Step 1: Consider the degree of a cubic polynomial, which is 3.

Step 2: Apply the Fundamental Theorem of Algebra, which states that a polynomial of degree \(n\) has exactly \(n\) roots.

Step 3: Conclude that a cubic polynomial has three roots, which are the zeroes of the polynomial. Quick Tip: The roots of a polynomial can be real or complex, and some roots may repeat. Always consider the possibility of complex roots when dealing with polynomials.

Step 1: Start with the given number 140.

Step 2: Divide 140 by the smallest prime number, which is 2. \[ 140 \div 2 = 70. \]

Step 3: Divide the quotient by 2 again. \[ 70 \div 2 = 35. \]

Step 4: Now divide 35 by the next smallest prime, which is 5. \[ 35 \div 5 = 7. \]

Step 5: The remaining quotient is 7, which is a prime number.

Step 6: Compile the results to express the prime factors of 140. \[ 140 = 2^2 \times 5 \times 7. \] Quick Tip: For efficient prime factorization, continuously divide by the smallest prime until reaching a prime number or 1.

Step 1: Recall the division algorithm, which states that any integer dividend can be expressed as the divisor times the quotient plus the remainder. Quick Tip: The division algorithm provides the fundamental basis for operations in modular arithmetic and number theory.

Step 1: Rewrite the equations in standard form:

First equation: \( x + 2y = -5 \)
Second equation: \( -3x - 6y = -1 \)


Step 2: Multiply the first equation by 3 to see if the second equation is proportional to the first: \[ 3(x + 2y) = 3(-5) \quad \Rightarrow \quad 3x + 6y = -15 \]

Step 3: Compare the two equations:

The modified first equation is \( 3x + 6y = -15 \).
The second equation is \( -3x - 6y = -1 \).


The second equation is not proportional to the first equation, meaning the system is inconsistent and does not represent the same line.

Conclusion: Since the system is inconsistent, there are no solutions. Therefore, the correct answer is (D) None of the above. Quick Tip: When two linear equations are inconsistent, they represent parallel lines and thus have no solution.

Step 1: Calculate the difference between consecutive terms in the sequence.

Step 2: The first term is \( -5 \) and the second term is \( -1 \). The difference is: \[ Difference = -1 - (-5) = -1 + 5 = 4 \]

Step 3: The second term is \( -1 \) and the third term is \( 3 \). The difference is: \[ Difference = 3 - (-1) = 3 + 1 = 4 \]

Step 4: The third term is \( 3 \) and the fourth term is \( 7 \). The difference is: \[ Difference = 7 - 3 = 4 \]

Step 5: Confirm the difference is consistent across all listed terms.


Conclusion: The common difference for the given Arithmetic Progression is \( 4 \). Therefore, the correct answer is (D) \(4\). Quick Tip: Always check multiple consecutive differences in an AP to ensure the common difference is consistent.

Step 1: Recall the formula for the discriminant of a quadratic equation, which is used to determine the nature of its roots. Quick Tip: The discriminant tells us the number and type of solutions: positive for two real solutions, zero for one real solution, negative for two complex solutions.

Step 1: Apply the distance formula, \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. Quick Tip: Always double-check calculations when applying the distance formula to avoid common errors like miscalculating squares or square roots.

Step 1: For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant (\( \Delta \)) must be zero. The discriminant is given by: \[ \Delta = b^2 - 4ac \]
So, for the equation \( 3x^2 - 12x + m = 0 \), we have: \[ a = 3, \quad b = -12, \quad c = m \]

Step 2: Set the discriminant to zero: \[ \Delta = b^2 - 4ac = 0 \]
Substitute the values of \(a\), \(b\), and \(c\) into the equation: \[ (-12)^2 - 4(3)(m) = 0 \] \[ 144 - 12m = 0 \]

Step 3: Solve for \(m\): \[ 12m = 144 \] \[ m = \frac{144}{12} = 12 \]

Conclusion: The value of \(m\) that ensures the quadratic equation has equal roots is \(m = 12\). Therefore, the correct answer is (D) \(12\). Quick Tip: Remember, equal roots in a quadratic equation imply the discriminant is exactly zero.

