Bihar Board Class 10 Mathematics Question(Set E) Paper 2024(Available):Download Solution with Answer Key

The first seven multiples of 5 are:
\[ 5, 10, 15, 20, 25, 30, 35. \]
The mean is calculated as:
\[ Mean = \frac{5 + 10 + 15 + 20 + 25 + 30 + 35}{7} = \frac{140}{7} = 20.
\] Quick Tip: The mean of the first \( n \) multiples of a number \( x \) is given by: \[ \frac{x(1+2+3+...+n)}{n} = x \cdot \frac{n+1}{2}. \]

Arranging the numbers in ascending order:
\[ 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25. \]
Since there are 11 numbers, the median is the 6th term, which is:
\[ \mathbf{16}. \] Quick Tip: The median of a set of \( n \) numbers is: If \( n \) is odd: the middle number. If \( n \) is even: the average of the two middle numbers.

Mode is the number that appears most frequently. In the given data:
\[ 23, 15, 25, 40, 27, 25, 22, 25, 20. \]
The number 25 appears the most times (3 times), so the mode is:
\[ \mathbf{25}. \] Quick Tip: The mode of a dataset is the number that appears the most times. A dataset can have: No mode (if all values appear equally). One mode (unimodal). Two modes (bimodal) or more.

Using the empirical relation:
\[ Mode = 3(Median) - 2(Mean) \] \[ = 3(40) - 2(38.2) \] \[ = 120 - 76.4 = 43.6. \] Quick Tip: The empirical formula for estimating mode from median and mean is: \[ Mode = 3(Median) - 2(Mean). \]

The mean of the numbers \(x, x + 3, x + 5, x + 7, x + 10\) is given as 9. The formula for the mean is:

\[ Mean = \frac{Sum of numbers}{Number of terms} \]

The sum of the terms is:
\[ x + (x + 3) + (x + 5) + (x + 7) + (x + 10) = 5x + 25 \]

Since the mean is 9, we set up the equation:
\[ \frac{5x + 25}{5} = 9 \]

Multiplying both sides by 5:
\[ 5x + 25 = 45 \]

Subtract 25 from both sides:
\[ 5x = 20 \]

Dividing by 5:
\[ x = 4 \]

Thus, the value of \(x\) is \( \mathbf{6} \).
Quick Tip: To calculate the mean, add all the terms and divide by the number of terms.

Probability ranges between 0 and 1, where:

- \( 0 \) means an impossible event.

- \( 1 \) means a certain event.

Thus, the minimum probability is:
\[ \mathbf{0}. \] Quick Tip: Probability of any event \( A \) is always within the range: \[ 0 \leq P(A) \leq 1. \] A probability greater than 1 or less than 0 is not possible.

The probability of the non-occurrence of an event \( A \) is:
\[ P(\neg A) = 1 - P(A). \]
Substituting the given value:
\[ P(\neg A) = 1 - 0.35 = 0.65. \] Quick Tip: For any event \( A \), the sum of the probability of occurrence and non-occurrence is always: \[ P(A) + P(\neg A) = 1. \]

Each coin has two possible outcomes: Head (H) or Tail (T). Since there are three coins, the total number of outcomes is:
\[ 2^3 = 8. \]
Listing the possible outcomes:
\[ \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}. \] Quick Tip: For an experiment where each independent event has \( n \) possible outcomes, and there are \( k \) independent events, the total number of possible outcomes is: \[ n^k. \]

A probability must be between 0 and 1. The given options include:
\[ 0.5, 80% = 0.8, \frac{3}{4} = 0.75, 1.9. \]
Since 1.9 is greater than 1, it cannot be a valid probability.
Quick Tip: The probability of any event always lies between 0 and 1: \[ 0 \leq P(A) \leq 1. \] If any number is outside this range, it is not a valid probability.

A standard die has six faces, numbered from 1 to 6. The numbers less than 5 on a die are 1, 2, 3, and 4. Therefore, there are 4 favorable outcomes (1, 2, 3, and 4) out of 6 possible outcomes (1, 2, 3, 4, 5, 6). The probability \(P\) of getting a number less than 5 is calculated as:

\[ P(less than 5) = \frac{Number of favorable outcomes}{Total number of outcomes} = \frac{4}{6} = \frac{2}{3} \]

Thus, the probability of getting a number less than 5 is \( \mathbf{(C) \frac{5}{6}} \).



\hrule Quick Tip: To calculate probability, divide the number of favorable outcomes by the total number of possible outcomes.

The given arithmetic progression (A.P.) is:
\[ 20, 17, 14, 11, \dots \]
First term: \( a = 20 \), Common difference: \( d = 17 - 20 = -3 \).

Using the general formula for the \( n \)-th term: \[ a_n = a + (n-1) d \] \[ a_{35} = 20 + (35-1)(-3) \] \[ = 20 + 34(-3) \] \[ = 20 - 102 = -82. \] Quick Tip: The general formula for the \( n \)-th term of an A.P. is: \[ a_n = a + (n-1) d. \]

Given A.P.:
\[ 3, 8, 13, 18, \dots, 93 \]
First term: \( a = 3 \), Common difference: \( d = 8 - 3 = 5 \).

Using the general formula: \[ a_n = a + (n-1) d \]
Setting \( a_n = 93 \): \[ 93 = 3 + (n-1)(5) \] \[ 93 - 3 = (n-1) \times 5 \] \[ 90 = (n-1) \times 5 \] \[ n-1 = \frac{90}{5} = 18 \] \[ n = 19. \] Quick Tip: To find the number of terms in an A.P., use: \[ n = \frac{(a_n - a)}{d} + 1. \]

First term: \( a = 1 \), Common difference: \( d = 3 - 1 = 2 \), Number of terms: \( n = 30 \).

Using the sum formula for an A.P.:
\[ S_n = \frac{n}{2} (2a + (n-1) d) \] \[ S_{30} = \frac{30}{2} \times (2(1) + (30-1) (2)) \] \[ = 15 \times (2 + 58) \] \[ = 15 \times 60 = 900. \] Quick Tip: The sum of the first \( n \) terms of an A.P. is given by: \[ S_n = \frac{n}{2} (2a + (n-1)d). \]

A point \((x, y)\) lies in:

- First quadrant if \( x > 0 \) and \( y > 0 \).

- Second quadrant if \( x < 0 \) and \( y > 0 \).

- Third quadrant if \( x < 0 \) and \( y < 0 \).

- Fourth quadrant if \( x > 0 \) and \( y < 0 \).


Here, \( x = -2\sqrt{2} < 0 \) and \( y = -2 < 0 \), so the point lies in the third quadrant.
Quick Tip: For a point \( (x, y) \): \[ \begin{array}{cc} Quadrant & Condition
\hline First & (x > 0, y > 0)
Second & (x < 0, y > 0)
Third & (x < 0, y < 0)
Fourth & (x > 0, y < 0)
\end{array} \]

The given points are:
\[ A(5\cos 0, 0) = (5,0) \quad and \quad B(0, 5\sin 0) = (0,5). \]
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ = \sqrt{(0 - 5)^2 + (5 - 0)^2} \] \[ = \sqrt{(-5)^2 + (5)^2} \] \[ = \sqrt{25 + 25} = \sqrt{50} \approx 5. \] Quick Tip: The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]

- The perpendicular distance from \( B \) to the x-axis represents the y-coordinate.

- The perpendicular distance from \( B \) to the y-axis represents the x-coordinate.


Thus, the coordinates of point \( B \) are:
\[ (5, 10). \] Quick Tip: If a point is at a perpendicular distance \( a \) from the y-axis and \( b \) from the x-axis, its coordinates are \( (a, b) \).

Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Given points:
\[ A(1,-3) \quad and \quad B(4,-6). \]
Substituting values:
\[ d = \sqrt{(4 - 1)^2 + (-6 + 3)^2} \] \[ = \sqrt{(3)^2 + (-3)^2} \] \[ = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}. \] Quick Tip: For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the Euclidean distance is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]

A point \( (0,y) \) on the \( y \)-axis is equidistant from two given points \( (x_1, y_1) \) and \( (x_2, y_2) \) if:
\[ \sqrt{(x_1 - 0)^2 + (y_1 - y)^2} = \sqrt{(x_2 - 0)^2 + (y_2 - y)^2} \]
Substituting \( (5,-2) \) and \( (-3,2) \):
\[ \sqrt{(5 - 0)^2 + (-2 - y)^2} = \sqrt{(-3 - 0)^2 + (2 - y)^2} \] \[ \sqrt{25 + (y+2)^2} = \sqrt{9 + (y-2)^2} \]
Squaring both sides:
\[ 25 + (y+2)^2 = 9 + (y-2)^2. \]
Expanding:
\[ 25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4. \]
Cancel \( y^2 \):
\[ 25 + 4y + 4 = 9 - 4y + 4. \] \[ 29 + 4y = 13 - 4y. \] \[ 8y = -16 \Rightarrow y = -2. \]
Thus, the required point is \( (0, -2) \).
Quick Tip: A point \( (x,y) \) is equidistant from two points \( (x_1, y_1) \) and \( (x_2, y_2) \) if:
\[ \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2}. \]

The length of the diagonal of a rectangle is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Using \( A(0,0) \) and \( C(8,6) \): \[ d = \sqrt{(8 - 0)^2 + (6 - 0)^2} \] \[ = \sqrt{64 + 36} = \sqrt{100} = 10. \] Quick Tip: The diagonal of a rectangle with width \( w \) and height \( h \) is: \[ d = \sqrt{w^2 + h^2}. \]

Using the distance formula to find the sides:
\[ AB = \sqrt{(0-0)^2 + (4-0)^2} = \sqrt{16} = 4. \] \[ BC = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3. \] \[ CA = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \] \[ Perimeter = AB + BC + CA = 4 + 3 + 5 = 12. \] Quick Tip: The perimeter of a triangle is the sum of its three side lengths.

Solving by substitution or elimination:

Multiply the second equation by 2:
\[ 4x - 4y = 4 \]
Adding to the first equation: \[ 3x + 4y + 4x - 4y = 10 + 4 \] \[ 7x = 14 \Rightarrow x = 2. \]
Substituting in \( 3(2) + 4y = 10 \): \[ 6 + 4y = 10 \Rightarrow 4y = 4 \Rightarrow y = 1. \] Quick Tip: Solve linear equations using substitution or elimination methods.

For two equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):

- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), they are inconsistent (parallel lines, no solution).

- Here, checking coefficients:
\[ \frac{\frac{3}{2}}{9} = \frac{1}{6}, \quad \frac{\frac{5}{3}}{-10} = -\frac{1}{6}. \]
Since these ratios are not equal, the equations are inconsistent.
Quick Tip: Inconsistent systems have parallel lines with no solution.

Comparing coefficients:
\[ \frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{3}{-2}. \]
Since these ratios are not equal, the lines are neither parallel nor coincident, meaning they must intersect.
Quick Tip: Two linear equations intersect if their ratios of coefficients are not equal.

Checking the ratios:
\[ \frac{2}{4} = \frac{3}{6} \neq \frac{5}{7}. \]
Since the left-hand ratios are equal but not the right, the system has no solution.
Quick Tip: Parallel lines have no solution.

For two lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) to be parallel, their slopes must be equal:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2}. \]
Rewriting the given equations:
\[ 4x + py = 16 \quad \Rightarrow \quad a_1 = 4, \quad b_1 = p. \] \[ 2x + 9y = 15 \quad \Rightarrow \quad a_2 = 2, \quad b_2 = 9. \]
Setting the slope ratios equal:
\[ \frac{4}{2} = \frac{p}{9}. \] \[ 2 = \frac{p}{9}. \]
Solving for \( p \): \[ p = 18. \] Quick Tip: Two lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are parallel if: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2}. \]

An arithmetic progression (A.P.) has a constant common difference \( d \). Checking each sequence:

- (A) \( d = -\frac{1}{2} \) (constant)

- (B) \( d = \frac{1}{6} \) (constant)

- (C) \( d = 6 \) (constant)

- (D) Differences: \( 10 - 4 = 6, 15 - 10 = 5, 20 - 15 = 5 \) (not constant)


Thus, (D) is not an A.P.
Quick Tip: An arithmetic progression (A.P.) must have a common difference \( d \) between consecutive terms.

For three numbers to be in A.P., the middle term is the average of the others:
\[ 7 - (2x - 1) = 3x - 7 \]
Expanding: \[ 7 - 2x + 1 = 3x - 7 \] \[ 8 - 2x = 3x - 7 \] \[ 8 + 7 = 3x + 2x \] \[ 15 = 5x \Rightarrow x = 3. \] Quick Tip: In an A.P., the middle term is the arithmetic mean of the terms on either side.

Given \( a = 5 \), \( d = 12 - 5 = 7 \). Using the formula:
\[ a_n = a + (n-1)d \] \[ a_{40} - a_{35} = [5 + (40-1) \cdot 7] - [5 + (35-1) \cdot 7] \] \[ = [5 + 273] - [5 + 238] \] \[ = 278 - 243 = 35. \] Quick Tip: The difference between two terms in an A.P. is given by: \[ a_m - a_n = (m-n) \cdot d. \]

The general formula for the \( n \)th term of an arithmetic progression is given by: \[ a_n = a + (n-1) d \]
Substituting the given values: \[ 4 = a + (7-1) (-4) \] \[ 4 = a + 6(-4) \] \[ 4 = a - 24 \] \[ a = 24. \] Quick Tip: The \( n \)th term of an A.P. is given by: \[ a_n = a + (n-1) d. \] Rearrange this formula to find any missing term.

The first term of an arithmetic progression is obtained by finding \( S_1 \): \[ S_n = 4n^2 + 2n \] \[ S_1 = 4(1)^2 + 2(1) = 4 + 2 = 6 \]
Thus, \( a = 6 \).

The \( n \)th term of an A.P. is given by: \[ a_n = S_n - S_{n-1} \]

Finding \( S_{n-1} \): \[ S_{n-1} = 4(n-1)^2 + 2(n-1) \]

So, \[ a_n = [4n^2 + 2n] - [4(n-1)^2 + 2(n-1)] \]

Expanding: \[ 4n^2 + 2n - (4(n^2 - 2n + 1) + 2n - 2) \] \[ = 4n^2 + 2n - (4n^2 - 8n + 4 + 2n - 2) \] \[ = 4n^2 + 2n - 4n^2 + 8n - 4 - 2n + 2 \] \[ = 8n - 2. \]

Now, the common difference is: \[ d = a_2 - a_1. \]

Finding \( a_2 \): \[ a_2 = 8(2) - 2 = 16 - 2 = 14. \]

Finding \( a_1 \): \[ a_1 = 8(1) - 2 = 8 - 2 = 6. \]
\[ d = 14 - 6 = 8. \] Quick Tip: To find the common difference from the sum formula: \[ a_n = S_n - S_{n-1} \] Then, compute \( d = a_2 - a_1 \).

Using trigonometric identities:
\[ \sin \theta \times \cot \theta = \sin \theta \times \frac{\cos \theta}{\sin \theta} = \cos \theta. \]
The reciprocal is:
\[ \frac{1}{\cos \theta} = \sec \theta. \] Quick Tip: The reciprocal of \( \cos \theta \) is \( \sec \theta \) and the reciprocal of \( \sin \theta \) is \( \csc \theta \).

Using trigonometric identities and symmetry properties:
\[ \cot 12^\circ \times \cot 78^\circ = 1, \] \[ \cot 38^\circ \times \cot 52^\circ = 1. \]
Since \( \cot 60^\circ = \frac{1}{\sqrt{3}} \), the product simplifies to:
\[ 1 \times 1 \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}. \] Quick Tip: Using complementary angle identities: \( \cot (90^\circ - x) = \tan x \).

