MCQ 11 Mark
The total surface area of a cube is $96\ cm^2.$ The volume of the cube is:
- A
$8\ cm^3$
- B
$27\ cm^3$
- ✓
$64\ cm^3$
- D
$512\ cm^3$
AnswerCorrect option: C. $64\ cm^3$
We know that,
Total surface area of a cube $= 6a^2$
$\Rightarrow 96 = 6a^2$
$\Rightarrow\text{a}^2=\frac{96}{6}$
$\Rightarrow a^2 = 16$
$\Rightarrow a = 4\ cm$
Now,
Volume of the cube $= a^3$
$= 4^3$
$= 64\ cm^3$
View full question & answer→MCQ 21 Mark
The volume of a right circular cone of height $24\ cm$ is $1232\ cm^3.$ Its curved surface area is:
- A
$1254\ cm^2$
- B
$704\ cm^2$
- ✓
$550\ cm^2$
- D
$462\ cm^2$
AnswerCorrect option: C. $550\ cm^2$
We know that,
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow1232=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow1232=\frac{1}{3}\times\frac{22}{7}\times\text{r}^2\times24$
$\Rightarrow\text{r}^2=\frac{7\times3\times1232}{24\times22}$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}=7\text{cm}$
$\Rightarrow r = 7\ cm$ and $h = 24\ cm$
$l^2 = r^2 + h^2$
$\Rightarrow l^2 = 7^2 + 24^2$
$\Rightarrow l^2 = 49 + 576$
$\Rightarrow l^2 = 625$
$\Rightarrow l = 25$
Curved surface area $=\pi\text{rl}$
$=\frac{22}{7}\times7\times25$
$=550\text{cm}^2$
View full question & answer→MCQ 31 Mark
The diameter of the base of a cylinder is $6\ cm$ and its height is $14\ cm.$ The volume of the cylinder is:
- A
$198\ cm^3$
- ✓
$396\ cm^3$
- C
$495\ cm^3$
- D
$297\ cm^3 $
AnswerCorrect option: B. $396\ cm^3$
Given, $d = 6\ cm$
$r = 3\ cm$
$h = 14\ cm$
Now,
Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times3\times3\times14$
$=396\text{cm}^3$
View full question & answer→Question 41 Mark
A beam 9m long, 40cm wide and 20cm high is made up of iron which weighs 50kg per cubic metre. The weight of the beam is:
- 27kg
- 48kg
- 36kg
- 56kg
Answer
- 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 $=\frac{18}{25}\times50$
$=36\text{kg}$ View full question & answer→MCQ 51 Mark
The length, breadth and height of a cuboid are $15\ cm, 12\ cm$ and $4.5\ cm$ respectively. Its volume is:
- A
$243\ cm^3$
- B
$405\ cm^3$
- ✓
$810\ cm^3$
- D
$603\ cm^3$
AnswerCorrect option: C. $810\ cm^3$
Volume of a cuboid $=$ length $\times$ breadth $\times$ height
$= 15 \times 12 \times 4.5$
$= 810\ cm^3$
View full question & answer→MCQ 61 Mark
The volume of a sphere of radius $10.5\ cm$ is:
- A
$9702\ cm^3$
- ✓
$4851\ cm^3$
- C
$19404\ cm^3$
- D
$14553\ cm^3$
AnswerCorrect option: B. $4851\ cm^3$
Given radius $=10.5=\frac{21}{2}\ cm$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$
$=11\times21\times21$
$=4851\ cm^3$
View full question & answer→Question 71 Mark
How many planks of dimensions (5m × 25m × 10cm) can be stored in a pit which is 20m long, 6m wide and 50cm deep?