Step 1: Note the formula for the altitude \( h \) of an equilateral triangle with side length \( s \) is given by: \[ h = \frac{\sqrt{3}}{2} \times s \]

Step 2: Substitute \( s = 2a \) into the formula: \[ h = \frac{\sqrt{3}}{2} \times 2a = a\sqrt{3} \]

Step 3: Thus, the length of each altitude in \( \triangle ABC \) is \( a\sqrt{3} \). Quick Tip: Remember, the altitude of an equilateral triangle splits it into two 30-60-90 right triangles, where the ratios of sides are consistent.

Step 1: Compute the midpoints (\( x_i \)) of each class interval using the formula: \[ x_i = \frac{Lower Bound + Upper Bound}{2} \]


% Table for Calculation

\begin{tabular{|c|c|c|c|
\hline
Class Interval & Frequency (f) & Midpoint (x) & f \(\times\) x

\hline
1 -- 3 & 3 & \( \frac{1+3}{2} = 2 \) & \(3 \times 2 = 6\)

3 -- 5 & 2 & \( \frac{3+5}{2} = 4 \) & \(2 \times 4 = 8\)

5 -- 7 & 4 & \( \frac{5+7}{2} = 6 \) & \(4 \times 6 = 24\)

7 -- 9 & 2 & \( \frac{7+9}{2} = 8 \) & \(2 \times 8 = 16\)

9 -- 11 & 3 & \( \frac{9+11}{2} = 10 \) & \(3 \times 10 = 30\)

\hline
Total & \( \sum f = 14 \) & & \( \sum f x = 84 \)

\hline
\end{tabular



Step 2: Compute the mean using the formula: \[ Mean = \frac{\sum f x}{\sum f} \]

Substituting the values: \[ Mean = \frac{84}{14} = 6 \]

Thus, the mean of the given data is 6. Quick Tip: To find the mean of grouped data, use \( \frac{\sum f x}{\sum f} \), where \( x \) is the class midpoint and \( f \) is the frequency.

Step 1: Using the trigonometric identity: \[ \sin x = \cos (90^\circ - x) \]
we substitute \( \sin 27^\circ \) as follows: \[ \sin 27^\circ = \cos (90^\circ - 27^\circ) = \cos 63^\circ \]

Step 2: Substituting in the given expression: \[ \frac{\sin 27^\circ}{\cos 63^\circ} = \frac{\cos 63^\circ}{\cos 63^\circ} = 1 \] Quick Tip: Use the co-function identity \( \sin x = \cos (90^\circ - x) \) for simplifications.

Step 1: Using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \]
Substitute \( \cos A = \frac{\sqrt{3}}{2} \): \[ \sin^2 A = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4} \] \[ \sin A = \frac{1}{2} \]

Step 2: Use the double angle identity for sine: \[ \sin 2A = 2 \sin A \cos A \]
Substitute the values for \( \sin A \) and \( \cos A \): \[ \sin 2A = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \] Quick Tip: Use the identity \( \sin 2A = 2 \sin A \cos A \) for double-angle simplifications.

Step 1: Using the Pythagorean identities: \[ 1 + \tan^2 A = \sec^2 A, \quad 1 + \cot^2 A = \csc^2 A \]

Step 2: Substituting in the given expression: \[ \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} \]

Step 3: Expressing in terms of sine and cosine: \[ \sec^2 A = \frac{1}{\cos^2 A}, \quad \csc^2 A = \frac{1}{\sin^2 A} \] \[ \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \] Quick Tip: Use the identities \( \sec^2 A = 1 + \tan^2 A \) and \( \csc^2 A = 1 + \cot^2 A \) for solving such problems.

Step 1: Given equation: \[ \sin 2A = 2 \sin A \]

Step 2: Using the double angle formula: \[ \sin 2A = 2 \sin A \cos A \]

Step 3: Equating both sides: \[ 2 \sin A \cos A = 2 \sin A \]

Step 4: Dividing by \( 2 \sin A \) (provided \( \sin A \neq 0 \)): \[ \cos A = 1 \]

Step 5: The value of \( A \) that satisfies \( \cos A = 1 \) is: \[ A = 0^\circ \] Quick Tip: The identity \( \sin 2A = 2 \sin A \cos A \) helps in solving trigonometric equations.

