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Maths 3 Mark Question

Volume and Surface Area of Solids · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 38 questions.

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Question 13 Marks
How many persons can be accommodated in a hall of length 16 m . Breadth 12.5 m and height 4.5 m , assuming that $3.6 m^3$ of air is required for each person?
Answer
Length of hall $(1)=16 m$
$\text { Breadth }(b)=12.5 m$
$\text { Height }(h)=4.5 m$
$\text { Volume of air in it }=I \times b \times h$
$=16 \times 12.5 \times 4.5 m^3$
$=900 m^3$
Air for one person is required $=3.6 m^3$
Number of person which can be accommodated in the hall $=\frac{900}{3.6}$
$=\frac{900 \times 10}{36}$
$=250$
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Question 23 Marks
A particular brand of talcum powder is available in two packs, a plastic can with a square base of side 5cm and of height 14cm, or one with a circular base of radius 3.5cm and of height 12cm. Which of them has greater capacity and by how much?
Answer
In frist case,
Side of square base (a) $=5 cm$.
And height $( h )=14 cm$.
Volume $=5 \times 5 \times 14=350 cm^3$
In second case,
Radius of the circular base $( r )=3.5 cm$.
Height $(h)=12 cm$.
Volume $\pi r ^2 h$
$=\frac{22}{7} \times 3.5 \times 3.5 \times 12 cm^3$
$=462 cm^2$
Hence second type of circular plastic can has greater capacity.
Difference $=462$ - 350
$=112 cm^3 .$
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Question 33 Marks
The curved surface area of a cylindrical pillar is $264 m^2$ and its volume is $924 m^3$. Find the diameter and height of the pillar.
Answer
Curved surface area $=2\pi\text{rh}=264\text{m}^2$
$\therefore\text{r}=\frac{264}{2\pi\text{h}}=\frac{132}{\pi\text{h}}\text{m}$
Volume $=\pi\text{r}^2\text{h}=\pi\times\frac{132}{\pi\text{h}}\times\frac{132}{\pi\text{h}}\times\text{h}=924\text{m}^3$
$\therefore\text{h}=-\frac{132\times132\times7}{22\times924}=6\text{m}$
Now, $\text{r}=\frac{132}{\pi\text{h}}=\frac{13 2\times7}{22\times6}=7\text{m}$
i.e., deameter of the piller, d = 7 × 2 = 14m.
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Question 43 Marks
A swimming pool is 260m long and 140m wide. If 54600 cubic metres of water is pumped into it, find the height of the water level in it.
Answer
Length of pool $=260 m$,
and width $=140 m$.
Volume of water $=54600 m^3$
$\therefore$ Height of water $=\frac{\text{Volume}}{\text {Length}\times\text{breadth}}$
$=\frac{546000}{260\times140}\text{m}$
$=\frac{3}{2}=1.5\text{m}$
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Question 53 Marks
How many bricks, each of size 25cm × 13.5cm × 6cm, will be required to build a wall 8m long, 5.4m high and 33cm thick?
Answer
Length of wall $(I)=8 m=800 cm$
Height $(h)=5.4 m=540 cm$
Width $( b )=33 cm$
$\therefore$ Volume of wall $=1 \times b \times h$,
$=800 \times 540 \times 33 cm^3$
Volume of one brick $=25 \times 13.5 \times 6 cm^3$
$=2025 cm^3$
$\therefore$ No. of bricks $=\frac{14256000}{2025}=7040$
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Question 63 Marks
The rainfall recorded on a certain day was 5cm. Find the volume of water that fell on a 2-hectare field.
Answer
Area of field $=2$ hectare,
$=20000 m^2$
$\text { Rainfall }=5 cm .=005 m$
Volume of water of rainfall,
$=$ Area of field $\times$ height of rainfall water,
$=20000 \times 0.05 m^3$
$=1000 m^3$
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Question 73 Marks
Find the cost of painting 15 cylindrical pillars of a building at Rs. 2.50 per square metre if the diameter and height of each pillar are 48cm and 7 metres respectively.
Answer
Diameter of a cylindrical pillar = 48cm.
Radius (r) $=\frac{48}{2}=24\text{cm}.$
$\frac{24}{100}\text{m}$
Height (h) = 7m
$\therefore$ Lateral surface area of one pillar
$=2\pi\text{rh}=2\times\frac{22}{7}\times\frac{24}{100}\times7\text{m}^2$
$=\frac{1056}{100}\text{m}^2$
Surface area of 15 pillars
$=\frac{1056}{100}\times15\text{m}^2=\frac{15840}{100}\text{m}^2$
Rate of painting = Rs. 2.50 per sq. m.
$\therefore$ Total cost $=\text{Rs.}2.50\times\frac{15840}{100}$
$=\text{Rs}.396$
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Question 83 Marks
A box with a lid is made of wood which is 3cm thick. Its external length, breadth and height are 56cm, 39cm and 30cm respectively. Find the capacity of the box. Also find the volume of wood used to make the box.