- 480
- 450
- 320
- 360
Answer
- 480
Solution:
Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$
$=\frac{(20\times100)\times(6\times100)\times50}{(5\times100)\times25\times10}$ $...(1\text{m}=100\text{cm})$
$=\text{480}$ View full question & answer→MCQ 81 Mark
The ratio between the radius of the base and the height of a cylinder is $2 : 3.$ If its volume is $1617\ cm^3,$ then its total surface area is:
- A
$308\ cm^2$
- B
$462\ cm^2$
- C
$540\ cm^2$
- ✓
$770\ cm^2$
AnswerCorrect option: D. $770\ cm^2$
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 $=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{21}{2}+7\Big)$
$=44\Big(\frac{35}{2}\Big)$
$=770\text{cm}^2$
View full question & answer→MCQ 91 Mark
The height of a cylinder is $14\ cm$ and its curved surface area is $264\ cm^2.$ The volume of the cylinder is:
- A
$308\ cm^3$
- ✓
$396\ cm^3$
- C
$1232\ cm^3$
- D
$1848\ cm^3$
AnswerCorrect option: B. $396\ cm^3$
Given,
Height $= 14\ cm$
Curved surface area $= 264\ cm^2$
Now,
Curved surface area $=2\pi\text{rh}$
$\Rightarrow264=2\times\frac{22}{7}\times\text{r}\times14$
$\Rightarrow264=88\times\text{r}$
$\Rightarrow\text{r}=\frac{264}{88}$
$\Rightarrow\text{r}=3\ cm$
Volume $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times3\times3\times14$
$=396\ cm^3$
View full question & answer→MCQ 101 Mark
If the length of diagonal of a cube is $8\sqrt3\ cm$ then its surface area is:
- A
$192\ cm^2$
- ✓
$384\ cm^2$
- C
$512\ cm^2$
- D
$768\ cm^2$
AnswerCorrect option: B. $384\ cm^2$
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 $= 6a^2$
$= 6 \times (8)^2$
$= 6 \times 64$
$= 384\ cm^2$
View full question & answer→Question 111 Mark
A metallic sphere of radius 10.5cm is melted and then recast into small cones, each of radius 3.5cm and height 3cm. The number of such cones will be:
- 21
- 63
- 126
- 130
Answer
- 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. View full question & answer→Question 121 Mark
How many bricks will be required to construct a wall 8m long, 6m high and 22.5cm thick if each brick measures (25cm × 11.25 × 6cm)?
- 4800
- 5600
- 6400
- 5200
Answer
- 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 $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{wall}}{\text{Volume}\ \text{of}\ 1\ \text{brick}}$
$=\frac{800\times60\times225}{\frac{1125}{2}\times3}$
$=\frac{800\times60\times225\times2}{1125\times3}$
$=6400$ View full question & answer→Question 131 Mark
In a cylinder, if the radius is halved and the height is doubled, then the volume will be:
- The same.
- Doubled.
- Halved.
- Four times.
Answer
- 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. View full question & answer→Question 141 Mark
The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be:
- 1 : 3
- 3 : 1
- 4 : 3
- 3 : 4
Answer
- 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. View full question & answer→MCQ 151 Mark
The volume of a cube is $512\ cm^3.$ Its total surface area is:
- A
$256\ cm^2$
- ✓
$384\ cm^2$
- C
$512\ cm^2$
- D
$64\ cm^2$
AnswerCorrect option: B. $384\ cm^2$
Volume of the cube $= a^3$
$\Rightarrow 512 = a^3$
$\Rightarrow a = 8\ cm$
Now,
Total surface area of a cube $= 6a^2$
$= 6 \times (8)^2$
$= 6 \times 64$
$= 384\ cm^2$
View full question & answer→Question 161 Mark
A solid metal ball of radius 8cm is melted and cast into smaller balls, each of radius 2cm. The number of such balls is:
- 8
- 16
- 32
- 64
Answer
- 64
Solution:
Let the required number of balls be n.