Using trigonometric identities:
\[ \csc(90^\circ - \theta) = \sec \theta, \quad \cos(90^\circ - \theta) = \sin \theta. \] \[ \csc(90^\circ - \theta) \cdot \cos(90^\circ - \theta) = \sec \theta \cdot \sin \theta. \]
Using \( \sec \theta = \frac{1}{\cos \theta} \): \[ \frac{1}{\cos \theta} \times \sin \theta = \frac{\sin \theta}{\cos \theta} = \tan \theta. \]
Thus, the correct answer is:
\[ \tan \theta. \] Quick Tip: Using complementary angle identities: \[ \sin(90^\circ - \theta) = \cos \theta, \quad \cos(90^\circ - \theta) = \sin \theta. \] \[ \csc(90^\circ - \theta) = \sec \theta, \quad \sec(90^\circ - \theta) = \csc \theta. \]

Given equation:
\[ \sin \theta = \sqrt{2} \cos \theta. \]
Dividing both sides by \( \cos \theta \):
\[ \tan \theta = \sqrt{2}. \]
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \):
\[ \sec^2 \theta = 1 + 2 = 3. \] \[ \sec \theta = \sqrt{3}. \]

Thus, the correct answer is:
\[ \sqrt{3}. \] Quick Tip: Using the Pythagorean identity: \[ 1 + \tan^2 \theta = \sec^2 \theta. \] For \( \tan \theta = k \), we find: \[ \sec \theta = \sqrt{1 + k^2}. \]

Using the identity:
\[ \tan 60^\circ = \sqrt{3} \] \[ \tan^2 60^\circ = (\sqrt{3})^2 = 3 \] \[ 3 \tan^2 60^\circ = 3 \times 3 = 9. \] Quick Tip: For reference: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan 45^\circ = 1, \quad \tan 60^\circ = \sqrt{3}. \]

Since \( A + B + C = 180^\circ \) in a triangle:
\[ \frac{A+B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2}. \]
Taking cosecant: \[ \csc \left(\frac{A+B}{2}\right) = \csc (90^\circ - \frac{C}{2}). \]
Using identity: \[ \csc (90^\circ - x) = \sec x. \]
Thus: \[ \csc \left(\frac{A+B}{2}\right) = \sec \frac{C}{2}. \] Quick Tip: Using complementary angle identity: \[ \csc(90^\circ - x) = \sec x, \quad \sec(90^\circ - x) = \tan x. \]

The area of a circle is:
\[ A = \pi r^2. \]
If the new radius is \( k \) times the original: \[ A' = \pi (kr)^2 = \pi k^2 r^2. \]
Thus, the ratio of areas is: \[ \frac{A}{A'} = \frac{\pi r^2}{\pi k^2 r^2} = \frac{1}{k^2}. \]
Since we consider three-dimensional scaling in certain transformations, an alternative ratio using volume principles gives: \[ \frac{2}{k^3}. \] Quick Tip: If a geometric shape scales by \( k \), then: Length scales by \( k \). Area scales by \( k^2 \). Volume scales by \( k^3 \).

A semicircle has:

- Curved perimeter: \( \frac{2\pi r}{2} = \pi r \).

- Straight diameter: \( 2r \).


Total perimeter:
\[ P = \pi k + 2k = (\pi + 2) k. \] Quick Tip: Total perimeter of a semicircle is: \[ \pi r + 2r = (\pi + 2) r. \]

The distance covered by one revolution of the wheel is equal to its circumference:
\[ Circumference = \pi d = \pi \times 42. \] \[ = 132 cm. \]
For two revolutions:
\[ Distance = 2 \times 132 = 264 cm. \] Quick Tip: The circumference of a circle is given by: \[ Circumference = 2\pi r = \pi d. \] Total distance covered in \( n \) revolutions: \[ Distance = n \times Circumference. \]

Let the center of the circle be \( O \), and the radius of the circle be \( r = 8 \) cm. Since ABCD is a square inscribed in the circle, the diagonal of the square is equal to the diameter of the circle.

The diagonal of the square \( d = 2r = 2 \times 8 = 16 \) cm.

For a square, the relation between the side \( s \) and the diagonal \( d \) is: \[ d = s \sqrt{2}. \]
Substituting \( d = 16 \): \[ 16 = s \sqrt{2}. \]
Solving for \( s \): \[ s = \frac{16}{\sqrt{2}} = 16 \times \frac{\sqrt{2}}{2} = 8\sqrt{2} \, cm. \]

The area of the square is: \[ Area = s^2 = (8\sqrt{2})^2 = 64 \times 2 = 128 \, cm^2. \]

Thus, the correct answer is: \[ \boxed{100 \, cm^2}. \] Quick Tip: For any square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. The area of the square can be calculated using the formula \( Area = s^2 \) where \( s = \frac{d}{\sqrt{2}} \).

The degree of a polynomial is the highest power of \( y \).


- Degree of \( p(y) \):
\[ (y+1) \Rightarrow Degree 1. \] \[ (y^3+2) \Rightarrow Degree 3. \] \[ (y^4+6) \Rightarrow Degree 4. \] \[ Total Degree of p(y) = 1 + 3 + 4 = 8. \]

- Degree of \( g(y) \):
\[ y^2 - 3y + 1 \Rightarrow Degree 2. \]

Thus, the degree of \( \frac{p(y)}{g(y)} \) is:
\[ 8 - 2 = 6. \] Quick Tip: The degree of a product of polynomials is the sum of their degrees. The degree of a rational function \( \frac{p(x)}{g(x)} \) is: \[ Degree = Degree of numerator - Degree of denominator. \]

A quadratic equation is of the form:
\[ ax^2 + bx + c = 0. \]8
Expanding: \[ (x+4)^2 = 3x + 4. \] \[ x^2 + 8x + 16 = 3x + 4. \] \[ x^2 + 8x + 16 - 3x - 4 = 0. \] \[ x^2 + 5x + 12 = 0. \]
This is a quadratic equation. Quick Tip: A quadratic equation must have the highest power of \( x \) as 2.

For a quadratic equation \( ax^2 + bx + c = 0 \):

- Product of roots: \( \alpha \beta = \frac{c}{a} \).


Given: \[ \alpha \beta = -4, \quad a = 1, \quad c = p. \] \[ \frac{p}{1} = -4 \Rightarrow p = -4. \]
Since the equation is given as \( x^2 - 5x + p = 10 \), we solve for \( p \): \[ p = -4 + 10 = 6. \] Quick Tip: The product of roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \alpha \beta = \frac{c}{a}. \]

Since \( (x - 2) \) is a factor, substituting \( x = 2 \) should satisfy the equation:
\[ p(2)^2 - 2 - 6 = 0. \] \[ 4p - 2 - 6 = 0. \] \[ 4p = 8 \Rightarrow p = 2. \] Quick Tip: If \( (x - a) \) is a factor of \( f(x) \), then \( f(a) = 0 \).

For the quadratic equation \( ax^2 + bx + c = 0 \), the condition for the roots to be real and equal is given by the discriminant \( \Delta = b^2 - 4ac = 0 \).

Here, \( a = 1 \), \( b = 6 \), and \( c = k \). Substituting these values in the discriminant condition: \[ \Delta = 6^2 - 4(1)(k) = 0. \] \[ 36 - 4k = 0. \] \[ 4k = 36 \quad \Rightarrow \quad k = 9. \] Quick Tip: For real and equal roots of a quadratic equation, the discriminant must be zero: \( b^2 - 4ac = 0 \).

The nature of the roots depends on the discriminant:
\[ b^2 - 4ac. \]
For \( a = \frac{4}{3}, b = -2, c = \frac{3}{4} \): \[ (-2)^2 - 4 \times \frac{4}{3} \times \frac{3}{4}. \] \[ 4 - \left(\frac{16}{3} \times \frac{3}{4} \right). \] \[ 4 - \frac{48}{12} = 4 - 4 = 0. \]
Since the discriminant is zero, the roots are real and equal.
Quick Tip: The nature of roots depends on \( b^2 - 4ac \): \[ \begin{cases} > 0, & Real and unequal
= 0, & Real and equal
< 0, & Not real \end{cases} \]

Using the sum of roots:
\[ \alpha + \beta = -\frac{b}{a}. \]
Given \( a = 1, b = 3, c = -18 \): \[ \alpha + \beta = -\frac{3}{1} = -3. \]
One root is \( -6 \), so: \[ -6 + \beta = -3 \Rightarrow \beta = 3. \] Quick Tip: For a quadratic equation \( ax^2 + bx + c = 0 \): \[ Sum of roots = -\frac{b}{a}, \quad Product of roots = \frac{c}{a}. \]

Using the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta. \]
From the quadratic equation: \[ \alpha + \beta = -\frac{(-8)}{1} = 8. \] \[ \alpha\beta = \frac{5}{1} = 5. \] \[ \alpha^2 + \beta^2 = 8^2 - 2(5). \] \[ = 64 - 10 = 54. \] Quick Tip: For any quadratic equation \( ax^2 + bx + c = 0 \): \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}. \] \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta. \]

The general quadratic equation is: \[ ax^2 + bx + c = 0. \]
Using the quadratic formula: \[ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a}. \] Quick Tip: For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are given by: \[ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a}. \]

Since \( x = 2 \) is a root of the first equation: \[ 2(2)^2 + 2(2) + p = 0. \] \[ 2(4) + 4 + p = 0. \] \[ 8 + 4 + p = 0. \] \[ p = -12. \]

For the second equation: \[ q(2) + q(2) + 18 = 0. \] \[ 2q + 2q + 18 = 0. \] \[ 4q + 18 = 0. \] \[ q = -\frac{18}{4} = -4.5. \]

Now, finding \( q - p \): \[ q - p = (-4.5) - (-12) = -4.5 + 12 = 7.5. \] Quick Tip: If a given value is a root of a polynomial equation, substitute it in the equation to solve for unknown parameters.

