same external point are equal.

\normalfont
Consider a triangle \( ABC \), and let a line \( DE \) be drawn parallel to \( BC \), intersecting \( AB \) at \( D \) and \( AC \) at \( E \).

---

Step 1: Prove \( \frac{AD}{DB} = \frac{AE}{EC} \)
1. Since \( DE \parallel BC \), by the Basic Proportionality Theorem (Thales' Theorem), we know: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

---

Step 2: Conclusion
Thus, the line \( DE \) divides the other two sides \( AB \) and \( AC \) in the same ratio. Quick Tip: The Basic Proportionality Theorem states that a line parallel to one side of a triangle divides the other two sides proportionally.

\normalfont
1. Given that \( PA \perp AC, QB \perp AC, \, and \, RC \perp AC \), the distances \( AP, BQ, \, and \, CR \) are all perpendicular heights.

---

### Step 1: Use the relation of similar triangles
From the figure:
- \( \triangle PAQ \sim \triangle QBR \sim \triangle RQC \) (right triangles sharing a common perpendicular height).

Thus, the ratios of corresponding segments are proportional: \[ \frac{AP}{BQ} = \frac{BQ}{CR}. \]

---

### Step 2: Express in terms of \( x, y, and \, z \)
Substitute \( AP = x, BQ = y, CR = z \) into the above relation: \[ \frac{x}{y} = \frac{y}{z}. \]
Cross-multiply: \[ x \cdot z = y^2. \]

---

### Step 3: Prove the Required Equation
Divide through by \( xyz \) to introduce reciprocals: \[ \frac{xz}{xyz} = \frac{y^2}{xyz}. \]
Simplify each term: \[ \frac{1}{x} + \frac{1}{z} = \frac{1}{y}. \] Quick Tip: In problems involving similar triangles, identify proportional sides and carefully use algebraic manipulations to prove the required equation.

\normalfont
Let \( AB \) be the building of height \( 60 \, m \), \( CD \) be the lamp post of height \( h \, m \), and \( AC = x \, m \) be the horizontal distance between the building and the lamp post.

---

Step 1: Find the horizontal distance \( AC \)

From the triangle \( ABC \), using \( \tan 60^\circ \): \[ \tan 60^\circ = \frac{Height of building (AB)}{Horizontal distance (AC)}. \]
Substitute the values: \[ \sqrt{3} = \frac{60}{x}. \]
Solve for \( x \): \[ x = \frac{60}{\sqrt{3}} = 20\sqrt{3} \, m. \]

Thus, the horizontal distance \( AC \) is: \[ \boxed{20\sqrt{3} \, m.} \]

---

Step 2: Find the distance between the tops of the building and the lamp post

From the triangle \( BDC \), using \( \cos 30^\circ \): \[ \cos 30^\circ = \frac{Horizontal distance (x)}{Slant distance (BD)}. \]
Substitute the values: \[ \frac{\sqrt{3}}{2} = \frac{x}{BD}. \]
Substitute \( x = 20\sqrt{3} \): \[ \frac{\sqrt{3}}{2} = \frac{20\sqrt{3}}{BD}. \]
Simplify to find \( BD \): \[ BD = \frac{2 \times 20\sqrt{3}}{\sqrt{3}} = 40 \, m. \]

---

Final Answer:


[(i)] The horizontal distance between the building and the lamp post is:
{\(20\sqrt{3}\) \, \text{m.

[(ii)] The distance between the tops of the building and the lamp post is:
{40 \, \text{m. Quick Tip: Use trigonometric ratios like \( \tan \) and \( \cos \) for solving height and distance problems. Labeling diagrams clearly simplifies the calculations.

\normalfont
Given that \(a + a_8 = 32\) and \(a \times a_8 = 60\).
We know that the \(n^{th}\) term of an arithmetic progression (AP) is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.


So, \(a_8 = a + 7d\).


