4 Mark Question

Question 14 Marks

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
Length = 24m, breadth = 25cm and height = 6m

Answer

Length of cuboid (l) = 24m
Breadth (b) = 25cm $=\frac{1}{4}\text{m}$
Height (h) = 6m

Volume = lbh $=24\times\frac{1}{4}\times6=36\text{cm}^3$
Lateral surface area = 2[l + b] × h,

$=2\Big(24+\frac{1}{4}\Big)\times6\text{m}^2$

$=2\times\frac{97}{4}\times6=291\text{m}^2$

Total surface area = 2(lb + bh + hb)

$=2\Big(24\times\frac{1}{4}+\frac{1}{4}\times6+6\times24\Big)\text{m}^2$

$=2\Big(6+\frac{3}{2}+144\Big)\text{m}^2$

$=2\Big(\frac{12+3+288}{2}\Big)\text{m}^2$

$=2\times\frac{203}{2}=303\text{m}^2$

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Question 24 Marks

A wooden cylindrical pole is 7m high and its base radius is 10cm. Find its weight if the wood weighs 225kg per cubic metre.

Answer

Radius of the base of pole (r)
$=10\text{dm}$
$=\frac{10}{100}\text{m}$
$=\frac{1}{10}\text{m}$
And height (h) = 7m.
$\therefore$ Volume of wood $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\frac{1}{10}\times\frac{1}{10}\times7\text{m}^3$
$=\frac{22}{100}\text{m}^3$
Weight of 1 cubic metre = 225kg.
$\therefore$ Total weight $=\frac{22}{100}\times225=\frac{4950}{100}$
$=49.5\text{kg.}$

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Question 34 Marks

A milk tank is in the form of a cylinder whose radius is 1.5m and height is 10.5m. Find the quantity of milk in litres that can be stored in the tank.

Answer

Radius of cylindrical tank (r) = 1.5m
And height (h) = 10.5m

$\therefore$ Volume of the tank $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times1.5\times1.5\times1.5\text{m}^3$
$=\frac{22\times15}{10}\times\frac{15}{100}\times\frac{15}{10}\times\frac{15}{10}=\frac{297}{4}\text{m}^3$
$=74.25\text{m}^3$
$\therefore$ Quantity of milk in litre,
= 74.25 × 1000 litre
= 74250 litre

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Question 44 Marks

Find the volume, curved surface area and total surface area of the cylinders whose dimensions are:
Radius of the base = 14dm and height = 15m

Answer

Radius of the base of a cylinder (r),
= 14dm = 1.4m.
Height (h) = 15m

Volume:

$=\pi\text{r}^2\text{h}=\frac{22}{7}\times(1.4)^2\times15\text{m}^3$

$=\frac{22}{7}\times1.4\times1.4\times15\text{m}^3=92.4\text{m}^3$

Lateral surface area $=2\pi\text{rh}$

$=2\times\frac{22}{7}\times1.4\times15\text{m}^2=132\text{m}^2$

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

$=2\times\frac{22}{7}\times1.4\times15\text{m}^2=132\text{m}^2$

$=8.8(16.4)\text{m}^2=144.32\text{m}^2$

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Question 54 Marks

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
Length $=15 m$, breadth $=6 m$ and height $=9 dm$.

Answer

Length of cuboid $( l )=15 m$
Breadth $(b)=6 m$
Height $(h)=9 dm =0.9 m$
a. Volume $= lbh =15 \times 6 \times 0.9=81 m^3$
b. Lateral surface area $=2(l+b) \times h$
$=2(15+6) \times 0.9 m^2$
$=2 \times 21 \times 0.9=37.8 m^2$
c. Total surface area $=2[ lb + bh + hl ]$
$=2(15 \times 6+6 \times 0.9+0.9 \times 15] m ^2$
$=2[90+5.4+13.5] m ^2$
$=2 \times 108.9=217.8 m^2$

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Question 64 Marks

A cylinder is open at both ends and is made of 1.5 cm -thick metal. Its external diameter is 12 cm and height is 84 cm . What is the volume of metal used in making the cylinder? Also, find the weight of the cylinder if $1 cm^3$ of the metal weighs 7.5 g .