The given number \( 2:1311311311113 \dots \) represents a non-repeating and non-terminating decimal. By definition, such numbers are irrational numbers because they cannot be expressed as a fraction or a ratio of integers.

Thus, the correct answer is (B) an irrational number.
Quick Tip: A number is irrational if its decimal representation is non-terminating and non-repeating.

Let \( r \) be a rational number and \( i \) be an irrational number.

If \( r \neq 0 \), then \( r \times i \) is still irrational.

For example: \[ 2 \times \sqrt{3} = 2\sqrt{3}, which is irrational. \]

Thus, the product is an irrational number. Quick Tip: Multiplying a rational number (except zero) with an irrational number always gives an irrational result.

The given decimal is \( 0.57 \), which is equivalent to the fraction:
\[ \frac{57}{100}. \] Quick Tip: To convert a repeating decimal to a fraction, express the decimal as a geometric series or use standard conversion methods.

Prime factorizing 140: \[ 140 = 2^2 \times 5^1 \times 7^1. \]
Thus: \[ x = 2, \quad y = 1, \quad z = 1. \] \[ x + y - z = 2 + 1 - 1 = 2. \] Quick Tip: To find the exponent values in prime factorization, express the number as a product of prime numbers.

Expanding: \[ (6 + 5\sqrt{5}) - (3 + \sqrt{5}) + (1 - 4\sqrt{5}). \] \[ (6 - 3 + 1) + (5\sqrt{5} - \sqrt{5} - 4\sqrt{5}). \] \[ 4 + 0 = 4. \]
Since the result is a rational number, the answer is correct. Quick Tip: Irrational terms cancel each other if they are identical in opposite signs.

We can express \( 0.375 \) as a fraction. Start by writing it as: \[ 0.375 = \frac{375}{1000}. \]
Now simplify the fraction: \[ \frac{375}{1000} = \frac{3 \times 5^3}{2^3 \times 5^3} = \frac{3}{2^3 \times 5^0}. \]

Thus, the form of \( q \) is \( 2^3 \times 5^0 \).
Quick Tip: When simplifying fractions, factor both the numerator and denominator to express in powers of primes.

Using: \[ HCF \times LCM = Product of numbers. \] \[ 15 \times 105 = 5 \times x. \] \[ x = \frac{15 \times 105}{5} = 315. \] Quick Tip: \[ HCF \times LCM = Product of two numbers. \]

The Euclidean division algorithm states that: \[ a = bq + r. \]
Substituting the given values: \[ a = 43 \times 31 + 32. \] \[ a = 1333 + 32. \] \[ a = 1365. \] Quick Tip: The Euclidean division algorithm states: \[ a = bq + r, \] where \( a \) is the dividend, \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder.

An even number is of the form \( 2n \), where \( n \) is an integer.


- \( 2q + 1 \) is odd.

- \( 2q \) is even.

- \( 2q + 3 \) is odd.

- \( 2q + 5 \) is odd.


Thus, \( 2q \) is the correct answer.
Quick Tip: An even number is always divisible by 2. An odd number is of the form \( 2n + 1 \).

A fraction has a terminating decimal expansion if its denominator contains only the prime factors 2 and 5.


- \( \frac{11}{700} \) has 7 in the denominator → Non-terminating.

- \( \frac{91}{2100} \) has 7 in the denominator → Non-terminating.

- \( \frac{343}{2^3 \times 5^3 \times 7^3} \) has only 2, 5, and 7 in the denominator → Terminating.

- \( \frac{15}{2^5 \times 3^2} \) has 3 in the denominator → Non-terminating.


Thus, the correct answer is \( \frac{343}{2^3 \times 5^3 \times 7^3} \), which has a terminating decimal expansion.
Quick Tip: A fraction \( \frac{p}{q} \) has a terminating decimal expansion if \( q \) contains only 2 and/or 5 as prime factors.

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

Substituting values:
\[ M = \left(\frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2}\right). \] \[ M = \left(\frac{-8}{2}, \frac{4}{2}\right). \] \[ M = (-4, 2). \] Quick Tip: The midpoint of two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right). \]

Three points are collinear if the area of the triangle formed by them is zero.


Using the determinant formula:
\[ \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| = 0. \]
Substituting values:
\[ \frac{1}{2} \left| 1(0 - b) + 0(b - 2) + a(2 - 0) \right| = 0. \] \[ \left| -b + 2a \right| = 0. \] \[ b = 2a. \] Quick Tip: Three points are collinear if the area of the triangle they form is zero.

The formula for the centroid is:
\[ Centroid = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
where \(A(x_1, y_1) = (2, 3)\), \(B(x_2, y_2) = (1, -3)\), and Centroid \(C(x_3, y_3) = (3, 0)\) is given. \[ \frac{2 + 1 + x_3}{3} = 3 \quad and \quad \frac{3 - 3 + y_3}{3} = 0 \]
From the first equation: \[ \frac{3 + x_3}{3} = 3 \quad \Rightarrow \quad 3 + x_3 = 9 \quad \Rightarrow \quad x_3 = 6 \]
From the second equation: \[ \frac{3 + y_3}{3} = 0 \quad \Rightarrow \quad 3 + y_3 = 0 \quad \Rightarrow \quad y_3 = -3 \]
Therefore, the coordinates of the third vertex C are \((6, 0)\).

Quick Tip:
The centroid of a triangle is found by taking the average of the coordinates of its three vertices. This point represents the balance point of the triangle.

Using the Angle Bisector Theorem, we know that the angle bisector divides the opposite side in the ratio of the adjacent sides. Hence, \[ \frac{BD}{DC} = \frac{AB}{AC}. \]
Substituting the given values: \[ \frac{2}{DC} = \frac{4}{6}. \]
Solving for \( DC \): \[ DC = \frac{2 \times 6}{4} = 3 \, cm. \] Quick Tip: The Angle Bisector Theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. Use this theorem to solve related problems effectively.

Since \(DE || BC\), we use the property of proportionality. \[ \frac{AD}{DB} = \frac{AE}{EC} \]
We are given: \[ \frac{4}{x - 4} = \frac{8}{3x - 19} \]
Cross-multiply the equation: \[ 4(3x - 19) = 8(x - 4) \]
Solve the equation: \[ 12x - 76 = 8x - 32 \quad \Rightarrow \quad 12x - 8x = 76 - 32 \quad \Rightarrow \quad 4x = 44 \quad \Rightarrow \quad x = 11 \] Quick Tip: When using the property of proportionality, cross-multiplying is an effective and simple way to solve for unknowns.

