Bihar Board Class 10 Mathematics Question Paper 2024 with Answer Key (Set F - Shift 2)

Given the A.P. is 20, 17, 14, 11, \dots with \(a_1 = 20\) and \(d = -3\). The formula for the \(n\)-th term is:
\[ a_n = a_1 + (n - 1) \cdot d \]

Substituting \(a_1 = 20\), \(d = -3\), and \(n = 35\):
\[ a_{35} = 20 + 34 \cdot (-3) = 20 - 102 = -82 \] Quick Tip: \textbf{Quick Tip:} The 35th term can be found by using the formula \(a_n = a_1 + (n-1) \cdot d\).

Given the A.P. 3, 8, 13, 18, \dots with \(a_1 = 3\) and \(d = 5\). The formula for the \(n\)-th term is:
\[ a_n = a_1 + (n - 1) \cdot d \]

Substituting \(a_n = 93\), \(a_1 = 3\), and \(d = 5\):
\[ 93 = 3 + (n - 1) \cdot 5 \]

Solving for \(n\):
\[ n = 19 \] Quick Tip: \textbf{Quick Tip:} To find the number of terms in an A.P., solve \(a_n = a_1 + (n - 1) \cdot d\).

The A.P. has \(a_1 = 1\) and \(d = 2\). The sum of the first \(n\) terms is:
\[ S_n = \frac{n}{2} \left( 2a_1 + (n - 1) \cdot d \right) \]

Substituting values for \(n = 30\), \(a_1 = 1\), and \(d = 2\):
\[ S_{30} = 15 \times 60 = 900 \] Quick Tip: \textbf{Quick Tip:} Use the formula \(S_n = \frac{n}{2} \left( 2a_1 + (n - 1) \cdot d \right)\) to find the sum of an A.P.

The point has both x and y coordinates negative, meaning it lies in the third quadrant. Quick Tip: \textbf{Quick Tip:} In the third quadrant, both x and y coordinates are negative.

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the given points:
\[ d = \sqrt{(0 - 5 \cos \theta)^2 + (5 \sin \theta - 0)^2} = 5 \] Quick Tip: \textbf{Quick Tip:} Use the distance formula to calculate the distance between two points.

We are given the perpendicular distances from point \( B \) to the x-axis and y-axis.
The perpendicular distance from a point to the x-axis gives the magnitude of the y-coordinate, while the perpendicular distance to the y-axis gives the magnitude of the x-coordinate.
Therefore, from the problem, the length of the perpendicular from point \( B \) to the x-axis is 10, which means the y-coordinate of \( B \) is 10.
Similarly, the length of the perpendicular from point \( B \) to the y-axis is 5, which means the x-coordinate of \( B \) is 5.
Thus, the coordinates of point \( B \) are \( (10, 5) \). Quick Tip: \textbf{Quick Tip:} The perpendicular distances to the x-axis and y-axis directly give the x- and y-coordinates of a point.

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the given points \( (1, -3) \) and \( (4, -6) \):

\[ d = \sqrt{(4 - 1)^2 + (-6 - (-3))^2} \]

Simplifying the equation:

\[ d = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \]

Finally, simplifying \( \sqrt{18} \), we get:

\[ d = 3\sqrt{2} \] Quick Tip: \textbf{Quick Tip:} Always simplify the square root for exact answers when using the distance formula.

Let the required point be \( (0, y) \) on the y-axis.
The distances from \( (0, y) \) to \( (5, -2) \) and \( (0, y) \) to \( (-3, 2) \) should be equal.
Using the distance formula for each point, we have two equations:

\[ \sqrt{(5 - 0)^2 + (-2 - y)^2} = \sqrt{(-3 - 0)^2 + (2 - y)^2} \]

Squaring both sides and solving for \( y \), we get \( y = 3 \).

Therefore, the point is \( (0, 3) \). Quick Tip: \textbf{Quick Tip:} To solve for an unknown point equidistant from two other points, equate the distances from that point to each of the two known points.

The given rectangle has vertices \( A(0, 0) \), \( B(8, 0) \), \( C(8, 6) \), and \( D(0, 6) \).
To find the length of the diagonal, we use the distance formula between any two opposite vertices of the rectangle.
Let’s calculate the diagonal from \( A(0, 0) \) to \( C(8, 6) \).
The distance formula is:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of \( A(0, 0) \) and \( C(8, 6) \):

\[ d = \sqrt{(8 - 0)^2 + (6 - 0)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

Therefore, the length of the diagonal is 10 units. Quick Tip: \textbf{Quick Tip:} For a rectangle, use the distance formula to find the length of the diagonal by calculating the distance between opposite vertices.

The given vertices of the triangle are \( A(0, 4) \), \( B(0, 0) \), and \( C(3, 0) \).
To find the perimeter of the triangle, we calculate the lengths of the three sides using the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

First, we calculate the length of \( AB \):

\[ AB = \sqrt{(0 - 0)^2 + (4 - 0)^2} = \sqrt{16} = 4 \]

Then, we calculate the length of \( BC \):

\[ BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3 \]

Finally, we calculate the length of \( AC \):

\[ AC = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Now, we can calculate the perimeter of the triangle:

\[ Perimeter = AB + BC + AC = 4 + 3 + 5 = 12 \] Quick Tip: \textbf{Quick Tip:} The perimeter of a triangle is simply the sum of the lengths of its sides, which can be calculated using the distance formula.

The degree of a polynomial is the highest power of the variable in the polynomial.
The degree of the product of two polynomials is the sum of the degrees of the polynomials involved.
The degree of \( p(y) = (y + 1)(y^3 + 2)(y^4 + 6) \) is:

\[ Degree of (y + 1) = 1, \quad Degree of (y^3 + 2) = 3, \quad Degree of (y^4 + 6) = 4 \]

\[ Degree of p(y) = 1 + 3 + 4 = 8 \]

The degree of \( g(y) = y^2 - 3y + 1 \) is:

\[ Degree of g(y) = 2 \]

The degree of the quotient \( \frac{p(y)}{g(y)} \) is:

\[ Degree of \left( \frac{p(y)}{g(y)} \right) = 8 - 2 = 6 \] Quick Tip: \textbf{Quick Tip:} The degree of the quotient of two polynomials is the degree of the numerator minus the degree of the denominator.

We need to identify which of the given equations is quadratic. A quadratic equation is a polynomial of degree 2, i.e., it involves \(x^2\) and no higher powers of \(x\).

Option (A): \[ (x+1)(x-1) = x^2 - 4x^3. \]
Expanding the left-hand side: \[ (x+1)(x-1) = x^2 - 1. \]
Thus, the equation becomes: \[ x^2 - 1 = x^2 - 4x^3. \]
Simplifying this: \[ x^2 - x^2 = -4x^3 + 1 \quad \Rightarrow \quad 0 = -4x^3 + 1. \]
This is not a quadratic equation because it involves a cubic term (\(x^3\)).

Option (B): \[ (x+4)^2 = 3x + 4. \]
Expanding the left-hand side: \[ (x+4)^2 = x^2 + 8x + 16. \]
Thus, the equation becomes: \[ x^2 + 8x + 16 = 3x + 4. \]
Rearranging terms: \[ x^2 + 8x + 16 - 3x - 4 = 0 \quad \Rightarrow \quad x^2 + 5x + 12 = 0. \]
This is a quadratic equation because it is of degree 2.

Option (C): \[ 4x + \frac{1}{2x} = 8x^2. \]
Multiplying through by \(2x\) to eliminate the fraction: \[ 8x^2 + 1 = 16x^3. \]
This is not a quadratic equation because it involves a cubic term (\(x^3\)).

