Question 14 Marks
Find the capacity of a closed rectangular cistern whose length is $8\ m$, breadth $6\ m$ and depth $2.5\ m$. Also, find the area of the iron sheet required to make the cistern.
AnswerLength of the cistern,$ l = 8m$
Breadth of the cistern, $b = 6m$
Height (or depth) of the cistern, $h = 2.5m$
$\therefore$ Capacity of the cistern
= Volume of the cistern
$= l \times b \times h$
$= 8 \times 6 \times 2.5$
$= 120m^3$
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
$= 2(lb + bh + hl)$
$= 2(8 \times 6 + 6 \times 2.5 + 2.5 \times 8)$
$= 2 \times 83$
$= 166m^2$
View full question & answer→Question 24 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are: Length $= 24\ m$, breadth $= 25\ cm$ and height $= 6\ m$
AnswerHere,$ l = 24m, b = 25cm = 0.25m, h = 6m$
Volume of the cuboid $= l \times b \times h = (24 \times 0.25 \times 6)m^3 = 36m^3$
Total surface area $= 2(lb \times lh \times bh)$
$= 2(24 \times 0.25 + 24 \times 6 + 0.25 \times 6)m^2$
$= 2(6 + 144 + 1.5)m^2$
$= 2 \times 151.5m^2$
$= 303m^2$
Lateral surface area $= 2(l + b) \times h$
$= [2(24 + 0.25) \times 6]m^2$
$= [2 \times 24 \times 6]m^2$
$= 291m^2$
View full question & answer→Question 34 Marks
It is required to make a closed cylindrical tank of height $1\ m$ and base diameter $140\ cm$ from a metal sheet. How many square metres of the sheet are required for the same?
AnswerDiameter of a cylinder $= 140\ cm$
$\Rightarrow $ Radius,$r = 70\ cm$
Height $(h)$ of a cylinder $= 1m = 100\ cm$
Now,
Area of sheet required = Total surface area of cylinder
$=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times70(100+70)\Big]\text{cm}^2$
$=\big[2\times22\times10\times170\big]\text{cm}^2$
$=74800\text{cm}^2$
$=7.48\text{m}^2$
View full question & answer→Question 44 Marks
A cylindrical water tank of diameter $1.4m$ and height $2.1m$ is being fed by a pipe of diameter $3.5cm$ through which water flows at the rate of $2m$ per second. In how much time will the tank be filled?
AnswerSuppose the tank is filled in $x$ minutes. Then,
Volume of the water that flows out through the pipe in $x$ minutes
= Volume of the tank
$\Rightarrow\pi\times\Big(\frac{3.5}{2\times100}\Big)^2\times(2\times60\text{x})=\pi\times(0.7)\times2.1$
$\Rightarrow\Big(\frac{35}{2000}\Big)^2\times120\text{x}=\Big(\frac{7}{10}\Big)^2\times\Big(\frac{21}{10}\Big)$
$\Rightarrow\frac{35}{2000}\times\frac{35}{2000}\times120\text{x}=\frac{7}{10}\times\frac{7}{10}\times\frac{21}{10}$
$\Rightarrow\text{x}=\frac{7\times7\times21\times2000\times2000}{10\times10\times10\times35\times35\times120}$
$\Rightarrow\text{x}=28$
Hence, the tank will be flled in $28$ minutes.
View full question & answer→Question 54 Marks
A man uses a piece of canvas having an area of $551\ m^2$, to make a conical tent of base radius $7\ m$ . Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately $1\ m^2$, find the volume of the tent that can be made with it. (Use $\left.\pi=\frac{22}{7}\right)$.
AnswerRadius of a conical tent, $r = 7m$ Area of $49$ canvas used in making conical tent $= (551 - 1)m^2 = 550m^2$
$\Rightarrow $ Curved surface areaof a conical tent$ = 550m^2$
$\Rightarrow\pi\text{rl}=550$
$\Rightarrow\frac{22}{7}\times7\times\text{l}$
$\Rightarrow\text{l}=\frac{550}{22}=25\text{m}=\text{slant}\ \text{height}$
Now, $l^2 = r^2 + h^2$
$\Rightarrow 25^2 = 7^2 + h^2$
$\Rightarrow h^2 = 25^2 - 7^{2$
$} = 625 - 49 = 576$
$\Rightarrow h = 24m =$ height
$\therefore$ Volume of conical tent $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times7\times7\times24\Big)\text{m}^3$
$=1232\text{m}^3$
View full question & answer→Question 64 Marks
The sum of length, breadth and depth of a cuboid is $19\ cm$ and the length of its diagonal is $11\ cm$. The surface area of the cuboid.