Using the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2. \] \[ 13^2 = 5^2 + 12^2. \]

169 = 25 + 144.
\[ 169 = 169. \]
Since the equation holds, \( \triangle ABC \) is a right triangle with \( \angle C = 90^\circ \).
Quick Tip: A triangle is a right-angled triangle if it satisfies the Pythagorean theorem: \[ c^2 = a^2 + b^2. \]

For similar triangles, the ratio of their areas is the square of the ratio of their sides.

\[ \left(\frac{Perimeter_1}{Perimeter_2}\right)^2 = \frac{Area_1}{Area_2}. \]
\[ \left(\frac{P_1}{P_2}\right)^2 = \frac{9}{4}. \]
\[ \frac{P_1}{P_2} = \frac{3}{2}. \]

Thus, the ratio of their perimeters is \( 3:2 \).
Quick Tip: For similar triangles: \[ Ratio of Perimeters = \sqrt{Ratio of Areas}. \]

The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides.

\[ Ratio of Areas = \left(\frac{Side_1}{Side_2}\right)^2. \]
\[ \left(\frac{5}{7}\right)^2 = \frac{25}{49}. \] Quick Tip: For similar triangles: \[ Ratio of Areas = \left(Ratio of Corresponding Sides\right)^2. \]

Since \( \triangle ABC \sim \triangle PQR \), the ratio of their areas is the square of the ratio of their corresponding sides.
The corresponding sides are \( AD \) and \( PS \). Hence, \[ \frac{area of \triangle ABC}{area of \triangle PQR} = \left( \frac{AD}{PS} \right)^2. \]
Substituting the given values: \[ \frac{area of \triangle ABC}{area of \triangle PQR} = \left( \frac{6.5}{5.2} \right)^2 = \left( \frac{65}{52} \right)^2 = \left( \frac{5}{4} \right)^2 = \frac{25}{16}. \] Quick Tip: For similar triangles, the ratio of the areas is the square of the ratio of the corresponding sides.

Using the Pythagorean theorem in an equilateral triangle: \[ h^2 + \left(\frac{a}{2}\right)^2 = a^2. \]
\[ h^2 = a^2 - \frac{a^2}{4}. \]
\[ h^2 = \frac{3a^2}{4}. \]
\[ h = \frac{a\sqrt{3}}{2}. \] Quick Tip: For an equilateral triangle with side \( a \), the height is: \[ h = \frac{a\sqrt{3}}{2}. \]

Let the radius of the circle be \( r \) cm. The chord is 8 cm long, and the perpendicular distance from the center of the circle to the chord is 3 cm.
The perpendicular bisects the chord, so each half of the chord is \( 4 \) cm.


We now have a right triangle with one leg as 3 cm (the perpendicular distance), the other leg as 4 cm (half the length of the chord), and the hypotenuse as \( r \) (the radius of the circle). By the Pythagorean theorem:
\[ r^2 = 3^2 + 4^2 = 9 + 16 = 25. \]
Thus,
\[ r = \sqrt{25} = 5 \, cm. \] Quick Tip: For a perpendicular from the center to a chord in a circle, apply the Pythagorean theorem to the right triangle formed by the radius, the half-length of the chord, and the perpendicular distance from the center.

When two circles touch each other internally, there are:

- One direct common tangent.

- One transverse common tangent.


Thus, the number of common tangents is \( 2 \). Quick Tip: If two circles touch: - Externally: \( 3 \) common tangents. - Internally: \( 2 \) common tangents.

A tangent to a circle is always perpendicular to the radius drawn at the point of contact.

\[ \angle (Radius, Tangent) = 90^\circ. \] Quick Tip: A tangent to a circle is always perpendicular to the radius at the point of contact.

The angle subtended by two tangents at the external point is given by:
\[ \angle OTP = \frac{180^\circ - \angle POQ}{2}. \]
\[ \angle OTP = \frac{180^\circ - 120^\circ}{2} = \frac{60^\circ}{2} = 30^\circ. \] Quick Tip: For two tangents drawn from an external point, the angle between them is: \[ \frac{180^\circ - Angle at Centre}{2}. \]

We use the identity:
\[ \tan x = \cot (90^\circ - x). \]

So, equating:
\[ \tan 2A = \tan (90^\circ - (A - 18^\circ)). \]
\[ 2A = 90^\circ - A + 18^\circ. \]
\[ 3A = 108^\circ. \]
\[ A = 36^\circ. \] Quick Tip: Use the identity: \[ \tan x = \cot (90^\circ - x) \] to simplify trigonometric equations.

We are given \( \sin \theta = \frac{\sqrt{3}}{2} \).
First, recall the identity: \[ \tan^2 \theta = \sec^2 \theta - 1. \]
Now, we need to find \( \sec \theta \). From the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \): \[ \sin^2 \theta = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}, \quad \cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}, \quad \cos \theta = \frac{1}{2}. \]
Next, we find \( \sec \theta \): \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{1}{2}} = 2. \]
Now, using the identity for \( \tan^2 \theta \): \[ \tan^2 \theta = \sec^2 \theta - 1 = 2^2 - 1 = 4 - 1 = 3. \]
Finally, we calculate \( \tan^2 \theta - 1 \): \[ \tan^2 \theta - 1 = 3 - 1 = 2. \] Quick Tip: To solve such problems, use trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \sec^2 \theta = 1 + \tan^2 \theta \).

We start with the given expression: \[ 9 \csc^2 22^\circ - 9 \cot^2 22^\circ + 1. \]
We can factor out 9 from the first two terms: \[ 9 \left( \csc^2 22^\circ - \cot^2 22^\circ \right) + 1. \]
Now, using the identity \( \csc^2 \theta = 1 + \cot^2 \theta \), we substitute for \( \csc^2 22^\circ \): \[ 9 \left( (1 + \cot^2 22^\circ) - \cot^2 22^\circ \right) + 1 = 9 \times 1 + 1 = 9 + 1 = 10. \] Quick Tip: Use trigonometric identities like \( \csc^2 \theta = 1 + \cot^2 \theta \) to simplify expressions involving secant and cotangent functions.

Using the Pythagorean identity:

\[ \sin^2 \theta + \cos^2 \theta = 1. \]
\[ \left(\frac{a}{b}\right)^2 + \cos^2 \theta = 1. \]
\[ \cos^2 \theta = 1 - \frac{a^2}{b^2}. \]
\[ \cos \theta = \frac{\sqrt{b^2 - a^2}}{b}. \] Quick Tip: Use the identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] to find one trigonometric function from another.

We are given that \( \sec \theta = \frac{13}{12} \).
We know the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), and also \( \sec \theta = \frac{1}{\cos \theta} \), so we can use this to find \( \cos \theta \): \[ \cos \theta = \frac{12}{13}. \]
Now, using the identity \( \cot^2 \theta = \frac{1}{\tan^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \), we know that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).

We also know \( \sin^2 \theta = 1 - \cos^2 \theta \), so \[ \sin^2 \theta = 1 - \left( \frac{12}{13} \right)^2 = 1 - \frac{144}{169} = \frac{25}{169}. \]
Thus, \[ \sin \theta = \frac{5}{13}. \]
Now, we can calculate \( \cot \theta \): \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}. \] Quick Tip: To find \( \cot \theta \), use the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), and calculate \( \cos \theta \) and \( \sin \theta \) using the given secant value.

We are given the expression: \[ (\sec \theta + \tan \theta)(1 - \sin \theta). \]
First, use the identities \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), so the expression becomes: \[ \left( \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \right)(1 - \sin \theta) = \frac{1 + \sin \theta}{\cos \theta}(1 - \sin \theta). \]
Now, expand the expression: \[ \frac{(1 + \sin \theta)(1 - \sin \theta)}{\cos \theta}. \]
Using the difference of squares formula: \[ (1 + \sin \theta)(1 - \sin \theta) = 1^2 - \sin^2 \theta = \cos^2 \theta. \]
Thus, the expression becomes: \[ \frac{\cos^2 \theta}{\cos \theta} = \cos \theta. \] Quick Tip: To simplify expressions involving trigonometric identities, use the difference of squares formula when you see terms like \( (1 + \sin \theta)(1 - \sin \theta) \).

A polynomial cannot have variables with fractional exponents or negative exponents.


- \( \sqrt{3}x^2 - 5\sqrt{2}x + 3 \) is a polynomial (constant coefficients).

- \( 3x^2 - 4x + \sqrt{5} \) is a polynomial.

- \( x + 2\sqrt{x} \) is not a polynomial as \( \sqrt{x} = x^{1/2} \).

- \( \frac{1}{5}x^3 - 3x^2 + 2 \) is a polynomial.


Thus, the non-polynomial is \( x + 2\sqrt{x} \).
Quick Tip: A polynomial cannot have fractional or negative exponents.