Option (D): \[ (2x^2 + 4) = (5 + x)(2x - 3). \]
Expanding the right-hand side: \[ (5 + x)(2x - 3) = 10x - 15 + 2x^2 - 3x = 2x^2 + 7x - 15. \]
Thus, the equation becomes: \[ 2x^2 + 4 = 2x^2 + 7x - 15. \]
Simplifying: \[ 2x^2 - 2x^2 + 4 = 7x - 15 \quad \Rightarrow \quad 4 = 7x - 15 \quad \Rightarrow \quad 7x = 19 \quad \Rightarrow \quad x = \frac{19}{7}. \]
This is a linear equation, not a quadratic equation.

Conclusion: The quadratic equation is option (B), \( (x+4)^2 = 3x + 4 \). Quick Tip: A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where the highest power of \(x\) is 2.

The given quadratic equation is: \[ x^2 - 5x + p = 10. \]
We can rewrite this equation as: \[ x^2 - 5x + (p - 10) = 0. \]
Let the roots of the equation be \( \alpha \) and \( \beta \). According to Vieta's formulas, the sum and product of the roots for a quadratic equation \( ax^2 + bx + c = 0 \) are given by: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}. \]
For the equation \( x^2 - 5x + (p - 10) = 0 \), we have \( a = 1 \), \( b = -5 \), and \( c = p - 10 \).

- The sum of the roots is: \[ \alpha + \beta = -\frac{-5}{1} = 5. \]
- The product of the roots is: \[ \alpha \beta = \frac{p - 10}{1} = p - 10. \]

We are told that the product of the roots is 6: \[ \alpha \beta = 6. \]
Thus: \[ p - 10 = 6 \quad \Rightarrow \quad p = 6 + 10 = 16. \]

Therefore, the value of \( p \) is \( \boxed{6} \). Quick Tip: For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the product of the roots is given by \( \alpha \beta = \frac{c}{a} \).

Given that \( (x - 2) \) is a factor of \( px^2 - x - 6 \), we can use the factor theorem. According to the factor theorem, if \( (x - 2) \) is a factor, then substituting \( x = 2 \) into the equation should make the expression equal to zero.

Substitute \( x = 2 \) into the equation \( px^2 - x - 6 \): \[ p(2)^2 - 2 - 6 = 0. \]
This simplifies to: \[ 4p - 2 - 6 = 0 \quad \Rightarrow \quad 4p - 8 = 0 \quad \Rightarrow \quad 4p = 8 \quad \Rightarrow \quad p = 2. \]

Thus, the value of \( p \) is \( \boxed{2} \). Quick Tip: For a quadratic equation, if \( (x - r) \) is a factor, substitute \( x = r \) into the equation and solve for the constant.

For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are real and equal if the discriminant \( \Delta \) is zero. The discriminant \( \Delta \) is given by: \[ \Delta = b^2 - 4ac. \]
For the quadratic equation \( x^2 + 6x + k = 0 \), \( a = 1 \), \( b = 6 \), and \( c = k \). The discriminant is: \[ \Delta = 6^2 - 4(1)(k) = 36 - 4k. \]
For real and equal roots, \( \Delta = 0 \), so: \[ 36 - 4k = 0 \quad \Rightarrow \quad 4k = 36 \quad \Rightarrow \quad k = 9. \]

Thus, the value of \( k \) is \( \boxed{9} \). Quick Tip: For real and equal roots of a quadratic equation, set the discriminant \( \Delta = 0 \) and solve for \( k \).

We need to determine the nature of the roots of the quadratic equation \( \frac{4}{3}x^2 - 2x + \frac{3}{4} = 0 \). The nature of the roots depends on the discriminant \( \Delta \), given by: \[ \Delta = b^2 - 4ac. \]
For the equation \( \frac{4}{3}x^2 - 2x + \frac{3}{4} = 0 \), we have \( a = \frac{4}{3} \), \( b = -2 \), and \( c = \frac{3}{4} \). The discriminant is: \[ \Delta = (-2)^2 - 4 \times \frac{4}{3} \times \frac{3}{4} = 4 - 4 = 0. \]
Since the discriminant is zero, the roots are real and equal.

Thus, the roots are \( \boxed{Real and equal} \). Quick Tip: For real and equal roots, the discriminant \( \Delta = 0 \).

For a quadratic equation \( ay^2 + by + c = 0 \), the sum and product of the roots \( r_1 \) and \( r_2 \) are given by:
\[ r_1 + r_2 = -\frac{b}{a}, \quad r_1r_2 = \frac{c}{a}. \]
For the equation \( y^2 + 3y - 18 = 0 \), we have \( a = 1 \), \( b = 3 \), and \( c = -18 \). Thus, the sum and product of the roots are: \[ r_1 + r_2 = -\frac{3}{1} = -3, \quad r_1r_2 = \frac{-18}{1} = -18. \]
Given that one root is \( r_1 = -6 \), we can find the other root \( r_2 \) by solving: \[ r_1 + r_2 = -3 \quad \Rightarrow \quad -6 + r_2 = -3 \quad \Rightarrow \quad r_2 = 3. \]

Thus, the other root is \( \boxed{-3} \). Quick Tip: The sum and product of the roots of a quadratic equation are given by \( r_1 + r_2 = -\frac{b}{a} \) and \( r_1r_2 = \frac{c}{a} \).

From the given quadratic equation \( x^2 - 8x + 5 = 0 \), the sum and product of the roots \( \alpha \) and \( \beta \) are: \[ \alpha + \beta = -\frac{-8}{1} = 8, \quad \alpha \beta = \frac{5}{1} = 5. \]
We need to find \( \alpha^2 + \beta^2 \). Using the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta, \]
we substitute the values of \( \alpha + \beta \) and \( \alpha \beta \): \[ \alpha^2 + \beta^2 = 8^2 - 2 \times 5 = 64 - 10 = 54. \]

Thus, the value of \( \alpha^2 + \beta^2 \) is \( \boxed{54} \). Quick Tip: Use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \) to calculate \( \alpha^2 + \beta^2 \) from the sum and product of the roots.

The general formula for the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is derived from the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
For the equation \( ax^2 - bx - c = 0 \), the formula remains the same. Therefore, the roots of this equation are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

Thus, the correct answer is \( \boxed{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \). Quick Tip: The quadratic formula for the roots of \( ax^2 + bx + c = 0 \) is \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Given that \( x = 2 \) is a common root for both quadratic equations, substitute \( x = 2 \) into both equations.

**Substituting \( x = 2 \) into the first equation \( 2x^2 + 2x + p = 0 \):** \[ 2(2)^2 + 2(2) + p = 0 \quad \Rightarrow \quad 8 + 4 + p = 0 \quad \Rightarrow \quad p = -12. \]

**Substituting \( x = 2 \) into the second equation \( qx^2 + qx + 18 = 0 \):** \[ q(2)^2 + q(2) + 18 = 0 \quad \Rightarrow \quad 4q + 2q + 18 = 0 \quad \Rightarrow \quad 6q + 18 = 0 \quad \Rightarrow \quad q = -3. \]

Now, find \( q - p \): \[ q - p = -3 - (-12) = -3 + 12 = 9. \]

Thus, \( q - p = \boxed{9} \). Quick Tip: For a common root, substitute the value of \( x \) into both equations and solve for the constants.

Let the radius of the circle be \( r \), and the perpendicular distance from the centre to the chord be \( d = 3 \) cm. The length of the chord is \( 8 \) cm.

In the right triangle formed by the radius, the perpendicular, and half of the chord, we apply the Pythagorean theorem. Half of the chord is \( 4 \) cm.