Step 1: Circumference of a circle is given by the formula: \[ 2 \pi r = 22 \]

Step 2: Solve for the radius \( r \): \[ r = \frac{22}{2\pi} = \frac{22}{2 \times \frac{22}{7}} = \frac{7}{2} \]

Step 3: Area of a full circle is given by: \[ \pi r^2 \]
Substitute \( r = \frac{7}{2} \): \[ \pi r^2 = \pi \times \left(\frac{7}{2}\right)^2 = \pi \times \frac{49}{4} = \frac{49\pi}{4} \]
Now substitute \( \pi = \frac{22}{7} \) into the equation: \[ \frac{49\pi}{4} = \frac{49}{4} \times \frac{22}{7} = \frac{49 \times 22}{4 \times 7} = \frac{1078}{28} = \frac{77}{4} \]

Step 4: Area of a quadrant of the circle is: \[ \frac{1}{4} \times \frac{77}{4} = \frac{77}{8} \]

Thus, the area of the quadrant is \( \frac{77}{8} \, cm^2 \). Quick Tip: For quadrant area, use \( \frac{1}{4} \pi r^2 \) after determining the radius from the circumference formula.

A capsule consists of a cylindrical middle section and two hemispherical ends.

Step 1: The middle section is a cylinder, which has a curved surface area and volume formula: \[ Curved Surface Area = 2\pi r h \]

Step 2: The two ends are hemispheres, which contribute to the total surface area and volume. The volume formula for one hemisphere is: \[ V = \frac{2}{3} \pi r^3 \]

Thus, a capsule is best represented by one cylinder and two hemispheres. Quick Tip: The combination of shapes helps in finding the total surface area and volume of real-life objects.

Step 1: Using the empirical formula: \[ Mode = 3 \times Median - 2 \times Mean \]

Step 2: Substituting the given values: \[ 35 = 3 \times Median - 2 \times 32 \]

Step 3: Solving for Median: \[ 35 = 3 \times Median - 64 \] \[ 3 \times Median = 99 \] \[ Median = \frac{99}{3} = 33 \] Quick Tip: The relation \( Mode = 3 \times Median - 2 \times Mean \) is useful in finding the missing measure of central tendency.

Step 1: The modal class is the class interval with the highest frequency.
From the frequency table, the maximum frequency is 19 (corresponding to the class interval 15-20).

Step 2: Since 19 is the highest frequency, the modal class is 15-20. Quick Tip: The modal class is the class interval with the highest frequency in a frequency distribution table.

Step 1: The total number of outcomes when rolling two dice is: \[ 6 \times 6 = 36 \]

Step 2: The possible dice pairs that give a sum of 10 are: \[ (4,6), (5,5), (6,4) \]

Step 3: There are 3 favorable outcomes, so the probability is: \[ \frac{Favorable Outcomes}{Total Outcomes} = \frac{3}{36} = \frac{1}{12} \] Quick Tip: To find the probability of a sum with two dice, list all possible pairs and count the favorable ones.

Step 1: The sample space for rolling a single die is: \[ \{1,2,3,4,5,6\} \]

Step 2: The even numbers are: \[ \{2,4,6\} \]

Step 3: The probability of rolling an even number is: \[ \frac{Number of favorable outcomes}{Total outcomes} = \frac{3}{6} = \frac{1}{2} \] Quick Tip: A fair six-sided die has equal chances of landing on any number; half of them are even.

Step 1: The volume of a cube is given by the formula: \[ V = s^3 \]
where \( s \) is the length of a side.

Step 2: Substituting the given volume: \[ 1331 = s^3 \]

Step 3: Taking the cube root on both sides: \[ s = \sqrt[3]{1331} = 11 \]

Thus, the side of the cube is 11 cm. Quick Tip: To find the side of a cube from its volume, take the cube root of the given volume.

Step 1: Given the quadratic equation: \[ x^2 + 3x - p = 0 \]
One root is given as \( x = 2 \).

Step 2: Using the property of quadratic equations, if \( x = 2 \) is a root, it satisfies the equation: \[ (2)^2 + 3(2) - p = 0 \]

Step 3: Solving for \( p \): \[ 4 + 6 - p = 0 \] \[ p = 10 \]

Thus, the value of \( p \) is 10. Quick Tip: To find an unknown coefficient in a quadratic equation, substitute the given root and solve for the unknown.