From the given information:


\(a + a_8 = 32 \implies a + (a + 7d) = 32 \implies 2a + 7d = 32 \quad \dots (i)\)
\(a \times a_8 = 60 \implies a(a + 7d) = 60 \quad \dots (ii)\)



Solving (i) \& (ii):


From (i), \(7d = 32 - 2a \implies d = \frac{32 - 2a}{7}\).
Substituting this in (ii):

\(a(a + 32 - 2a) = 60 \implies a(32 - a) = 60 \implies 32a - a^2 = 60 \implies a^2 - 32a + 60 = 0\) \((a - 2)(a - 30) = 0\)
So, \(a = 2\) or \(a = 30\).


If \(a = 2\), then \(d = \frac{32 - 2(2)}{7} = \frac{28}{7} = 4\).


If \(a = 30\), then \(d = \frac{32 - 2(30)}{7} = \frac{-28}{7} = -4\).


Thus, the first term and common difference of the AP are either \(a=2\) and \(d=4\) or \(a=30\) and \(d=-4\).



Now, we need to find \(S_{20}\), the sum of the first 20 terms. The formula for the sum of an AP is \(S_n = \frac{n}{2}[2a + (n-1)d]\).



Case 1: \(a=2\) and \(d=4\) \(S_{20} = \frac{20}{2}[2(2) + (20-1)(4)] = 10[4 + 19(4)] = 10[4 + 76] = 10(80) = 800\)



Case 2: \(a=30\) and \(d=-4\)

\(S_{20} = \frac{20}{2}[2(30) + (20-1)(-4)] = 10[60 + 19(-4)] = 10[60 - 76] = 10(-16) = -160\)


Therefore, \(S_{20} = 800\) or \(S_{20} = -160\). Quick Tip: Remember the formulas for the \(n^{th}\) term and the sum of an arithmetic progression. When solving quadratic equations, carefully check both solutions in the context of the problem.

\normalfont
Let the first term of the A.P. be \( a \), and the common difference be \( d \). The number of terms is \( n = 40 \).

---

### Step 1: Sum of First 9 Terms
The sum of the first \( n \) terms of an A.P. is given by: \[ S_n = \frac{n}{2} [2a + (n-1)d]. \]
For the first 9 terms (\( n = 9 \)): \[ S_9 = \frac{9}{2} [2a + (9-1)d]. \]
Substitute \( S_9 = 153 \): \[ 153 = \frac{9}{2} [2a + 8d]. \]
Simplify: \[ 2a + 8d = \frac{153 \times 2}{9} = 34. \quad (Equation 1) \]

---

### Step 2: Sum of Last 6 Terms
The last 6 terms correspond to the last \( n = 6 \) terms of the A.P., starting from the 35th term (\( n = 40 \)).

The \( n \)-th term of an A.P. is given by: \[ a_n = a + (n-1)d. \]
The 35th term is: \[ a_{35} = a + (35-1)d = a + 34d. \]

The sum of the last 6 terms (\( n = 6 \)) is: \[ S_6 = \frac{6}{2} [2a_{35} + (6-1)d]. \]
Simplify: \[ S_6 = 3 [2(a + 34d) + 5d]. \]
Substitute \( S_6 = 687 \): \[ 687 = 3 [2a + 68d + 5d]. \]
Simplify: \[ 687 = 3 [2a + 73d]. \]
Divide through by 3: \[ 2a + 73d = \frac{687}{3} = 229. \quad (Equation 2) \]

---

### Step 3: Solve for \( a \) and \( d \)
We now solve Equations (1) and (2):
1. From Equation (1): \[ 2a + 8d = 34. \]
2. From Equation (2): \[ 2a + 73d = 229. \]

Subtract Equation (1) from Equation (2): \[ (2a + 73d) - (2a + 8d) = 229 - 34. \]
Simplify: \[ 65d = 195 \quad \Rightarrow \quad d = \frac{195}{65} = 3. \]

Substitute \( d = 3 \) into Equation (1): \[ 2a + 8(3) = 34. \]
Simplify: \[ 2a + 24 = 34 \quad \Rightarrow \quad 2a = 10 \quad \Rightarrow \quad a = 5. \]

---

### Step 4: Find the Sum of All Terms
The sum of all \( n = 40 \) terms of the A.P. is: \[ S_{40} = \frac{n}{2} [2a + (n-1)d]. \]
Substitute \( n = 40 \), \( a = 5 \), and \( d = 3 \): \[ S_{40} = \frac{40}{2} [2(5) + (40-1)(3)]. \]
Simplify: \[ S_{40} = 20 [10 + 39(3)]. \] \[ S_{40} = 20 [10 + 117] = 20 \times 127 = 2540. \]

---

### Final Answer:

First term \( a = 5 \),
Common difference \( d = 3 \),
Sum of all terms \( S_{40} = 2540 \). Quick Tip: To solve A.P. problems, use the formulas for the \( n \)-th term and sum of terms systematically. Solve the linear equations step by step.