Answer
Outer length of the box $=56 cm$
$\text { Width }=39 cm$
$\text { And height }=30 cm$
$\text { Volume }=56 \times 39 \times 30$
$=65520 cm^3$
Thickness of wood used $=3 cm$.
$\therefore \text { Inner length }=56-2 \times 3$
$=56-6$
$=50 cm$
$\text { Width }=39-2 \times 3$
$=39-6$
$=33 cm$
$\text { And height }=30-2 \times 3$
$=30-6$
$=24 cm$
$\therefore$ Inner volume of the box $=50 \times 33 \times 24 cm^3$
$=39600 cm^3$
And volume of wood used = Outer volume - Inner volume
$=(65520-39600) cm^3$
$=25920 cm^3$
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Question 93 Marks
The dimensions of a rectangular water tank are 2m 75cm by 1m 80cm by 1m 40cm. How many litres of water does it hold when filled to the brim?
Answer
Length of water tank $(1)=2 m$
$75 cm=2.75 m$
Breadth (b) $=1 m 80 cm=1.80 m$,
and height $(h)=1 m 40 cm=1.40 m$
Volume of water filled in it $=$ l.b.h $=2.75 \times 1.80 \times 1.40 m^3$
$=6.93 m^3$
Water in litres $=6.93 \times 1000$
$=6930$ litres $\left(1 m^3=1000\right.$ litres $)$
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Question 103 Marks
Find the number of coins, 1.5cm in diameter and 0.2cm thick, to be melted to form a right circular cylinder with a height of 10cm and a diameter of 4.5cm.
Answer
Volume of the coin $=\pi\text{r}^2\text{h}=\frac{22}{7}\times0.75\times0.75\times0.2$
Volume of the cylinder $=\pi\text{r}^2\text{h}=\frac{22}{7}\times2.25\times2.25\times10$
No. of coins $=\frac{\text{Volume of cylinder}}{\text{Volume of coin}}$
$=\frac{2.25\times2.25\times10}{0.75\times0.75\times0.2}=450\text{ coins}$
$\therefore$ 450 coins must be melted to form the required cylinder.
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Question 113 Marks
The circumference of the base of a cylinder is 88cm and its height is 60cm. Find the volume of the cylinder and its curved surface area.
Answer
Circumference of the base of cylinder = 88cm.
$\because\text{radius}=\frac{\text{Circumference}}{2\pi}=\frac{88\times7}{2\times22}=14\text{cm}$
Height (h) = 60cm.
$\therefore\text{Volume}=\pi\text{r}^2\text{h}=\frac{22}{7}\times14\times14\times60\text{cm}^3$
$=36960\text{cm}^3$
And curved surface area $=2\pi\text{rh}$
$=88 \times 60=5280 cm^2$
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Question 123 Marks
The lateral surface area of a cylinder of length 14 m is $220 m^2$. Find the volume of the cylinder.
Answer
Lateral surface of cylinder $=220 m^2$
Height $(h)=14 m$
Let radius of a cylinder $=r$
Then $2 \pi rh =220$
$\Rightarrow 2 \times \frac{22}{7} \times r \times 14=220$
$\Rightarrow r=\frac{220 \times 7}{2 \times 22 \times 14}=\frac{5}{2}=2.5 m$
$\therefore \text { Volume }=\pi r^2 h$
$=\frac{22}{7} \times(2.5)^2 \times 14 m^3$
$=\frac{22}{7} \times 2.5 \times 2.5 \times 14 m^3$
$=22 \times 6.25 \times 2=275 m^3$
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Question 133 Marks
If the length of each edge of a cube is doubled, how many times does its volume become? How many times does its surface area become?
Answer
Let edge of given cube $= a$
Volume $=a^3$
And surface area $=6 a ^2$
By doubing the edge of cube 3 the side of new cube $= a \times 2=2 a$
Volume $(2 a)^3=8 a^3$
And surface area $=6(2 a)^2=6 \times 4 a^2$
$=24 a^2=4 \times 6 a^2$
It is clear from the above that,
Volume is increased 8 times and surface area is 4 times.
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Question 143 Marks
A rectangular vessel 22cm by 16cm by 14cm is full of water. If the total water is poured into an empty cylindrical vessel of radius 8cm, find the height of water in the cylindrical vessel.
Answer
Length of rectangular vessel $(l)=22 cm$.
Breadth $(A)=16 cm$.
And height $(A)=14 cm$.
$\therefore$ Volume of vessel $= lbh$
$=22 \times 16 \times 14 cm^3=4928 cm^3$
Volume of water in the cylindrical vessel $=4928 cm^3$
Radius $( r )=8 cm$.
Let height of water in the vessel $= h$
$\therefore \pi r^2 h=4928$
$\Rightarrow \frac{22}{7} \times 8 \times 8 \times h=4928$
$\Rightarrow h=\frac{4928 \times 7}{22 \times 8 \times 8} \Rightarrow h=24.5$
Hence height of water $=24.5 cm$.
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Question 153 Marks
Find the volume, lateral surface area and the total surface area of a cube whose edges measures:
$5.6cm.$
Answer
Edge of cube $(a)=5.6 cm$.

