Step 1: Calculate the selling price.

\[ \text{Selling Price} = \text{Cost Price} + \text{Profit} \]

\[ = 8,000 + (15\% \times 8,000) \]

\[ = 8,000 + \frac{15}{100} \times 8,000 \]

\[ = 8,000 + 1,200 \]

\[ = 9,200 \]

Step 2: Calculate the invested amount.

\[ \text{Invested Amount} = \text{Selling Price} - \text{Amount Spent} \]

\[ = 9,200 - 4,550 \]

\[ = 4,650 \]

Step 3: Use the simple interest formula.

\[ \text{Simple Interest} = \frac{P \times R \times T}{100} \]

\[ P = 4,650, R = 7\%, T = 4 \text{ years} \]

\[ \text{Interest} = \frac{4,650 \times 7 \times 4}{100} \]

\[ = \frac{130,200}{100} \]

\[ = 1,302 \]

Final Answer: The interest earned in 4 years is \$1,302.00.

Step 1: Use the logarithmic property: \( \log a + \log b = \log (a \cdot b) \).

\[ \log(3x - 1) + \log 2 = \log [2(3x - 1)] \]

Step 2: Rewrite the right-hand side using the logarithmic exponent rule.

\[ \log [2(3x - 1)] = 2\log y \]

Step 3: Remove the logarithms by rewriting the equation in exponential form.

\[ 2(3x - 1) = y^2 \]

Step 4: Solve for \( x \).

\[ 3x - 1 = \frac{y^2}{2} \]

\[ 3x = \frac{y^2}{2} + 1 \]

\[ x = \frac{y^2 + 2}{6} \]

Final Answer: \( x = \frac{y^2 + 2}{6} \).

Step 1: Let the original cost of a chair be x and a table be y.

From the first purchase:

\[ 2x + 3y = 1800 \quad \text{(Equation 1)} \]

Step 2: After the 20% increase:

The new cost of a chair = 1.2x, the new cost of a table = 1.2y

From the second purchase:

\[ 6(1.2x) + 2(1.2y) = 4800 \]

\[ 7.2x + 2.4y = 4800 \quad \text{(Equation 2)} \]

Step 3: Solve the two equations:

Multiply Equation 1 by 2.4: \[ 4.8x + 7.2y = 4320 \] Subtract:

\[ (7.2x + 2.4y) - (4.8x + 7.2y) = 480 \]

New chair cost: 1.2 × 600 = 720.00

New table cost: 1.2 × 200 = 240.00

Step 1: Calculate arc length of the sector.

\[ \text{Arc length} = \frac{\theta}{360^\circ} \times 2 \pi r \] \[= \frac{126}{360} \times 2 \times \frac{22}{7} \times 10 \]

\[ = \frac{7}{20} \times \frac{440}{7} = 22 \, \text{cm} \]

Step 2: Arc length becomes the circumference of the cone’s base.

\[ 2 \pi r_{\text{base}} = 22 \quad \Rightarrow \] \[ \quad r_{\text{base}} = \frac{22}{2 \times \frac{22}{7}} \] \[= \frac{22 \times 7}{44} = 3.5 \, \text{cm} \]

Step 3: Surface area of the cone.

The surface area is the lateral area of the cone (no base).

\[ \text{Surface Area} = \pi r_{\text{base}} l \] \[= \frac{22}{7} \times 3.5 \times 10 = 110 \, \text{cm}^2 \]

Final Answer: The surface area is 110 cm².

Step 1: Find the height using Pythagoras' Theorem.

\[ h = \sqrt{l^2 - r_{\text{base}}^2} \] \[= \sqrt{10^2 - 3.5^2} \] \[= \sqrt{100 - 12.25} \] \[= \sqrt{87.75} \approx 9.37 \, \text{cm} \]

Step 2: Calculate the volume.

\[ V = \frac{1}{3} \pi r_{\text{base}}^2 h \] \[= \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 9.37 \]

\[ = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 9.37 \] \[ \approx \frac{1}{3} \times 38.5 \times 9.37 \] \[ \approx \frac{1}{3} \times 360.745 \] \[\approx 120.25 \, \text{cm}^3 \]

Final Answer: The volume of the cone is approximately 120.25 cm³.

Step 1: Use the sum of interior angles formula:

\[ S = (n - 2) \times 180 \]

Let the number of sides of the first polygon be \( n \), and the second polygon be \( 2n - 3 \).

From the given sum:

\[ (n - 2) \times 180 + ((2n - 3) - 2) \times 180 = 2520 \]

Step 2: Expand the equation:

\[ 180(n - 2) + 180(2n - 5) = 2520 \]

\[ 180n - 360 + 360n - 900 = 2520 \]

\[ 540n - 1260 = 2520 \]

Step 3: Solve for \( n \):

\[ 540n = 3780 \]

\[ n = \frac{3780}{540} = 7 \]

Step 4: Find the number of sides of the second polygon:

\[ 2n - 3 = 2(7) - 3 = 11 \]

Final Answer: The polygons have 7 sides and 11 sides, respectively.

