Maths M.C.Q

2. $16\pi\text{cm}^3$

Solution:

So here, radius would be half of edge of cube, or diameter of rod = edge of cube

r = 2cm

h = 4cm

So, volume $=\pi\text{r}^2\text{h}$

$=\pi\times2\times2\times4$

$=16\pi\text{cm}^2$

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. 3.5cm

Solution:

Let a, b, c. be the sides of three cubes

Then a³ $=\frac{1}{8}\Rightarrow\text{a}=\frac{1}{2}$

b³ = 1 ⇒ b = 1

c³ = 8 ⇒ c = 2,

Now height of resulting cube

0.5 + 1 + 2 = 3.5cm

2. 1920

Solution:

Volume of pit = (16 × 12 × 4)m³

Volume of a plank = (4 × 0.5 × 0.2)m³

Required number of planks $=\frac{\text{Volume of pit}}{\text{Volume of a plank}}$

$=\frac{16\times12\times4}{4\times0.5\times0.2}=1920$

Hence, (b) is the correct answer.

1. $2\pi\text{r}(\text{r}+\text{h})$

Solution:

Let S be the total surface area of the closed cylinder with radius r and height h, then

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

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

2. $\frac{4}{3}\pi(\text{R}^3-\text{r}^3)$

Solution:

The volume of a spherical shell is given by $\frac{4}{3}\pi(\text{R}^3-\text{r}^3)$ where R = Larger radius and r = smaller radius.

3. 15cm

Solution:

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

$2310=\frac{22}{7}\times7\times7\times\text{h}$

$\text{h}=\frac{2310}{22\times7}$

$\text{h}=15\text{cm}$

$=15\text{cm}$

3. 770cm².

Solution:

$\frac{\text{r}}{\text{h}}=\frac{2}{3}$

$\text{r}=\frac{2}{3}\text{h}$

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

$1617=\frac{22}{7}\cdot\frac{2}{3}\text{h}\cdot\frac{2}{3}\text{h}\cdot\frac{2}{3}\text{h}\cdot\text{h}$

$\text{h}^3=\frac{1617.7.9}{22.4}$

$\text{h}=\frac{21}{2}\text{cm}\text{r}=7\text{cm}$

Total surface area of cylinder $=2\pi\text{rh}+2\pi\text{r}^2$

$=2\cdot\frac{22}{7}\big(7\cdot\frac{21}{2}+7\cdot7\big)$

$=770\text{cm}^2$

1. 10%

Solution:

The formula of the curved surface area of a cone with base radius 'r' and slant height '' is given as

Curved Surface Area $=\pi\text{rl}$

Now, it is said that the slant height has increase by 10%. So the new slant height '1.1 l'

So, now

New Curved Surface Area $=1.1\pi\text{rl}$

We see that the percentage increase of the Curved Surface Area is 10%

2. $\pi\text{r}\Big(\text{l+}\frac{\text{r}}{4}\Big)$

Solution:

Total surface area of cone = Area of the base + Curved Surface area of cone

$=\pi\Big(\frac{\text{r}}{2}\Big)^2+\pi\Big(\frac{\text{r}}{2}\Big)\times2\text{l}=\frac{\pi\text{r}}{2}\Big(\frac{\text{r}}{2}+2\text{l}\Big)$

$=\frac{\pi\text{r}}{4}(\text{r}+4\text{l})=\pi\text{r}\Big(\text{l}+\frac{\text{r}}{4}\Big)$

Hence, (b) is the correct answer.

1. 37-47

Solution:

Let the lower limit be x.

Here,

Class size = 10

$\therefore$ Upper limit = Class size + Lower limit

Upper limit = (x + 10)

Mid value of the class interval = 42

$\therefore\frac{\text{x}+\text{x}+10}{2}=42$

$\Rightarrow\frac{2\text{x}+10}{2}=42$

$\Rightarrow2\text{x}+10=84$

$\Rightarrow2\text{x}=74$

$\Rightarrow\text{x}=37$

Thus, we have:

Lower limit = 37

Upper limit = 37 + 10 = 47

4. 125%

Solution:

Let the original side be a then

Original surface area S = 6a²

When side is increased by 50%, side becomes $\frac{3}{2\text{a}}$ than

Surface area $\text{s}=\frac{27}{2}{\text{a}^2}$

Hence,

Surface area is increases by $\text{x}=\frac{\frac{27}{2}\text{a}^2-6\text{a}^2}{6\text{a}^2}\times100=125\%$

3. 450

Solution:

Diameter of a coin = 1.5cm

Radius $=\frac{1.5}{2}\text{cm}.$

Thickness of coin = 0.2cm

Volume of one coin $=\pi\text{r2h}=\Big(\frac{1.5}{2}\Big)^2\times0.2\text{cm}^3$