Answer

Thickness of the metal $=1.5 cm$.
External diameter $=12 cm$.
$\therefore$ External radius $(R)=\frac{12}{2}=6 cm$.
And then internal radius $( r )=6-1.5$
$=4.5 cm$.
Height $(h)=84 cm$.
$\therefore$ Volume of metal used $=\pi r^2 h-\pi r^2 h$
$=\pi h \left( R ^2- r ^2\right)$
$=\pi h ( R + r )( R - r )$
$=\frac{22}{7} \times 84(6+4.5)(6-4.5) cm ^2$
$=264(10.5)(1.5) cm ^3=4158 cm^3$
Weight of $1 cm^3=7.5 g$
$\therefore$ Total weight $=7.5 \times 4158 g$
$=31185 g=31.185 kg$

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Question 74 Marks

Find the volume of wood used to make a closed box of outer dimensions $60 cm \times 45 cm \times 32 cm$, the thickness of wood being 2.5 cm all around.

Answer

Outer length of wooden box $( L )=60 cm$
Width $(B)=45 cm$
And height $( H )=32 cm$
Thickness of wood used $=2.5 cm$.
Inner length $( l )=60-2 \times 2.5$
$=60-5$
$=55 cm$.
Width $(b)=45-2 \times 2.5=45-5=40 cm$
And height $( h )=32-2 \times 2.5=32-5=27 cm$
Now outer volume $= L \times B \times H =60 \times 45 \times 32 cm^3$
Inner volume $= I \times b \times h =55 \times 40 \times 27$
$\therefore$ Volume of wood $=60 \times 45 \times 32-55 \times 40 \times 27$
$=86400-59400=27000 cm^3$

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Question 84 Marks

The length of a metallic tube is 1 metre, its thickness is 1 cm and its inner diameter is 12 cm . Find the weight of the tube if the density of the metal is 7.7 grams per cubic centimetre.

Answer

Inner diameter of tube $=12 cm$.
Inner radius $( r )=\frac{12}{2}=6 cm$.
Thickness of metal $=1 m$.
$\therefore$ Outer radius $(R)=6+1=7 cm$.
Length of the tube $( h )=1 m=100 cm$.
$\therefore$ Volume of metal used $=\pi R^2-\pi r^2 h$
$=\pi h \left( R ^2- r ^2\right)=\pi h ( R + r )( R - r )$
$=\frac{22}{7} \times 100(7+6)(7-6) cm ^3$
$=\frac{2200}{7} \times 13 \times 1 cm^3=\frac{28600}{7} cm^3$
Weight of $1 cm^3$ metal $=7.7 g$.
$\therefore$ Total weight,
$=\frac{28600}{7} \times 7.7 g=31460 g$
$=31.46 kg$.

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Question 94 Marks

The external dimensions of a closed wooden box are 62cm, 30cm and 18cm. If the box is made of 2cm-thick wood, find the capacity of the box.

Answer

Outer length of box = 62cm.
Outer width = 30cm.
Outer height = 18cm.
Thickness of wood = 2cm.
$\therefore$ Inernal length = 62 - 2 × 2
= 58cm.
Internal width = 30 - 2 × 2
= 26cm.
Inernal height = 18 - 2 × 2
= 14cm.
Capacity of the box = lbh
$=58 \times 26 \times 14 cm^3$
$=21112 cm^3$

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Question 104 Marks

Find the volume, curved surface area and total surface area of the cylinders whose dimensions are:
Radius of the base = 5.6m and height = 1.25m.

Answer

Radius of the base of the cylunder (r),
= 5.6m
Height (h) = 1.25m

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

$=\frac{22}{7}\times(5.6)^2\times1.25\text{cm}^3$

$=\frac{22}{7}\times5.6\times5.6\times1.25\text{m}^3=123.2\text{m}^3$

Lateral surface area $=2\pi\text{rh}$

$=2\times\frac{22}{7}\times5.6\times1.25\text{m}^2=44\text{m}^2$

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

$=2\times\frac{22}{7}\times5.6(1.25+5.6)$

$=44\times0.8(6.85)\text{m}^2=241.12\text{m}^2$

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Question 114 Marks

Find the volume, curved surface area and total surface area of the cylinders whose dimensions are:
Radius of the base = 7cm and height = 50cm.

Answer

Radius of the base of the cylinder (r) = 7cm.
Height (h) = 50cm.