AnswerLet the length and height (or depth) of the cuboid be $l\ cm, b\ cm$ and $h\ cm$, respectively.
$\therefore$$ l + b + h = 19 ....(1)$
Also, Length of the diagonal $= 11cm$
$\Rightarrow\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}=11$
$\Rightarrow\text{l}^2+\text{b}^2+\text{h}^2=121\ ...(2)$ Squaring $(1)$
we get $(l + b + h)^2 = 19^2$
$\Rightarrow l^2 + b^2 + h^2 + 2(lb + bh + hl) = 361$
$\Rightarrow 121 + 2(lb + bh + hl) = 361 [Using (2)]$
$\Rightarrow 2(lb + bh + hl) = 361 - 121 = 240cm^2$
Thus, the surface area of the cuboid is $240cm^2.$
View full question & answer→Question 74 Marks
A wall $15\ m$ long, $30\ cm$ wide and $4\ m$ high is made of bricks, each measuring $(22cm \times 12.5cm \times 7.5cm)$. If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?
AnswerLength of the wall $=15 m=1500 cm$
Breadth of the wall $=30 cm$
Height of the wall $=4 m=400 cm$
Volume of wall $=1500 \times 30 \times 400 cm^3=18000000 cm^3$
Now, volume of each brick $= 22 \times 12.5 \times 7.5cm^3$
$= 2062.5cm^3$
Also, volume of the morter $=\frac{1}{12}\times\text{Volume}\ \text{of}\ \text{the}\ \text{wall}$
$=\frac{18000000}{12}=1500000\text{cm}^3$
Total volume of the bricks in the wall
= Volume of the wall - Volume of the mortar
$= (18000000 - 1500000)cm^3 = 16500000cm^3$
$\therefore$ Number of briks
$=\frac{\text{Volume}\ \text{of}\ \text{bricks}}{\text{Volume}\ \text{of}\ \text{one}\ \text{brick}}$
$=\frac{16500000}{2062.5}=8000\ \text{bricks}$
View full question & answer→Question 84 Marks
A cylindrical bucket, $28\ cm$ in diameter and $72\ cm$ and high, is full of water. The water is emptied into a rectangular tank, $66\ cm$ long and $28\ cm$ wide. Find the height of the water level in the tank.
AnswerHere, cylindrical bucket has diameter $= 28\ cm$.
$\therefore$ radius $=\Big(\frac{28}{2}\Big)\text{cm}=14\text{cm}$ and height $= 72\ cm$.
Length of the tank $= 66\ cm$
Breadth of the tank $= 28\ cm$
$\therefore$ Volume of the tank = Volume of cylindrical bucket
$\Rightarrow\text{l}\times\text{b}\times\text{h}=\pi\text{r}^2\text{h}$
$\Rightarrow66\times28\times\text{h}=\frac{22}{7}\times14\times14\times72$
$\Rightarrow\text{h}=\Big(\frac{22\times2\times14\times72}{66\times28}\Big)\text{cm}$
$\Rightarrow\text{h}=24\text{cm}.$
$\therefore$ The height of the water level in the tank $= 24\ cm$.
View full question & answer→Question 94 Marks
The curved surface area of a cone is $308\ cm^2$ and its slant height is $14\ cm$. Find the radius of the base and total surface area of the cone. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerLet $r$ be the radius of a cone. Slant height of a cone, $l = 14cm$
Curved surface area of a cone $= 308cm^2$
$\Rightarrow\pi\text{rl}=308$
$\Rightarrow\frac{22}{7}\times\text{r}\times14=308$
$\Rightarrow22\times\text{r}\times2=308$
$\Rightarrow\text{r}=\frac{308}{22\times2}=7\text{cm}$
Total surface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times7(14+7)\Big]\text{cm}^2$
$=(22\times21)\text{cm}^2$
$=462\text{cm}^2$
View full question & answer→Question 104 Marks
Each edge of a cube is increased by $50\%$. Find the percentage increase in the surface area of the cube.