The mean is given by:

\[ \text{Mean} = \frac{\sum \text{Scores}}{\text{Total Count}} \]

\[ = \frac{2 + 6 + 4 + 7 + 6 + 8 + 9 + 5 + 3 + 5 + 4 + 2 + 8 + 7 + 6}{15} \]

\[ = \frac{82}{15} = 5.47 \]

Final Answer: The mean of the scores is 5.47.

List the even numbers in the set: 2, 6, 4, 6, 8, 4, 2, 8, 6.

Number of even numbers = 9.

\[ P(\text{even}) = \frac{9}{15} = \frac{3}{5} \]

Final Answer: \( P(\text{even}) = \frac{3}{5} \).

List the scores that are at least 7: 7, 8, 9, 7, 8.

Number of scores at least 7 = 5.

\[ P(\text{at least } 7) = \frac{5}{15} = \frac{1}{3} \]

Final Answer: \( P(\text{at least } 7) = \frac{1}{3} \).

Step 1: Relationship in a G.P.

In a G.P., the ratio of consecutive terms is constant. Let the first term be \( a \) and the common ratio be \( r \).
Second term: \( ar = 2x - 4 \)
Fourth term: \( ar^3 = 13x + 4 \)
Sixth term: \( ar^5 = 122x - 4 \)

Step 2: Form ratios.

\[ \frac{ar^3}{ar} = r^2 = \frac{13x + 4}{2x - 4} \] \[ \frac{ar^5}{ar^3} = r^2 = \frac{122x - 4}{13x + 4} \]

Step 3: Equate both expressions for \( r^2 \).

\[ \frac{13x + 4}{2x - 4} = \frac{122x - 4}{13x + 4} \]

Step 4: Cross-multiply.

\[ (13x + 4)^2 = (2x - 4)(122x - 4) \]

Step 5: Expand both sides.

\[ (169x^2 + 104x + 16) \] \[= (244x^2 - 8x - 488x + 16) \] \[ 169x^2 + 104x + 16 \] \[= 244x^2 - 496x + 16 \]

Step 6: Bring all terms to the left.

\[ 169x^2 + 104x + 16 - 244x^2 + 496x - 16 = 0 \]

\[ -75x^2 + 600x = 0 \] \[ -75x(x - 8) = 0 \]

Step 7: Solve for x.

\[ x = 0 \quad \text{or} \quad x = 8 \] x = 0 is not meaningful in this context, so:
Final Answer: \( x = 8 \)

Step 1: Substitute \( x = 8 \) into second and fourth terms.

Second term: \( 2(8) - 4 = 12 \)
Fourth term: \( 13(8) + 4 = 108 \)

Step 2: Calculate common ratio \( r \).

\[ r^2 = \frac{108}{12} = 9 \quad \Rightarrow \quad r = 3 \]

Final Answer: The common ratio is 3.

Step 1: Use the second term formula.

\[ ar = 12 \quad \Rightarrow \] \[\quad a \times 3 = 12 \quad \Rightarrow \quad a = 4 \]

Final Answer: The first term is 4.

Step 1: Calculate the cost per bottle.

\[ \text{Cost per bottle} = \frac{\text{Total cost}}{\text{Number of bottles}} \]

\[ = \frac{240}{60} = 4.00 \]

Step 2: Find the profit per bottle.

\[ \text{Profit per bottle} = \text{Selling price} - \text{Cost price} \]

\[ = 5.50 - 4.00 = 1.50 \]

Step 3: Calculate the percentage profit.

\[ \text{Percentage Profit} = \frac{\text{Profit per bottle}}{\text{Cost price}} \times 100 \]

\[ = \frac{1.50}{4.00} \times 100 \]

\[ = 37.5\% \]

Final Answer: The percentage profit on each bottle is 37.5%.

Step 1: Find revenue from 45 bottles sold at full price.

\[ \text{Revenue} = 45 \times 5.50 = 247.50 \]

Step 2: Find revenue from 15 discounted bottles.

Discounted price per bottle:

\[ \text{Discounted price} = 5.50 \times (1 - 0.20) \]

\[ = 5.50 \times 0.80 = 4.40 \]

\[ \text{Revenue from discounted bottles} \] \[= 15 \times 4.40 = 66.00 \]

Step 3: Find total revenue.

\[ \text{Total Revenue} = 247.50 + 66.00 = 313.50 \]

Step 4: Find total profit.

\[ \text{Total Profit} = \text{Total Revenue} - \text{Total Cost} \]

\[ = 313.50 - 240.00 \]

\[ = 73.50 \]

Final Answer: The total profit made on the 60 bottles is \$73.50.