Step 1: Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]

Step 2: Substituting \( \cos \theta = \frac{15}{17} \): \[ \sin^2 \theta + \left(\frac{15}{17}\right)^2 = 1 \]

Step 3: Solving for \( \sin \theta \): \[ \sin^2 \theta + \frac{225}{289} = 1 \] \[ \sin^2 \theta = 1 - \frac{225}{289} = \frac{289}{289} - \frac{225}{289} = \frac{64}{289} \]

Step 4: Taking the square root: \[ \sin \theta = \pm \frac{8}{17} \]

Thus, \( \sin \theta \) can be \( \frac{8}{17} \) or \( -\frac{8}{17} \), depending on the quadrant of \( \theta \). Quick Tip: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find one trigonometric ratio from another.

Step 1: Find the class midpoints (\( x_i \)): \[ x_i = \frac{Lower Bound + Upper Bound}{2} \]
\[ x_i = \{1, 3, 5, 7, 9\} \]

Step 2: Compute the sum of \( f_i x_i \): \[ \sum f_i x_i = (1 \times 1) + (2 \times 3) + (6 \times 5) + (8 \times 7) + (3 \times 9) \]
\[ = 1 + 6 + 30 + 56 + 27 = 120 \]

Step 3: Compute the total frequency: \[ \sum f_i = 1 + 2 + 6 + 8 + 3 = 20 \]

Step 4: Find the mean using the formula: \[ Mean = \frac{\sum f_i x_i}{\sum f_i} = \frac{120}{20} = 6 \]

Thus, the mean is 6. Quick Tip: For grouped frequency data, the mean is found using \( \frac{\sum f_i x_i}{\sum f_i} \).

Step 1: The midpoint formula is: \[ M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Step 2: Substituting given points \( (-3,10) \) and \( (5,4) \): \[ M = \left( \frac{-3 + 5}{2}, \frac{10 + 4}{2} \right) \]

Step 3: Solving: \[ M = \left( \frac{2}{2}, \frac{14}{2} \right) = (1,7) \]

Thus, the midpoint is (1,7). Quick Tip: The midpoint of a segment joining two points is the average of their coordinates.

Step 1: Using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substituting given values \( (-1,-3) \) and \( (x,9) \): \[ 13 = \sqrt{(x + 1)^2 + (9 + 3)^2} \]

Step 3: Squaring both sides: \[ 169 = (x+1)^2 + 144 \]

Step 4: Solving for \( x \): \[ (x+1)^2 = 169 - 144 \] \[ (x+1)^2 = 25 \]
\[ x+1 = \pm 5 \]

Step 5: Finding \( x \): \[ x = -1 + 5 = 4 \quad or \quad x = -1 - 5 = -6 \]

Thus, \( x \) can be 4 or -6. Quick Tip: Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find unknown coordinates.

Step 1: Let the two consecutive odd integers be \( x \) and \( x+2 \).

Step 2: Given condition: \[ x^2 + (x+2)^2 = 290 \]

Step 3: Expanding the equation: \[ x^2 + x^2 + 4x + 4 = 290 \]

Step 4: Simplifying: \[ 2x^2 + 4x + 4 = 290 \] \[ 2x^2 + 4x - 286 = 0 \]

Step 5: Dividing by 2: \[ x^2 + 2x - 143 = 0 \]

Step 6: Solving using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-143)}}{2(1)} \]
\[ x = \frac{-2 \pm \sqrt{4 + 572}}{2} \]
\[ x = \frac{-2 \pm \sqrt{576}}{2} \]
\[ x = \frac{-2 \pm 24}{2} \]

Step 7: Finding possible values of \( x \): \[ x = \frac{-2 + 24}{2} = \frac{22}{2} = 11 \]
or \[ x = \frac{-2 - 24}{2} = \frac{-26}{2} = -13 \quad (not positive) \]

Step 8: The two consecutive odd positive integers are 11 and 13. Quick Tip: For problems involving consecutive numbers, define them in terms of a variable and use algebraic equations to solve.

Step 1: Given that \( LM \parallel CB \) and \( LN \parallel CD \), we will use the Basic Proportionality Theorem (Thales' Theorem).