AnswerLet the initial edge of the cube be a units.
$\therefore$ Initial surface area of the cube $=6 a^2$
square units New edge of the cube $=a+50 \%$ of $a=a+\frac{50}{100} a=1.5 a$ units
$\therefore$ New surface of the cube $=6(1.5 a)^2=13.5 a^2$
square units Increase in surface area of the cube
$=13.5 a ^2-6 a ^2=7.5 a ^2$ square units
$\therefore$ Percentage increase in the surface area of the cube
$=\frac{\text{Increase}\ \text{in}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}{\text{Initial}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}\times100\%$
$=\frac{7.5\text{a}^2}{6\text{a}^2}\times100\%$
$=125\%$
View full question & answer→Question 114 Marks
There are $20$ cylindrical pillars in a building, each having a diameter of $50\ cm$ and height $4\ m$. Find the cost of cleaning them at ₹ $14\ per\ m^2.$
AnswerRadius $(r)$ of $1$ pillar $=\frac{50}{100\times2}=\frac{1}{4}\text{m}$
Height $(h)$ of $1$ pillar $= 4m$
$\therefore$ Lateral surface area of $1$ pillar $=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times\frac{1}{4}\times4\Big)\text{m}^2$
$=\frac{44}{7}\text{m}^2$
$\Rightarrow$ Lateral surface area of $20$ such pillars $=20\times\frac{44}{7}=\frac{880}{7}\text{m}^2$
Cost of cleaning $= ₹ 14/m^2$
$\Rightarrow$ Cost of cleaning $\frac{880}{7}\text{m}^2=₹\ \Big(14\times\frac{880}{7}\Big)$
$=₹\ 1760$
View full question & answer→Question 124 Marks
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is $10.4\ cm$ and its length is $25\ cm.$ The thickness of the metal is $8mm$ everywhere. Calculate the volume of the metal.
AnswerInternal diameter of the tube $= 10.4cm$
Internal radius$=\Big(\frac{10.4}{2}\Big)\text{cm}=5.2\text{cm}$ and
length $= 25\ cm$ and external radius
$= (5.2 + 0.8)\text{cm}= 6\text{cm}$
Required volume $=\big[\pi\times(6)^2\times25-\pi\times(5.2)^2\times25\big]\text{cm}^2$
$=\pi\times25\big[(6)^2-(5.2)^2\big]\text{cm}^3$
$=\frac{22}{7}\times25\big[36-27.04\big]\text{cm}^3$
$=\Big(\frac{22}{7}\times25\times8.96\Big)\text{cm}^3$
$=704\text{cm}^3$
$\therefore$ the volume of the metal $= 704\ cm^3.$
View full question & answer→Question 134 Marks
A classroom is $10\ m$ long, $6.4\ m$ wide and $5\ m$ high. If each student be given $1.6\ m^2$ of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?
AnswerLength of the classroom $=10 m$
Breadth of the classroom $=6.4 m$
Height of the classroom $=5 m$
Area of the floor $=$ length $\times$ breadth $=10 \times 6.4 m^2$
No. of students
$=\frac{\text{Area}\ \text{of}\ \text{the}\ \text{floor}}{\text{Area}\ \text{given}\ \text{to}\ \text{student}\ \text{on}\ \text{the}\ \text{floor}}$
$=\frac{640}{16}=40\ \text{students}$
Now, volume of the classroom $= 10 \times 6.4 \times 5m^3$
$\therefore$ Air required by each student $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{room}}{\text{Number}\ \text{of}\ \text{students}}$
$=\frac{10\times6.4\times5}{40}\text{m}^3$
$=8\text{m}^3$
View full question & answer→Question 144 Marks
The dimension of a room are $(9 m \times 8 m \times 6.5 m)$. It has one door of dimension $(2 m \times 1.5 m)$ and two window, each of dimension $(1.5 \times 1 m)$. Find the cost of whitewashing the walls at $₹ 25$ per square metre.