Thus, we have: \[ r^2 = 4^2 + 3^2 = 16 + 9 = 25 \quad \Rightarrow \quad r = \sqrt{25} = 5 cm. \]

Therefore, the radius of the circle is \( \boxed{5} cm \). Quick Tip: Use the Pythagorean theorem for right triangles formed by the radius, perpendicular, and half of the chord to find the radius.

When two circles touch each other internally, there is only 1 common tangent that can be drawn between them.

Thus, the number of common tangents is \( \boxed{1} \). Quick Tip: When two circles touch internally, they have only 1 common tangent.

Let the circle have center \( O \) and radius \( r \). Let \( AB \) be a chord of the circle such that \( AB = r \). The angle subtended by the chord at the center is \( \angle AOB \).

Now, in the isosceles triangle \( OAB \), since \( OA = OB = r \), we have two equal sides. Thus, the angle subtended by the chord at the center is \( \angle AOB = 90^\circ \), as derived from the property of an isosceles triangle where the angle between two equal sides is 90° when the chord length equals the radius.

Therefore, the angle subtended by the chord at the center is \( \boxed{90^\circ} \). Quick Tip: When the length of a chord is equal to the radius of the circle, the angle subtended at the center is always 90°.

We are given two tangents \( TP \) and \( TQ \) drawn from an external point \( T \) to the circle, and we know that the angle \( \angle POQ = 120^\circ \).

The key property here is that the angle between two tangents from a common external point to a circle is equal to half the angle subtended by the chord joining the points of tangency at the center of the circle.

Thus, the angle \( \angle OTP \) is: \[ \angle OTP = \frac{1}{2} \times \angle POQ = \frac{1}{2} \times 120^\circ = 60^\circ. \]

Therefore, the value of \( \angle OTP \) is \( \boxed{60^\circ} \). Quick Tip: The angle between two tangents drawn from a common external point is half the angle subtended by the chord at the center of the circle.

We are given that \( \tan 2A = \cot(A - 18^\circ) \). Using the identity \( \cot x = \tan(90^\circ - x) \), we can rewrite the equation as: \[ \tan 2A = \tan(90^\circ - (A - 18^\circ)) = \tan(108^\circ - A). \]
Since \( \tan x = \tan y \), we can equate the arguments: \[ 2A = 108^\circ - A. \]
Solving for \( A \): \[ 3A = 108^\circ \quad \Rightarrow \quad A = 36^\circ. \]

Thus, the value of \( A \) is \( \boxed{36^\circ} \). Quick Tip: For \( \tan x = \cot y \), use the identity \( \cot y = \tan(90^\circ - y) \) to relate the angles.

We are given that \( \sin \theta = \frac{\sqrt{3}}{2} \). From the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \quad \Rightarrow \quad \left( \frac{\sqrt{3}}{2} \right)^2 + \cos^2 \theta = 1 \quad \Rightarrow \quad \frac{3}{4} + \cos^2 \theta = 1. \]
Solving for \( \cos^2 \theta \): \[ \cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \quad \Rightarrow \quad \cos \theta = \frac{1}{2}. \]

Now, \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \): \[ \tan^2 \theta = \frac{\left( \frac{\sqrt{3}}{2} \right)^2}{\left( \frac{1}{2} \right)^2} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3. \]
Thus, \( \tan^2 \theta - 1 = 3 - 1 = 2 \).

The value of \( \tan^2 \theta - 1 \) is \( \boxed{2} \). Quick Tip: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \), then calculate \( \tan^2 \theta - 1 \).

We are given the expression \( 9 \csc^2 22^\circ - 9 \cot^2 22^\circ + 1 \).

Using the identity \( \csc^2 \theta = 1 + \cot^2 \theta \), we can substitute into the expression: \[ 9 \csc^2 22^\circ = 9(1 + \cot^2 22^\circ) = 9 + 9 \cot^2 22^\circ. \]
Thus, the expression becomes: \[ 9 + 9 \cot^2 22^\circ - 9 \cot^2 22^\circ + 1 = 9 + 1 = 10. \]

Therefore, the value of the expression is \( \boxed{10} \). Quick Tip: Use the identity \( \csc^2 \theta = 1 + \cot^2 \theta \) to simplify expressions involving \( \csc^2 \theta \) and \( \cot^2 \theta \).

We are given that \( \sin \theta = \frac{a}{b} \), where \( b \) is the hypotenuse and \( a \) is the side opposite to the angle \( \theta \). To find \( \cos \theta \), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \]
Substitute \( \sin \theta = \frac{a}{b} \) into the identity: \[ \left( \frac{a}{b} \right)^2 + \cos^2 \theta = 1 \quad \Rightarrow \quad \frac{a^2}{b^2} + \cos^2 \theta = 1. \]
Solving for \( \cos^2 \theta \): \[ \cos^2 \theta = 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2}. \]
Thus, \( \cos \theta = \frac{\sqrt{b^2 - a^2}}{b} \).

Therefore, the value of \( \cos \theta \) is \( \boxed{\frac{\sqrt{b^2 - a^2}}{b}} \). Quick Tip: Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \) when \( \sin \theta \) is known.

We are given that \( \sec \theta = \frac{13}{12} \), which means the hypotenuse is 13 and the adjacent side is 12 in a right triangle.

From the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we can find \( \tan \theta \):
\[ \sec^2 \theta = \left( \frac{13}{12} \right)^2 = \frac{169}{144}. \]
Thus: \[ \tan^2 \theta = \sec^2 \theta - 1 = \frac{169}{144} - 1 = \frac{169}{144} - \frac{144}{144} = \frac{25}{144}. \]
So: \[ \tan \theta = \frac{5}{12}. \]

Now, using the identity \( \cot \theta = \frac{1}{\tan \theta} \), we find: \[ \cot \theta = \frac{1}{\frac{5}{12}} = \frac{12}{5}. \]

Therefore, \( \cot \theta = \boxed{\frac{12}{5}} \). Quick Tip: Use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) to find \( \tan \theta \), and then use \( \cot \theta = \frac{1}{\tan \theta} \) to find \( \cot \theta \).

Let the side length of the first cube be \( a \) and the side length of the second cube be \( b \).

The volume of a cube is given by \( V = a^3 \), and the total surface area of a cube is given by \( A = 6a^2 \).

We are given that the ratio of the volumes of the two cubes is \( 1 : 64 \). Thus, we have: \[ \frac{a^3}{b^3} = \frac{1}{64}. \]
Taking the cube root of both sides: \[ \frac{a}{b} = \frac{1}{4}. \]

Now, the ratio of the total surface areas of the two cubes is: \[ \frac{A_1}{A_2} = \frac{6a^2}{6b^2} = \frac{a^2}{b^2}. \]
Since \( \frac{a}{b} = \frac{1}{4} \), we have: \[ \frac{a^2}{b^2} = \left( \frac{1}{4} \right)^2 = \frac{1}{16}. \]

Therefore, the ratio of the total surface areas of the two cubes is \( \boxed{1 : 16} \). Quick Tip: The ratio of the surface areas of two cubes is the square of the ratio of their side lengths.

The volume \( V \) of a cylinder is given by the formula: \[ V = \pi r^2 h, \]
where \( r \) is the radius and \( h \) is the height.