Step 1: Identify the middle terms.

Since there are 10 values, the median is the average of the 5th and 6th terms.

The 5th term is \( x + 4 \) and the 6th term is \( 2x^2 - 1 \).

Median: \[ \frac{(x + 4) + (2x^2 - 1)}{2} = 6.5 \]

Step 2: Solve for x.

\[ \frac{x + 4 + 2x^2 - 1}{2} = 6.5 \] \[ \frac{2x^2 + x + 3}{2} = 6.5 \] \[ 2x^2 + x + 3 = 13 \] \[ 2x^2 + x - 10 = 0 \]

Step 3: Solve the quadratic equation.

\[ (2x - 4)(x + 2.5) = 0 \] \[ x = 2 \quad \text{or} \quad x = -2.5 \]

Since the values must make sense in the ordered data, we accept x = 2.

Final Answer: x = 2.

Step 1: Substitute x = 2 into all variable terms.

The scores become: 3, 4, 4, 5, 6, 7, 8, 8, 9, 10.

Step 2: Calculate the total sum.

\[ \text{Sum} = 3 + 4 + 4 + 5 + 6 + 7 + 8 + 8 + 9 + 10 = 64 \]

Step 3: Calculate the mean.

\[ \text{Mean} = \frac{64}{10} = 6.4 \]

Final Answer: The mean mark is 6.4.

Step 1: Identify the scores that are ≥ 8.

Pass marks: 8, 8, 9, 10 (4 students)

Step 2: Calculate the percentage.

\[ \text{Percentage Passed} = \frac{4}{10} \times 100\% \] \[= 40\% \]

Final Answer: 40% of the applicants passed the interview.

Step 1: Define variables and set up two right-angled triangles.

• Let the height of the object above the ground be \( h \).
• Let the initial horizontal distance from P to the point directly beneath K be \( x \) m.

At Point P:

\[ \tan 32^\circ = \frac{h - 1.62}{x} \] \[ h - 1.62 = x \times \tan 32^\circ \quad \] \[\Rightarrow \quad h = x \times \tan 32^\circ + 1.62 \]

At Point Q:

Distance to the object is now \( x - 32 \) m. \[ \tan 42^\circ = \frac{h - 1.62}{x - 32} \] \[ h - 1.62 = (x - 32) \times \tan 42^\circ \quad \]

\[ \quad h = (x - 32) \times \tan 42^\circ + 1.62 \]

Step 2: Set both expressions for h equal.

\[ x \times \tan 32^\circ + 1.62 = (x - 32) \times \tan 42^\circ + 1.62 \]

\[ x \times \tan 32^\circ = (x - 32) \times \tan 42^\circ \]

\[ x \times \tan 32^\circ = x \times \tan 42^\circ - 32 \times \tan 42^\circ \]

\[ x ( \tan 32^\circ - \tan 42^\circ ) = -32 \times \tan 42^\circ \]

\[ x = \frac{ -32 \times \tan 42^\circ }{ \tan 32^\circ - \tan 42^\circ } \]

Step 3: Calculate the values.

\[ \tan 32^\circ \approx 0.6249 \quad \] \[\text{and} \quad \tan 42^\circ \approx 0.9004 \] \[ x = \frac{ -32 \times 0.9004 }{ 0.6249 - 0.9004 } \] \[= \frac{ -28.8128 }{ -0.2755 } \approx 104.6 \, \text{m} \]

Step 4: Find h.

\[ h = 104.6 \times 0.6249 + 1.62 \] \[\approx 65.42 + 1.62 \] \[\approx 67.04 \, \text{m} \]

Final Answer: The perpendicular distance from K to the ground is 67 m (to the nearest whole number).

Step 1: Use the perimeter to find the radius.

\[ \text{Perimeter} = OA + OB + AB \] \[= 2r + 24.2 = 52.2 \] \[ 2r = 52.2 - 24.2 = 28 \quad \] \[\Rightarrow \quad r = 14 \, \text{cm} \]

Step 2: Use the cosine rule to find \( \angle AOB \).

\[ \cos(\angle AOB) = \frac{OA^2 + OB^2 - AB^2}{2 \cdot OA \cdot OB} \]

\[ = \frac{14^2 + 14^2 - 24.2^2}{2 \cdot 14 \cdot 14} \] \[ = \frac{392 - 585.64}{392} \] \[= \frac{-193.64}{392} \approx -0.493 \]

Step 3: Calculate the angle using inverse cosine:

\[ \angle AOB = \cos^{-1}(-0.493) \] \[\approx 118.5^\circ \Rightarrow \boxed{119^\circ} \]

Final Answer: \( \angle AOB \approx \boxed{119^\circ} \)

Step 1: Find the slope (gradient) of the line passing through (1, 4) and (3, 0).