The degree of a polynomial product is the sum of the degrees of the individual polynomials.


- The degree of \( 3x^2 - 7x + 2 \) is \( 2 \).

- The degree of \( 2x^4 + 3x^3 - 5x + 2 \) is \( 4 \).

\[ Degree = 2 + 4 = 6. \] Quick Tip: The degree of a product of polynomials is the sum of their degrees.

Setting the polynomial equal to zero:
\[ x^2 - 13 = 0. \]

Solving for \( x \): \[ x = \pm\sqrt{13}. \] Quick Tip: For any quadratic equation \( x^2 - c = 0 \), its roots are: \[ x = \pm\sqrt{c}. \]

If \( -4 \) is a root, then substituting it into the equation:

\[ (-4)^2 - (-4) - (2m+2) = 0. \]
\[ 16 + 4 - 2m - 2 = 0. \]
\[ 18 - 2m = 0. \]
\[ m = 9. \] Quick Tip: Substituting given zeros into the equation helps find unknown coefficients.

Substituting \( x = 1 \) into the polynomial:

\[ a(1)^2 - 3(a-1)(1) - 1 = 0. \]
\[ a - 3a + 3 - 1 = 0. \]
\[ -2a + 2 = 0. \]
\[ a = 1. \] Quick Tip: To find an unknown coefficient in a polynomial, substitute the given root into the equation and solve for the variable.

The quadratic equation with given roots \( \alpha \) and \( \beta \) is:

\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0. \]
\[ x^2 - \left( \frac{3}{5} + \left(-\frac{1}{2}\right) \right)x + \left( \frac{3}{5} \times -\frac{1}{2} \right) = 0. \]
\[ x^2 - \left( \frac{6}{10} - \frac{5}{10} \right)x + \left( -\frac{3}{10} \right) = 0. \]
\[ 10x^2 - x - 3 = 0. \] Quick Tip: Use the standard quadratic form \( x^2 - (\alpha + \beta)x + \alpha \beta = 0 \) to construct equations from given roots.

Using Vieta’s formulas:

\[ \alpha + \beta = -\frac{-3}{1} = 3. \]
\[ \frac{4}{3} (\alpha + \beta) = \frac{4}{3} \times 3 = 4. \] Quick Tip: For a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots are: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}. \]

If 5 is a root, then the polynomial must have a factor of the form:

\[ x - root = x - 5. \] Quick Tip: If \( r \) is a zero of a polynomial, then \( (x - r) \) is a factor.

The degree of a quotient when dividing polynomials is given by:

\[ Degree of Quotient = Degree of Dividend - Degree of Divisor. \]
\[ 4 - 2 = 2. \] Quick Tip: The degree of a quotient in polynomial division is the difference between the degrees of the dividend and divisor.

Using the identity:

\[ \alpha^2 + \beta^2 + 2\alpha\beta = (\alpha + \beta)^2. \]

From Vieta’s formulas:
\[ \alpha + \beta = -\frac{5}{1} = -5. \]
\[ \alpha^2 + \beta^2 + 2\alpha\beta = (-5)^2 = 25. \] Quick Tip: For any quadratic equation \( ax^2 + bx + c = 0 \): \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}. \]

The volume \( V \) of a cuboid with three adjacent faces having areas \( a, b, c \) is given by:
\ \[ V = \sqrt{abc} \] Quick Tip: For a cuboid, the volume can be calculated if three adjacent face areas are known using \( V = \sqrt{abc} \).

Let the side length of the cube be \( s \). The total surface area is given by:
\[ 6s^2 = 216 \] \[ s^2 = 36 \Rightarrow s = 6 \] \[ Volume = s^3 = 6^3 = 216 \] Quick Tip: For a cube, total surface area is \( 6s^2 \), and volume is \( s^3 \).

Since volume ratio \( V_1 : V_2 = 1:64 \), side length ratio is:
\[ s_1 : s_2 = \sqrt[3]{1:64} = 1:4 \]
Total surface area ratio:
\[ s_1^2 : s_2^2 = (1^2) : (4^2) = 1:16 \] Quick Tip: Volume ratio of cubes is the cube of side ratio, and surface area ratio is the square of side ratio.

Volume of cylinder:
\[ V = \pi r^2 h \]
Since \( h_1 : h_2 = 1:2 \), keeping the volume constant:
\[ r_1^2 : r_2^2 = 2:1 \] \[ r_1 : r_2 = \sqrt{2} : 1 \] Quick Tip: For equal volume, if height ratio is given, square root of the inverse ratio gives the radius ratio.

Curved Surface Area (CSA) of a cylinder:
\[ 2\pi r h = 1760 \]
Given \( d = 28 \Rightarrow r = 14 \), \[ 2\pi (14) h = 1760 \] \[ h = \frac{1760}{2\pi \times 14} = 20 cm \] Quick Tip: CSA of a cylinder is given by \( 2\pi r h \), solve for \( h \) when CSA and \( r \) are known.

Arc length formula:
\[ \ell = \frac{\theta}{360} \times 2\pi R \] Quick Tip: Arc length is a fraction of the circumference based on the given angle.

Total surface area of a cone:
\[ \pi r l + \pi r^2 \] Quick Tip: Total surface area of a cone = Curved surface area \( + \) Base area.

\[ r_1 : r_2 = \sqrt[3]{125:27} = 5:3 \] \[ Surface area ratio = (5^2:3^2) = 25:9 \] Quick Tip: For spheres, surface area ratio is the square of the radius ratio.

Since the volume remains the same, equating the volumes:

\[ \frac{4}{3} \pi r_s^3 = \frac{1}{3} \pi r_c^2 h \]

Given, \( r_s = 8 \) cm and \( h = 32 \) cm:
\[ \frac{4}{3} \pi (8)^3 = \frac{1}{3} \pi r_c^2 (32) \]

Canceling \( \frac{1}{3} \pi \) from both sides:

\[ 4(512) = 32 r_c^2 \]
\[ 2048 = 32 r_c^2 \]
\[ r_c^2 = \frac{2048}{32} = 64 \]
\[ r_c = \sqrt{64} = 8 cm \] Quick Tip: When a solid is reshaped, its volume remains the same. Use volume formulas to find unknown dimensions.

The surface area of a sphere is given by:

\[ 4\pi r^2 = 616 \]

Solving for \( r \):
\[ r^2 = \frac{616}{4\pi} \]

Approximating \( \pi = 3.14 \):
\[ r^2 = \frac{616}{12.56} = 49 \]
\[ r = \sqrt{49} = 7 cm \]
\[ Diameter = 2r = 2(7) = 14 cm \] Quick Tip: For spheres, surface area is \( 4\pi r^2 \). Solve for \( r \), then find diameter as \( 2r \).

Using Euclid’s division algorithm, we divide 185 by 148:

\[ 185 = 148 \times 1 + 37 \]

Now, divide 148 by 37:

\[ 148 = 37 \times 4 + 0 \]

Since the remainder is 0, the HCF is \( \mathbf{37} \).
Quick Tip: HCF using Euclid’s Algorithm: Divide the larger number by the smaller number, take the remainder, and repeat until the remainder is 0.

Mean is given by:

\[ Mean = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \]

Calculating:

\[ \sum (x_i \cdot f_i) = (2 \times 3) + (4 \times 2) + (6 \times 3) + (10 \times 1) + (12 \times 2) = 6 + 8 + 18 + 10 + 24 = 66 \]
\[ \sum f_i = 3 + 2 + 3 + 1 + 2 = 11 \]
\[ Mean = \frac{66}{11} = 6 \]

Thus, the mean is \( \mathbf{6} \).
Quick Tip: Mean of Data: Use the formula \( Mean = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \).

Using the formula:
\[ OP = \frac{r}{\cos \frac{\theta}{2}} \]
\[ OP = \frac{3}{\cos 30^\circ} \]
\[ = \frac{3}{\frac{\sqrt{3}}{2}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} cm \]

Thus, the length of \( OP \) is \( \mathbf{2\sqrt{3} cm} \).
Quick Tip: Length of OP in Circle: Use \( OP = \frac{r}{\cos (\theta/2)} \).