$\Rightarrow\text{n}\times\frac{4}{3}\pi\times(2)^3=\frac{4}{3}\pi\times(8)^3$
$\Rightarrow\text{n}=\frac{8^3}{2^3}$
$\Rightarrow\text{n}=\frac{8\times8\times8}{8}$
$\Rightarrow\text{n}=64$ View full question & answer→MCQ 171 Mark
The curved surface area of one cone is twice that of the other while the slant height of the latter is twice of the former. The ratio of their radii is:
- A
$2 : 1$
- ✓
$4 : 1$
- C
$8 : 1$
- D
$1 : 1$
AnswerCorrect option: B. $4 : 1$
Let the slant height be $l$ and $2l$ and their radii be $r_1$ and $r_2$
The ratio of the curved surface area is given by $=\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}$
$\Rightarrow\frac{\pi\text{r}_1\text{l}}{\pi\text{r}_2(2\text{l})}=\frac{2}{1}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2(2)}=\frac{2}{1}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{1}$
$\Rightarrow$ Ratio of their radii is $4 : 1.$
View full question & answer→MCQ 181 Mark
The ratio between the curved surface area and the total surface area of a right circular cylinder is $1 : 2.$ If the total surface area is $616\ cm^2,$ then the volume of the cylinder is:
- ✓
$1078\ cm^3$
- B
$1232\ cm^3$
- C
$1848\ cm^3$
- D
$924\ cm^3$
AnswerCorrect option: A. $1078\ cm^3$
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$=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times7\times7\times7$
$=1078\text{cm}^3$
View full question & answer→Question 191 Mark
The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio:
- 9 : 8
- 8 : 9
- 3 : 4
- 4 : 3
Answer
- 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. View full question & answer→MCQ 201 Mark
A conical tent is to accommodate $11$ persons such that each person occpies $4m^2$ of space on the ground. They have $220m^3$ of air to breathe. The height of the cone is:
AnswerArea of the ground $=$ number of person $\times$ the amount of space each person occupies
$= 11 \times 4$
$\Rightarrow\pi\text{r}^2=44\ m^2\ ...(\text{i})$
Given that the volume $= 220m^3$
$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=220$
$\Rightarrow\frac{1}{3}\times44\times\text{h}=220 ...($from$ (i))$
$\Rightarrow\text{h}=\frac{220\times3}{44}$
$\Rightarrow\text{h}=15m$
$\Rightarrow$ Height of the cone $= 15m.$
View full question & answer→MCQ 211 Mark
A solid ball od radius $6\ cm$ is melted and then drawn into a wire of diameter $0.2\ cm.$ The length of wire is:
- A
$272m$
- ✓
$288m$
- C
$292m$
- D
$296m$
AnswerCorrect option: B. $288m$
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 $r_1$ 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
$\Rightarrow288\pi=\pi(0.1)^2\text{h}$
$\Rightarrow\text{h}=28800\text{cm}=288\text{m}$
View full question & answer→Question 221 Mark
A solid metallic cylinder of base radius 3cm and height 5cm is melted to make n solid cones of height 1cm and base radius 1mm. The value of n is:
- 450
- 1350
- 4500
- 13500
Answer
- 13500
Solution:
n = number of cones $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{cylinder}}{\text{Volume}\ \text{of}\ \text{one}\ \text{cone}}$
$=\frac{\pi(3)^2\times5}{\frac{1}{3}\pi\Big(\frac{1}{10}\Big)^2\times1}$
$=\frac{9\times5}{\frac{1}{3}\times\frac{1}{100}}$
$=\frac{45}{\frac{1}{300}}$
$=45\times300$
$=13500$ View full question & answer→MCQ 231 Mark
The volume of a cone is $1570\ cm^3$ and its height is $15\ cm.$ What is the radius of the cone? $($Use pi $=3.14).$
- ✓
$10\ cm$
- B
$9\ cm$
- C
$12\ cm$
- D
$8.5\ cm$
AnswerCorrect option: A. $10\ cm$
Let r be the radius of the cone.Volume $= 1570\ cm^3$
$1570=\frac{1}{3}\times3.14\times\text{r}^2\times15$
$\Rightarrow1570=3.14\times\text{r}^2\times15$
$\Rightarrow\text{r}^2=100$
$\Rightarrow\text{r}=10\text{ cm}$
View full question & answer→Question 241 Mark
If the ratio of the volumes of two spheres is 1 : 8 then the ratio of their surface area is:
- 1 : 2
- 1 : 4
- 1 : 8
- 1 : 16
Answer
- 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. View full question & answer→Question 251 Mark
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is:
- 1 : 2 : 3
- 2 : 1 : 3
- 2 : 3 : 1
- 3 : 2 : 1
Answer
- 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
$=\frac{1}{3}\pi\text{r}^2(\text{r}):\frac{2}{3}\pi\text{r}^3:\pi\text{r}^2(\text{r})$
$=\frac{1}{3}:\frac{2}{3}:1$
$=1:2:3$ View full question & answer→MCQ 261 Mark
How many persons can be accommodated in a dining hall of dimensions $(20m \times 15m \times 4.5m),$ assuming that each person requires $5m^3$ of air?
