Maths M.C.Q

3. 64cm³

Solution:

We know that,

Total surface area of a cube = 6a²

⇒ 96 = 6a²

$\Rightarrow\text{a}^2=\frac{96}{6}$

⇒ a² = 16

⇒ a = 4cm

Now,

Volume of the cube = a³

= 4³

= 64cm³

3. 550cm²

Solution:

We know that,

Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$

$\Rightarrow1232=\frac{1}{3}\pi\text{r}^2\text{h}$

$\Rightarrow1232=\frac{1}{3}\times\frac{22}{7}\times\text{r}^2\times24$

$\Rightarrow\text{r}^2=\frac{7\times3\times1232}{24\times22}$

$\Rightarrow\text{r}^2=49$

$\Rightarrow\text{r}=7\text{cm}$

⇒ r = 7cm and h = 24cm

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = 25

Curved surface area $=\pi\text{rl}$

$=\frac{22}{7}\times7\times25$

$=550\text{cm}^2$

3. 36kg

​​​​​​​Solution:

Volume of the beam = length × breadth × height

$=9\times\frac{40}{100}\times\frac{20}{100}$

$...\Big(1\text{cm}=\frac{1}{100}\text{m}\\\Rightarrow40\text{cm}=\frac{40}{100}\text{m}\ \text{and}\ 20\text{cm}=\frac{20}{100}\text{m}\Big)$

$=\frac{18}{25}\text{cm}^3$

Weight of the beam $=\frac{18}{25}\times50$

$=36\text{kg}$

2. 396cm³

Solution:

Given, d = 6cm

r = 3cm

h = 14cm

Now,

Volume of the cylinder $=\pi\text{r}^2\text{h}$

$=\frac{22}{7}\times3\times3\times14$

$=396\text{cm}^3$

2. 4851cm³

Solution:

Given radius $=10.5=\frac{21}{2}\text{cm}$

Volume of sphere $=\frac{4}{3}\pi\text{r}^3$

$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$

$=11\times21\times21$

$=4851\text{cm}^3$

1. 480

​​​​​​​Solution:

Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$

$=\frac{(20\times100)\times(6\times100)\times50}{(5\times100)\times25\times10}$ $...(1\text{m}=100\text{cm})$

$=\text{480}$

4. 770cm²

Solution:

Let the radius be 2x and the height be 3x cm.

We know that,

Volume of the cylinder $=\pi\text{r}^2\text{h}$

$=\pi\times(2\text{x})^2\times3\text{x}$

$=\frac{22}{7}\times12\times\text{x}^3$

$\Rightarrow1617=\frac{22}{7}\times12\text{x}^3$

$\Rightarrow\text{x}^3=\frac{7\times1617}{22\times12}$

$\Rightarrow\text{x}^3=\frac{7\times49}{2\times4}$

$\Rightarrow\text{x}^3=\frac{343}{8}$

$\Rightarrow\text{x}^3=\Big(\frac{7}{2}\Big)^3$

$\Rightarrow\text{x}=\frac{7}{2}$

$\therefore$ radius $=2\times\frac{7}{2}=7\text{cm}$

Height $=3\times\frac{7}{2}=\frac{21}{2}\text{cm}$

$\therefore$ total surface area $=2\pi\text{r}(\text{h}+\text{r})$

$=2\times\frac{22}{7}\times7\Big(\frac{21}{2}+7\Big)$

$=44\Big(\frac{21}{2}+7\Big)$

$=44\Big(\frac{35}{2}\Big)$

$=770\text{cm}^2$

2. 396cm³

Solution:

Given,

Height = 14cm

Curved surface area = 264cm²

Now,

Curved surface area $=2\pi\text{rh}$

$\Rightarrow264=2\times\frac{22}{7}\times\text{r}\times14$

$\Rightarrow264=88\times\text{r}$

$\Rightarrow\text{r}=\frac{264}{88}$

$\Rightarrow\text{r}=3\text{cm}$

Volume $=\pi\text{r}^2\text{h}$

$=\frac{22}{7}\times3\times3\times14$

$=396\text{cm}^3$

2. 384cm²

Solution:

We know that,

Length of the longest diagonal $=\sqrt{3}\text{a}$

$\Rightarrow8\sqrt3=\sqrt{3}\text{a}$

$\Rightarrow\text{a}=8$

Now,

Total surface area = 6a²

= 6 × (8)²

= 6 × 64

= 384cm²

3. 126

Solution:

Let the number of cones be n.