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

$=\frac{22}{7}\times(7)^2\times50\text{cm}^3$

$=\frac{22}{7}\times7\times7\times50\text{cm}^3=7700\text{cm}^3$

Lateral surface area $=2\pi\text{rh}$

$=2\times\frac{22}{7}\times7\times50=2200\text{cm}^2$

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

$=2\times\frac{22}{7}\times7(50+7)\text{cm}^2$

$=44\times57=2508\text{cm}^2$

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Question 124 Marks

The volume of a cylinder of height 8 cm is $1232 cm^3$. Find its curved surface area and the total surface area.

Answer

Volume of cylinder $=1232 cm^3$
Height $(h)=8 cm$
Let $r$ be the radius, then
$\pi\text{r}^2\text{h}=1232$
$=\frac{22}{7}\times\text{r}^2\times8=1232$
$\Rightarrow\text{r}^2=\frac{1232\times7}{22\times8}$
$\Rightarrow\text{r}^2=49=(7)^2$
$\Rightarrow\text{r}=7\text{cm}$
$\therefore$ Curved surface area $=2\pi\text{rh}$
$=2\times\frac{22}{7}\times(7+8)\text{cm}^2$
$=44\times15=660\text{cm}^2$

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Question 134 Marks

How many planks of size $2 m \times 25 cm \times 8 cm$ can be prepared from a wooden block 5 m long, 70 cm broad and 32 cm thick, assuming that there is no wastage?

Answer

Length of one plank $=2 m$
$=200 cm$
Breadth $(b)=25 cm$,
Thickesss $( h )=8 cm$
$\therefore$ Volume of plank $= I \times b \times h =200 \times 25 \times 8 cm^3$
$=40000 cm^3$
Length of block $( l )=5 m=500 cm$,
Width (b) $=70 cm$
and thickness $( h )=32 cm$
$\therefore$ Volume of block $=500 \times 70 \times 32 cm^3$
Number of planks $=\frac{\text { Volume of black }}{\text { Volume of plank }}$
$=\frac{500 \times 70 \times 32}{40000}$
$=\frac{1120000}{40000}=28$

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Question 144 Marks

Find the volume of iron required to make an open box whose external dimensions are $36 cm \times 25 cm \times 16.5 cm$, the box being 1.5 cm thick throughout. If $1 cm^3$ of iron weighs 8.5 grams, find the weight of the empty box in kilograms.

Answer

Outer length of open box $=36 cm$
Breadth $=25 cm$
And height $=16.5 cm$
Thickness of iron $=1.5 cm$.
$\therefore \text { Inner length }=36-2 \times 1.5=36-3$
$=33 cm,$
Breadth $=25-2 \times 1.5$
$=25-3$
$=22 cm$
And height $=16.5-1.5$
$=15 cm$
$\therefore$ Volume of iron used in it = Out volume - inner volume
$=36 \times 25 \times 16.5 cm^3-33 \times 22 \times 15 cm^3$
$=14850-10890$
$=3960 cm^3$
Weight of $1 cm^3=8.5$ gram
$\therefore \text { Total weight }=3960 \times 8.5 g$
$=33660 g$
$=33.660 kg$
$=33.66 kg$

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Question 154 Marks

A solid cube of metal each of whose sides measures 2.2cm is melted to form a cylindrical wire of radius 1mm. Find the length of the wire so obtained.

Answer

Side of a solid cube $=2.2 cm$.
Volume $=(\text { side })^3$
$=(2.2)^3$
$=10.648 cm^3$

$\therefore$ Volume of cylinderdrical wire $=10.648 cm^3$
Radius = 1mm $=\frac{1}{10}\text{cm}$
$\therefore\text{Length}=\frac{\text{Volume}}{\pi\text{r}^2}$
$=\frac{10.648\times7\times10\times10}{22\times1\times1}$
$=\frac{10648\times7\times100}{1000\times22}=\frac{484\times7}{10}\text{cm}$
$=\frac{3388}{10}=338.8\text{cm}$

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Question 164 Marks

A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring $22 cm \times 12.5 cm \times 7.5 cm$. If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer

Length of wall (l) = 15m.
Width (b) = 30cm = 0.3m
Height (h) = 4m
$\therefore$ Volume of the wall = lbh
$=15 \times 0.3 \times 4 m^3=18 m^3$
$\therefore \text { Volume of bricks }=18-1.5=16.5 m^3$
Volume of one brick,
$=22 cm . \times 12.5 cm . \times 7.5 cm \text {. }$
$=22 \times 12.5 \times 7.5 cm^3$
$=2062.5\text{cm}^3=\frac{2062.5}{100\times100\times100}\text{m}^3$
$=0.0020625\text{m}^3$
No. of bricks $=\frac{\text{Total volume of bricks}}{\text{Volume of one brick}}$
$=\frac{16.5}{0.0020625}=\frac{165\times10000000}{10\times20625}$
$=8000$

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Question 174 Marks

Find the capacity of a rectangular cistern in litres whose dimensions are $11.2 m \times 6 m \times 5.8 m$. Find the area of the iron sheet required to make the cistern.

