A model of a building is constructed using a scale of 1:500. The height of a room in the model is 5 cm long. What is the actual height of the room in metres?

$$\text{Actual height (cm)} = \text{Model height} \times \text{Scale factor}$$ $$\text{Actual height (cm)} = 5 \text{ cm} \times 500$$ $$\text{Actual height (cm)} = 2500 \text{ cm}$$

Since $1 \text{ m} = 100 \text{ cm}$:

$$\text{Actual height (m)} = \frac{2500 \text{ cm}}{100 \text{ cm/m}}$$ $$\text{Actual height (m)} = 25 \text{ m}$$

The correct option is D.
The actual height of the room is $\boxed{\text{25 m}}$.

:
For similar solids, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions (like height).
Let $V_1$ and $h_1$ be the volume and height of the smaller glass, and $V_2$ and $h_2$ be for the larger glass.

$$\frac{V_1}{V_2} = \left(\frac{h_1}{h_2}\right)^3$$

Given: $V_1 = 54 \text{ ml}$, $V_2 = 250 \text{ ml}$, $h_1 = 6 \text{ cm}$.

$$\sqrt[3]{\frac{27}{125}} = \frac{6}{h_2}$$ $$\frac{3}{5} = \frac{6}{h_2}$$

$$3 \times h_2 = 5 \times 6$$ $$3h_2 = 30$$ $$h_2 = \frac{30}{3}$$ $$h_2 = 10 \text{ cm}$$

The correct option is C.
The height of the larger glass is $\boxed{\text{10 cm}}$.

:
The party hats are shaped like a cone. The volume of a cone is given by the formula $V = \frac{1}{3}\pi r^2 h$.
Given: height $h = 21 \text{ cm}$, radius $r = 10 \text{ cm}$.

$$V = \frac{1}{3} \times \pi \times 100 \times 21$$ $$V = \pi \times 100 \times \frac{21}{3}$$ $$V = \pi \times 100 \times 7$$ $$V = 700\pi \text{ cm}^3$$

$$V \approx 700 \times 3.14159$$ $$V \approx 2199.113 \text{ cm}^3$$

Rounding to the nearest whole number, $V \approx 2199 \text{ cm}^3$.
The correct option is B.
The maximum volume of soil each hat could hold is $\boxed{\text{2199 cm}^3}$.

:
This problem forms a right-angled triangle where the flagpole is the opposite side to the angle of elevation, and the shadow is the adjacent side.
Height of flagpole (opposite) = $6 \text{ m}$
Length of shadow (adjacent) = $8 \text{ m}$

$$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$ $$\tan \theta = \frac{6 \text{ m}}{8 \text{ m}}$$ $$\tan \theta = \frac{3}{4}$$ $$\tan \theta = 0.75$$

$$\theta = \arctan(0.75)$$ $$\theta \approx 36.86989^\circ$$

$$\theta \approx 36.9^\circ$$

The correct option is A.
The angle of elevation of the sun is $\boxed{\text{36.9}^\circ}$.

:
The box is a cuboid with dimensions: Length = 70 cm, Width = 30 cm, Height = 40 cm.
The cylindrical tins have: Height = 20 cm, Diameter = 15 cm.

$$\text{Cylinders along length} = \frac{\text{Cuboid length}}{\text{Cylinder diameter}} = \frac{70 \text{ cm}}{15 \text{ cm}} = 4.66...$$

Since we can only fit whole cylinders, this means 4 cylinders can fit along the length.

$$\text{Total cylinders} = (\text{Cylinders along length}) \times (\text{Cylinders along width}) \times (\text{Number of layers})$$ $$\text{Total cylinders} = 4 \times 2 \times 2$$ $$\text{Total cylinders} = 16$$

The correct option is C.
The maximum number of cylinders that can completely fit inside the box is $\boxed{\text{16}}$.

3 zimekamilika, zimebaki 2 leo. Unapiga hatua.