Volume of N coins $=\pi\text{r2h}=\pi\Big(\frac{1.5}{2}\Big)^2\times0.2\times\text{ N cm}^3$

Volume of N coins = Volume of the right circular cylinder to be formed

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

$\Rightarrow\text{N}=\frac{4.5\times4.5\times10\times2\times2}{2\times2\times1.5\times1.5\times0.2}=450$

Hence, the required number of coins = 450

1. 10%

Solution:

C.S.A of a cone $=\pi\text{rl}$

If l' = 1 + 10% of l

$=\text{l}+\frac{10}{100}\times\text{l}$

$=\text{l}+\frac{\text{l}}{10}$

And, r' = r

$\text{C.S.A.}=\pi\text{r}\Big(\text{l}+\frac{\text{l}}{10}\Big)=\frac{11}{10}\pi\text{rl}$

So, increase in C.S.A. $=\frac{\frac{11}{10}\pi\text{rl}-\pi\text{rl}}{\pi\text{rl}}\times100\%=10\%$

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 $=\pi\text{rl}$

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

$=550\text{cm}^2$

1. $100\pi$

Solution:

Volume of solid sphere $=\frac{4}{3}\pi(10)^3=\frac{4000\pi}{3}\text{cm}^3$

Vomule 8 solid sphere of radius (say) $\text{r}=8\times\frac{4}{3}\pi\text{r}^3=\frac{32\pi\text{r}^3}{3}\text{cm}^3$

Now, $\frac{32\pi\text{r}^3}{3}=\frac{4000\pi}{3}$

$\Rightarrow\text{r}=\Big(\frac{1000}{8}\Big)^\frac{1}{3}=\frac{10}{2}=5\text{cm}$

Surface Area of each small ball $=4\pi\text{r}^2=4\pi(5)^2=100\pi\text{ cm}^2 $

Hence, correct option is (a).

3. 2cm

Solution:

CSA of cylinder $=2\pi\text{rh}$

$88=2\times\frac{22}{7}\times\text{r}\times14$

$\text{r}=\frac{88\times7}{2\times22\times14}$

$\text{r}=1\text{cm}$

Hence Diameter of base = 2cm.

2. 2.1cm

Solution:

Given, height of a cone = 8.4cm

Radius of the base of cone = 2.1cm

Volume of a cone = $\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\times\pi(2.1)^2\times8.4$

$=\frac{1}{3}\times\pi\times2.1\times2.1\times8.4=\pi\times4.41\times2.8\text{cm}^3$

Since, cone is melted and recast into a sphere.

let the radius of a sphere be R.

Then, Volume of a sphere = Volume of a cone

$\Rightarrow\frac{4}{3}\pi\text{r}^3=\pi\times4.41\times2.8$

$\Rightarrow\text{r}^3=\frac{4.41\times2.8\times3}{4}$

$\Rightarrow\text{r}^3=4.41\times0.7\times3$

$\Rightarrow\text{r}^3=4.41\times2.1$

$\therefore\text{r}=2.1$

Hence, the radius of the sphere is 2.1cm.

2. $0$

Solution:

For a cone,

$\text{V}=\frac{1}{3}\pi\text{R}^2\text{h}$

S = curved Surface Area $=\pi\text{R}\text{L}$

$\text{L}=\sqrt{\text{h}^2+\text{R}^2}$

$3\pi\text{V}\text{h}^3-\text{S}^2\text{h}^2+\text{9V}^2$

$=3\pi\Big(\frac{1}{3}\pi\text{R}^2\text{h}\Big)\text{h}^3-\pi^2\text{R}^2\big(\text{h}^2+\text{R}^2\big)\text{h}^2\\+9\times\frac{1}{9}\pi^2\text{R}^4\text{h}^2$

$=\pi^2\text{R}^2\text{h}^4-\pi^2\text{R}^2\text{h}^4-\pi^2\text{R}^4\text{h}^2+\pi^2\text{R}^4\text{h}^2$

$=0$

3. 13500

Solution:

Radius of cylinder = 3cm.