AnswerLength of the room, $I =9 m$
Breadth of the room, $b =8 m$
Height of the room, $h =6.5 m$
Now, Area of the walls to be whitewashed
$=$ Curved surface area of the room - Area of the door $-2 \times$ Area of each window
$=2 h( I + b )-2 m \times 1.5 m-2 \times 1.5 m \times 1 m=2 \times 6.5 \times(9+8)-3-3=221-6=215 m^2$
$\therefore$ Cost of whitewashing the walls at $₹ 25$ per square metre $=$ Area of the walls to be whitewashed $\times ₹ 25$ per square metre $=215 \times 25=₹ 5,375$
View full question & answer→Question 154 Marks
How many metres of cloth, $2.5m$ wide, will be required to make a conical tent whose base radius is $7m$ and height $24m?$ $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerHere, radius $= 7m$ and height $(h) = 24m$
$\therefore$ slant height $\text{l}=\sqrt{\text{h}^2+\text{r}^2}$
$=\sqrt{(24)^2+(7)^2}$ $\text{l}=\sqrt{576+49}$
$\text{l}=\sqrt{625}$ $\text{l}=25\text{m}$
Now, area of cloth $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{m}^2$
$=550\text{m}^2$
$\therefore$ length of cloth $=\frac{\text{Area}\ \text{of}\ \text{cloth}}{\text{Width}\ \text{of}\ \text{cloth}}$
$=\Big(\frac{550}{2.5}\Big)\text{m}$
$=220\text{m}$
$\therefore$ Length of cloth required to make conical tent $= 220m$.
View full question & answer→Question 164 Marks
Water in a canal, $30\ dm$ wide and $12dm$ deep, is flowing with a velocity of $20\ km$ per hour. How much area will it irrigate, if $9\ cm$ of standing water is desired?
AnswerWidth of the canal $=30 dm =3 m(1 m=10 dm )$
Depth of the canal $=12\ dm =1.2 m$
Speed of the water flow $=$ $20 km / h =20000 m / h$
$\therefore$ Volume of water flowing out of the canal in $1 h=3 \times 1.2 \times 20000=72000 m^3$
Height of standing water on field $-9 cm-0.09 m(1 m-100 cm)$
Assume that water flows of the canal for $1 h$ . Then,
Area of the field irrigated $=\frac{\text{Volume}\ \text{of}\ \text{water}\ \text{flowing}\ \text{out}\ \text{of}\ \text{the}\ \text{canal}}{\text{Height}\ \text{of}\ \text{standing}\ \text{water}\ \text{on}\ \text{the}\ \text{field}}$
$=\frac{72000}{0.09}$
$=800000\text{m}^2$
$=\frac{800000}{10000}$
$\big(1\ \text{hectare}=1000\text{m}^2\big)$
$=80\ \text{hectare}$
Thus, the area of the field irrigated is $=80\ \text{hectare}$.
Disclaimer: In this question time is not given,
so the question is solved assuming that the water flows out of the canal for i hour.
View full question & answer→Question 174 Marks
The curved surface area of a cylinder is $4400\ cm^2$ and the circumference of its base is $110\ cm$ . Find the height and the volume of the cylinder.
AnswerLet base radius be $r$ and height be $h$ Then, $2\pi\text{rh}=4400\text{cm}^2$ And, $2\pi\text{r}=110\text{cm}$
$\Rightarrow\frac{2\pi\text{rh}}{2\pi\text{r}}=\frac{4400}{110}$
$\Rightarrow\text{h}=40\text{cm}$
$\therefore2\times\frac{22}{7}\times\text{r}\times\text{h}\times40=4400\text{cm}.$
$\Rightarrow\text{r}=\Big(\frac{4400\times7}{44\times40}\Big)\text{cm}=\frac{35}{2}\text{cm}.$
$\therefore$ Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\times40\Big)\text{cm}^3$
$=38500\text{cm}^3.$
View full question & answer→Question 184 Marks
Find the weight of a solid cylinder of radius $10.5\ cm$ and height $60\ cm$ if the material of the cylinder weighs $5$ g per $cm ^3$.