AnswerRequired number of persons $=\frac{\text{length}\times\text{breadth}\times\text{height}}{\text{amount}\ \text{of}\ \text{air}\ \text{each}\ \text{person}\ \text{requires}}$
$=\frac{20\times15\times4.5}{5}$
$=270$
View full question & answer→MCQ 271 Mark
The lateral surface area of a cube is $256m^2.$ The volume of the cube is:
- A
$64m^3$
- B
$216m^3$
- C
$256m^3$
- ✓
$512m^3$
AnswerCorrect option: D. $512m^3$
We know that,
Lateral surface area of a cube $= 4a^2$
$\Rightarrow 256 = 4a^2$
$\Rightarrow\text{a}^2=\frac{256}{4}$
$\Rightarrow a^2 = 64$
$\Rightarrow a = 8m$
Now,
Volume of the cube $= a^3$
$= 8^3$
$= 512m^3$
View full question & answer→MCQ 281 Mark
The volumes of two spheres are in the ratio $64 : 27$ and the sum of their radii is $7\ cm.$ The difference in their total surface areas is:
- A
$38\ cm^2$
- B
$58\ cm^2$
- C
$78\ cm^2$
- ✓
$88\ cm^2$
AnswerCorrect option: D. $88\ cm^2$
Let the radii be $x \ cm$ and $(7 - x)\ cm.$
Volume of the two spheres are in the ratio $64 : 27.$
$\Rightarrow\frac{\frac{4}{3}\pi\text{x}^3}{\frac{4}{3}\pi(7-\text{x})^3}=\frac{64}{27}$
$\Rightarrow\Big(\frac{\text{x}}{7-\text{x}}\Big)^3=\Big(\frac{4}{3}\Big)^3$
$\Rightarrow\frac{\text{x}}{7-\text{x}}=\frac{4}{3}$
$\Rightarrow\text{x}=4\ cm$
So, their radii are $4\ cm$ and $3\ cm.$
Difference of their tital surface areas
$=4\pi(4)^2-4\pi(3)^2$
$=4\times\frac{22}{7}(16-9)$
$=88\ cm^2$
View full question & answer→Question 291 Mark
A hemispherical bowl of radius 9cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3cm and height 4cm. How many bottles will be needed to empty the bowl?