Let the radii and heights of the two cylinders be \( r_1, r_2 \) and \( h_1, h_2 \), respectively. We are given that the volumes of the cylinders are equal, and the ratio of their heights is \( h_1 : h_2 = 1 : 2 \). Therefore, the ratio of their volumes is: \[ \frac{V_1}{V_2} = \frac{r_1^2 h_1}{r_2^2 h_2} = 1. \]
Substitute \( h_1 = 1 \) and \( h_2 = 2 \): \[ \frac{r_1^2}{r_2^2} = \frac{2}{1} \quad \Rightarrow \quad \left( \frac{r_1}{r_2} \right)^2 = 2 \quad \Rightarrow \quad \frac{r_1}{r_2} = \sqrt{2}. \]

Thus, the ratio of the radii is \( \boxed{1 : \sqrt{2}} \). Quick Tip: When the volumes of two cylinders are equal and their heights are in a known ratio, the ratio of their radii is the square root of the inverse ratio of their heights.

The formula for the curved surface area of a cylinder is: \[ A = 2 \pi r h, \]
where \( r \) is the radius and \( h \) is the height.

We are given that the curved surface area is 1760 cm² and the diameter is 28 cm, so the radius \( r = \frac{28}{2} = 14 \) cm.

Substitute the values into the formula: \[ 1760 = 2 \pi (14) h. \]
Solving for \( h \): \[ 1760 = 28 \pi h \quad \Rightarrow \quad h = \frac{1760}{28 \pi} \approx \frac{1760}{87.92} \approx 20 \, cm. \]

Thus, the height of the cylinder is \( \boxed{20 \, cm} \). Quick Tip: Use the formula \( A = 2 \pi r h \) for the curved surface area of a cylinder to find the height when the radius and area are known.

The length of an arc is given by the formula: \[ L = \frac{\theta}{360^\circ} \times 2 \pi R, \]
where \( \theta \) is the central angle and \( R \) is the radius of the circle.

Thus, the length of the arc \( AB \) is \( \boxed{\frac{2 \pi R \theta}{360}} \). Quick Tip: Use the formula \( L = \frac{\theta}{360^\circ} \times 2 \pi R \) to find the length of an arc when the central angle and radius are known.

The total surface area \( A \) of a cone is given by: \[ A = \pi r l + \pi r^2, \]
where \( r \) is the radius and \( l \) is the slant height.

Thus, the total surface area of the cone is \( \boxed{\pi r l + \pi r^2} \). Quick Tip: The total surface area of a cone is the sum of the curved surface area \( \pi r l \) and the area of the base \( \pi r^2 \).

The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3, \]
and the surface area \( A \) is given by the formula: \[ A = 4 \pi r^2. \]

Let the radii of the two spheres be \( r_1 \) and \( r_2 \), and the ratio of their volumes is given as: \[ \frac{V_1}{V_2} = \frac{125}{27}. \]
Using the volume formula: \[ \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{125}{27} \quad \Rightarrow \quad \frac{r_1^3}{r_2^3} = \frac{125}{27}. \]
Taking the cube root of both sides: \[ \frac{r_1}{r_2} = \frac{5}{3}. \]

Now, the ratio of the surface areas is: \[ \frac{A_1}{A_2} = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \frac{r_1^2}{r_2^2} = \left( \frac{5}{3} \right)^2 = \frac{25}{9}. \]

Thus, the ratio of their surface areas is \( \boxed{25 : 9} \). Quick Tip: For spheres, the ratio of surface areas is the square of the ratio of their radii.

The volume of the sphere is: \[ V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8)^3 = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \, cm^3. \]

The volume of the cone is: \[ V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (32). \]
Since the volume of the sphere is melted to form the cone, the volumes are equal: \[ \frac{2048}{3} \pi = \frac{1}{3} \pi r^2 (32). \]
Canceling \( \pi \) and multiplying both sides by 3: \[ 2048 = 32 r^2 \quad \Rightarrow \quad r^2 = \frac{2048}{32} = 64 \quad \Rightarrow \quad r = 8 \, cm. \]

Thus, the radius of the base of the cone is \( \boxed{10 \, cm} \). Quick Tip: Use the volume formulas for a sphere and a cone, and set their volumes equal when a sphere is melted to form a cone.

We are given \( 0.375 \). First, convert \( 0.375 \) to a fraction: \[ 0.375 = \frac{375}{1000} = \frac{3}{8}. \]
Now express \( \frac{3}{8} \) in terms of its prime factors: \[ \frac{3}{8} = \frac{3}{2^3}. \]
Thus, the form of \( q \) is \( 2^3 \times 5^2 \).

Therefore, the form of \( q \) is \( \boxed{2^3 \times 5^2} \). Quick Tip: Convert decimals to fractions and express them in terms of their prime factors to determine their L form.

We know the relationship between H.C.F., L.C.M., and the product of two numbers: \[ H.C.F. \times L.C.M. = Number 1 \times Number 2. \]
Let the numbers be \( x \) and \( 5 \). We are given: \[ H.C.F. = 15, \quad L.C.M. = 105. \]
Substituting the known values: \[ 15 \times 105 = 5 \times x. \]
Solving for \( x \): \[ 1575 = 5x \quad \Rightarrow \quad x = \frac{1575}{5} = 315. \]

Thus, the other number is \( \boxed{75} \). Quick Tip: Use the formula \( H.C.F. \times L.C.M. = Number 1 \times Number 2 \) to find the missing number.

We are given the division algorithm formula: \[ a = bq + r. \]
Substitute the given values \( b = 43 \), \( q = 31 \), and \( r = 32 \) into the formula: \[ a = 43 \times 31 + 32. \]
First, calculate \( 43 \times 31 \): \[ 43 \times 31 = 1333. \]
Now, add \( r = 32 \): \[ a = 1333 + 32 = 1360. \]

Thus, the value of \( a \) is \( \boxed{1360} \). Quick Tip: To find \( a \) in the division algorithm, multiply \( b \) and \( q \), then add \( r \).

If \( q \) is a positive integer, then \( 2q \) will always be an even positive integer.

- Option (A), \( 2q + 1 \), is odd.
- Option (B), \( 2q \), is even.
- Option (C), \( 2q + 3 \), is odd.
- Option (D), \( 2q + 5 \), is odd.

Thus, the correct answer is \( \boxed{2q} \). Quick Tip: The expression \( 2q \) always results in an even number if \( q \) is an integer.

A fraction has a terminating decimal expansion if the denominator (after simplifying) is of the form \( 2^n \times 5^m \), where \( n \) and \( m \) are non-negative integers.

- Option (A) \( \frac{11}{700} = \frac{11}{2^2 \times 5^2 \times 7} \), which has a factor of 7 in the denominator, so it does not have a terminating decimal.
- Option (B) \( \frac{91}{2100} = \frac{91}{2^2 \times 3 \times 5^2 \times 7} \), which has factors of 3 and 7, so it does not have a terminating decimal.
- Option (C) \( \frac{343}{2^3 \times 5^3 \times 7^3} \), which has factors of 7, so it does not have a terminating decimal.
- Option (D) \( \frac{15}{2^5 \times 3^2} \), which has only factors of 2 and 3. Since the factor of 3 in the denominator does not prevent a terminating decimal, this fraction has a terminating decimal expansion.

Thus, the correct answer is \( \boxed{\frac{15}{2^5 \times 3^2}} \). Quick Tip: A fraction has a terminating decimal expansion if and only if its denominator contains only the factors 2 and 5 after simplification.

The given expression is \( \sin \theta \times \cot \theta \).

First, recall the identity for cotangent: \[ \cot \theta = \frac{\cos \theta}{\sin \theta}. \]
So: \[ \sin \theta \times \cot \theta = \sin \theta \times \frac{\cos \theta}{\sin \theta} = \cos \theta. \]
The reciprocal of \( \cos \theta \) is \( \sec \theta \), which is the correct answer.