AnswerHere, radius $( r )=10.5\ cm$ and height $=60\ cm$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10.5\times10.5\times60\Big)\text{cm}^3$
$=20790\text{cm}^3$
$\therefore$ Weight of the solid cylinder if the material of the cylinder.
Weighs $5g$ per $cm3 = (20790 \times 5) = 103950g$
$=\frac{103950}{1000}$
$\big[\therefore1000\text{g}=1\text{kg}\big]$
$=103.95\text{kg}$
View full question & answer→Question 194 Marks
$1\ cm^3$ of gold is drawn into a wire $0.1\ mm$ in diameter. Find the length of the wire.
Answer$1 cm^3=1 cm \times 1 cm \times 1 cm=0.01 m$
Therefore, Volume of the gold $=0.01 m \times 0.01 m \times 0.01 m=0.000001 m^3 \ldots(1)$
Diameter of the wire drawn $=0.1 mm$
Radius of the wire drawn
$=\frac{0.1}{2} mm=0.05 mm r =0.00005 m \ldots...(2)$
Length of the wire $= h m \ldots...(3)$
Volume pf the wire drawn $=$ Volume of the gold
$\Rightarrow\pi\text{r}^2\text{h}=0.000001$
$\Rightarrow\pi\times0.00005\times0.00005\times\text{h}=0.000001$ $[$from equations $(1), (2)$ and $(3)]$
$\text{h}=\frac{0.00000\times7}{0.00005\times0.00005\times22}$
$\therefore\text{h}=127.27\text{m}$
$\therefore$ the length of the wire is $127.27m.$
View full question & answer→Question 204 Marks
The curved surface area of a cylinder is $1210 cm^2$ and its diameter is $20\ cm$ . Find its height and volume.
AnswerHere, curved surface area $= 1210\ cm^2$
Diameter $= 20\ cm$
$\Rightarrow\text{radius}=\frac{20}{2}=10\text{cm}$
$\therefore$ Curved surface area of the cylinder $=2\pi\text{rh}$
$\Rightarrow1210=2\times\frac{22}{7}\times10\times\text{h}$
$\Rightarrow\text{h}=\Big(\frac{1210\times7}{2\times22\times10}\Big)\text{cm}=19.25\text{cm}$
$\therefore\text{Height}=19.25\text{cm}$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10^2\times19.25\Big)\text{cm}^3$
$=\Big(\frac{22}{7}\times10\times10\times19.25\Big)\text{cm}^3$
$=6050\text{cm}^3$
$\therefore$ Volume of the cylinder $= 6050cm^3.$
View full question & answer→Question 214 Marks
Water flows at the rate of $10$ metres per minute through a cylindrical pipe $5\ mm$ in diameter. How long would it take to fill a conical vessel whose diameter art the surface $40\ cm$ and depth $24\ cm$?
AnswerDiameter of the pipe $= 5\ mm = 0.5\ cm$
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe $= 10$ metres $= 1000\ cm$
Volume that flows in $1$ min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$
$=51\ \text{min}\ 12\ \text{sec}$
View full question & answer→Question 224 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are: Length $= 26\ m,$ breadth $= 14\ m$ and height $= 6.5\ m$
AnswerHere, $I =26 m, b =14 m, h =6.5 m$
Volume of the cuboid $= I \times b \times h =(26 \times 14 \times 6.5) m ^3=2366 m^3$
Total surface area $=2( lb \times lh \times bh )=2(26 \times 14+26 \times 6.5+14 \times 6.5) m ^2$ $=2(364+169+91) m ^2=2 \times 624 m^2=1248 m^2$
Lateral surface area $=2(1+b) \times h=[2(26+14) \times 6.5] m ^2$
$=[2 \times 40 \times 6.5] m^2=520 m^2$
View full question & answer→Question 234 Marks
It costs $₹ 3300$ to paint the inner curved surface of a cylindrical vessel $10\ m$ deep at the rate of ₹ $30$ per $m^2$. Find the:
$i.$ Inner curved surface area of the vessel.
$ii.$ Inner radius of the base.
$iii.$ Capacity of the vessel.