- 27
- 35
- 54
- 63
Answer
- 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. View full question & answer→MCQ 301 Mark
The diameter of a roller, $1m$ long, is $84\ cm.$ If it takes $500$ complete revolutions to level a playground, the area of the playground is:
- A
$1440m^2$
- ✓
$1320m^2$
- C
$1260m^2$
- D
$1550m^2$
AnswerCorrect option: B. $1320m^2$
The diameter of the roller $=84\ cm=\frac{84}{100}\ m$
So, the radius $=\frac{84}{200}\text{m}$
The area covered by the roller in $1$ revolution
$=2\pi\text{rh}$
$=2\times\frac{22}{7}\times\frac{84}{200}\times1$
$=2.64\ m^2$
$\therefore$ Area covered in $500$ complete revolution $= 500 \times 2.64 = 1320m^2$
Thus, the area of the playground is $1320m^2.$
View full question & answer→MCQ 311 Mark
If the curved surface area of a cylinder is $1760\ cm^2$ and its base radius is $14\ cm,$ then its height is:
- A
$10\ cm$
- B
$15\ cm$
- ✓
$20\ cm$
- D
$40\ cm$
AnswerCorrect option: C. $20\ cm$
Given that, $r = 14\ cm$
Curved surface area $= 1760\ cm^2$
Now,
Curved surface area $=2\pi\text{rh}$
$\Rightarrow1760=2\times\frac{22}{7}\times14\times\text{h}$
$\Rightarrow1760=88\times\text{h}$
$\Rightarrow\text{h}=\frac{1760}{88}$
$\Rightarrow\text{h}=20\ cm$
View full question & answer→Question 321 Mark
The number of planks of dimension (4m × 5m × 2m) that can be stored in a pit which is 40m long, 12m wide and 16m deep, is:
- 190
- 192
- 184
- 180
Answer
- 192
Solution:
Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}}{\text{Volume}\ \text{of}\ 1\ \text{plank}}$
$=\frac{40\times12\times16}{4\times5\times2}$
$=192$ View full question & answer→MCQ 331 Mark
The height of a cone is $21\ cm$ and its slant height is $28\ cm.$ The volume of the cone is:
- A
$7356\ cm^3$
- ✓
$7546\ cm^3$
- C
$7506\ cm^3$
- D
$7564\ cm^3$
AnswerCorrect option: B. $7546\ cm^3$
Let r be the radius of the cone.
$l^2 = h^2 + r^2$
$\Rightarrow r^2 = l^2 - h^2$
$= 28^2 - 21^2$
$= 49 \times 7$
$\Rightarrow\text{r}=7\sqrt{7}\ cm$
Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow$ Volume of the cone $=\frac{1}{3}\times\frac{22}{7}\times(7\sqrt{7})^2\times21$
$\Rightarrow$ Volume of the cone $=7546\ cm^3$
View full question & answer→MCQ 341 Mark
The height of a cone is $24\ cm$ and the diameter of its base is $14\ cm.$ The curved surface area of the cone is:
- A
$528\ cm^2$
- ✓
$550\ cm^2$
- C
$616\ cm^2$
- D
$704\ cm^2$
AnswerCorrect option: B. $550\ cm^2$
The height of the cone is $24\ cm$ and the diameter of its base is $14\ cm.$
So, its radius $= 7\ cm.$
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{7^2+24^2}$
$\text{l}=\sqrt{49+576}$
$\text{l}=25\ cm$
So, curved surface area of the cone $=\pi\text{r}=\frac{22}{7}\times7\times25=550\ cm^2$
View full question & answer→Question 351 Mark
If the surface area of a sphere is $(144\pi)\text{m}^2$ then its volume is:
- $(288\pi)\text{m}^3$
- $(188\pi)\text{m}^3$
- $(300\pi)\text{m}^3$
- $(316\pi)\text{m}^3$
Answer
- $(288\pi)\text{m}^3$
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 $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\pi\times6\times6\times6$
$=4\times\pi\times2\times6\times6$
$=288\pi\text{m}^3$ View full question & answer→Question 361 Mark
How many lead shots, each 0.3cm in diameter, can be made from a cuboid of dimension 9cm × 11cm × 12cm?
- 7200
- 8400
- 72000
- 84000
Answer
- 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. View full question & answer→Question 371 Mark
The number of coins 1.5cm in diameter and 0.2cm thick to be melted to form a right circular cylinder of height 10cm and diameter 4.5cm is:
- 540
- 450
- 380
- 472
Answer
- 450
Solution:
Let the required number of coins be n.