Volume of the metallic sphere = n × volume of each cone

$\Rightarrow\frac{4}{3}\pi(10.5)^3=\text{n}\times(3.5)^2(3)$

$\Rightarrow4(10.5)^3=\text{n}(3.5)^2(3)$

$\Rightarrow\text{n}=126$

Thus, the number of such cones is 126.

3. 6400

Solution:

Volume of the wall = length × breadth × height

= (8 × 100) × (6 × 100) × 22.5 ...(1m = 100cm)

$=800\times600\times\frac{225}{10}$

$=800\times60\times225$

Volume of 1 brick = length × breadth × height

$=25\times\frac{1125}{100}\times6$

$=\frac{1125}{2}\times3$

Required number of bricks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{wall}}{\text{Volume}\ \text{of}\ 1\ \text{brick}}$

$=\frac{800\times60\times225}{\frac{1125}{2}\times3}$

$=\frac{800\times60\times225\times2}{1125\times3}$

$=6400$

3. Halved.

Solution:

Let original radius be r cm and the original height be h cm.

⇒ Original volume $=\pi\text{r}^2\text{h}$

Given that the new radius $=\frac{\text{r}}{2}$ and new height = 2h

⇒ New volume $=\pi\times\Big(\frac{\text{r}}{2}\Big)^2\times2\text{h}$

$=\pi\times\frac{\text{r}^2}{4}\times2\text{h}$

$=\frac{1}{2}\pi\text{r}^2\text{h}$

so, the new volume is halved.

2. 3 : 1

Solution:

Th ratio of the volume of a right circular cylinder and a right circular cone is given by

$\frac{\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}=\frac{1}{\frac{1}{3}}$ ...(Same base and same height)

$=\frac{3}{1}$

⇒ Ratio of the volumes is 3 : 1. 

4. 64

Solution:

Let the required number of balls be n.

$\Rightarrow\text{n}\times\frac{4}{3}\pi\times(2)^3=\frac{4}{3}\pi\times(8)^3$

$\Rightarrow\text{n}=\frac{8^3}{2^3}$

$\Rightarrow\text{n}=\frac{8\times8\times8}{8}$

$\Rightarrow\text{n}=64$

2. 4 : 1

Solution:

Let the slant height be l and 2l and their radii be r₁ and r₂

The ratio of the curved surface area is given by $=\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}$

$\Rightarrow\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}=\frac{2}{1}$

$\Rightarrow\frac{\text{r}_1}{\text{r}_2(2)}=\frac{2}{1}$

$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{1}$

⇒ Ratio of their radii is 4 : 1.

1. 1078cm³

Solution:

The ratio between the curved surface area and total surface area given by,

$\frac{2\pi\text{rh}}{2\pi\text{rh}+2\pi\text{r}^2}=\frac{1}{2}$

$\Rightarrow\frac{2\pi\text{r}(\text{h})}{2\pi\text{r}(\text{h}+\text{r})}=\frac{1}{2}$

$\Rightarrow\frac{\text{h}}{\text{h}+\text{r}}=\frac{1}{2}$

$\Rightarrow\text{h}+\text{r}=2\text{h}$

$\Rightarrow\text{h}=\text{r}$

Given total surface area $=2\pi\text{rh}+2\pi\text{r}^2=2\pi\text{r}(\text{h}+\text{r})=616$