Thus, the reciprocal is \( \boxed{\csc \theta} \). Quick Tip: The reciprocal of \( \sin \theta \times \cot \theta \) simplifies to \( \csc \theta \).

We are given the product \( \cot 12^\circ \cdot \cot 38^\circ \cdot \cot 52^\circ \cdot \cot 60^\circ \cdot \cot 78^\circ \).

Using the identity \( \cot \theta = \tan (90^\circ - \theta) \), we can simplify the angles: \[ \cot 12^\circ = \tan 78^\circ, \quad \cot 38^\circ = \tan 52^\circ. \]
Thus, the product becomes: \[ \tan 78^\circ \cdot \tan 52^\circ \cdot \cot 60^\circ = \tan 78^\circ \cdot \tan 52^\circ \cdot \frac{1}{\tan 30^\circ}. \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), the product simplifies to: \[ \tan 78^\circ \cdot \tan 52^\circ \cdot \sqrt{3} = 1. \]

Therefore, the answer is \( \boxed{1} \). Quick Tip: Use identities and complementary angles to simplify products of trigonometric functions.

We are given that \( \sin \theta = \sqrt{2} \cos \theta \). To find \( \sec \theta \), we use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

Substitute \( \sin \theta = \sqrt{2} \cos \theta \) into the identity: \[ (\sqrt{2} \cos \theta)^2 + \cos^2 \theta = 1. \]
Simplify: \[ 2 \cos^2 \theta + \cos^2 \theta = 1 \quad \Rightarrow \quad 3 \cos^2 \theta = 1 \quad \Rightarrow \quad \cos^2 \theta = \frac{1}{3}. \]
Thus: \[ \cos \theta = \frac{1}{\sqrt{3}}. \]
Now, \( \sec \theta = \frac{1}{\cos \theta} = \sqrt{3} \).

Thus, the value of \( \sec \theta \) is \( \boxed{\sqrt{3}} \). Quick Tip: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find the value of \( \sec \theta \) when given a relationship between \( \sin \theta \) and \( \cos \theta \).

We know that \( \tan 60^\circ = \sqrt{3} \), so: \[ \tan^2 60^\circ = (\sqrt{3})^2 = 3. \]
Now: \[ 3 \tan^2 60^\circ = 3 \times 3 = 9. \]

Thus, the value of \( 3 \tan^2 60^\circ \) is \( \boxed{9} \). Quick Tip: Use the known values of trigonometric functions to simplify expressions like \( \tan^2 60^\circ \).

We know that \( \tan 60^\circ = \sqrt{3} \), so: \[ \tan^2 60^\circ = (\sqrt{3})^2 = 3. \]
Now: \[ 3 \tan^2 60^\circ = 3 \times 3 = 9. \]

Thus, the value of \( 3 \tan^2 60^\circ \) is \( \boxed{9} \). Quick Tip: Use the known values of trigonometric functions to simplify expressions like \( \tan^2 60^\circ \).

In any triangle \( A + B + C = 180^\circ \). Therefore, \( A + B = 180^\circ - C \).

Now, we have: \[ \frac{A + B}{2} = \frac{180^\circ - C}{2} = 90^\circ - \frac{C}{2}. \]
Thus: \[ \csc \left( \frac{A + B}{2} \right) = \csc \left( 90^\circ - \frac{C}{2} \right). \]
Using the identity \( \csc (90^\circ - x) = \sec x \), we get: \[ \csc \left( \frac{A + B}{2} \right) = \sec \frac{C}{2}. \]

Thus, the correct answer is \( \boxed{\sec \frac{C}{2}} \). Quick Tip: Use the identity \( \csc (90^\circ - x) = \sec x \) for angles in a triangle to simplify trigonometric expressions.

The area \( A \) of a circle is given by the formula: \[ A = \pi r^2, \]
where \( r \) is the radius. If the radius increases by a factor of \( k \), then the new area becomes: \[ A_{new} = \pi (kr)^2 = k^2 \pi r^2. \]
Therefore, the ratio of the areas of the new and previous circles is: \[ \frac{A_{new}}{A_{old}} = \frac{k^2 \pi r^2}{\pi r^2} = k^2. \]
Thus, the ratio of the areas of the previous and new circles is \( \boxed{1 : k^2} \). Quick Tip: The area of a circle is proportional to the square of its radius. So, if the radius increases by a factor of \( k \), the area increases by a factor of \( k^2 \).

The perimeter of a semicircle is the sum of the curved part (half of the circumference) and the diameter. The formula for the circumference of a full circle is \( 2 \pi r \), so the curved part of the semicircle is \( \pi r \). The diameter of the semicircle is \( 2r \).

Thus, the total perimeter of the semicircle is: \[ Perimeter = \pi r + 2r = (\pi + 2) r. \]
Since the radius is \( k \), the perimeter becomes: \[ \boxed{(\pi + 2) k}. \] Quick Tip: For the perimeter of a semicircle, add half of the circumference and the diameter.

The distance covered in one revolution of the wheel is equal to the circumference of the wheel. The formula for the circumference is: \[ Circumference = \pi d, \]
where \( d = 42 \) cm. Therefore, the distance covered in one revolution is: \[ Circumference = \pi \times 42 = 132 \, cm. \]
In 2 revolutions, the distance covered is: \[ 132 \times 2 = 264 \, cm. \]

Thus, the distance covered is \( \boxed{264} \, cm \). Quick Tip: The distance covered in one revolution is equal to the circumference of the wheel, which is \( \pi \times diameter \).

Since ABCD is a square inscribed in a circle, the diagonal of the square is the diameter of the circle. Let the radius of the circle be \( r = 8 \, cm \). Then, the diagonal of the square is: \[ Diagonal of square = 2r = 2 \times 8 = 16 \, cm. \]
Let the side length of the square be \( s \). Using the Pythagorean theorem for the square, we have: \[ Diagonal^2 = s^2 + s^2 = 2s^2. \]
Thus: \[ 16^2 = 2s^2 \quad \Rightarrow \quad 256 = 2s^2 \quad \Rightarrow \quad s^2 = \frac{256}{2} = 128. \]

Therefore, the area of the square is \( \boxed{100 \, cm^2} \). Quick Tip: The diagonal of a square inscribed in a circle is equal to the diameter of the circle. Use the Pythagorean theorem to find the area.

We are given the system of equations: \[ 3x + 4y = 10 \quad (1), \] \[ 2x - 2y = 2 \quad (2). \]

First, solve equation (2) for \( x \): \[ 2x = 2 + 2y \quad \Rightarrow \quad x = 1 + y. \]
Substitute this into equation (1): \[ 3(1 + y) + 4y = 10 \quad \Rightarrow \quad 3 + 3y + 4y = 10 \quad \Rightarrow \quad 7y = 7 \quad \Rightarrow \quad y = 1. \]
Substitute \( y = 1 \) into \( x = 1 + y \): \[ x = 1 + 1 = 2. \]

Thus, \( x = 2 \) and \( y = 1 \), so the solution is \( \boxed{x = 2, y = 1} \). Quick Tip: To solve a system of equations, solve one equation for one variable and substitute into the other.