Since the n coins are melted form a right circular cylinder,
$\text{n}\times\pi\times\Big(\frac{1.5}{2}\Big)^2\times0.2=\pi\times\Big(\frac{4.5}{2}\Big)^2\times10$
$\Rightarrow\text{n}\times\Big(\frac{15}{10}\Big)^2\times\frac{2}{10}=\Big(\frac{45}{20}\Big)^2\times10$
$\Rightarrow\text{n}\times\frac{9}{16}\times\frac{2}{10}=\frac{81}{16}\times10$
$\Rightarrow\text{n}=\frac{81\times10\times16\times10}{16\times9\times2}$
$\Rightarrow\text{n}=450$ View full question & answer→Question 381 Mark
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is:
- 27 : 20
- 20 : 27
- 4 : 9
- 9 : 4
Answer
- 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 $=\frac{\pi\text{r}_1\text{h}_1}{\pi\text{r}_2\text{h}_2}$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)\times\Big(\frac{\text{h}_1}{\text{h}_2} \Big)$
$=\Big(\frac{2}{3}\Big)^2\times\Big(\frac{5}{3}\Big)^2$
$=\frac{4}{9}\times\frac{5}{3}$
$=\frac{20}{27}$
$=20:27$ View full question & answer→MCQ 391 Mark
If each edge of a cube is increased by $50\%,$ then the percentage increase in its surface area is:
- A
$50\%$
- B
$75\%$
- C
$100\%$
- ✓
$125\%$
AnswerCorrect option: D. $125\%$
Let each edge be $a$
$\Rightarrow$ its surface area $= 6a^2$
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
$=\frac{\text{Increase}\ \text{in}\ \text{the}\ \text{surface}\ \text{area}}{\text{original}\ \text{surface}}\times100$
$=\frac{15\text{a}^2}{2}\times\frac{1}{6\text{a}^2}\times100$
$=125\%$
View full question & answer→MCQ 401 Mark
If the diameter of a cylinder is $28\ cm$ and its height is $20\ cm$ then its curved surface area is:
- A
$880\ cm^2$
- ✓
$1760\ cm^2$
- C
$3520\ cm^2$
- D
$2640\ cm^2$
AnswerCorrect option: B. $1760\ cm^2$
Given$, d = 628$
$r = 14\ cm$
$h = 20\ cm$
Now,
Curved surface area $=2\pi\text{rh}$
$=2\times\frac{22}{7}\times14\times20$
$=1760\ cm^2$
View full question & answer→MCQ 411 Mark
The surface area of a sphere is $1386\ cm^2.$ Its volume is:
- A
$1617\ cm^3$
- B
$3234\ cm^3$
- ✓
$4851\ cm^3$
- D
$9702\ cm^3$
AnswerCorrect option: C. $4851\ cm^3$
Given surface area of a sphere $= 1386\ cm^2$
$\Rightarrow4\pi\text{r}^2=1386$
$\Rightarrow4\times\frac{22}{7}\times\text{r}^2=1386$
$\Rightarrow\frac{88}{7}\times\text{r}^2=1386$
$\Rightarrow\text{r}^2=\frac{1386\times7}{88}$
$\Rightarrow\text{r}^2=\frac{441}{88}$
$\Rightarrow\text{r}=\sqrt{\frac{441}{4}}$
$\Rightarrow\text{r}=\frac{21}{2}\ cm$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}$
$=11\times21\times21$
$=4851\ cm^3$
View full question & answer→Question 421 Mark
A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is:
- 1 : 2
- 2 : 1
- 4 : 1
- $\sqrt2:1$
Answer
- 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. View full question & answer→MCQ 431 Mark
If each side of a cube is doubled, then its volume:
- A
- B
Becomes $4$ times.
- C
Becomes $6$ times.
- ✓
becomes $8$ times.
AnswerCorrect option: D. becomes $8$ times.
Let the original side be $'x\ ' \ cm$
$\therefore$ Original volume $= x^3 \ cm^3$
Now,
New side $= 2x \ cm$
$\therefore$ New volume $= (2x)^3$
$= 8x^3 \ cm^3$
So, the volume becomes $8$ times.