$\Rightarrow2\pi\text{r}(\text{h}+\text{r})=616$

$\Rightarrow2\pi\text{r}(\text{r}+\text{r})=616$ $(\because\text{h}=\text{r})$

$\Rightarrow2\pi\text{r}(2\text{r})=616$

$\Rightarrow4\pi\text{r}^2=616$

$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=616$

$\Rightarrow\text{r}^2=\frac{616\times7}{88}$

$\Rightarrow\text{r}^2=49$

$\Rightarrow\text{r}=7\text{cm}$

$\Rightarrow\text{h}=7\text{cm}$ $(\because\text{h}=\text{r})$

Volume of the cylinder$=\pi\text{r}^2\text{h}$

$=\frac{22}{7}\times7\times7\times7$

$=1078\text{cm}^3$

1. 9 : 8

Solution:

Let the radii of the bases of a cylinder and a cone be 3x cm and 4x cm respectively and let their heights be 2y cm and 3y cm respectively.

⇒ Ratio of the volumes $=\frac{\pi(3\text{x})^2\times2\text{y}}{\frac{1}{3}\pi(4\text{x})^2\times3\text{y}}$

$=\frac{9\times2}{16}$

$=\frac{9}{8}$

⇒ Ratio of the volume is 9 : 8.

2. 15m

Solution:

Area of the ground = number of person × the amount of space each person occupies

= 11 × 4

$\Rightarrow\pi\text{r}^2=44\text{m}^2\ ...(\text{i})$

Given that the volume = 220m³

$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=220$

$\Rightarrow\frac{1}{3}\times44\times\text{h}=220$ ...(from (i))

$\Rightarrow\text{h}=\frac{220\times3}{44}$

$\Rightarrow\text{h}=15\text{m}$

⇒ Height of the cone = 15m.

2. 288m

Solution:

Volume of the solid lead ball $=\frac{4}{3}\pi\times\text{r}^3$

$=\frac{4}{3}\pi\times6^3=288\pi\text{cm}^3$

Let the length of the wire be h.

Its radius $=\text{r}_1=\frac{0.2}{2}=0.1\text{cm}$

Volume of the wire $=\pi\text{r}_1^2\text{h},$ where r₁ is the radius of the wire

Since the wire is drawn into a solid lead ball,

Volume of the solid lead ball = volume of the wire

$\Rightarrow288\pi=\pi(0.1)^2\text{h}$

$\Rightarrow\text{h}=28800\text{cm}=288\text{m}$

4. 13500

Solution:

n = number of cones $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cylinder}}{\text{Volume}\ \text{of}\ \text{one}\ \text{cone}}$

$=\frac{\pi(3)^2\times5}{\frac{1}{3}\pi\Big(\frac{1}{10}\Big)^2\times1}$

$=\frac{9\times5}{\frac{1}{3}\times\frac{1}{100}}$

$=\frac{45}{\frac{1}{300}}$

$=45\times300$

$=13500$

1. 10cm

Solution:

Let r be the radius of the cone.

Volume = 1570cm³

$1570=\frac{1}{3}\times3.14\times\text{r}^2\times15$

$\Rightarrow1570=3.14\times\text{r}^2\times15$

$\Rightarrow\text{r}^2=100$

$\Rightarrow\text{r}=10\text{cm}$

2. 1 : 4

Solution:

The ratio of the volumes of two sphere is given by $\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}.$

$\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}=\frac{1}{8}$

$\Rightarrow\frac{\text{r}^3}{\text{R}^3}=\frac{1}{8}$

$\Rightarrow\Big(\frac{\text{r}}{\text{R}}\Big)^3=\frac{1}{8}$

$\Rightarrow\frac{\text{r}}{\text{R}}=\frac{1}{2}\ ...(\text{i})$

⇒ Ratio of the volumes = 1 : 2

Now,

Ratio of their surface area $=\frac{4\pi\text{r}^2}{4\pi\text{R}^2}$

$=\frac{\text{r}^2}{\text{R}^2}$

$=\Big(\frac{\text{r}}{\text{R}}\Big)^2$

$=\Big(\frac{1}{2}\Big)^2$ ...(from (i))

$=\frac{1}{4}$

⇒ Ratio of their surface area = 1 : 4.

1. 1 : 2 : 3

Solution:

Let the radius of each be r.

Height of the hemisphere = its radius = r cm

So, height of each is r cm.