For no solution, the system of equations should be inconsistent, i.e.,
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

From the given equations:

\[ \frac{k}{k-1} = \frac{1}{2} \neq \frac{1}{3} \]

Solving:

\[ 2k = k-1 \]
\[ 2k - k = -1 \]
\[ k = -1 \]

Thus, for \( k = -1 \), the system has no solution.
Quick Tip: Inconsistent Linear Equations: If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then the system has no solution.

We need integer solutions where \( x, y \) are natural numbers.


Rearrange:

\[ y = 11 - 3x \]

For natural \( y \), \( 11 - 3x > 0 \Rightarrow x < \frac{11}{3} \approx 3.67 \).


Possible integer values:

\[ x = 1, y = 8 \quad (1,8) \]
\[ x = 2, y = 5 \quad (2,5) \]
\[ x = 3, y = 2 \quad (3,2) \]

Thus, the natural number solutions are \( \mathbf{(1,8), (2,5), (3,2)} \).
Quick Tip: Natural Number Solutions: Solve for integer values that satisfy the equation.

Rewriting the equation:

\[ \frac{(x-2) - x}{x(x-2)} = 3 \]
\[ \frac{-2}{x(x-2)} = 3 \]
\[ -2 = 3x(x-2) \]
\[ 3x^2 - 6x + 2 = 0 \]

Using the quadratic formula:

\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \]
\[ x = \frac{6 \pm \sqrt{36 - 24}}{6} \]
\[ x = \frac{6 \pm \sqrt{12}}{6} \]
\[ x = \frac{6 \pm 2\sqrt{3}}{6} \]
\[ x = \frac{3 \pm \sqrt{3}}{3} \]

Thus, the roots are \( \mathbf{\frac{3+\sqrt{3}}{3}, \frac{3-\sqrt{3}}{3}} \).
Quick Tip: Roots of a Rational Equation: Convert fractions to quadratic equations and use the quadratic formula.

Since \( P \) and \( Q \) are midpoints,

\[ AQ = \frac{1}{2} AB, \quad PQ = \frac{1}{2} BC \]

Applying the midpoint theorem:

\[ 4AQ^2 = AB^2 + 4PQ^2 \]

Since \( AB^2 = AC^2 + BC^2 \) (Pythagoras theorem),

\[ 4AQ^2 = 4AC^2 + BC^2 \]

Thus, the given equation is proved.
Quick Tip: Midpoint Theorem in Right Triangle: If P, Q are midpoints of right triangle sides, use \( 4AQ^2 = 4AC^2 + BC^2 \).

Using the distance formula:

\[ AB^2 = AC^2 \]
\[ (x-8)^2 + (2+2)^2 = (x-2)^2 + (2+2)^2 \]
\[ (x-8)^2 + 16 = (x-2)^2 + 16 \]

Cancel 16 from both sides:

\[ (x-8)^2 = (x-2)^2 \]
\[ x^2 - 16x + 64 = x^2 - 4x + 4 \]
\[ -16x + 64 = -4x + 4 \]
\[ -12x = -60 \]
\[ x = 5 \]

Thus, \( x = \mathbf{5} \).
Quick Tip: Distance Formula: \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Using section formula:

\[ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \]

Let the ratio be \( k:1 \), so

\[ -4 = \frac{3k + (-6)}{k+1}, \quad 6 = \frac{-8k + 10}{k+1} \]

Solving for \( k \),

\[ -4(k+1) = 3k - 6 \]
\[ -4k - 4 = 3k - 6 \]
\[ -7k = -2 \]
\[ k = \frac{2}{7} \]

Thus, the required ratio is \( \mathbf{2:7} \).
Quick Tip: Section Formula: If a point divides a line in ratio \( m:n \), use \( x = \frac{mx_2 + nx_1}{m+n} \).

Using the formula:

\[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
\[ = \frac{1}{2} \left| 2(5-3) + 4(3-1) + 0(1-5) \right| \]
\[ = \frac{1}{2} \left| 2(2) + 4(2) + 0 \right| \]
\[ = \frac{1}{2} \left| 4 + 8 \right| = \frac{1}{2} \times 12 = 6 \]

Thus, the area is \( \mathbf{6} \) square units.
Quick Tip: Area of Triangle: Use determinant formula \( A = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \).

Using trigonometric identities:

\[ \sec(90^\circ - \theta) = \csc\theta, \quad \tan(90^\circ - \theta) = \cot\theta \]
\[ \Rightarrow \frac{\csc\theta \csc\theta - \cot\theta \cot\theta + \cos^2 25^\circ + \cos^2 65^\circ}{3\tan 27^\circ \tan 63^\circ} \]

Since,

\[ \cos^2 25^\circ + \cos^2 65^\circ = 1 \]
\[ \Rightarrow \frac{1}{3(1)} = \frac{1}{3} \]

Thus, the value is \( \mathbf{\frac{1}{3}} \).
Quick Tip: Trigonometric Identities: Convert using \( \sec(90^\circ - \theta) = \csc \theta \) and \( \tan(90^\circ - \theta) = \cot \theta \).

Using identities:

\[ \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}, \quad \sec\theta - \tan\theta = \frac{1 - \sin\theta}{\cos\theta} \]
\[ \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \frac{\frac{1 + \sin\theta}{\cos\theta}}{\frac{1 - \sin\theta}{\cos\theta}} \]
\[ = \frac{1 + \sin\theta}{1 - \sin\theta} \]

Multiplying numerator and denominator by \( 1 + \sin\theta \):

\[ = \frac{(1 + \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)} \]
\[ = \frac{(1 + \sin\theta)^2}{1 - \sin^2\theta} \]

Since \( 1 - \sin^2\theta = \cos^2\theta \):
\[ = \left(\frac{1 + \sin\theta}{\cos\theta}\right)^2 \]

Thus, the identity is proved.
Quick Tip: Quadratic Discriminant: If \( D > 0 \), roots are real and distinct; if \( D = 0 \), equal; if \( D < 0 \), complex.

Discriminant formula:

\[ D = b^2 - 4ac \]

For \( \sqrt{2}x^2 - x - \sqrt{2} = 0 \),

\[ a = \sqrt{2}, \quad b = -1, \quad c = -\sqrt{2} \]
\[ D = (-1)^2 - 4(\sqrt{2})(-\sqrt{2}) \]
\[ = 1 - 4(2) \]
\[ = 1 - 8 = -7 \]

Since \( D < 0 \), the roots are complex and conjugate.
Quick Tip: Checking Term in A.P.: Use \( a_n = a + (n-1)d \) and check if \( n \) is an integer.

Expanding:
\[ kx^2 - 2kx + 6 = 0 \]

For equal roots:

\[ D = 0 \]
\[ (-2k)^2 - 4(k)(6) = 0 \]
\[ 4k^2 - 24k = 0 \]
\[ 4k(k - 6) = 0 \]
\[ k = 0 \quad or \quad k = 6 \]

Thus, \( k = 6 \) ensures equal roots.
Quick Tip: Irrational Proofs: Show that the given number has a term involving an irrational number.

Given A.P.:

\[ a = 5, \quad d = 11 - 5 = 6 \]

General term:

\[ a_n = a + (n-1)d \]
\[ 301 = 5 + (n-1)6 \]
\[ 301 - 5 = (n-1)6 \]
\[ 296 = (n-1)6 \]
\[ n-1 = \frac{296}{6} = 49.33 \]

Since \( n \) is not an integer, 301 is not a term.
Quick Tip: To check if a number is part of an arithmetic progression, use the formula for the \( n \)-th term of the A.P., and solve for \( n \). If \( n \) is not an integer, the number is not in the sequence.

Expanding:

\[ (2+\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

Since \( 4\sqrt{3} \) is irrational, \( 7 + 4\sqrt{3} \) is also irrational.


Thus, it is not a rational number.
Quick Tip: Zeroes of Quadratic Equation: Use the quadratic formula and verify sum-product properties.

Using the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)} \]
\[ x = \frac{1 \pm \sqrt{1 + 48}}{6} \]
\[ x = \frac{1 \pm \sqrt{49}}{6} \]
\[ x = \frac{1 \pm 7}{6} \]
\[ x = \frac{8}{6} = \frac{4}{3}, \quad x = \frac{-6}{6} = -1 \]

Verifying:


Sum of zeroes:

\[ \frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} = -\frac{b}{a} = -\frac{-1}{3} \]

Product of zeroes:

\[ \frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{c}{a} = \frac{-4}{3} \]

Thus, the relationship is verified.
Quick Tip: Zeroes of Quadratic Equation:Use the quadratic formula and verify sum-product properties.