Height = 5cm

Radius of cone= 1mm = 0.1cm

Height = 1cm

Volume of cylinder = number of cones × Volume of cone

$\pi\text{R}^2\text{h}=\text{n}\times\frac{1}{3}\pi\text{r}^2\text{h}$

$\pi32\times5=\text{n}\times\frac{1}{3}\pi\times0.1^2\times1$

$45=\text{n}\times\frac{1}{3}\times(.1)^2\times1$

$45=\text{n}\times\frac{1}{100}$

$\text{n}=13500$

$13500$

3. 192

Solution:

Volume of a cuboid = Length × Breadth × Height

Volume of pit = 40m × 12m × 16m = 7680m³

Volume of plank = 4m × 5m × 2m = 40m³

No. of planks $=\frac{\text{Volume of pit }}{\text{Volume of plank }}=\frac{7680}{40}=192$

4. $36\pi\text{cm}^2$

Solution:

AB = 12cm

Area of circular Base $=\pi\text{r}^2=64\pi$

⇒ r = 8cm

AD = 9cm

Consider $\triangle\text{ADE}$ and $\triangle\text{ABC},$

$\angle\text{DAE}=\angle\text{BAC}$ (common)

$\angle\text{ADE}=\angle\text{ABC}$ (each 90º)

$\angle\text{AED}=\angle\text{ACB}$ (Third angle will also be same)

Hence $\triangle\text{ADE}\sim\triangle\text{ABC}$

So $\frac{\text{AD}}{\text{AB}}=\frac{\text{DE}}{\text{BC}}$

$\Rightarrow\frac{9}{12}=\frac{\text{DE}}{8}$

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

Radius of base of new cone = 6cm

⇒ Area $=\pi(6)^2=36\pi\text{cm}^2$

3. Same.

Solution:

curved surface area of a cylinder of radius 'r' and heightn 'h' is given by $\text{A}=2\pi\text{rh}$

Now, if r' = 2r and $\text{h}'=\frac{\text{h}}{2}$

Then $\text{A}'=2\pi\times(2\text{r})\times\frac{\text{h}}{2}$

$=2\pi\text{rh}=\text{A}$

⇒ C.S.A. remains the same.

3. 550cm²

Solution:

Radius of cone = r cm

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

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

$\Rightarrow\text{r}^2=\frac{1232\times3\times7}{22\times24}=49$

$\Rightarrow\text{r}=\sqrt{49}=7\text{cm}$

Area of curved surface area $=\pi\text{rl}$

$\text{l}=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25$

$\therefore\frac{22}{7}\times7\times25=550\text{cm}^2$

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 $=\pi\text{r}^2\text{h}$

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

$=396\text{cm}^3$

1. 1000 cubic cm

Solution:

Perimeter of one face of a cube = 4a = 40cm (where a = side of the cube) sides of face of a cube = 4

Therefore, side of the cube $=\frac{4\text{a}}{4}=\frac{40}{4}=10\text{cm}$

Therefore, volume of the cube = a³ = 10³ = 1000 cubic cm

2. 54

Solution:

Volume of a hemisphere $=\frac{2}{3}\pi\text{r}^3=\frac{2}{3}\pi(9)^3\text{cm}^3$

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

Volume of a cylinderical bottle $=\pi(1.5)^2\times4$

No. of bottles required $=\frac{\text{Volume of a hemisphere}}{\text{Volume of a cylinderical bottle}}$

$=\frac{\frac{2}{3}\pi(9)^2}{\pi(1.5)^2\times4}$

$=\frac{81\times3}{2.25\times1.5\times2}=54$

Thus, total 54 bottals are required.

4. $288\pi$

Solution:

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

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

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

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

Volume of the sphere $=\big(\frac{4}{3}\big)\times\pi\times6^3$

$\Rightarrow$ Volume of the sphere $=288\pi\text{m}^3$

4. 25.344 kg wt.

Solution:

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

$=\frac{1}{3}\times\frac{22}{7}\times24\times24\times42$

= 25344cm³

= 25.344 dm³ (1dm³ =1000cm³)

= 25.344 Kg wt. (1 kg - wt = 1dm³)

3. 18

​​​​​​​Solution:

We have the cuboid of dimensions 10cm × 9cm × 6cm.

We are to find how many cubs with edge 3cm can be cut from the given cuboid

Let us cut this cuboid into following two cuboids

9cm × 9cm × 6cm

And

1cm × 9cm × 6cm

So, the number of cubes of sides 3cm, that can be cut from the firest cuboid,

$=\frac{9\text{cm}\times9\text{cm}\times6\text{cm}}{3\text{cm}\times3\text{cm}\times3\text{cm}}$

= 18

We can not cut single cube of side 3cm from the second cuboid of dimension 1cm × 9cm × 6cm

Hence, this much volume is useless for us.

So, we can cut maximum 18 cubes of side 3cm from the cuboid of dimensions 10cm × 9cm × 6cm.

Hence, the correct option is (c).