Volume of the cone : Volume of the hemisphere : Volume of the cylinder

$=\frac{1}{3}\pi\text{r}^2(\text{r}):\frac{2}{3}\pi\text{r}^3:\pi\text{r}^2(\text{r})$

$=\frac{1}{3}:\frac{2}{3}:1$

$=1:2:3$

2. 270

Solution:

Required number of persons $=\frac{\text{length}\times\text{breadth}\times\text{height}}{\text{amount}\ \text{of}\ \text{air}\ \text{each}\ \text{person}\ \text{requires}}$

$=\frac{20\times15\times4.5}{5}$

$=270$

4. 512m³

Solution:

We know that,

Lateral surface area of a cube = 4a²

⇒ 256 = 4a²

$\Rightarrow\text{a}^2=\frac{256}{4}$

⇒ a² = 64

⇒ a = 8m

Now,

Volume of the cube = a³

= 8³

= 512m³

4. 88cm²

Solution:

Let the radii be x cm and (7 - x)cm.

Volume of the two spheres are in the ratio 64 : 27.

$\Rightarrow\frac{\frac{4}{3}\pi\text{x}^3}{\frac{4}{3}\pi(7-\text{x})^3}=\frac{64}{27}$

$\Rightarrow\Big(\frac{\text{x}}{7-\text{x}}\Big)^3=\Big(\frac{4}{3}\Big)^3$

$\Rightarrow\frac{\text{x}}{7-\text{x}}=\frac{4}{3}$

$\Rightarrow\text{x}=4\text{cm}$

So, their radii are 4cm and 3cm.

Difference of their tital surface areas

$=4\pi(4)^2-4\pi(3)^2$

$=4\times\frac{22}{7}(16-9)$

$=88\text{cm}^2$

3. 54

Solution:

Let the number of bottles be n.

The number of bottles needed to empty the bowl = n × volume of each cylinder

that is, volume of the hemispherical bowl = n × volume of each cylinder

$\Rightarrow\frac{2}{3}\pi(9)^2=\text{n}\times\pi(1.5)^2(4)$

$\Rightarrow\text{n}=54$

Thus, there are 54 bottles.

2. 1320m²

Solution:

The diameter of the roller $=84\text{cm}=\frac{84}{100}\text{m}$

So, the radius $=\frac{84}{200}\text{m}$

The area covered by the roller in 1 revolution

$=2\pi\text{rh}$

$=2\times\frac{22}{7}\times\frac{84}{200}\times1$

$=2.64\text{m}^2$

$\therefore$ Area covered in 500 complete revolution = 500 × 2.64 = 1320m²

Thus, the area of the playground is 1320m².

3.  20cm

Solution:

Given that, r = 14cm

Curved surface area = 1760cm²

Now,

Curved surface area $=2\pi\text{rh}$

$\Rightarrow1760=2\times\frac{22}{7}\times14\times\text{h}$

$\Rightarrow1760=88\times\text{h}$

$\Rightarrow\text{h}=\frac{1760}{88}$

$\Rightarrow\text{h}=20\text{cm}$ 

2. 192

​​​​​​​Solution:

Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$

$=\frac{40\times12\times16}{4\times5\times2}$

$=192$

2. 7546cm³

Solution:

Let r be the radius of the cone.

l² = h² + r²

⇒ r² = l² - h²

= 28² - 21²

= 49 × 7

$\Rightarrow\text{r}=7\sqrt{7}\text{cm}$

Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$

⇒ Volume of the cone $=\frac{1}{3}\times\frac{22}{7}\times(7\sqrt{7})^2\times21$

⇒ Volume of the cone $=7546\text{cm}^3$

2. 550cm²

Solution:

The height of the cone is 24cm and the diameter of its base is 14cm.

So, its radius = 7cm.