Step 2: According to Thales' Theorem, if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Since \( LM \parallel CB \), in \( \triangle ABM \), we get: \[ \frac{AM}{BM} = \frac{AL}{LC} \]

Similarly, since \( LN \parallel CD \), in \( \triangle AND \), we get: \[ \frac{AN}{DN} = \frac{AL}{LC} \]

Step 3: Equating both ratios, we obtain: \[ \frac{AM}{BM} = \frac{AN}{DN} \]

Thus, the given statement is proved. Quick Tip: Thales' theorem states that a line parallel to one side of a triangle divides the other two sides in the same ratio.

Step 1: The given quadratic polynomial is: \[ x^2 + 7x + 10 = 0 \]

Step 2: Factorizing the quadratic equation: \[ x^2 + 5x + 2x + 10 = 0 \] \[ x(x+5) + 2(x+5) = 0 \]
\[ (x+5)(x+2) = 0 \]

Step 3: Setting each factor to zero: \[ x+5 = 0 \quad \Rightarrow \quad x = -5 \] \[ x+2 = 0 \quad \Rightarrow \quad x = -2 \]

Thus, the zeroes are \( -5 \) and \( -2 \).


Step 4: Verifying the relationship between zeroes and coefficients:

From the general quadratic equation \( ax^2 + bx + c = 0 \), we have:
- Sum of zeroes:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{7}{1} = -7 \]
\[ (-5) + (-2) = -7 \quad (Verified) \]

- Product of zeroes:
\[ \alpha \beta = \frac{c}{a} = \frac{10}{1} = 10 \]
\[ (-5) \times (-2) = 10 \quad (Verified) \]

Thus, the relationships are verified. Quick Tip: For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( -\frac{b}{a} \), and the product of the roots is \( \frac{c}{a} \).

Step 1: Compute the cumulative frequency:
\[ \begin{array}{|c|c|c|} \hline \textbf{Class Interval} & \textbf{Frequency (f)} & \textbf{Cumulative Frequency (CF)}
\hline 0 - 10 & 2 & 2
10 - 20 & 4 & 6
20 - 30 & 7 & 13
30 - 40 & 3 & 16
40 - 50 & 2 & 18
\hline \end{array} \]

Step 2: Find the median class.
The total frequency \( n = 18 \), so \[ \frac{n}{2} = \frac{18}{2} = 9 \]

The cumulative frequency just greater than 9 is 13, corresponding to the class interval 20-30.
Thus, the median class is 20-30.

Step 3: Apply the median formula: \[ Median = L + \left( \frac{\frac{n}{2} - CF}{f} \right) \times h \]

where:
- \( L = 20 \) (lower boundary of median class)
- \( n = 18 \) (total frequency)
- \( CF = 6 \) (cumulative frequency before median class)
- \( f = 7 \) (frequency of median class)
- \( h = 10 \) (class width)

Step 4: Substituting values: \[ Median = 20 + \left( \frac{9 - 6}{7} \right) \times 10 \]
\[ = 20 + \left( \frac{3}{7} \times 10 \right) \]
\[ = 20 + 4.29 \]
\[ = 24.29 \]

Thus, the median is approximately 24.3. Quick Tip: The median class is the first class interval where the cumulative frequency exceeds \( n/2 \).

Step 1: Let the number of blue balls be \( x \).
Total number of balls = \( 5 + x \).

Step 2: Probability of drawing a red ball: \[ P(Red) = \frac{5}{5 + x} \]

Step 3: Probability of drawing a blue ball: \[ P(Blue) = \frac{x}{5 + x} \]

Step 4: Given that \( P(Blue) = 2 \times P(Red) \), \[ \frac{x}{5 + x} = 2 \times \frac{5}{5 + x} \]

Step 5: Solving for \( x \): \[ x = 2 \times 5 \]
\[ x = 10 \]

Thus, the number of blue balls is 10. Quick Tip: If one probability is given as a multiple of another, use proportion equations to solve for the unknown.

Step 1: Let the side lengths of the two squares be \( x \) and \( y \).

