The box is a closed right square prism, so first find its total surface area.

Area of the two bases:

\(2\cdot 40^2=2\cdot 1600=3200\text{ cm}^2\).

The lateral surface area of the prism equals the perimeter of the base times the height:

\(P_{\text{base}}\cdot h=4\cdot 40\cdot 3=160\cdot 3=480\text{ cm}^2\).

Total surface area of the box:

\(3200+480=3680\text{ cm}^2\).

Now add \(10\%\) of cardboard for fastenings:

\(3680\cdot 1.1=4048\text{ cm}^2\).

Answer: \(4048\text{ cm}^2\).

The volume of water in the first aquarium is:

\(V=60\cdot 40\cdot 30=72000\text{ cm}^3\).

In the second aquarium, the base area is:

\(80\cdot 50=4000\text{ cm}^2\).

Let the water height in the second aquarium be \(h\). Then

\(4000h=72000\).

Hence

\(h=\dfrac{72000}{4000}=18\text{ cm}\).

Answer: \(18\text{ cm}\).

The front surface does not need painting because it consists of glass doors.

So we paint the back, the two side faces, the top, and the bottom.

Back face:

\(2\cdot 1=2\text{ m}^2\).

Two side faces:

\(2\cdot (2\cdot 0.5)=2\cdot 1=2\text{ m}^2\).

Top and bottom:

\(2\cdot (1\cdot 0.5)=2\cdot 0.5=1\text{ m}^2\).

Total painted area:

\(2+2+1=5\text{ m}^2\).

Amount of paint needed:

\(5\cdot 100=500\text{ g}\).

Answer: \(500\text{ g}\).

One wrap along the length of the box forms a rectangle with sides \(30\text{ cm}\) and \(10\text{ cm}\). Its perimeter is:

\(2(30+10)=80\text{ cm}\).

There are two such wraps, so this requires:

\(2\cdot 80=160\text{ cm}\).

One wrap along the width forms a rectangle with sides \(25\text{ cm}\) and \(10\text{ cm}\). Its perimeter is:

\(2(25+10)=70\text{ cm}\).

There are two such wraps, so this requires:

\(2\cdot 70=140\text{ cm}\).

The bow requires an additional \(25\text{ cm}\).

Total:

\(160+140+25=325\text{ cm}=3.25\text{ m}\).

Answer: \(3.25\text{ m}\).

After cutting squares from the corners, the height of the box will be equal to the side of the cut square:

\(h=20\text{ cm}\).

The side of the base decreases by \(20\text{ cm}\) from each side, that is by \(40\text{ cm}\) in total:

\(a=80-20-20=40\text{ cm}\).

The base of the box is a square with side \(40\text{ cm}\), so its area is:

\(40\cdot 40=1600\text{ cm}^2\).

The volume of the box is:

\(V=1600\cdot 20=32000\text{ cm}^3\).

Answer: \(32000\text{ cm}^3\).

The water depth must be \(10\text{ cm}=0.1\text{ m}\) less than the pool depth:

\(0.5-0.1=0.4\text{ m}\).

The volume of water is:

\(V=2\cdot 1.5\cdot 0.4=1.2\text{ m}^3\).

Convert to liters: \(1\text{ m}^3=1000\text{ l}\), hence

\(1.2\text{ m}^3=1200\text{ l}\).

One bucket holds \(10\text{ l}\), so the number of buckets needed is:

\(\dfrac{1200}{10}=120\).

Answer: \(120\).

The thickness of the glaze is \(2\text{ mm}=0.2\text{ cm}\).

First, find the area of the surface covered with glaze.

The base of the prism is an isosceles triangle with sides \(10,10,12\). The altitude of this triangle to the side \(12\) is:

\(h_{\triangle}=\sqrt{10^2-6^2}=\sqrt{100-36}=8\text{ cm}\).

Area of one triangular base:

\(S_{\triangle}=\dfrac{12\cdot 8}{2}=48\text{ cm}^2\).

Two triangular bases give:

\(2\cdot 48=96\text{ cm}^2\).

The lateral rectangular faces have areas \(10\cdot 15\), \(10\cdot 15\), and \(12\cdot 15\). The bottom is not covered with glaze, so the rectangle \(12\cdot 15\) is excluded.

This leaves:

\(10\cdot 15+10\cdot 15=300\text{ cm}^2\).

Thus the coated area is:

\(96+300=396\text{ cm}^2\).

The volume of glaze is approximately the coated area multiplied by the thickness:

\(V=396\cdot 0.2=79.2\text{ cm}^3\).

Answer: \(79.2\text{ cm}^3\).

Consider the cross-section of the greenhouse — an isosceles triangle with base \(6\text{ m}\) and height \(4\text{ m}\).

Half of the base is \(3\text{ m}\), so the equal side of the triangle is:

\(\sqrt{3^2+4^2}=5\text{ m}\).

The transparent roof consists of two equal rectangles, each with one side \(5\text{ m}\) and the other side equal to the length of the greenhouse \(L\).