Answer$i.$ Cost of painting inner curved surface area of vassel
$=$ Cost of painting per $m^2 \times $ Inner curved surface of vessel
$\Rightarrow $ ₹ $3300 = ₹ 30 \times $ Inner curved surface of vessel
$\Rightarrow $ Inner curved surface of vessel $= 110\ m^2$
$ii.$ Let inner radius of the base $= r$
Depth, $h = 10\ m$
Inner curved surface of vessel $=2\pi\text{rh}$
$\Rightarrow10=2\times\frac{22}{7}\times\text{r}\times10$
$\Rightarrow\text{r}=\frac{110\times7}{2\times22\times10}=1.75\text{ m}$
$iii.$ Capacity of the vessel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times1.75\times1.75\times10\Big)\text{m}^3$
$=96.25\text{ m}^3$
View full question & answer→Question 244 Marks
The capacity of a closed cylindrical vessel of height $1m$ is $15.4$ litres. Find the area of the metal sheet needed to make it.
AnswerVolume of a cylinder $= 15.4$ litres $= 15400cm^3$
Height $(h)$ of a cylinder $= 1\ m = 100\ cm$
Volume of a cylinder $=\pi\text{r}^2\text{h}$
$\Rightarrow15400=\Big(\frac{22}{7}\times\text{r}^2\times100\Big)$
$\Rightarrow\text{r}^2=\frac{15400\times7}{22\times100}=49$
$\Rightarrow\text{r}=7\text{cm}$
Now, Area of metal sheet needed = Total surface area of cylinder $=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times7(100+7)\Big]\text{cm}^2$
$=\big[2\times22\times107\big]\text{cm}^2$
$=4708\text{cm}^2$
View full question & answer→Question 254 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are: Length $= 15\ m$, breadth $= 6\ m$ and height $= 5\ dm$
AnswerHere, $l = 15\ m, b = 64\ m, h = 5 dm = 0.5$
$3$Volume of the cuboid $= l \times b \times h = (15 \times 6 \times 0.5)m^3 = 45m^3$
Total surface area $= 2(lb \times lh \times bh) = 2(15 \times 6 + 15 \times 0.5 + 6 \times 0.5)m^2$
$= 2(90 + 7.5 + 3)m^2 = 2 \times 100m^2 = 201m^2$
Lateral surface area $= 2(l + b) \times h = [2(15 + 6) \times 0.5]m^2$
$= [2 \times 21 \times 0.5]m^2 = 21m^2$
View full question & answer→Question 264 Marks
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are $6\ cm$ and $10\ cm$ respectively. $\big(\text{Take}\ \pi=3.14\big)$
AnswerHere, height $(h) = 6\ cm$ and slant height $(l) = 10\ cm$
$\therefore$ radius $(r)$ $=\sqrt{\text{l}^2-\text{h}^2}$
$\text{r}=\sqrt{10^2-6^2}$ $\text{r}=\sqrt{100-36}$
$\text{r}=\sqrt{64}$ $\text{r}=8\text{cm}$
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times3.14\times8\times8\times8\Big)\text{cm}^3$
$=401.92\text{cm}^3$
$\therefore$ Curved surface area $=\pi\text{rl}$
$=\big(3.14\times8\times10\big)\text{cm}^2$
$=251.2\text{cm}^2$
$\therefore$ Total surface area $=\pi\text{r}(\text{l}+\text{r})$
$=\pi\text{r}(10+8)$ $=(3.14\times8\times18\big)\text{cm})^2$
$=452.16\text{cm}^2$
View full question & answer→Question 274 Marks
How many planks of dimensions $(5\ m \times 25\ cm \times 10\ cm)$ can be stored in a pit which is $20\ m$ long, $6\ m$ wide and $80\ cm$ deep?
AnswerNumber of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}\ \text{in}\ \text{cm}^3}{\text{Volume}\ \text{of}\ 1\ \text{plank}\ \text{in}\ \text{cm}^3}$
Volume of one plank $= (l \times b \times h)cm^3 = 500 \times 25 \times 10cm^3 = 125000cm^3$
Volume of the pit $= (l \times b \times h)cm^3$
Here, $l = 20\ m = 2000\ cm, b = 6\ m = 600\ cm, h = 80\ cm$
i.e., volume of the pit $= 2000 \times 600 \times 80cm^3 = 96000000\ cm^3$
$\therefore$ Number of planks $=\frac{96000000}{125000}$
$=\frac{96000}{125}=768$
View full question & answer→Question 284 Marks
Water flows at the rate of $10$ metres per minute through a cylindrical pipe $5\ mm$ in diameter. How long would it take to fill a conical vessel whose diameter art the surface $40\ cm$ and depth $24\ cm$?