a. Volume $=a^3=(5.6)^3 cm^3$

$=5.6 \times 5.6 \times 5.6 cm^2=175.616 cm^3$

b. Lateral surface area $=4 a ^2$

$=4(5.6)^2=4 \times 5.6 \times 5.6 cm^2$

$=125.44 cm^2$

c. Total surface area $=6 a ^2=6 \times(5.6)^2 cm^2$

$=6 \times 5.6 \times 5.6 cm^2=188.16 cm^2$
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Question 163 Marks
A pit 5 m long and 3.5 m wide is dug to a certain depth. If the volume of earth taken out of it is $14 m^3$, what is the depth of the pit?
Answer
Length of pit $(1)=5 m$
Width $(b)=3.5 m$
Let depth of pit $=h$
Then volume of earth dug out,
$=\text { I.b.h }=5 \times 3.5 \times h=17.5 h m^3$
But volume of earth $=14 m^3$
$17.5 h=14$
$h=\frac{14}{17.5}=\frac{140}{175}$
$\Rightarrow h=\frac{4}{5} m$
$=\frac{4}{5} \times 100$
$=80 cm$
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Question 173 Marks
The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm . It is melted and recast into cubes, each of side 45 cm . How many cubes are formed?
Answer
Length of metal block $( l )=2.25 m=225 cm$
Width (b) $=1.5 m=150 cm$
And height ( $h$ ) $=27 cm$
Volume of block $=1 \times b \times n$
$=225 \times 150 \times 27 cm^2$
$=911250 cm^3$
Side of each cube $=45 cm$.
Volume of each cube $=a^2$
$=45 \times 45 \times 45$
$=94125 cm^3$
Number of cubes $=\frac{911250}{91125}$
$=10$
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Question 183 Marks
Find the volume, lateral surface area and the total surface area of a cube whose edges measures:
7m.
Answer
Edge of cube (a) $=7 m$