View full question & answer→MCQ 441 Mark
The surface area of a sphere of radius $21\ cm$ is:
- A
$2772\ cm^2$
- B
$1386\ cm^2$
- C
$4158\ cm^2$
- ✓
$5544\ cm^2$
AnswerCorrect option: D. $5544\ cm^2$
Surface area of a sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=4\times22\times3\times21$
$=5544\ cm^2$
View full question & answer→MCQ 451 Mark
$2.2\ dm^3$ of lead is to be drawn into a cylindrical wire $0.50\ cm$ in diameter. The length of the wire is:
- A
$110m$
- ✓
$112m$
- C
$98m$
- D
$124m$
AnswerCorrect option: B. $112m$
We know that $1\ dm = 10\ cm.$
Given that $2.2\ dm^3$ is drawn into a cylindrical wire.
That is, $(2.2 \times 1000) = 2200\ cm^3$ 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 $= 0.50\ cm)$
$\Rightarrow\text{h}=11200\ cm$
$\Rightarrow\text{h}=112\ m$
So, the length of the wire is $112m.$
View full question & answer→Question 461 Mark
The radius of a hemispheical balloon increase from 6cm to 12cm as air is being pumped into it. The ratio of the surface areas of the ballons in two cases is:
- 1 : 4
- 1 : 3
- 2 : 3
- 1 : 2
Answer
- 1 : 4
Solution:
Ratio of the surface areas of the ballons.
$\Rightarrow\frac{3\pi(6)^2}{3\pi(12)^2}$
$=\frac{1}{4}$ View full question & answer→MCQ 471 Mark
A river $1.5m$ deep and $30m$ wide is flowing at the rate of $3\ km$ per hour. The volume of water that runs into the sea per minute is:
- A
$2000m^3$
- ✓
$2250m^3$
- C
$2500m^3$
- D
$2750m^3$
AnswerCorrect option: B. $2250m^3$
Volume of the water running into the sea per hour $= 1.5 \times 30 \times 3000$
$...(1\ km = 1000m)$
$= 45 \times 3000$
Volume of the water running into the sea per minute $=\frac{45\times3000}{60}$
$=45\times50$
$=2250\ m^3$
View full question & answer→MCQ 481 Mark
In a shower, $5\ cm$ of rain falls. What is the volume of water that falls on $2$ hectares of ground?
- A
$500m^3$
- B
$750m^3$
- C
$800m^3$
- ✓
$1000m^3$
AnswerCorrect option: D. $1000m^3$
Volume of water that falls on ground $=2\times10000\times\frac{5}{100}$
$...(1$ hector $=10000m$ and $1\ cm =\frac{1}{100}m)$
$=1000\ m^3$
View full question & answer→MCQ 491 Mark
The curved surface area of a cylindrical pillar is $264m^2$ and its volume is $924m^3.$ The height of the pillar is:
AnswerGiven that,
Curved surface area $= 264m^2$
Volume $= 924m^3$
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 $=\pi\text{r}^2\text{h}$
$\Rightarrow924=\frac{22}{7}\times\frac{42}{\text{h}}\times\frac{42}{\text{h}}\times\text{h}$
$\Rightarrow924=22\times\frac{6}{\text{h}}\times42$
$\Rightarrow\frac{924}{22\times42\times6}=\frac{1}{\text{h}}$
$\Rightarrow\frac{924}{5544}=\frac{1}{\text{h}}$
$\Rightarrow\text{h}=\frac{5544}{924}$
$\Rightarrow\text{h}=6\ m$
View full question & answer→MCQ 501 Mark
The length, breadth and height of a cuboid are $15m, 6m$ and $5\ dm$ respectively. The lateral area of the cuboid is:
- A
$45m^2$
- ✓
$21m^2$
- C
$201m^2$
- D
$90m^2$
AnswerCorrect option: B. $21m^2$
Lateral surface area of the cuboid $= 2(l + b) \times h$
$=2(15+6)\times\frac{5}{10}$ $...\Big(1\ dm=\frac{1}{10}m \Rightarrow5\ dm=\frac{5}{10}m\Big)$
$=2(21)\times\frac{1}{2}$
$=21\ m^2$
View full question & answer→