AnswerDiameter of the pipe $= 5\ mm = 0.5\ cm$
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe $= 10$ metres $= 1000\ cm$
Volume that flows in $1$ min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$ $=51\ \text{min}\ 12\ \text{sec}$
View full question & answer→Question 294 Marks
The inner diameter of a cylindrical wooden pipe is $24\ cm$ and its outer diameter is $28\ cm$.
The length of the pipe is $35\ cm$. Find the mass of the pipe, if $1cm^3$ of wood has a mass of $0.6g.$
AnswerInternal diameter of a cylinder $= 24cm$
$\Rightarrow$ Internal radius of a cylinder,$ r = 12cm$
External diameter of a cylinde$r = 28cm$
$\Rightarrow$ External radius of a cylinder, $R = 14cm$
Length of the pipe, i.e. height, $h = 35cm$
Now, Volume of pipe = Volume of cylinder $=\pi(\text{R}^2-\text{r}^2)\text{h}$
$=\Big[\frac{22}{7}(14^2-12^2)\times35\Big]\text{cm}^3$
$=\big[22\times(196-144)\times5\big]\text{cm}^3$
$=(22\times52\times5)\text{cm}^3$
$=5720\text{cm}^3$ Given, $1cm3 = 0.6g$
$\therefore$ Mass of pipe $= (5720 \times 0.6)g = 3432g = 3.432kg$
View full question & answer→Question 304 Marks
The barrel of a fountain pen, cylindrical in shape, is $7cm$ long and $5mm$ in diameter. A full barrel of ink in the pen will be used up on writing $330$ words on an average. How many words would use up a bottle of ink containing one fifth of a litre?
AnswerLength $= 7\ cm$
= (height) Diameter $= 5mm$
$\Rightarrow\text{radius}=\Big(\frac{5}{2}\Big)\text{mm}=2.5\text{mm}$ $=0.25\text{cm}$
$\therefore$ Volume of the barrel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times0.25\times0.25\times7\Big)\text{cm}^3$
$=\frac{11}{8}\text{cm}^3$ $\frac{11}{8}\text{cm}^3$ isused for writing $330$ words.
So, $\Big(\frac{1}{5}\times1000\Big)\text{cm}^3$ will be used for writing $\Big(330\times\frac{8}{11}\times\frac{1}{5}\times1000\Big)\text{words}$
$=48000\ \text{words}$
View full question & answer→Question 314 Marks
A bus stop is barricated from the remaining part of the road by using $50$ hollow cones made of recycled cardboard. Each one has a base diameter of $40\ cm$ and height $1\ m$. If the outer side of each of the cones is to be painted and the cost of painting is $₹ 25$ per $m^2$, what will be the cost of painting all these cones? $\big(\text{Use}\ \pi=3.14\ \text{and}\ \sqrt{1.04}=1.02\big)$
AnswerRadius of a cone, $r = 20cm$
$=\frac{20}{100}\text{m}=\frac{1}{5}\text{m}$
Height of a cone, $h = 1m$
$\therefore$ Slant height of a cone, $\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{\Big(\frac{1}{5}\Big)^2+1}$
$\text{l}=\sqrt{\frac{26}{25}}$
$\text{l}=\sqrt{1.04}$
$\text{l}=1.02$
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(3.14\times\frac{1}{5}\times1.02\Big)\text{m}^2$
$=\frac{3.2028}{5}\text{m}^2$
$\Rightarrow$ Curved surface area of $50$ cones
$=\Big(50\times\frac{3.2028}{5}\Big)\text{m}^2$
$=32.028\text{m}^2$
Cost of painting $= ₹ 25 per m^2$
$\Rightarrow$ Cost of painting $32.028m^2$
area $= ₹ (25 \times 32.028) = ₹ 800.70$
View full question & answer→Question 324 Marks
A box made of sheet metal costs ₹ $6480$ at ₹ $120$ per square metre. If the box is $5m$ long and $3\ m$ wide, find its height.