a. Volume $=a^3=(7)^3$

$=7 \times 7 \times 7 m^3$

$=343 m^3$

b. Leteral surface area $=4 a ^2$

$=4(7)^2=4 \times 49 m^2=196 m^2$

c. Total surface area $=6 a^2=6(7)^2 m^2$

$=6 \times 49=294 m^2$
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Question 193 Marks
Find the volume of a cube whose total surface area is $384 cm^2$.
Answer
Total surface area $=6 a ^2$
$\Rightarrow6\text{a}^2=384$
$\Rightarrow\text{a}=\sqrt{\frac{384}{6}}=8\text{cm}$
$\therefore$ Volume $=\text{a}^2=512\text{cm}^3$
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Question 203 Marks
The size of a matchbox is $4 cm \times 2.5 cm \times 1.5 cm$. What is the volume of a packet containing 144 matchboxes? How many such packets can be placed in a carton of size $1.5 m \times 84 cm \times 60 cm$ ?
Answer
Volume of one matchbox $=4 \times 2.5 \times 1.5 cm^3=15 cm^3$
Volume of 144 matchbox $=15 \times 144 cm^3$
or volume of one packet $=2160 cm^3$
Length of catron $( I )=1.5 m=150 cm$.
Breadth (b) $=84 cm$.
Volume of one carton $=1 \times b \times h$,
$=1502 \times 84 \times 60 cm^3$
$=756000 cm^3$
No. of packets $=\frac{756000}{2160}$
$=350$
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Question 213 Marks
The area of a courtyard is $3750 m^2$. Find the cost of covering it with gravel to a height of 1 cm if the gravel costs Rs. 6.40 per cubic metre.
Answer
Area of courtyard $=3750 m^2$
Height of gravel $=1 cm$.
Volume of gravel $=3750 \times \frac{1}{100} m^3$
$=37.50 m^3$
Cost of $1 m^3$ gravel $=$ Rs. 6.40
Total cost $=$ Rs. $6.40 \times 37.50$
$=\text { Rs. } 240$
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Question 223 Marks
A box with a lid is made of wood which is 3cm thick. Its external length, breadth and height are 56cm, 39cm and 30cm respectively. Find the capacity of the box. Also find the volume of wood used to make the box.
Answer
Outer length of the box $=56 cm$
$\text { Width }=39 cm$
$\text { And height }=30 cm$
$\text { Volume }=56 \times 39 \times 30$
$=65520 cm^3$
Thickness of wood used $=3 cm$.
\$1therefore $\$$ Inner length = $56-2 \times 3$
$=56-6$
$=50 cm$
$\text { Width }=39-2 \times 3$
$=39-6$
$=33 cm$
And height $=30-2 \times 3$
$=30-6$
$=24 cm$
\$1therefore $\$$ Inner volume of the box $=50 \times 33 \times 24 cm^3$
$=39600 cm^3$
And volume of wood used = Outer volume - Inner volume
$=(65520-39600) cm^3$
$=25920 cm^3$
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Question 233 Marks
Find the volume, lateral surface area and the total surface area of a cube whose edges measures:
8dm, 5cm.
Answer
Edge of cube $( a )=8 dm 5 cm=85 cm$.