4. 21m²

Solution:

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

$=\big(2(15+6)\times\frac{5}{10}\big)\text{m}^2=21\text{m}^2$

3. 3 : 1

Solution:

Let V₁ and V₂ be the volume of the two cylinders with radius r₁ and height h₁, and radius r₂ and height h₂,

Where, $\frac{2\text{r}_1}{2\text{r}_2}=\frac{3}{1},\frac{\text{h}_1}{\text{h}_2}=\frac{1}{3}$

So,

$\text{V}_1=\pi\text{r}^2_1\text{h}_1\ ...(\text{i})$

Now,

$\text{V}_2=\pi\text{r}^2_2\text{h}_2\ ...(\text{ii})$

From equation (i) and (ii), we have

$\frac{\text{V}_1}{\text{V}_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$

$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\Big(\frac{\text{2r}_1}{\text2{r}_2}\Big)^2\Big(\frac{\text{h}_1}{\text{h}_2}\Big)$

$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\big(3\big)^2\Big(\frac{1}{3}\Big)=\frac{3}{1}$

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 $=\frac{18}{25}\times50$

$=36\text{kg}$

3. $10\sqrt{3}\text{cm}$

Solution:

The longest rod that can be fitted in the cubical vessel is its diagonal.

side of the cube (l) = 10cm

So, the diagonal of the cube,

$=\sqrt{3}\text{l}$

$=10\sqrt{3}\text{cm}$

So, the length of the longest rod that can be fitted in the cubical box is $10\sqrt{3}\text{cm}.$

Hence, the correct choice is (c).​​​​​​

4. $2\pi\text{rh}$

Solution:

The lateral surface area of a cylinder is equal to its curved surface area.

$\therefore$ Lateral surface area of a cylinder $=2\pi\text{rh}$

2. 314cm²

Solution:

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

$1570=\frac{1}{3}\times\pi\times\text{r}^2\times15$

$\pi\text{r}^2=314\text{cm}^2$

Base area is 314cm³

4. $\pi\text{rl}$

Solution:

The formula of the curved surface area of a cone with base radius 'r' and slant height '1' us given as

Curved Surface Area $=\pi\text{rl}$

Hence the base radius is given as '2r' and the slant height is given as $\frac{1}{2}$

Substituting these values in the above equation we have

Curved Surface Area $=\frac{(\pi)(2)(\text{r})(l))}{2}$

$=\pi\text{rl}$

1. a₂v₁ = a₁v₂

Solution:

Let the radius of the base of the cylinder be r and R.

Now, let sheet with length a₁ be used to form a cylinder with volume v₁

So,

$2\pi\text{r}=\text{a}_1$

$\Rightarrow\text{r}=\frac{\text{a}_1}{2\pi}$

Volume $\text{V}_1=\pi\text{r}^2\text{h}=\pi\Big(\frac{\text{a}_1}{2\pi}\Big)^2\text{a}_2=\frac{\text{a}^2_1\text{a}_2}{74\pi}$

Similarly, let sheet with length a₂ be used to form cylinder with volume v₂

So,

$2\pi\text{r}=\text{a}_1$

$\Rightarrow\text{r}=\frac{\text{a}_1}{2\pi}$

Volume $\text{V}_1=\pi\text{r}^2\text{h}=\pi\Big(\frac{\text{a}_1}{2\pi}\Big)^2\text{a}_2=\frac{\text{a}^2_1\text{a}_2}{74\pi}$

Now, $\frac{\text{V}_1}{\text{V}_2}=\frac{\text{a}^2_1\text{a}_2}{\text{a}^2_2\text{a}_2}=\frac{\text{a}_1}{\text{a}_2}$

$\Rightarrow\text{V}_1\text{a}_2=\text{V}_2\text{a}_1$

3. 220m

Solution:

Let the length of the required cloth be L m and its breadth be B m.

Here, B = 2.5m

Radius of the conical tent = 7m

Height of the tent = 24m

Area of cloth required = curved surface area of the conical tent

$\Rightarrow\text{L}\times\text{B}=\pi\text{rl}​​$

$\Rightarrow\text{L}\times\text{B}=\pi\text{rl}​​$

$\Rightarrow\text{L}\times2.5=\frac{22}{7}\times7\times\sqrt{7^2+24^2}$

$\Rightarrow2.5\text{L}=22\times\sqrt{49}+576$

$\therefore\text{L}=\frac{22\times25}{2.5}=220\text{m}$

2. 384cm²

Solution:

Diagonal $=\text{s}\sqrt{3}$ Where s = side

Here s = 8

Surface area of cube = 6a² = 6 × 8 × 8 = 384cm²

2. 1 : 4

Solution:

Volume of reduced cylinder : Volume of original cylinder

$\pi\big(\frac{\text{r}}{2}\big)^2\text{h}:\pi(\text{r})^2\text{h}$

$1:4$

1. $\frac{\text{h}^3}{4\pi}$

Solution:

Circumference of cylinder $=2\pi\text{r}$

Height = h

Volume $\pi\text{r}^2\text{h}=\pi\Big(\frac{\text{h}^2}{4\pi^2}\Big)\text{h}=\frac{\text{h}^3}{4\pi}$