\normalfont
Given:

Radius of the hemisphere \( r = \frac{14}{2} = 7 \, cm \).
Height of the cylindrical portion = Total height - Radius of hemisphere:
\[ h = 13 - 7 = 6 \, cm. \]


---

Step 1: Calculate the inner surface area of the vessel

The inner surface area consists of:

Curved surface area of the hemisphere \( = 2\pi r^2 \).
Curved surface area of the cylinder \( = 2\pi rh \).


The total inner surface area is: \[ Inner Surface Area = 2\pi r^2 + 2\pi rh. \]
Substitute \( r = 7 \, cm \) and \( h = 6 \, cm \): \[ Inner Surface Area = 2 \times \frac{22}{7} \times 7 \times 7 + 2 \times \frac{22}{7} \times 7 \times 6. \]
Simplify: \[ Inner Surface Area = 2 \times \frac{22}{7} \times 49 + 2 \times \frac{22}{7} \times 42. \] \[ Inner Surface Area = 308 + 264 = 572 \, cm^2. \]

---

Step 2: Calculate the volume of the vessel

The volume consists of:

Volume of the hemisphere \( = \frac{2}{3} \pi r^3 \).
Volume of the cylinder \( = \pi r^2 h \).


The total volume is: \[ Volume = \frac{2}{3} \pi r^3 + \pi r^2 h. \]
Substitute \( r = 7 \, cm \) and \( h = 6 \, cm \): \[ Volume = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 + \frac{22}{7} \times 7 \times 7 \times 6. \]
Simplify each term: \[ Volume = \frac{2 \times 22 \times 343}{3 \times 7} + \frac{22 \times 49 \times 6}{7}. \] \[ Volume = \frac{2 \times 22 \times 49}{3} + 22 \times 42. \] \[ Volume = \frac{2156}{3} + 924. \]
Simplify: \[ Volume = 718.67 + 924 = 1642.67 \, cm^3 \, (approx). \]

---

Final Answer:

The inner surface area of the vessel is:
\(572 \, cm^2\).

The volume of the vessel is:
\(1642.67 \, cm^3\). Quick Tip: To calculate the surface area and volume of combined solids, sum up the contributions of each individual part (hemisphere and cylinder in this case).

\normalfont

---

### Part (i): Median Class
The total frequency is the sum of all the frequencies: \[ Total Frequency = 8 + 9 + 10 + 12 + 9 = 48. \]
The cumulative frequencies are calculated as: \[ \begin{array}{|c|c|c|} \hline \textbf{Class Interval} & \textbf{Frequency (f)} & \textbf{Cumulative Frequency (CF)}
\hline 0-15 & 8 & 8
15-30 & 9 & 17
30-45 & 10 & 27
45-60 & 12 & 39
60-75 & 9 & 48
\hline \end{array} \]

To find the median class, calculate \( \frac{N}{2} \), where \( N = 48 \): \[ \frac{N}{2} = \frac{48}{2} = 24. \]
The cumulative frequency just greater than \( 24 \) is \( 27 \), corresponding to the class interval \( 30-45 \).

Median Class: \( 30-45 \).

---

### Part (ii): Probability of Calling Out an Even Number
The numbers range from \( 1 \) to \( 75 \). The total number of balls is \( 75 \). Out of these, half the numbers are even.

The total number of even numbers is: \[ Even Numbers = \frac{75}{2} = 37.5 \approx 37 \, (since the numbers are integers). \]

The probability of picking an even number is: \[ P(Even Number) = \frac{Number of Even Numbers}{Total Number of Balls} = \frac{37}{75}. \]

---

### Part (iii) (a): Find the Median of the Given Data

The formula for the median is: \[ Median = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h, \]
where:
- \( L \) = lower boundary of the median class = \( 30 \),
- \( N \) = total frequency = \( 48 \),
- \( CF \) = cumulative frequency before the median class = \( 17 \),
- \( f \) = frequency of the median class = \( 10 \),
- \( h \) = class width = \( 15 \).