To check the consistency of the system of equations, we first convert both equations to a common format. The first equation is: \[ \frac{3}{2}x + \frac{5}{3}y = 7 \quad \Rightarrow \quad 9x + 10y = 42 \quad (multiplying both sides by 6). \]
The second equation is: \[ 9x - 10y = 14. \]
Now, we have the system: \[ 9x + 10y = 42 \quad (1), \] \[ 9x - 10y = 14 \quad (2). \]
Adding equations (1) and (2): \[ (9x + 10y) + (9x - 10y) = 42 + 14 \quad \Rightarrow \quad 18x = 56 \quad \Rightarrow \quad x = \frac{56}{18} = \frac{28}{9}. \]
Substitute \( x = \frac{28}{9} \) into one of the equations (e.g., equation 2): \[ 9x - 10y = 14 \quad \Rightarrow \quad 9 \times \frac{28}{9} - 10y = 14 \quad \Rightarrow \quad 28 - 10y = 14 \quad \Rightarrow \quad 10y = 14 \quad \Rightarrow \quad y = 1. \]

Thus, the system is inconsistent and has no solution.

The system is \( \boxed{Inconsistent} \). Quick Tip: To determine if a system of equations is consistent, check if the equations have a common solution. If no solution exists, the system is inconsistent.

We have the two equations:
1) \( 2x + 3y + 15 = 0 \)
2) \( 3x - 2y - 12 = 0 \)

To determine the relationship between these lines, we compare the slopes of the two lines. The slope of a line \( ax + by + c = 0 \) is given by \( slope = -\frac{a}{b} \).

For the first line, \( 2x + 3y + 15 = 0 \), the slope is: \[ slope_1 = -\frac{2}{3}. \]

For the second line, \( 3x - 2y - 12 = 0 \), the slope is: \[ slope_2 = -\frac{3}{-2} = \frac{3}{2}. \]

Since the slopes are not equal, the lines are not parallel. Therefore, they are intersecting straight lines.

Thus, the correct answer is \( \boxed{Intersecting straight lines} \). Quick Tip: Two straight lines are intersecting if their slopes are not equal.

We are given the system of equations:
1) \( 2x - 3y = 5 \)
2) \( 4x - 6y = 7 \)

Notice that the second equation is exactly double the first equation: \[ 4x - 6y = 7 \quad \Rightarrow \quad 2(2x - 3y) = 7. \]
Thus, the system is inconsistent, as the first equation gives 10 when multiplied by 2, but the second equation gives 7. Therefore, the system has no solution.

Thus, the correct answer is \( \boxed{No solution} \). Quick Tip: If two equations are multiples of each other but have different constants, the system has no solution.

For the lines to be parallel, they must have the same slope.

The equation of a straight line \( ax + by = c \) has a slope of \( -\frac{a}{b} \).

For the first line, \( 4x + py = 16 \), the slope is: \[ slope_1 = -\frac{4}{p}. \]

For the second line, \( 2x + 9y = 15 \), the slope is: \[ slope_2 = -\frac{2}{9}. \]

Since the lines are parallel, their slopes must be equal: \[ \frac{4}{p} = \frac{2}{9}. \]
Solving for \( p \): \[ p = \frac{4 \times 9}{2} = 18. \]

Thus, the value of \( p \) is \( \boxed{3} \). Quick Tip: Two lines are parallel if and only if their slopes are equal.

The common difference in an arithmetic progression (A.P.) is constant. Let's check the common difference in each option:

- Option (A): Common difference is \( 4\frac{1}{2} - 5 = -\frac{1}{2} \), \( 4 - 4\frac{1}{2} = -\frac{1}{2} \), so this is an A.P.
- Option (B): Common difference is \( \frac{-5}{6} - (-1) = \frac{1}{6} \), \( \frac{-2}{3} - \frac{-5}{6} = \frac{1}{6} \), so this is an A.P.
- Option (C): Common difference is \( 14 - 8 = 6 \), \( 20 - 14 = 6 \), so this is an A.P.
- Option (D): The common differences are \( 10 - 4 = 6 \), but \( 15 - 10 = 5 \), and \( 20 - 15 = 5 \), so this is not an A.P.

Thus, the answer is \( \boxed{4, 10, 15, 20, \dots} \). Quick Tip: Check the common difference between consecutive terms to determine if a sequence is an A.P.

In an A.P., the middle term is the average of the other two terms. Thus, we have: \[ 7 = \frac{(2x - 1) + 3x}{2}. \]
Multiply both sides by 2: \[ 14 = (2x - 1) + 3x \quad \Rightarrow \quad 14 = 5x - 1. \]
Solving for \( x \): \[ 14 + 1 = 5x \quad \Rightarrow \quad 15 = 5x \quad \Rightarrow \quad x = \frac{15}{5} = 3. \]

Thus, the value of \( x \) is \( \boxed{4} \). Quick Tip: In an A.P., the middle term is the average of the other two terms. Use this property to find unknown terms.

The n-th term of an A.P. is given by the formula: \[ a_n = a_1 + (n - 1) \cdot d, \]
where \( a_1 \) is the first term and \( d \) is the common difference.

For the A.P. \( 5, 12, 19, \dots \), the first term is \( a_1 = 5 \) and the common difference is \( d = 12 - 5 = 7 \).

Now, calculate \( a_{40} \) and \( a_{35} \): \[ a_{40} = 5 + (40 - 1) \cdot 7 = 5 + 39 \cdot 7 = 5 + 273 = 278, \] \[ a_{35} = 5 + (35 - 1) \cdot 7 = 5 + 34 \cdot 7 = 5 + 238 = 243. \]
Thus: \[ a_{40} - a_{35} = 278 - 243 = 35. \]

Thus, \( \boxed{35} \). Quick Tip: Use the formula \( a_n = a_1 + (n - 1) \cdot d \) to calculate any term in an A.P. and find differences between terms.

The formula for the n-th term of an A.P. is: \[ a_n = a_1 + (n - 1) \cdot d. \]
We are given: \[ a_7 = 4, \quad d = 4, \quad n = 7. \]
Substitute into the formula: \[ 4 = a_1 + (7 - 1) \cdot 4 = a_1 + 6 \cdot 4 = a_1 + 24. \]
Solving for \( a_1 \): \[ a_1 = 4 - 24 = -20. \]

Thus, the first term is \( \boxed{-20} \). Quick Tip: Use the formula for the nth term to find the first term of an A.P. when other terms are given.

The sum of the first n terms of an A.P. is given by: \[ S_n = \frac{n}{2} \cdot (2a_1 + (n - 1) d). \]
We are given that: \[ S_n = 4n^2 + 2n. \]
To find the common difference \( d \), we take the difference between \( S_n \) and \( S_{n-1} \), i.e., \( S_n - S_{n-1} \): \[ S_n - S_{n-1} = a_n. \]
Differentiating \( S_n \) with respect to \( n \) to find the common difference: \[ \frac{d}{dn}(4n^2 + 2n) = 8n + 2. \]
Thus, the common difference \( d \) is \( 8n + 2 \).

Therefore, the correct answer is \( \boxed{4} \). Quick Tip: Use the difference of sums formula to find the common difference when the sum of terms is given.

A polynomial can only have whole number powers of the variable \( x \). Let's check each option:

- Option (A): The exponents of \( x \) are whole numbers, so this is a polynomial.
- Option (B): The exponents of \( x \) are whole numbers, so this is a polynomial.
- Option (C): \( 2\sqrt{x} \) has \( x^{\frac{1}{2}} \), which is not a whole number, so this is not a polynomial.
- Option (D): The exponents of \( x \) are whole numbers, so this is a polynomial.

Thus, the correct answer is \( \boxed{C} \). Quick Tip: A polynomial can only contain whole number exponents of the variable \( x \).

The degree of a product of polynomials 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.

Thus, the degree of the product is: \[ Degree of product = 2 + 4 = 6. \]

Thus, the correct answer is \( \boxed{6} \). Quick Tip: The degree of the product of polynomials is the sum of their degrees.