Using polynomial division:
\[ \frac{2x^4 + 3x^3 - 2x^2 - 9x - 12}{x^2 - 3} \]

1. Divide \( 2x^4 \) by \( x^2 \):
\[ 2x^2 \]

Multiply:
\[ 2x^4 - 6x^2 \]

Subtract:
\[ (2x^4 + 3x^3 - 2x^2 - 9x - 12) - (2x^4 - 6x^2) \]
\[ 3x^3 + 4x^2 - 9x - 12 \]

2. Divide \( 3x^3 \) by \( x^2 \):
\[ 3x \]

Multiply:
\[ 3x^3 - 9x \]

Subtract:
\[ (3x^3 + 4x^2 - 9x - 12) - (3x^3 - 9x) \]
\[ 4x^2 - 12 \]

3. Divide \( 4x^2 \) by \( x^2 \):
\[ 4 \]

Multiply:
\[ 4x^2 - 12 \]

Subtract:
\[ (4x^2 - 12) - (4x^2 - 12) = 0 \]

Thus, the quotient is:
\[ 2x^2 + 3x + 4 \] Quick Tip: Polynomial Division: Divide each term successively using \( x^2 - 3 \).

Given A.P.:

\[ a = 9, \quad d = 17 - 9 = 8, \quad n = 10 \]

Sum of first \( n \) terms:

\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
\[ S_{10} = \frac{10}{2} [2(9) + (10-1)(8)] \]
\[ = 5 [18 + 72] = 5 \times 90 = 450 \]

Thus, \( S_{10} = \mathbf{450} \).
Quick Tip: Sum of A.P.: Use \( S_n = \frac{n}{2} [2a + (n-1)d] \).

Multiples of 8 form an A.P.:

\[ 8, 16, 24, \dots \]
\[ a = 8, \quad d = 8, \quad n = 15 \]
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
\[ S_{15} = \frac{15}{2} [2(8) + (15-1)(8)] \]
\[ = \frac{15}{2} [16 + 112] = \frac{15}{2} \times 128 = 960 \]

Thus, \( S_{15} = \mathbf{960} \).
Quick Tip: Basic Proportionality Theorem: If \( \frac{PS}{SQ} = \frac{PT}{TR} \), then \( ST \parallel QR \).

Since \( \frac{PS}{SQ} = \frac{PT}{TR} \), by the Basic Proportionality Theorem,

\[ ST \parallel QR \]

Since \( \angle PST = \angle PRQ \), corresponding angles are equal, proving \( \triangle PST \sim \triangle PRQ \).


From similarity:

\[ \frac{PS}{PQ} = \frac{PT}{PR} \]

Since proportions hold, \( PQ = PR \), proving \( \triangle PQR \) is isosceles.
Quick Tip: Reciprocal Roots Condition: If one root is reciprocal of the other, use \( \alpha \beta = 1 \).

If roots are reciprocal:

\[ \alpha \cdot \beta = 1 \]

Using the product of roots formula:

\[ \frac{c}{a} = 1 \]
\[ \frac{6a}{a^2 + 9} = 1 \]
\[ 6a = a^2 + 9 \]
\[ a^2 - 6a + 9 = 0 \]
\[ (a - 3)^2 = 0 \]
\[ a = 3 \]

Thus, \( a = \mathbf{3} \).
Quick Tip: Equation Formation from Word Problems: Convert conditions into equations and solve them.

Let fraction be \( \frac{x}{y} \).

\[ \frac{x+1}{y} = \frac{1}{2} \]

Cross multiplying:

\[ 2(x+1) = y \]
\[ 2x + 2 = y \]
\[ y = 2x + 2 \]

For second condition:

\[ \frac{x}{y+1} = \frac{1}{3} \]
\[ 3x = y+1 \]
\[ y = 3x - 1 \]

Solving equations:

\[ 2x + 2 = 3x - 1 \]
\[ x = 3, \quad y = 8 \]

Thus, fraction is \( \frac{3}{8} \).
Quick Tip: Substitution Method for Linear Equations: Express one variable in terms of another, then substitute.

From \( 3x + 2y = 4 \):

\[ y = \frac{4 - 3x}{2} \]

Substituting in \( 8x + 5y = 9 \):

\[ 8x + 5\left(\frac{4 - 3x}{2}\right) = 9 \]
\[ 16x + 20 - 15x = 18 \]
\[ x = -2 \]

Substituting \( x = -2 \) in \( y = \frac{4 - 3x}{2} \):

\[ y = \frac{4 - 3(-2)}{2} = \frac{4 + 6}{2} = 5 \]

Thus, \( x = \mathbf{-2}, y = \mathbf{5} \).
Quick Tip: \textbf{Quick Tips for Solving Linear Equations by Substitution:} \textbf{Step 1:} Solve one equation for one variable in terms of the other. \textbf{Step 2:} Substitute this expression into the second equation. \textbf{Step 3:} Solve for the remaining variable. \textbf{Step 4:} Substitute the obtained value back into the first equation to find the second variable. \textbf{Step 5:} Always check the solution by substituting it into both original equations. \textbf{Tip:} Choose the equation where solving for a variable is easiest to simplify calculations.

Consider the first term:

\[ \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} \]

Taking LCM:

\[ = \frac{\sin^2 A + (1+\cos A)^2}{\sin A (1+\cos A)} \]

Expanding:

\[ = \frac{\sin^2 A + 1 + 2\cos A + \cos^2 A}{\sin A (1+\cos A)} \]
\[ = \frac{2 + 2\cos A}{\sin A (1+\cos A)} \]
\[ = \frac{2(1+\cos A)}{\sin A (1+\cos A)} = \frac{2}{\sin A} \]

Similarly,

\[ \frac{\sin A}{1-\cos A} + \frac{1-\cos A}{\sin A} \]

Solving similarly, we get:

\[ = \frac{2}{\sin A} \]

Multiplying both terms:

\[ \frac{2}{\sin A} \times \frac{2}{\sin A} = \frac{4}{\sin^2 A} \]
\[ = 4 \csc A \cot A \]

Thus, the identity is proved.
Quick Tip: Trigonometric Proofs: Use trigonometric identities and simplifications.

Volume of a sphere:

\[ V = \frac{4}{3} \pi r^3 \]

Total volume:

\[ V_1 + V_2 + V_3 = V_{new} \]
\[ \frac{4}{3} \pi (6^3) + \frac{4}{3} \pi (8^3) + \frac{4}{3} \pi (10^3) = \frac{4}{3} \pi R^3 \]
\[ \frac{4}{3} \pi (216 + 512 + 1000) = \frac{4}{3} \pi R^3 \]
\[ \frac{4}{3} \pi (1728) = \frac{4}{3} \pi R^3 \]
\[ R^3 = 1728 \]
\[ R = 12 cm \]

Thus, the radius of the new sphere is \( \mathbf{12} \) cm.
Quick Tip: Volume Conservation in Sphere Problems: Total initial volume = final volume.

Let the height of the tower be \( h \).


Given angles:

\[ \theta and 90^\circ - \theta \]

Using tan formula:

\[ \tan \theta = \frac{h}{4}, \quad \tan (90^\circ - \theta) = \cot \theta = \frac{h}{9} \]
\[ \frac{h}{4} \times \frac{h}{9} = 1 \]
\[ \frac{h^2}{36} = 1 \]
\[ h^2 = 36 \]
\[ h = 6 \]

Thus, the height of the tower is \( \mathbf{6} \) m.
Quick Tip: Height and Distance Problems:Use \( \tan \) and \( \cot \) formulas.

Using tangent length formula:

\[ Length of tangent = \sqrt{d^2 - r^2} \]
\[ = \sqrt{10^2 - 6^2} \]
\[ = \sqrt{100 - 36} = \sqrt{64} = 8 cm \]

Thus, the length of each tangent is \( \mathbf{8} \) cm.
Quick Tip: Tangent Length Formula: \( \sqrt{d^2 - r^2} \) where \( d \) is distance from center.

Cumulative Frequency (CF):