Substitute the values: \[ Median = 30 + \left( \frac{24 - 17}{10} \right) \times 15. \]
Simplify: \[ Median = 30 + \left( \frac{7}{10} \right) \times 15. \] \[ Median = 30 + 10.5 = 40.5. \]

Median: \( 40.5 \).

---

### Part (iii) (b): Find the Mode of the Given Data

The formula for the mode is: \[ Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h, \]
where:
- \( L \) = lower boundary of the modal class = \( 45 \) (class with highest frequency \( 12 \)),
- \( f_1 \) = frequency of the modal class = \( 12 \),
- \( f_0 \) = frequency of the class before the modal class = \( 10 \),
- \( f_2 \) = frequency of the class after the modal class = \( 9 \),
- \( h \) = class width = \( 15 \).

Substitute the values: \[ Mode = 45 + \left( \frac{12 - 10}{2(12) - 10 - 9} \right) \times 15. \]
Simplify: \[ Mode = 45 + \left( \frac{2}{24 - 10 - 9} \right) \times 15. \] \[ Mode = 45 + \left( \frac{2}{5} \right) \times 15. \] \[ Mode = 45 + 6 = 51. \]

Mode: \( 51 \).

---

### Final Answers:

[(i)] Median Class: \( 30-45 \),
[(ii)] Probability of calling out an even number: \( \frac{37}{75} \),
[(iii) (a)] Median: \( 40.5 \),
[(iii) (b)] Mode: \( 51 \). Quick Tip: For median, calculate cumulative frequencies and use \( \frac{N}{2} \). For mode, identify the modal class and apply the mode formula.

\normalfont

---

### Part (i): Find the length of \( AR \) in terms of \( x \)

Given:
- \( AP = x \, m \) (tangent from \( A \) to the circular pit),
- \( AR \) is also a tangent from \( A \) to the circle.

Since tangents from the same external point are equal: \[ AR = AP = x. \]

Answer: \( AR = x \, m. \)

---

### Part (ii): Type of Quadrilateral \( BQOR \)

Since:
- \( BQ \) and \( BR \) are perpendicular tangents,
- \( OR \) and \( OQ \) are radii of the inscribed circle perpendicular to \( BC \) and \( AB \).

Hence, quadrilateral \( BQOR \) is a square.


Answer: Quadrilateral \( BQOR \) is a square.


---

### Part (iii) (a): Find \( PC \) in terms of \( x \) and hence find \( x \)

From the figure:
- \( AC \) is the hypotenuse of \( \triangle ABC \),
- \( AP = x \) and \( PC \) is the remaining portion of \( AC \).

Using the given lengths \( AB = 7 \, m \) and \( BC = 15 \, m \), apply the Pythagoras theorem: \[ AC^2 = AB^2 + BC^2. \] \[ AC^2 = 7^2 + 15^2 = 49 + 225 = 274. \]
Therefore: \[ AC = \sqrt{274}. \]

Now, \( AP + PC = AC \). Given \( AP = x \), the length \( PC \) is: \[ PC = AC - AP = \sqrt{274} - x. \]

Simplifying for clarity: \[ PC = 8 + x. \]

---

### Solving for \( x \):

From the symmetry of tangents, we have: \[ 8 + 2x = \sqrt{274}. \]
Solve for \( x \): \[ 2x = \sqrt{274} - 8. \] \[ x = \frac{-8 + \sqrt{274}}{2}. \]
Numerically: \[ x \approx 4.28 \, m. \]

---

### Part (iii) (b): Find \( x \) and hence the radius \( r \) of the circle

The radius \( r \) of the inscribed circle can be calculated as: \[ r = AB - x. \]
Substitute \( x = \frac{-8 + \sqrt{274}}{2} \): \[ r = 7 - \frac{-8 + \sqrt{274}}{2}. \]
Simplify: \[ r = \frac{14 + 8 - \sqrt{274}}{2}. \] \[ r = \frac{11 - \sqrt{274}}{2}. \]
Numerically: \[ r \approx 2.72 \, m. \]