Therefore, the roof area is:

\(2\cdot 5\cdot L=200\).

Hence \(10L=200\Rightarrow L=20\text{ m}\).

The area of the triangular base is:

\(S_{\triangle}=\dfrac{6\cdot 4}{2}=12\text{ m}^2\).

The volume of the prism is:

\(V=12\cdot 20=240\text{ m}^3\).

Answer: \(240\text{ m}^3\).

Only the lateral surface of the prism must be painted.

First find the hypotenuse of the triangular base:

\(c=\sqrt{12^2+16^2}=\sqrt{144+256}=20\text{ cm}\).

The perimeter of the base is:

\(12+16+20=48\text{ cm}\).

The beam length is \(2\text{ m}=200\text{ cm}\).

The lateral surface area is:

\(S_{\text{lat}}=48\cdot 200=9600\text{ cm}^2\).

Convert to square meters:

\(9600\text{ cm}^2=0.96\text{ m}^2\).

Therefore, the amount of paint needed is:

\(0.96\cdot 100=96\text{ ml}\).

Answer: \(96\text{ ml}\).

Along the length of the crate, we can place:

\(\dfrac{60}{10}=6\) boxes.

Along the width:

\(\dfrac{40}{10}=4\) boxes.

Along the height:

\(\dfrac{40}{10}=4\) boxes.

So in total:

\(6\cdot 4\cdot 4=96\).

Answer: \(96\).

The water level is \(20\text{ cm}=0.2\text{ m}\) below the rim, so the actual water depths are:

\(1.2-0.2=1.0\text{ m}\) and \(2.2-0.2=2.0\text{ m}\).

Since the bottom changes uniformly, the volume can be found as the area of the pool surface multiplied by the average water depth.

Average depth:

\(\dfrac{1.0+2.0}{2}=1.5\text{ m}\).

Pool surface area:

\(16\cdot 8=128\text{ m}^2\).

Therefore, the volume of water is:

\(V=128\cdot 1.5=192\text{ m}^3\).

Answer: \(192\text{ m}^3\).

The base of the cake is a circle of radius \(10\text{ cm}\), so the diameter of the cake is \(20\text{ cm}\).

The box must be \(2\text{ cm}\) larger in each dimension, so the length and width of the box are:

\(20+2=22\text{ cm}\).

The height of the box is:

\(8+2=10\text{ cm}\).

Therefore, the volume of the rectangular prism is:

\(V=22\cdot 22\cdot 10=4840\text{ cm}^3\).

Answer: \(4840\text{ cm}^3\).

The volume of the small aquarium is:

\(V_{\text{small}}=40\cdot 20\cdot 20=16000\text{ cm}^3\).

According to the condition, the volume of the large aquarium is twice as large:

\(V_{\text{large}}=2\cdot 16000=32000\text{ cm}^3\).

Let its width be \(x\). Then

\(50\cdot x\cdot 32=32000\).

\(1600x=32000\Rightarrow x=20\text{ cm}\).

Answer: \(20\text{ cm}\).

Convert the dimensions to decimeters, since \(1\text{ l}=1\text{ dm}^3\):

\(5\text{ m}=50\text{ dm},\quad 80\text{ cm}=8\text{ dm},\quad 30\text{ cm}=3\text{ dm}\).

The volume of the flowerbed is:

\(V=50\cdot 8\cdot 3=1200\text{ dm}^3=1200\text{ l}\).

One bag contains \(20\text{ l}\), so the number of bags needed is:

\(\dfrac{1200}{20}=60\).

Answer: \(60\).

Ceiling area:

\(8\cdot 6=48\text{ m}^2\).

Total wall area:

\(2\cdot 8\cdot 3+2\cdot 6\cdot 3=48+36=84\text{ m}^2\).

So the ceiling and walls together give:

\(48+84=132\text{ m}^2\).

The window and door are not painted, their total area is:

\(4+2=6\text{ m}^2\).

Area to be painted for one coat:

\(132-6=126\text{ m}^2\).

Since there are two coats, the effective painted area for paint consumption is:

\(126\cdot 2=252\text{ m}^2\).

The amount of paint needed is:

\(252\cdot 100=25200\text{ g}=25.2\text{ kg}\).

Answer: \(25.2\text{ kg}\).

Floor area:

\(2\cdot 1.5=3\text{ m}^2\).

Wall area:

\(2\cdot 2\cdot 2.5+2\cdot 1.5\cdot 2.5=10+7.5=17.5\text{ m}^2\).

Total area of floor and walls:

\(3+17.5=20.5\text{ m}^2\).

The door of area \(2\text{ m}^2\) is not tiled, so the remaining area is:

\(20.5-2=18.5\text{ m}^2\).

Area of one tile:

\(20\text{ cm}\times 25\text{ cm}=0.2\text{ m}\times 0.25\text{ m}=0.05\text{ m}^2\).

Number of tiles:

\(\dfrac{18.5}{0.05}=370\).

Answer: \(370\).