AnswerLength of the box $= 5m$
Breadth of the box $= 3m$
Area of the sheet required
$=\frac{\text{Total}\ \text{cost}}{\text{Cost}\ \text{per}\ \text{metre}\ \text{square}}$
Let $h$ m be the height of the box.
Then area of the sheet = total surface area of the box
$= 2(lb + lh + bh)m^2 =2(5 \times 3 + 5 \times h + 3 \times h)m^2$
$= 2(15 + 8h) = (30 + 16h )m^2$
Now, $30 + 16\text{h}=\frac{6480}{120}$
$\Rightarrow 30 + 16h = 54$
$\Rightarrow 16h = 24$
$\Rightarrow h = 1.5m$
$\therefore$ The height of the box is $1.5m.$
View full question & answer→Question 334 Marks
The volume of a sphere is $38808cm^3$. Find its radius and hence its surface area. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerVolume of the sphere $= 38808cm^3$
Suppose that $r\ cm$ is the radius of the given sphere.
$\therefore\frac{4}{3}\pi\text{r}^3=38808$
$\Rightarrow\text{r}^3=\frac{38808\times3\times7}{4\times22}=9261$
$\Rightarrow\text{r}=\sqrt[3]{9261}=21\text{cm}$
$\therefore$ Surface area of the sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=5544\text{cm}^2$
View full question & answer→Question 344 Marks
A conical tent is $10m$ high and the radius of its base is $24\ m$. Find the slant height of the tent. If the cost of $1m^2$ canvas is $₹ 70$, find the cost of canvas required to make the tent. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius of a conical tent, $r = 24m$
Height of a conical tent, $h = 10m$
$\therefore$ Slant height of a conical tent,
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{24^2+10^2}$
$\text{l}=\sqrt{576+100}$
$\text{l}=\sqrt{676}$
$\text{l}=26\text{cm}$
Curved surface area of a conical tent $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times24\times26\Big)\text{m}^2$
$=\frac{13728}{7}\text{m}^2$ Cost of $1m^2$ canvas $= ₹ 70$
$\Rightarrow\text{Cost}\ \text{of}\ \frac{13728}{7}\text{m}^2\ \text{canvas}$
$=₹\ \Big(70\times\frac{13728}{7}\Big)=₹\ 137280$
View full question & answer→Question 354 Marks
The surface area of a cuboid is $758\ cm^2$. Its length and breadth are $14\ cm$ and $11\ cm$ respectively. Find its height.
AnswerLength of the cuboid $= 14\ cm$ Breadth of the cuboid $= 11\ cm$
Let the height of the cuboid be $x cm$.
Surface area of the cuboid $= 758\ cm^2$
Then $758 = 2(14 \times 11 + 14 \times x + 11 \times x)$
$\Rightarrow 758 = 2(154 + 14x + 11x)$
$\Rightarrow 758 = 2(154 + 25x)$
$\Rightarrow 758 = 308 + 50x$
$\Rightarrow 50x = 758 - 308 = 450$
$\Rightarrow\text{x}=\frac{450}{50}=9$
$\therefore$ The height of the cuboid is $9\ cm.$
View full question & answer→Question 364 Marks
Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are: Length $= 12cm,$ breadth $= 8\ cm$ and height $= 4.5\ cm$
AnswerHere, $l = 12cm, b = 8cm, h = 4.5cm$
Volume of the cuboid $= l \times b \times h = (12 \times 8 \times 4.5)cm^3 = 432cm^3$
Total surface area $= 2(lb \times lh \times bh) = 2(12 \times 8 + 12 \times 4.5 + 8 \times 4.5)cm^2$
$= 2(96 + 54 + 36)cm^2$
$= 2 \times 186cm^2 = 372cm^2$
Lateral surface area $= 2(l + b) \times h = [2(12 + 8) \times 4.5]cm^2$
$= [2(20) \times 4.5]cm^2 = 40 \times 4.5cm^2 = 180cm^2$
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