Answer

Length of rectangular cistern $( l )=11.2 m$
Breadth $(b)=6 m$
Height $(h)=5.8 m$
a. Volume $= lbh =11.2 \times 6 \times 5.8 m^3$
$=389.76 m^3$
$\therefore$ Capacity in litres,
$=389.76 \times 1000$ litres
$=389760$ litres.
b. Total surface area $=2[ lb + bh + hl ]$
$=2[11.2 \times 6+65.8+5.8 \times 11.2] m ^2$
$=2 \times 166.96=333.92 m^2$.

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Question 184 Marks

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
$Length = 22cm, breadth = 12cm and height = 7.5cm.$

Answer

Length of cuboid (l) = 22cm.
Breadth (b) = 12cm
and height (h) = 7.5cm.

Volume $= lbh =22 \times 12 \times 7.5 cm^3$

$= 1980 cm^3$

Lateral surface area = $=2(l+ b ) \times h$

$=2(22+12) \times 7.5 cm^2$

$=2 \times 34 \times 7.5=510 cm^2$

Total surface area = $=2[ lb + bh + hl ]$

$= 2[22 \times 12+12 \times 7.5+7.5 \times 22] cm ^2$

$= 2 \times 34 \times 7.5=510 cm^2$

$= 2 \times 519=1038 cm^2$

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Question 194 Marks

The curved surface area of a cylinder is $4400 cm^2$ and the circumference of its base is 110 cm . Find the volume of the cylinder.

Answer

Curved surface area $=4400 cm^3$
Circumference of base $=110 cm$
Let r be the radius,
$\Rightarrow2\pi\text{r}=110$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=110$
$\Rightarrow\text{r}=\frac{110\times7}{2\times22}=\frac{35}{2}\text{cm}$
Now, curved surface area $=2\pi\text{rh}$
$\therefore2\pi\text{rh}=4400$
$\Rightarrow2\times\frac{22}{7}\times\frac{35}{2}\times\text{h}=4400$
$110.\text{h}=4400$
$\Rightarrow\text{h}=\frac{4400}{110}$
$\Rightarrow\text{h}=40\text{cm}$
Hence volume of cylinder $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\times40\text{cm}^3$
$=38500\text{cm}^3$

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Question 204 Marks

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
Length $=48 cm$, breadth $=6 dm$ and height $=1 m$.

Answer

Length of cuboid (l) = 48cm.
Breadth (b) = 6dm = 60cm.
Height (h) = 1m = 100cm.

Volume = lbh = 48 × 60 × 100

$=288000 cm^3$

Lateral surface area = 2(l + b) × h

$=2(48+60) \times 100 cm^2$

$=2 \times 108 \times 100=21600 cm^2$

Total surface area = 2[lb + bh + hl]

$=2[48 \times 60+60 \times 100+100 \times 48] cm^2$

$=2[2880+6000+4800] cm^2$

$=2[13680]=27360 cm^2$

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Question 214 Marks

A piece of ductile metal is in the form of a cylinder of diameter 1cm and length 11cm. It is drawn out into a wire of diameter 1mm. What will be the length of the wire so obtained?

Answer

Diameter of cylindrical metal = 1cm.
Radius (r) $=\frac{1}{2}\text{cm}.$
Length. (A) = 11cm.
Volume $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\Big(\frac{1}{2}\Big)^2\times11\text{cm}^3=\frac{22}{7}\times\frac{1}{4}\times11$
$=\frac{22}{7}\times\frac{11}{4}=\frac{121}{14}\text{cm}^3$
$\therefore$ Volume of wire $=\frac{121}{14}\text{cm}^3$
Diameter of wire = 1mm
$\therefore\text{Radius}=\frac{1}{2}\text{mm}=0.5\text{mm}=0.05\text{cm}.$
Let lenght of wire = h
Then $\pi\text{r}^2\text{h}=\frac{22}{70}\times(0.05)^2\times\text{h}$
$\Rightarrow\frac{22}{7}\times0025\times\text{h}\frac{121}{14}$
$\Rightarrow\text{h}=\frac{121\times7}{14\times22\times0025}$
$=\frac{121\times7\times10000}{14\times22\times25}=1100\text{cm}$
$\therefore$ Length of wire = 11m.