$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$

$\text{l}=\sqrt{7^2+24^2}$

$\text{l}=\sqrt{49+576}$

$\text{l}=25\text{cm}$

So, curved surface area of the cone $=\pi\text{r}=\frac{22}{7}\times7\times25=550\text{cm}^2$

1. $(288\pi)\text{m}^3$

Solution:

Given that surface area of a sphere $=144\pi\text{m}^2$

$\Rightarrow4\pi\text{r}^2=144\pi$

$\Rightarrow4\times\text{r}^2=144$

$\Rightarrow\text{r}^2=\frac{144}{4}$

$\Rightarrow\text{r}^2=36$

$\Rightarrow\text{r}=6\text{cm}$

Volume of sphere $=\frac{4}{3}\pi\text{r}^3$

$=\frac{4}{3}\times\pi\times6\times6\times6$

$=4\times\pi\times2\times6\times6$

$=288\pi\text{m}^3$

4. 84000

Solution:

Let the number of lead shots be n.

Volume of the cuboid = n × volume of each lead shot

$\Rightarrow9\times11\times12=\text{n}\times\frac{4}{3}\pi\Big(\frac{0.3}{2}\Big)^3$ $...\Big(\text{since}\ \text{radius}=\frac{0.3}{2}\text{cm}\Big)$

$\Rightarrow9\times11\times12=\text{n}\times\frac{4}{3}\times\frac{22}{7}\times\frac{27}{8000}$

$\Rightarrow\text{n}=84000$

Thus, there are 84000 lesd shots.

2. 450

Solution:

Let the required number of coins be n.

Since the n coins are melted form a right circular cylinder,

$\text{n}\times\pi\times\Big(\frac{1.5}{2}\Big)^2\times0.2=\pi\times\Big(\frac{4.5}{2}\Big)^2\times10$

$\Rightarrow\text{n}\times\Big(\frac{15}{10}\Big)^2\times\frac{2}{10}=\Big(\frac{45}{20}\Big)^2\times10$

$\Rightarrow\text{n}\times\frac{9}{16}\times\frac{2}{10}=\frac{81}{16}\times10$

$\Rightarrow\text{n}=\frac{81\times10\times16\times10}{16\times9\times2}$

$\Rightarrow\text{n}=450$

2. 20 : 27

Solution:

 $\text{Let}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}\ \text{and}\ \frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$

Ratio of the surface area $=\frac{\pi\text{r}_1\text{h}_1}{\pi\text{r}_2\text{h}_2}$

$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)\times\Big(\frac{\text{h}_1}{\text{h}_2} \Big)$

$=\Big(\frac{2}{3}\Big)^2\times\Big(\frac{5}{3}\Big)^2$

$=\frac{4}{9}\times\frac{5}{3}$

$=\frac{20}{27}$

$=20:27$

4. 125%

Solution:

Let each edge be a

⇒ its surface area = 6a²

Now,

New edge = 150% of a

$=\frac{150}{100}\times\text{a}$

$=\frac{3\text{a}}{2}$

New surface area $=6\times\Big(\frac{3\text{a}}{2}\Big)^2$

$=\frac{27\text{a}^2}{2}$

Increase surface area $=\frac{27\text{a}^2}{2}-6\text{a}^2$

$=\frac{27\text{a}^2-12\text{a}^2}{2}$

$=\frac{15\text{a}^2}{2}$

Percentage increase in its surface area

$=\frac{\text{Increase}\ \text{in}\ \text{the}\ \text{surface}\ \text{area}}{\text{original}\ \text{surface}}\times100$

$=\frac{15\text{a}^2}{2}\times\frac{1}{6\text{a}^2}\times100$

$=125\%$

2. 1760cm²

Solution:

Given, d = 628

r = 14cm

h = 20cm

Now,

Curved surface area $=2\pi\text{rh}$

$=2\times\frac{22}{7}\times14\times20$

$=1760\text{cm}^2$

3. 4851cm³

Solution:

Given surface area of a sphere = 1386cm²

$\Rightarrow4\pi\text{r}^2=1386$

$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=1386$

$\Rightarrow\frac{88}{7}\times\text{r}^2=1386$

$\Rightarrow\text{r}^2=\frac{1386\times7}{88}$

$\Rightarrow\text{r}^2=\frac{441}{88}$

$\Rightarrow\text{r}=\sqrt{\frac{441}{4}}$

$\Rightarrow\text{r}=\frac{21}{2}\text{cm}$

Volume of sphere $=\frac{4}{3}\pi\text{r}^3$

$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$

$=11\times21\times21$

$=4851\text{cm}^3$

2. 2 : 1

Solution:

Let the radius of each be r.

