1. **Find the value of x in the given circle with center O and angle 50°**. In a circle, the angle at the center is twice the angle at the circumference subtended by the same arc. Formula: $$\text{Angle at center} = 2 \times \text{Angle at circumference}$$ Given angle at circumference = 50° So, $$x = 2 \times 50 = 100°$$ 2. **Find the quarterly rate to be paid if the annual rate is 8% on a house valued at 60000**. Annual rate = 8% Quarterly rate = $$\frac{8}{4} = 2\%$$ per quarter Quarterly amount = $$60000 \times \frac{2}{100} = 1200$$ 3. **Find the LCM of $$x(x+5)$$ and $$3x^2$$**. Factorize: $$x(x+5) = x(x+5)$$ $$3x^2 = 3 \times x \times x$$ LCM takes highest powers of each factor: LCM = $$3 \times x^2 \times (x+5) = 3x^2(x+5)$$ 4. **Find the value of y in the triangle with angles y° and 35° (isosceles triangle)**. Sum of angles in triangle = 180° Since two sides are equal, two angles are equal. Let the equal angles be y°. So, $$y + y + 35 = 180$$ $$2y = 180 - 35 = 145$$ $$y = \frac{145}{2} = 72.5°$$ 5. **Find the length of the train traveling at 72 km/h crossing a lamp post in 4 seconds**. Speed = 72 km/h = $$72 \times \frac{1000}{3600} = 20$$ m/s Length = speed \times time = $$20 \times 4 = 80$$ meters 6. **Find $$P(A \cap A')$$ given $$P(A) = \frac{2}{5}$$**. Since $$A'$$ is the complement of $$A$$, $$A \cap A' = \emptyset$$ (empty set) So, $$P(A \cap A') = 0$$ 7. **Solve $$x^2 - x - 90 = 0$$ for $$x < 0$$**. Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1, b=-1, c=-90$$ $$x = \frac{1 \pm \sqrt{1 + 360}}{2} = \frac{1 \pm 19}{2}$$ Solutions: $$x = 10$$ or $$x = -9$$ Since $$x < 0$$, $$x = -9$$ 8. **Find the value of x in triangle with sides 2x, 2x, and base 18 cm**. Since two sides are equal, triangle is isosceles. By triangle inequality: $$2x + 2x > 18 \Rightarrow 4x > 18 \Rightarrow x > 4.5$$ No other info given, so value of $$x$$ must satisfy $$x > 4.5$$. 9. **Find the common ratio of geometric progression 72, 48, 32...**. Common ratio $$r = \frac{48}{72} = \frac{2}{3}$$ 10. **Find curved surface area of cylinder made from square of perimeter 56 cm**. Square perimeter = 56 cm Side length $$s = \frac{56}{4} = 14$$ cm Cylinder height = side length = 14 cm Cylinder circumference = side length = 14 cm Radius $$r = \frac{14}{2\pi} = \frac{7}{\pi}$$ cm Curved surface area $$= 2\pi r h = 2\pi \times \frac{7}{\pi} \times 14 = 2 \times 7 \times 14 = 196$$ cm² 11. **Find length of BD in quadrilateral with triangles ABC and BDE**. Triangles ABC and BDE are similar (by angle-side-angle or given info). Sides AB=5, AC=3, BE=12, DE=y, BD=x From similarity: $$\frac{AB}{BE} = \frac{AC}{DE} = \frac{BC}{BD}$$ Using $$\frac{5}{12} = \frac{3}{y}$$, solve for $$y$$: $$y = \frac{3 \times 12}{5} = 7.2$$ Using $$\frac{5}{12} = \frac{x}{BC}$$, but BC not given, so insufficient info to find BD. 12. **Simplify $$\frac{5x}{2(x-1)} - \frac{3}{1-x}$$**. Note $$1-x = -(x-1)$$ Rewrite second term: $$\frac{3}{1-x} = \frac{3}{-(x-1)} = -\frac{3}{x-1}$$ Expression becomes: $$\frac{5x}{2(x-1)} - \left(-\frac{3}{x-1}\right) = \frac{5x}{2(x-1)} + \frac{3}{x-1}$$ Find common denominator $$2(x-1)$$: $$\frac{5x}{2(x-1)} + \frac{3 \times 2}{2(x-1)} = \frac{5x + 6}{2(x-1)}$$ 13. **Find equation of line passing through (0, -2) and parallel to line $$3y - 6x = 5$$**. Rewrite given line: $$3y = 6x + 5 \Rightarrow y = 2x + \frac{5}{3}$$ Slope $$m = 2$$ Equation of line parallel with slope 2 passing through (0, -2): $$y - (-2) = 2(x - 0) \Rightarrow y + 2 = 2x \Rightarrow y = 2x - 2$$ 14. **Find base diameter of cone made from sector of radius 25 cm with height 24 cm**. Slant height $$l = 25$$ cm (radius of sector) Height $$h = 24$$ cm Use Pythagoras to find radius $$r$$: $$r = \sqrt{l^2 - h^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7$$ cm Base diameter $$= 2r = 14$$ cm