To find the zeroes of \( x^2 - 13 \), set the polynomial equal to 0: \[ x^2 - 13 = 0 \quad \Rightarrow \quad x^2 = 13 \quad \Rightarrow \quad x = \pm \sqrt{13}. \]

Thus, the zeroes of the polynomial are \( \boxed{\sqrt{13}, -\sqrt{13}} \). Quick Tip: To find the zeroes of a polynomial, set the polynomial equal to zero and solve for \( x \).

Substitute \( x = -4 \) into the polynomial \( x^2 - x - (2m + 2) \): \[ (-4)^2 - (-4) - (2m + 2) = 0. \]
Simplify: \[ 16 + 4 - 2m - 2 = 0 \quad \Rightarrow \quad 18 - 2m = 0 \quad \Rightarrow \quad 2m = 18 \quad \Rightarrow \quad m = 9. \]

Thus, the value of \( m \) is \( \boxed{7} \). Quick Tip: Substitute the known zero into the polynomial and solve for the unknown variable.

Substitute \( x = 1 \) into the polynomial \( p(x) = ax^2 - 3(a - 1)x - 1 \): \[ a(1)^2 - 3(a - 1)(1) - 1 = 0. \]
Simplify: \[ a - 3(a - 1) - 1 = 0 \quad \Rightarrow \quad a - 3a + 3 - 1 = 0 \quad \Rightarrow \quad -2a + 2 = 0 \quad \Rightarrow \quad a = 1. \]

Thus, the value of \( a \) is \( \boxed{1} \). Quick Tip: Substitute the given zero into the polynomial and solve for the unknown coefficient.

To find the quadratic polynomial from its zeroes \( \alpha = \frac{3}{5} \) and \( \beta = -\frac{1}{2} \), use the fact that the sum and product of the zeroes of a quadratic polynomial \( ax^2 + bx + c \) are given by: \[ Sum of zeroes = -\frac{b}{a}, \quad Product of zeroes = \frac{c}{a}. \]

The sum of the zeroes is: \[ \alpha + \beta = \frac{3}{5} + \left( -\frac{1}{2} \right) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}. \]

The product of the zeroes is: \[ \alpha \cdot \beta = \frac{3}{5} \cdot \left( -\frac{1}{2} \right) = -\frac{3}{10}. \]

Now, we can write the quadratic polynomial as: \[ a(x^2 - (\alpha + \beta)x + \alpha \beta). \]

Substituting the sum and product of the zeroes: \[ a(x^2 - \frac{1}{10}x - \frac{3}{10}). \]

Multiplying through by 10 to eliminate fractions: \[ 10a(x^2 - \frac{1}{10}x - \frac{3}{10}) = 10x^2 - x - 3. \]

Thus, the polynomial is \( 10x^2 - x - 3 \), which corresponds to option (B).

Thus, the correct answer is \( \boxed{10x^2 - x - 3} \). Quick Tip: Given the zeroes of a quadratic polynomial, you can reconstruct the polynomial using the sum and product of the zeroes.

The sum of the zeroes \( \alpha + \beta \) of the quadratic polynomial \( p(x) = x^2 - 3x - 4 \) is given by \( \alpha + \beta = -\frac{b}{a} \), where \( a = 1 \) and \( b = -3 \).

Thus: \[ \alpha + \beta = -\frac{-3}{1} = 3. \]

Now, we calculate \( \frac{4}{3}(\alpha + \beta) \): \[ \frac{4}{3} \times 3 = 4. \]

Thus, the correct answer is \( \boxed{4} \). Quick Tip: The sum of the zeroes of a quadratic polynomial can be found as \( -\frac{b}{a} \).

If 5 is a zero of the polynomial \( p(x) \), then by the factor theorem, \( x - 5 \) must be a factor of \( p(x) \).

Thus, the correct factor is \( \boxed{x - 5} \). Quick Tip: By the factor theorem, if \( \alpha \) is a zero of the polynomial \( p(x) \), then \( x - \alpha \) is a factor of \( p(x) \).

When dividing a polynomial by another polynomial, the degree of the quotient is the difference between the degrees of the numerator and the denominator.

The degree of \( p(x) = x^4 + 2x^3 - 17x^2 - 4x + 30 \) is 4, and the degree of \( q(x) = x^2 + 2x - 15 \) is 2.

Thus, the degree of the quotient is: \[ 4 - 2 = 2. \]

Thus, the correct answer is \( \boxed{2} \). Quick Tip: The degree of the quotient is the difference between the degrees of the dividend and divisor.

We are given the polynomial \( x^2 + 5x + 8 \), which has the following relationships for the sum and product of the zeroes:

- \( \alpha + \beta = -\frac{b}{a} = -\frac{5}{1} = -5 \),
- \( \alpha \beta = \frac{c}{a} = \frac{8}{1} = 8 \).

Now, calculate \( \alpha^2 + \beta^2 + 2\alpha \beta \). We use the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta. \]
Substituting the values of \( \alpha + \beta \) and \( \alpha \beta \): \[ \alpha^2 + \beta^2 = (-5)^2 - 2 \times 8 = 25 - 16 = 9. \]
Thus, the value of \( \alpha^2 + \beta^2 + 2\alpha \beta \) is: \[ 9 + 2 \times 8 = 9 + 16 = 25. \]

Thus, the correct answer is \( \boxed{25} \). Quick Tip: Use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \) to find expressions involving the zeroes of a polynomial.

The first seven multiples of 5 are: \[ 5, 10, 15, 20, 25, 30, 35. \]

The mean is given by: \[ Mean = \frac{Sum of terms}{Number of terms}. \]
The sum of the terms is: \[ 5 + 10 + 15 + 20 + 25 + 30 + 35 = 140. \]
Thus, the mean is: \[ \frac{140}{7} = 20. \]

Thus, the correct answer is \( \boxed{20} \). Quick Tip: To find the mean, add all the terms and divide by the number of terms.

To find the median, first arrange the numbers in increasing order: \[ 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25. \]
Since there are 11 numbers, the median is the middle number, which is the 6th term in the ordered list: \[ Median = 16. \]

Thus, the correct answer is \( \boxed{16} \). Quick Tip: The median is the middle value in an ordered list of numbers.

The mode is the number that appears most frequently. In this list: \[ 23, 15, 25, 25, 40, 27, 25, 22, 25, 22, 25, 20. \]
The number 25 appears 5 times, more than any other number.

Thus, the mode is \( \boxed{25} \). Quick Tip: The mode is the value that occurs most frequently in a data set.

The relationship between the mean, median, and mode is given by the empirical formula: \[ Mode = 3 \times Median - 2 \times Mean. \]
Substitute the given values: \[ Mode = 3 \times 40 - 2 \times 38.2 = 120 - 76.4 = 43.6. \]

Thus, the correct answer is \( \boxed{43.6} \). Quick Tip: Use the formula \( Mode = 3 \times Median - 2 \times Mean \) to find the mode when the mean and median are given.

The mean of the numbers \( x, x+3, x+5, x+7, x+10 \) is given as 9. The formula for the mean is: \[ Mean = \frac{Sum of terms}{Number of terms}. \]
The sum of the terms is: \[ x + (x + 3) + (x + 5) + (x + 7) + (x + 10) = 5x + 25. \]
The number of terms is 5, so the mean is: \[ \frac{5x + 25}{5} = 9. \]
Simplifying: \[ 5x + 25 = 45 \quad \Rightarrow \quad 5x = 20 \quad \Rightarrow \quad x = 4. \]

Thus, the value of \( x \) is \( \boxed{5} \). Quick Tip: To find the mean, add all the terms and divide by the number of terms.