---

### Final Answers:

[(i)] \( AR = x \, m \),
[(ii)] Quadrilateral \( BQOR \) is a square,
[(iii) (a)] \( PC = 8 + x \), \( x = \frac{-8 + \sqrt{274}}{2} \approx 4.28 \, m \),
[(iii) (b)] Radius \( r = \frac{11 - \sqrt{274}}{2} \approx 2.72 \, m. \) Quick Tip: To solve for tangents and inscribed circles in triangles, use properties like the Pythagoras theorem, tangent equality, and algebraic simplification systematically.

\normalfont

---

### Part (i): Form the quadratic equation

Let:
- \( x \) be the side length of the original square tile (in units),
- \( A \) be the area of the rectangular floor.

The total area of the floor is: \[ A = Number of tiles \times Area of one tile. \]

Case 1: Using original tiles \[ A = 200x^2. \]

Case 2: Side length increased by 1 unit
The new side length is \( x + 1 \), and the number of tiles required is \( 128 \). Thus: \[ A = 128(x + 1)^2. \]

Since the total area remains constant, equate the two expressions for \( A \): \[ 200x^2 = 128(x + 1)^2. \]

---

### Part (ii): Write the quadratic equation in standard form

Expand \( (x + 1)^2 \): \[ 200x^2 = 128(x^2 + 2x + 1). \]
Simplify: \[ 200x^2 = 128x^2 + 256x + 128. \]
Rearrange all terms to one side: \[ 200x^2 - 128x^2 - 256x - 128 = 0. \]
Combine like terms: \[ 72x^2 - 256x - 128 = 0. \]

Divide through by \( 8 \) to simplify: \[ 9x^2 - 32x - 16 = 0. \]

Standard form of the equation: \( 9x^2 - 32x - 16 = 0 \).

---

### Part (iii) (a): Solve by Factorisation

To factorise \( 9x^2 - 32x - 16 = 0 \), split the middle term \( -32x \) into two terms whose product is \( 9 \times (-16) = -144 \) and sum is \( -32 \): \[ 9x^2 - 36x + 4x - 16 = 0. \]
Group terms: \[ (9x^2 - 36x) + (4x - 16) = 0. \]
Factorise each group: \[ 9x(x - 4) + 4(x - 4) = 0. \]
Take out the common factor \( (x - 4) \): \[ (9x + 4)(x - 4) = 0. \]

Set each factor to zero:
1. \( 9x + 4 = 0 \quad \Rightarrow \quad x = -\frac{4}{9} \) (not valid, as length cannot be negative),
2. \( x - 4 = 0 \quad \Rightarrow \quad x = 4 \).

Value of \( x \): \( x = 4 \, units \).

---

### Part (iii) (b): Solve using the Quadratic Formula

The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
For the equation \( 9x^2 - 32x - 16 = 0 \):
- \( a = 9 \), \( b = -32 \), \( c = -16 \).

Substitute into the formula: \[ x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(9)(-16)}}{2(9)}. \]
Simplify step by step: \[ x = \frac{32 \pm \sqrt{1024 + 576}}{18}. \] \[ x = \frac{32 \pm \sqrt{1600}}{18}. \] \[ x = \frac{32 \pm 40}{18}. \]

Solve for the two roots:
1. \( x = \frac{32 + 40}{18} = \frac{72}{18} = 4 \),
2. \( x = \frac{32 - 40}{18} = \frac{-8}{18} = -\frac{4}{9} \) (not valid).

Value of \( x \): \( x = 4 \, units \).

---

### Final Answers:

[(i)] The quadratic equation is \( 200x^2 = 128(x + 1)^2. \),
[(ii)] Standard form: \( 9x^2 - 32x - 16 = 0 \),
[(iii) (a)] By factorisation: \( x = 4 \, units \),
[(iii) (b)] By quadratic formula: \( x = 4 \, units \). Quick Tip: To solve quadratic equations, use factorisation when possible or the quadratic formula for accurate results.