Step 2: Using the given sum of areas: \[ x^2 + y^2 = 157 \]

Step 3: Using the given sum of perimeters: \[ 4x + 4y = 68 \]

Step 4: Simplifying the perimeter equation: \[ x + y = 17 \]

Step 5: Express \( y \) in terms of \( x \): \[ y = 17 - x \]

Step 6: Substituting in the area equation: \[ x^2 + (17-x)^2 = 157 \]

Step 7: Expanding: \[ x^2 + 289 - 34x + x^2 = 157 \]
\[ 2x^2 - 34x + 289 = 157 \]

Step 8: Simplifying: \[ 2x^2 - 34x + 132 = 0 \]

Step 9: Dividing by 2: \[ x^2 - 17x + 66 = 0 \]

Step 10: Solving using factorization: \[ (x - 11)(x - 6) = 0 \]

Step 11: Finding values of \( x \): \[ x = 11 \quad or \quad x = 6 \]

Step 12: Finding corresponding values of \( y \):
- If \( x = 11 \), then \( y = 6 \).
- If \( x = 6 \), then \( y = 11 \).

Thus, the sides of the squares are 11 m and 6 m. Quick Tip: For problems involving squares, use the formulas: - Area of a square = \( side^2 \), - Perimeter of a square = \( 4 \times side \).

Step 1: Let the speed of the current be \( x \) km/h.

Step 2: The effective speed of the boat:
- Downstream speed = \( (18 + x) \) km/h
- Upstream speed = \( (18 - x) \) km/h

Step 3: Using the time formula \( Time = \frac{Distance}{Speed} \), \[ Time taken downstream = \frac{24}{18 + x} \] \[ Time taken upstream = \frac{24}{18 - x} \]

Step 4: Given that the upstream journey takes one hour more than the downstream journey: \[ \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \]

Step 5: Taking LCM and solving for \( x \): \[ \frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1 \]
\[ \frac{432 + 24x - 432 + 24x}{324 - x^2} = 1 \]
\[ \frac{48x}{324 - x^2} = 1 \]
\[ 48x = 324 - x^2 \]

Step 6: Rearranging the quadratic equation: \[ x^2 + 48x - 324 = 0 \]

Step 7: Solving using the quadratic formula: \[ x = \frac{-48 \pm \sqrt{(48)^2 - 4(1)(-324)}}{2(1)} \]
\[ x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} \]
\[ x = \frac{-48 \pm \sqrt{3600}}{2} \]
\[ x = \frac{-48 \pm 60}{2} \]

Step 8: Finding values of \( x \): \[ x = \frac{-48 + 60}{2} = \frac{12}{2} = 6 \]
\[ x = \frac{-48 - 60}{2} = \frac{-108}{2} = -54 \quad (Not valid as speed cannot be negative) \]

Thus, the speed of the current is 6 km/h. Quick Tip: For boat and stream problems, use the formula: - Downstream speed = \( Boat speed + Current speed \), - Upstream speed = \( Boat speed - Current speed \).

Step 1: Let the height of the pedestal be \( h \) meters, and let the distance from the observation point to the base of the pedestal be \( d \) meters.

Step 2: Using the tangent formula in right-angled triangles: \[ \tan \theta = \frac{opposite side}{adjacent side} \]

For the pedestal (\( \angle 45^\circ \)): \[ \tan 45^\circ = \frac{h}{d} \] \[ 1 = \frac{h}{d} \] \[ h = d \]

For the statue + pedestal (\( \angle 60^\circ \)): \[ \tan 60^\circ = \frac{h + 1.6}{d} \] \[ \sqrt{3} = \frac{h + 1.6}{d} \] \[ h + 1.6 = \sqrt{3} d \]

Step 3: Substituting \( h = d \): \[ d + 1.6 = \sqrt{3} d \]

Step 4: Solving for \( d \): \[ 1.6 = (\sqrt{3} - 1) d \] \[ d = \frac{1.6}{\sqrt{3} - 1} \]

Rationalizing: \[ d = \frac{1.6 (\sqrt{3} + 1)}{(3 - 1)} = 0.8 (\sqrt{3} + 1) \]

Approximating \( \sqrt{3} \approx 1.732 \): \[ d = 0.8 (1.732 + 1) = 0.8 \times 2.732 = 2.19 m \]

Step 5: Since \( h = d \), the height of the pedestal is 2.19 m. Quick Tip: In height and distance problems, use \( \tan \theta = \frac{height}{base} \) to find unknown distances.

Step 1: Let the height of each pole be \( h \) meters.
Let the distances of the point from one pole be \( x \) meters, so the distance from the other pole is \( 80 - x \) meters.