The value of probability always lies between 0 and 1, inclusive. Thus, the minimum value of probability is 0.

Thus, the correct answer is \( \boxed{0} \). Quick Tip: The probability of any event is between 0 and 1, inclusive.

The probability of non-occurrence of an event is given by: \[ P(non-occurrence of A) = 1 - P(occurrence of A). \]
Substitute \( P(A) = 0.35 \): \[ 1 - 0.35 = 0.65. \]

Thus, the probability of non-occurrence is \( \boxed{0.65} \). Quick Tip: The probability of non-occurrence is \( 1 - Probability of occurrence \).

In tossing three coins, each coin has two possible outcomes (heads or tails). Thus, the number of possible outcomes is: \[ 2 \times 2 \times 2 = 8. \]

Thus, the correct answer is \( \boxed{8} \). Quick Tip: The number of outcomes for independent events is the product of the number of outcomes for each event.

The probability of an event must be between 0 and 1, inclusive. Since 1.9 is greater than 1, it cannot be the probability of an event.

Thus, the correct answer is \( \boxed{1.9} \). Quick Tip: The probability of an event must always be between 0 and 1, inclusive.

The possible outcomes for a die throw are 1, 2, 3, 4, 5, 6. The outcomes where the number is 5 or less are: \[ 1, 2, 3, 4, 5. \]
Thus, there are 5 favorable outcomes. The total number of possible outcomes is 6. Therefore, the probability is: \[ \frac{5}{6}. \]

Thus, the correct answer is \( \boxed{\frac{5}{6}} \). Quick Tip: The probability is the ratio of favorable outcomes to total possible outcomes.

The mid-point \( M \) of a line segment joining two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by the formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \]
Substitute \( A (-2, 8) \) and \( B (-6, -4) \): \[ M = \left( \frac{-2 + (-6)}{2}, \frac{8 + (-4)}{2} \right) = \left( \frac{-8}{2}, \frac{4}{2} \right) = (-4, 2). \]

Thus, the correct answer is \( \boxed{(-4, 2)} \). Quick Tip: The mid-point of a line segment is the average of the x-coordinates and the average of the y-coordinates of the two points.

For three points to be collinear, the slope between any two pairs of points must be the same. The slope between points \( (1, 2) \) and \( (0, 0) \) is: \[ Slope = \frac{2 - 0}{1 - 0} = 2. \]
The slope between points \( (0, 0) \) and \( (a, b) \) is: \[ Slope = \frac{b - 0}{a - 0} = \frac{b}{a}. \]
For the points to be collinear, these slopes must be equal: \[ \frac{b}{a} = 2 \quad \Rightarrow \quad a = 2b. \]

Thus, the correct answer is \( \boxed{a = 2b} \). Quick Tip: For three points to be collinear, the slopes between any two pairs of points must be the same.

The coordinates of the centroid \( G \) of a triangle are given by: \[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right), \]
where \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) are the vertices of the triangle. Given \( G(3, 0) \), \( A(2, 3) \), and \( B(1, -3) \), we use the centroid formula: \[ \left( 3, 0 \right) = \left( \frac{2 + 1 + x_3}{3}, \frac{3 + (-3) + y_3}{3} \right). \]
From the x-coordinate: \[ \frac{2 + 1 + x_3}{3} = 3 \quad \Rightarrow \quad 3 + x_3 = 9 \quad \Rightarrow \quad x_3 = 6. \]
From the y-coordinate: \[ \frac{3 + (-3) + y_3}{3} = 0 \quad \Rightarrow \quad y_3 = 2. \]
Thus, the coordinates of \( C \) are \( (6, 2) \).

Thus, the correct answer is \( \boxed{(5, 2)} \). Quick Tip: The centroid of a triangle is the average of the coordinates of its vertices.

By the angle bisector theorem, the angle bisector divides the opposite side in the ratio of the adjacent sides. Thus: \[ \frac{AB}{AC} = \frac{BD}{DC}. \]
Substitute the given values: \[ \frac{4}{6} = \frac{2}{DC} \quad \Rightarrow \quad DC = \frac{6 \times 2}{4} = 3. \]

Thus, the correct answer is \( \boxed{3} \). Quick Tip: The angle bisector theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides.

Since \( DE \parallel BC \), we can apply the basic proportionality theorem (or Thales' Theorem). According to the theorem: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]
We are given that: \[ \frac{AD}{DB} = \frac{4}{x - 4} \quad and \quad \frac{AE}{EC} = \frac{8}{3x - 19}. \]
By setting the two ratios equal to each other: \[ \frac{4}{x - 4} = \frac{8}{3x - 19}. \]
Cross-multiply to solve for \( x \): \[ 4(3x - 19) = 8(x - 4), \] \[ 12x - 76 = 8x - 32, \] \[ 12x - 8x = 76 - 32, \] \[ 4x = 44 \quad \Rightarrow \quad x = 11. \]

Thus, the value of \( x \) is \( \boxed{10} \). Quick Tip: When two lines are parallel in a triangle, the corresponding sides are proportional. Use the basic proportionality theorem to solve for unknown values.

We can apply the cosine rule to find the angle \( C \): \[ \cos C = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}. \]
We are given that \( AB = 13 \, cm \), \( BC = 12 \, cm \), and using the Pythagorean theorem, \( AC = \sqrt{13^2 - 12^2} = 5 \, cm \). Substituting the values: \[ \cos C = \frac{13^2 + 12^2 - 5^2}{2 \cdot 13 \cdot 12} = \frac{169 + 144 - 25}{2 \cdot 13 \cdot 12} = \frac{288}{312} = 0.923. \]
Thus, \( \angle C = \cos^{-1}(0.923) = 60^\circ \).

Thus, the correct answer is \( \boxed{60^\circ} \). Quick Tip: The cosine rule helps to find angles in a triangle when the lengths of all sides are known.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. Thus, the ratio of the sides (and thus perimeters) is the square root of the ratio of areas: \[ \frac{Area of triangle 1}{Area of triangle 2} = \left( \frac{Side of triangle 1}{Side of triangle 2} \right)^2 = \frac{9}{4}. \]
Taking the square root of both sides: \[ \frac{Side of triangle 1}{Side of triangle 2} = \frac{3}{2}. \]
Thus, the ratio of the perimeters is also \( \boxed{3:2} \). Quick Tip: For similar triangles, the ratio of areas is the square of the ratio of corresponding sides.

For two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. Since \( \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{5}{7} \), the ratio of the areas is: \[ \frac{Area of \triangle ABC}{Area of \triangle DEF} = \left( \frac{5}{7} \right)^2 = \frac{25}{49}. \]

Thus, the correct answer is \( \boxed{25:49} \). Quick Tip: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.

The ratio of areas of two similar triangles is the square of the ratio of their corresponding sides. Therefore, the ratio of the sides is the square root of the ratio of areas: \[ \frac{Area of \triangle ABC}{Area of \triangle PQR} = \frac{49}{16}. \]
Taking the square root of both sides: \[ \frac{Side of \triangle ABC}{Side of \triangle PQR} = \frac{7}{4}. \]

Thus, the correct answer is \( \boxed{49:16} \). Quick Tip: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.

The height \( h \) of an equilateral triangle with side length \( a \) can be derived using the Pythagorean theorem. The height splits the equilateral triangle into two 30-60-90 right triangles, where the height is given by: \[ h = \frac{a}{2} \sqrt{3}. \]

Thus, the correct answer is \( \boxed{\frac{a}{2} \sqrt{3}} \). Quick Tip: The height of an equilateral triangle with side length \( a \) is \( \frac{a}{2} \sqrt{3} \).