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \] \[= \frac{0 - 4}{3 - 1} = \frac{-4}{2} = -2 \]

Step 2: Use point-slope form to find the equation.

\[ y - y_1 = m(x - x_1) \Rightarrow \] \[y - 4 = -2(x - 1) \] \[ y = -2x + 2 + 4 = -2x + 6 \]

Final Answer: The equation of the line is \( \boxed{y = -2x + 6} \)

Step 1: Use triangle BED to find \( \angle EBD \).

Since \( BE = DE \), triangle \( BED \) is isosceles.
Therefore, \( \angle EBD = \angle EDB \)

\[ \angle EBD = \angle EDB \] \[ = \frac{180^\circ - 40^\circ}{2} = 70^\circ \]

Step 2: Find \( \angle DBC \) using angle subtraction.

\[ \angle DBC = \angle EBD - \angle CBD \] \[= 70^\circ - 50^\circ = 20^\circ \]

Step 3: Use the circle theorem:

\( \angle BAD \) is equal to \( \angle DBC \) because they subtend the same arc \( BD \).
So, \[ \angle BAD = \angle DBC = \boxed{20^\circ} \]

Final Answer: \( \angle BAD = \boxed{20^\circ} \)

Step 1: Let the length = 5x and the width = 2x.

Step 2: Use the perimeter formula:

\[ P = 2(\text{length} + \text{width}) = 70 \] \[ 2(5x + 2x) = 70 \] \[ 2(7x) = 70 \Rightarrow \] \[14x = 70 \Rightarrow x = 5 \]

Step 3: Calculate actual length and width:

\[ \text{Length} = 5x = 5 \times 5 = 25 \, \text{cm} \] \[ \text{Width} = 2x = 2 \times 5 = 10 \, \text{cm} \]

Step 4: Calculate area:

\[ \text{Area} = \text{Length} \times \text{Width} \] \[= 25 \times 10 = 250 \, \text{cm}^2 \]

Final Answer: The area of the rectangle is 250 cm².

Step 1: Use the probability of selecting fruit C.

\[ \text{P(C)} = \frac{3}{10} \Rightarrow \] \[\text{Number of C} = \frac{3}{10} \times 20 \] \[= 6 \]

Step 2: Let the number of fruit A be \( x \). Then fruit B is:

\[ \text{B} = 2x - 1 \]

Step 3: Total number of fruits:

\[ A + B + C = 20 \Rightarrow \] \[x + (2x - 1) + 6 = 20 \] \[ 3x + 5 = 20 \] \[ 3x = 15 \] \[ x = 5 \]

Step 4: Find each value:

\[ \text{A} = 5, \] \[\quad \text{B} = 2(5) - 1 = 9, \] \[\quad \text{C} = 6 \]

Final Answer: A = 5, B = 9, C = 6.

Step 1: Use log rules to combine the terms.

\[ 2\log_{10}x - \log_{10}9 = 4 \] \[ \log_{10}(x^2) - \log_{10}(9) = 4 \] \[ \log_{10}\left(\frac{x^2}{9}\right) = 4 \]

Step 2: Rewrite in exponential form.

\[ \frac{x^2}{9} = 10^4 = 10000 \] \[ x^2 = 10000 \times 9 = 90000 \]

Step 3: Take square root of both sides.

\[ x = \sqrt{90000} = 300 \]

Final Answer: \( x = \mathbf{300} \)

Step 1: Calculate \( y \) values for integer \( x \) from -3 to 3:

• \(x = -3, y = 17\)
• \(x = -2, y = 6\)
• \(x = -1, y = -1\)
• \(x = 0, y = -4\)
• \(x = 1, y = -3\)
• \(x = 2, y = 6\)
• \(x = 3, y = 11\)

Step 2:ploted Graph:

Final note: The graph is a parabola opening upwards with vertex near \(x = \frac{1}{4}\).

(i) Roots of the equation:

The roots are the x-values where the curve intersects the x-axis.

From the graph, these roots are approximately:

\[ x \approx -1.5 \quad \text{and} \quad x \approx 1.5 \]

(ii) Range where y decreases as x increases:

The graph decreases for values of \( x \) less than the vertex.

This happens when:

\[ x < 0.25 \]

(iii) Minimum point:

The minimum point is the vertex of the parabola, which appears to be:

\[ (0.25, -4.125) \]

Final Summary:

• Roots: \( x \approx -1.5 \) and \( x \approx 1.5 \)
• Decreasing Interval: \( x < 0.25 \)
• Minimum Point: \( (0.25, -4.125) \)