a. Volume $=a^3=(85)^3 cm^3$

$=85 \times 85 \times 85=614125 cm^3$

a. Lateral surface area $=4 a ^2$

$=4 \times(85)^2 cm^2$

$=4 \times 85 \times 85=28900 cm^2$

c. Total surface area $=6 a ^2=6(85)^2 cm^2$

$=6 \times 85 \times 85=43350 cm^2$
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Question 243 Marks
The volume of a circular iron rod of length 1 m is $3850 cm^3$. Find its diameter.
Answer
Volume of cylindrical rod $=3850 cm^3$
Length of $\operatorname{rod}( h )=1 m=100 cm$
Let radius of the base of the rod $=r$
Then volume $=\pi r ^2 h$
$\therefore\pi\text{r}^2\text{h}=3850$
$\Rightarrow\frac{22}{7}\times\text{r}^2\times100=3850$
$\Rightarrow\text{r}^2=\frac{3850\times7}{2200}=\frac{1225}{100}=\Big(\frac{35}{10}\Big)^2$
$=(3.5)^2$
$\therefore\text{r}=3.5\text{m}$
Hence diameter $=2 r =2 \times 3.5$
$= 7cm.$
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Question 253 Marks
A closed cylindrical tank of diameter 14m and height 5m is made from a sheet of metal. How much sheet of metal will be required?
Answer
Deameter of close cylinder = 14m
Radius $=\frac{14}{2}$
= 7m
Height = 5
$\therefore$ Total surface area of closed cylinder,
$=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{r}(\text{h}+\text{r})$
$=2\times\frac{22}{7}\times7\times(5+7)\text{m}^2$
$=44\times12=528\text{m}^2$
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Question 263 Marks
Find the height of the cylinder whose volume is $1.54 m^3$ and diameter of the base is 140 cm ?
Answer
Volume of cylinder $=1.54 m^3$
$=1540000 cm^3$
Diameter of its base $=140 cm$
Radius $( r )=70 cm$
$\therefore$ Height $=\frac{\text { Volume }}{\pi r ^2}=\frac{1540000 \times 7}{22 \times 70 \times 70} cm$
$=100 cm=1 m$
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Question 273 Marks
A solid rectangular piece of iron measures $1.05 m \times 70 cm \times 1.5 cm$. Find the weight of this piece in kilograms if $1 cm^3$ of iron weighs 8 grams.
Answer
$\text { Length of iron }(l)=1.05 m$
$=105 cm \text {, }$
$\text { Breadth }(b)=70 cm \text { and height }(h)=1.5 cm .$
$\text { Volume of iron }=1 \times b \times h=105 \times 70 \times 1.5 cm^3$
$=11025 cm^3$
$\text { Weight of } 1 cm^3 \text { iron }=8 \text { gram. }$
$\text { Total weight }=11025 \times g=88200 g$
$=\frac{88200}{1000} kg$
$88.2 kg \text {. }$
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Question 283 Marks
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84cm and its length is 1m.
Answer
No. of revolutions = 750
Diameter of road roller = 84cm
Length (h) = 1m
Radius $=\frac{84}{2}=42\text{cm}=0.42\text{m}$
$\therefore$ Surface area $=2\pi\text{rh}$
$=2\times\frac{22}{7}\times\frac{42}{100}\times1\text{m}^2$
$=\frac{264}{100}\text{m}^2$
$\therefore$ Area of road $=\frac{264}{100}\times750\text{m}^2$
$=1980\text{m}^2$
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Question 293 Marks
The surface area of a cube is $1176 cm^2$. Find its volume.
Answer
Surface area of a cube $=1126 cm^2$
Let edge of the cube $=a$
Then $6 a^2=1176 \Rightarrow a^2=\frac{1176}{6}=196$
$\Rightarrow a^2=(14)^2 \Rightarrow a=14 cm .$
$\therefore$ Volume of cube $= a ^3=(14)^3 cm^3$
$=14 \times 14 \times 14 cm^3=2744 cm^3$
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Question 303 Marks
The volume of a block of gold is $0.5 m^3$. If it is hammered into a sheet to cover an area of 1 hectare, find the thickness of the sheet.
Answer
$\text { Volume of block of gold }=0.5 m^3$
$=0.5 \times 1000000 cm^3$
$=500000 cm^3$
Area of sheet formed = 1 heckare
$=10000 m^2=10000 \times 10000 cm^2$
$\therefore \text { Thickness of sheet }=\frac{500000}{10000 \times 10000} cm$