Step 2: Using the tangent formula:
For the pole observed at \( 60^\circ \): \[ \tan 60^\circ = \frac{h}{x} \] \[ \sqrt{3} = \frac{h}{x} \] \[ h = \sqrt{3} x \]

For the pole observed at \( 30^\circ \): \[ \tan 30^\circ = \frac{h}{80 - x} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \] \[ h = \frac{(80 - x)}{\sqrt{3}} \]

Step 3: Equating both equations for \( h \): \[ \sqrt{3} x = \frac{(80 - x)}{\sqrt{3}} \]

Step 4: Cross-multiplying: \[ 3x = 80 - x \]

Step 5: Solving for \( x \): \[ 4x = 80 \] \[ x = 20 \]

Step 6: Substituting \( x = 20 \) into \( h = \sqrt{3} x \): \[ h = \sqrt{3} \times 20 \] \[ h = 20\sqrt{3} \]

Approximating \( \sqrt{3} \approx 1.732 \): \[ h \approx 20 \times 1.732 = 34.64 m \]

Thus, the height of the poles is 34.64 m and the distance of the point from one pole is 20 m. Quick Tip: When a point is between two objects, set up separate tangent equations and solve using substitution.

Step 1: Given data:
- Diameter of the glass = 5 cm, so radius \( r = \frac{5}{2} = 2.5 \) cm
- Height of the cylindrical part = 10 cm

Step 2: Volume of the cylindrical part: \[ V_{cylinder} = \pi r^2 h \]
\[ = 3.14 \times (2.5)^2 \times 10 \]
\[ = 3.14 \times 6.25 \times 10 \]
\[ = 196.25 cm^3 \]

Step 3: Volume of the hemispherical portion: \[ V_{hemisphere} = \frac{1}{2} \times \frac{4}{3} \pi r^3 \]
\[ = \frac{2}{3} \times 3.14 \times (2.5)^3 \]
\[ = \frac{2}{3} \times 3.14 \times 15.625 \]
\[ = \frac{98.12}{3} = 32.71 cm^3 \]

Step 4: Actual capacity of the glass: \[ V_{actual} = V_{cylinder} - V_{hemisphere} \]
\[ = 196.25 - 32.71 \]
\[ = 163.54 cm^3 \]

Thus,
- The apparent capacity of the glass = 196.25 cm\(^3\)
- The actual capacity of the glass = 163.54 cm\(^3\) Quick Tip: To find the effective volume of a container with a raised portion, subtract the displaced volume from the total volume.

Step 1: Given data:
- Diameter of the top = 3.5 cm, so radius \( r = \frac{3.5}{2} = 1.75 \) cm
- Total height of the top = 5 cm
- Radius of the hemisphere = 1.75 cm

Step 2: Find the height of the conical portion: \[ h_{cone} = 5 - 1.75 = 3.25 cm \]

Step 3: Slant height (\( l \)) of the cone using Pythagoras theorem: \[ l = \sqrt{h^2 + r^2} \]
\[ = \sqrt{(3.25)^2 + (1.75)^2} \]
\[ = \sqrt{10.56 + 3.06} \]
\[ = \sqrt{13.62} \approx 3.69 cm \]

Step 4: Curved surface area of the cone: \[ CSA_{cone} = \pi r l \]
\[ = \frac{22}{7} \times 1.75 \times 3.69 \]
\[ = \frac{22 \times 6.46}{7} \]
\[ = \frac{142.12}{7} \approx 20.3 cm^2 \]

Step 5: Curved surface area of the hemisphere: \[ CSA_{hemisphere} = 2 \pi r^2 \]
\[ = 2 \times \frac{22}{7} \times (1.75)^2 \]
\[ = 2 \times \frac{22}{7} \times 3.06 \]
\[ = \frac{134.64}{7} \approx 19.2 cm^2 \]

Step 6: Total surface area to be coloured: \[ Total area = CSA_{cone} + CSA_{hemisphere} \]
\[ = 20.3 + 19.2 \]
\[ = 39.5 cm^2 \]

Thus, the area Rasheed has to colour is 39.5 cm\(^2\). Quick Tip: The total surface area of a composite shape is found by adding the individual curved surface areas.