Let the length be $l$, width be $b$, and height be $h$. According to the question $l+b+h=x$. Squaring both sides, $l^2+b^2+h^2+2(lb+bh+hl) = x^2$---(i). Given diagonal $y = \sqrt{l^2+b^2+h^2} \implies y^2 = l^2+b^2+h^2$---(ii). Substituting equation (ii) into (i), we get $y^2 + 2(lb+bh+hl) = x^2 \implies 2(lb+bh+hl) = x^2 - y^2$.

Number of coins = Volume of cylinder / Volume of coin. Cylinder ($r=3, h=8$), Coin ($r=0.75=3/4, h=0.2=1/5$). Number = $\frac{\pi(3)^2(8)}{\pi(3/4)^2(1/5)} = \frac{72}{9/16 \times 1/5} = \frac{72 \times 80}{9} = 640$.

Explanation - Volume of water flowed in 30 minutes = $6 \times 1.5 \times 5000$ m = $45000$ cubic meters. Height of water in the field = $8$ cm or $8/100 = 2/25$ m. Area of the irrigated field = $45000 / (2/25) = 45000 \times 25/2 = 562500$ sq meters. $1$ hectare = $10000$ sq meters. Therefore, total area of the field irrigated = $562500 / 10000 = 56.25$ hectares.

Explanation - Volume of cylinder = $\pi r^2 h$. Increase in radius = $20$, so volume increase = $20 + 20 + (20^2/100) = 44$ and increase in height = $20$, so volume increase = $44$.

Radius of sphere $R=4$ cm, $r=2$ cm. Volume of sphere $= \frac{4}{3} \pi (R^3 - r^3) = \frac{4}{3} \pi (64 - 8) = \frac{4}{3} \pi (56)$. Volume of cone $= \frac{1}{3} \pi r_c^2 h = \frac{1}{3} \pi (4^2) h = \frac{16}{3} \pi h$. $\frac{4}{3} \pi (56) = \frac{16}{3} \pi h \Rightarrow 224 = 16h \Rightarrow h = 14$ cm.

Explanation - Curved surface area of cone = $\pi rl$. According to the question, $\pi r_1 l_1 = 3 \pi r_2 l_2$ (1st cone) (2nd cone) $\Rightarrow r_1 l_1 = 3 r_2 l_2 ... (1)$. And $l_2 = 3 l_1 ... (2)$. Putting $l_2 = 3 l_1$ in equation (1) $r_1 l_1 = 3 r_2 (3 l_1) \Rightarrow r_1 = 9 r_2 \Rightarrow r_1/r_2 = 9/1$. Area of cone base = $\pi r^2$. Thus $\pi r_1^2 / \pi r_2^2 = (r_1/r_2)^2 = (9)^2 = 81 : 1$.

Total area of field $= 588 \times 50 = 29400$ sq m. Area of tank $= 30 \times 20 = 600$ sq m. Area of remaining field $= 29400 - 600 = 28800$ sq m. Volume of soil taken out $= 30 \times 20 \times 12 = 7200$ cu m. Height $= 7200 / 28800 = 0.25$ meters $= 25$ cm.

Explanation - Number of cubes = $\frac{\text{Volume of spherical ball}}{\text{Volume of cube}}$ = $\frac{4/3 \pi (21/2)^3}{1^3} = 4/3 \times 22/7 \times (21/2) \times (21/2) \times (21/2) = 4851$

Explanation - Let the side of the cube = $x$ cm. Then its volume = $x^3$ cubic cm, and the diameter of the sphere that fits inside it = $x$ cm. Hence, radius of the sphere = $x/2$ cm. Hence, volume of the sphere = $4/3 \pi (x/2)^3 = 4/3 \pi (x^3/8) = 1/6 \pi x^3$. Therefore, the ratio of volumes of the cube and sphere = $x^3 : 1/6 \pi x^3 = 6 : \pi$

Volume of melted piece $=$ Volume of sphere. $49 \times 33 \times 24 = \frac{4}{3} \pi r^3$. $49 \times 33 \times 24 = \frac{4}{3} \times \frac{22}{7} \times r^3$. Simplifying gives $r^3 = 2744 = 14^3$. Therefore, $r=14$ cm.

Volume of rod ($r=0.5, h=4$) = Volume of wire ($h=900$ cm). $\pi (0.5)^2 \times 4 = \pi r^2 \times 900 \Rightarrow 0.25 \times 4 = 900 r^2 \Rightarrow 1 = 900 r^2 \Rightarrow r = 1/30$. Diameter = $2r = 2/30 = 1/15$ cm.

Explanation- Let side be 100, then after increase it will be 120. Volume = (side)^3. Old volume = 100^3 = 1000000. New volume = 120^3 = 1728000. Increase = 1728000 - 1000000 = 728000. Percentage increase = (728000/1000000) * 100 = 72.8%. Alternative method: 20 + 20 + (2020)/100 = 44%. Then 44 + 20 + (4420)/100 = 64 + 8.8 = 72.8%.

The volume of the given figure (cuboid) = Length $\times$ Breadth $\times$ Height = $3y \times y \times 2y = 6y^3$ cubic units.

Explanation - According to the question, the length of the rectangular paper is 22 cm and width is 4 cm. Since this has been folded to form a cylinder, the area of the rectangle will be the curved surface area of the cylinder. $22 \times 4 = 88$ cm^2 will be the answer.

Given: Volume of brass = 2.2 $dm^3$ = $2.2 \times 1000$ $cm^3$ = 2200 $cm^3$. Radius of cylindrical wire $(r) = \frac{0.50}{2} = 0.25$ cm. Volume of cylindrical wire = Volume of brass $\pi r^2 h = 2200 \implies \frac{22}{7} \times (0.25)^2 \times h = 2200 \implies h = \frac{2200 \times 7}{22 \times 0.0625} = 11200$ cm = 112 m.

Circumference of wheel $= \pi d = \frac{22}{7} \times 84 = 264$ cm. Speed of car $= 72 \times \frac{1000}{60} = 1200$ m/min $= 120000$ cm/min. Distance covered in 11 minutes $= 120000 \times 11 = 1320000$ cm. Number of revolutions $= \frac{1320000}{264} = 5000$.

Side of cube = $a$. Radius of sphere = $a/2$. Ratio = $a^3 : (4/3 \pi (a/2)^3) = a^3 : (\pi a^3 / 6) = 6 : \pi$.

Ratio of volumes $V_1:V_2 = 27:64 \Rightarrow a_1^3:a_2^3 = 27:64 \Rightarrow a_1:a_2 = 3:4$ (ratio of sides). Ratio of surface areas $= a_1^2:a_2^2 = 3^2:4^2 = 9:16$.

Number of cubes $= \frac{\text{Volume of big cube}}{\text{Volume of small cube}} = \frac{15^3}{5^3} = (15/5)^3 = 3^3 = 27$.

Volume = $\pi r^2 h = 1408 \Rightarrow \frac{22}{7} \times r^2 \times 7 = 1408 \Rightarrow 22 r^2 = 1408 \Rightarrow r^2 = 64 \Rightarrow r=8$ m.