$=\frac{5}{1000} cm=\frac{5}{1000} \times 10$
$=\frac{5}{1000} mm=0.05 mm$
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Question 313 Marks
The volume of a room is $378 m^3$ and the area of its floor is $84 m^2$. Find the height of the room.
Answer
Volume of a room $=378 m^3$
Area of its floor $=84 m^2$
Height $=\frac{\text { Volume }}{\text { Area }}$
$=\frac{378}{84} m$
$=4.5 m$
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Question 323 Marks
A cardboard box is 1.2 m long, 72 cm wide and 54 cm high. How many bars of soap can be put into it if each bar measures $6 cm \times 4.5 cm \times 4 cm$ ?
Answer
Length of cardboard box (1),
$=1.2 m=120 cm .$
Breadth (b) $=72 cm$.
Height $(h)=54 cm$.
Volume of box $= l \times b \times h$
$=120 \times 72 \times 54 cm^3$
$=466560 cm^3$
Volume of one soap bar $=6 \times 4.5 \times 4 cm^3$
$=108 cm^3$
No, of bars to be kept in it $=\frac{466560}{108}$
$=4320$
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Question 333 Marks
A solid cubical block of fine wood costs Rs. 256 at Rs. 500 per $m ^2$. Find its volume and the length of each side.
Answer
Total cost of wood = Rs. 256
Rate $=$ Rs. 500 per $m ^3$
Volume od wood $=\frac{256}{500}=0.512 m^3$
$=0.512 \times 100 \times 100 \times 100 cm^3$
$=512000 cm^3$
Let length of each side $=a$
Then $a^3=512000=(80)^3$
$a=80$
Hence length of each side $=80 cm$.
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Question 343 Marks
The radius and height of a cylinder are in the ratio $5: 7$ and its volume is $550 cm^3$. Find its radius and height.
Answer
$\frac{\text {Radius}}{\text{Height}}=\frac{\text{r}}{\text{h}}=\frac{5}{7}$
$\Rightarrow\text{r}=\frac{5}{7}\text{h}$
Now, volume $=\pi\text{r}^2\text{h}=\frac{22}{7}\times\frac{5}{7}\text{h}\times\text{h}=550\text{cm}^3$
$\therefore\text{h}=\sqrt[3]{\frac{550\times7\times7\times7}{22\times5\times5}}=7\text{cm}$
Also, $\text{r}=\frac{5}{7}\text{h}=5\text{cm}$
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Question 353 Marks
A river 2 m deep and 45 m wide is flowing at the rate of $3 km / h$. Find the quantity of water that runs into the sea per minute.
Answer
Speed of water = 3km/ h
Length of water flow in 1 minute
$=\frac{ 3\text{km}}{60\text {m}}$
$=\frac{3000}{60}$
$=50\text{m}$
Width of river = 45m
Depth of river = 2m
Volume of water in 1 minute
$=45 \times 2 \times 50 m^3$
$=4500 m^3$
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Question 363 Marks
The volume of a cube is $729 cm^3$. Find its surface area.
Answer
Volume of a cube $=729 cm^3$
Let edge of cube $=(9)^3$
$a=9 cm .$
Hence surface $=6 a ^2=6(9)^2 cm^2$
$=6 \times 81$
$=486 cm^2$
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Question 373 Marks
A rectangular water tank is 90cm wide and 40cm deep. If it can contain 576 litres of water, what is its length?
Answer
Width of tank = 90cm $=\frac {90}{100} \text{m}$
Depth = 40cm $=\frac{40}{100}\text{m}$
Water = 576 liter,
$\therefore$ Volume of tank $=\frac{576}{100}\text{cm}^3$
$\therefore$ Length $=\frac{\text{Volume}}{\text{Width}\times\text{Depth}}$
$=\frac{576\times100\times100}{90\times40\times1000}$
$=\frac{16}{10}\text{m}=1.6\text{m}$
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Question 383 Marks
A beam of wood is 5 m long and 36 cm thick. It is made of $1.35 m^3$ of wood. What is the width of the beam?
Answer
Volume of wood = $1.35 m^3$
Length of beam =$5m$
Thickness = 36cm $=\frac{36}{100}\text{m}.$
$\text{Width}=\frac{\text{Volume}}{\text{Length}\times\text{thickness}}$
$=\frac{1.35\times100}{5\times36}=\frac{135\times100}{100\times5\times36}$
$=\frac{3}{4}\text{m}=0.75\text{m}$
$=75\text{cm}$
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