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Question 224 Marks

A closed wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5 cm -thick wood. Find the capacity of the box and its weight if $100 cm^3$ of wood weighs 8 g .

Answer

Outer length $=80 cm$.
Outer width $=65 cm$.
Outer height $=45 cm$.
Total volume $=80 \times 65 \times 45 cm^3$
$=234000 cm^3$
Thickness of wood $=2.5 cm$.
$\therefore$ Inner length $=80-2 \times 2.5=75 cm$.
Inner width $=65-2 \times 2.5=60 cm$.
Inner height $=45-2 \times 2.5=40 cm$.
$\therefore$ Inner capacity (volume)
$=75 \times 60 \times 40 cm^3=180000 cm^3$
Volume of wood $=234000-180000$
$=54000 cm^3$
$\therefore$ Weight of wood $=\frac{54000 \times 8}{100} g$
$=4320 g=4.320 kg$.

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Question 234 Marks

How many cubic metres of earth must be dug out to sink a well which is 20m deep and has a diameter of 7 metres? If the earth so dug out is spread over a rectangular plot 28m by 11m, what is the height of the platform so formed?

Answer

Diameter of a well $=7 m$.
Radius $( r )=\frac{7}{2} m$
Depth $(h)=20 m$
$\therefore$ Volume of earth dug out $=\pi r ^2 h$
$=\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 20 m^3 \\
=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 m^3=770 m^3$
Now volume of platform $=770 m^3$
Length $( l )=28 m$
Breadth (b) $=11 m$
Let height $=h$
$\therefore lbh=770 \Rightarrow 11 \times h=770$
$\Rightarrow h=\frac{770}{28 \times 11}=\frac{5}{2}=2.5 m$
$\therefore$ Height of the platform $=2.5 m$.

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Question 244 Marks

The radius and height of a cylinder are in the ratio $7: 2$. If the volume of the cylinder is $8316 cm^3$, find the total surface area of the cylinder.

Answer

Ratio in radius and height of a cylinder = 7 : 2
Let radius = 7x
Then height = 2x
$\therefore\text{Volume}=\pi\text{r}^2\text{h}=\frac{22}{7}\times7\text{x}\times7\text{x}\times2\text{x}$
$=308\text{x}^3$
$\therefore308\text{x}^3=8316$
$\text{x}^3=\frac{8316}{308}=27=(3)^3$
$\therefore\text{x}=3$
$\therefore$ Radius of cylinder (r) = 7x = 7 × 3 = 21cm,
And height (h) = 2x = 2 × 6cm
Now, total surface area $=2\pi\text{r}(\text{r}+\text{h})$
$=2\times\frac{22}{7}\times21(21+6)\text{cm}^2$
$=132\times27\text{cm}^2$
$=3564\text{cm}^2$

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Question 254 Marks

A well of inner diameter 14m is dug to a depth of 12m. Earth taken out of it has been evenly spread all around it to a width of 7m to form an embankment. Find the height of the embankment so formed.

Answer

Inner diameter of well = 14m
Inner radius $=\frac{14}{2}=7\text{m}$
Depth (h) = 12m
$\therefore$ Volume of earth dig out $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times(7)^2\times12\text{m}^3$
$=\frac{22}{7}\times7\times7\times12\text{m}^3=1848\text{m}^3$
Width of the embankment of the well
= 7m
$\therefore$ Outer radius (R) = 7 + 7 = 14m.
Let height of the embankment = h
$\therefore$ Volume $\pi\text{r}^2\text{h}-\pi\text{r}^2\text{h}$
$=\pi\text{h}\big(\text{R}^2-\text{r}^2\big)=\frac{22}{7}\text{h}\big(\text{R}^2-\text{r}^2\big)$
$\therefore\frac{22}{7}\text{h}\big(14^2-7^2\big)=1848$
$\Rightarrow\frac {22}{7}\text{h}(14+7)(14-7)=1848$
$\Rightarrow\frac{22}{7}\text{h}\times21\times7=1848$
$\Rightarrow462\text{h}=1848$
$\therefore\text{h}=\frac{1848}{462}=4$
$\therefore$ Height of embankment = 4m.

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