Height of the hemisphere = its radius = r cm

Volume of the cone = volume of the hemisphere

$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=\frac{2}{3}\pi\text{r}^3$

$\Rightarrow\text{h}=2\text{r}$

$\Rightarrow\frac{\text{h}}{\text{r}}=\frac{2}{1}$

Thus, the ratio of their heights is 2 : 1.

4. becomes 8 times.

Solution:

Let the original side be 'x' cm

$\therefore$ Original volume = x³ cm³

Now,

New side = 2x cm

$\therefore$ New volume = (2x)³

= 8x³ cm³

So, the volume becomes 8 times.

4. 5544cm²

Solution:

Surface area of a sphere $=4\pi\text{r}^2$

$=4\times\frac{22}{7}\times21\times21$

$=4\times22\times3\times21$

$=5544\text{cm}^2$

2. 112m

Solution:

We know that 1dm = 10cm.

Given that 2.2dm³ is drawn into a cylindrical wire.

That is, (2.2 × 1000) = 2200cm³ is drawn into a cylindrical wire.

Let the radius of the wire be r and the height be h.

Volume of the cylindrical = 2200

$\Rightarrow\pi\text{r}^2\text{h}=2200$

$\Rightarrow\frac{22}{7}\times0.25\times0.25\times\text{h}=2200$ ...(Since the diameter = 0.50cm)

$\Rightarrow\text{h}=11200\text{cm}$

$\Rightarrow\text{h}=112\text{m}$

So, the length of the wire is 112m.

1. 1 : 4

Solution:

Ratio of the surface areas of the ballons.

$\Rightarrow\frac{3\pi(6)^2}{3\pi(12)^2}$

$=\frac{1}{4}$

4. 1000m³

Solution:

Volume of water that falls on ground $=2\times10000\times\frac{5}{100}$

$...\Big(1\ \text{hector}=10000\text{m}\ \text{and}\ 1\text{cm}=\frac{1}{100}\text{m}\Big)$

$=1000\text{m}^3$

2. 2250m³

Solution:

Volume of the water running into the sea per hour = 1.5 × 30 × 3000

...(1km = 1000m)

= 45 × 3000

Volume of the water running into the sea per minute $=\frac{45\times3000}{60}$

$=45\times50$

$=2250\text{m}^3$

3. 6m

Solution:

Given that,

Curved surface area = 264m²

Volume = 924m³

Now,

Curved surface area $=2\pi\text{rh}$

$\Rightarrow264=2\times\frac{22}{7}\times\text{r}\times\text{h}$

$\Rightarrow264=\frac{44}{7}\times\text{r}\times\text{h}$

$\Rightarrow\text{r}=\frac{264\times7}{44\times\text{h}}$

$\Rightarrow\text{r}=\frac{42}{\text{h}}$

Volume $=\pi\text{r}^2\text{h}$

$\Rightarrow924=\frac{22}{7}\times\frac{42}{\text{h}}\times\frac{42}{\text{h}}\times\text{h}$

$\Rightarrow924=22\times\frac{6}{\text{h}}\times42$

$\Rightarrow\frac{924}{22\times42\times6}=\frac{1}{\text{h}}$

$\Rightarrow\frac{924}{5544}=\frac{1}{\text{h}}$

$\Rightarrow\text{h}=\frac{5544}{924}$

$\Rightarrow\text{h}=6\text{m}$

2. 21m²

​​​​​​​Solution:

Lateral surface area of the cuboid = 2(l + b) × h

$=2(15+6)\times\frac{5}{10}$ $...\Big(1\text{dm}=\frac{1}{10}\text{m}\Rightarrow5\text{dm}=\frac{5}{10}\text{m}\Big)$

$=2(21)\times\frac{1}{2}$

$=21\text{m}^2$