Question 1: A cuboid water tank is $5 \ m$ long, $5 \ m$ wide and $4.5 \ m$ deep. How many liters of water can it hold?

Dimension of a cuboid tank:

Length $(l) = 6 \ m$     Breadth $(b) = 5 \ m$     Height $(h) = 4.5 \ m$

Therefore Volume $= lbh = 6 \times 5 \times 4.5 = 135 \ m^3$

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Question 2: A cubical vessel is $10 \ m$ long and $8 \ m$ wide. How high must it be made to hold $380$ cubic meters of a liquid?

Dimension of a cuboid tank:

Length $(l) = 10 \ m$     Breadth $(b) = 8 \ m$     Height $(h) = h \ m$

Therefore  $380 = lbh$

$\Rightarrow 380 = 10 \times 8 \times h$

$\Rightarrow h =$ $\frac{380}{10 \times 8}$ $= 4.75\ m$

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Question 3: Find the cost of digging a cuboid pit $8 \ m$ long, $6 \ m$ broad and $3 \ m$ deep at the rate of $Rs. 30 \ per \ m^3$.

Dimension of a cuboid pit:

Length $(l) = 8 \ m$     Breadth $(b) = 6 \ m$     Height $(h) = 3 \ m$

Cost of digging $= Volume \times Rate = 8 \times 6 \times 3 \times 30 = 4320 \ Rs.$

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Question 4: If $V$ is the volume of a cuboid of dimensions $a, \ b, \ c$ and $S$ is its surface area, then prove that $\frac{1}{V}$ $=$ $\frac{2}{S}$ $\Big($ $\frac{1}{a}$ $+$ $\frac{1}{b}$ $+$ $\frac{1}{c})$

Volume $= V$

Surface area $= S$

Dimensions are $a, \ b, \ c$

Therefore $V = abc$ … … … … … i)

$S = 2(ab + bc+ ac)$  … … … … … ii)

Dividing ii) by i) we get

$\frac{S}{V}$ $=$ $\frac{2(ab + bc+ ac) }{abc}$

$\Rightarrow \frac{1}{V}$ $=$ $\frac{2}{S}$ $\Big($ $\frac{1}{a}$ $+$ $\frac{1}{b}$ $+$ $\frac{1}{c})$

Hence Proved.

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Question 5: The areas of three adjacent faces of a cuboid are $x, \ y$ and $z$. If the volume is $V$, Prove that $V^2 = xyz$.

Area of adjacent faces $= x, y, z$

Let the dimensions be Length $(l) = a$     Breadth $(b) = b$     Height $(h) = c$

Therefore $V = abc$

Also $ab = x$ … … … … … i)

$bc = y$ … … … … … ii)

$ac = z$ … … … … … iiii)

Multiplying i), ii) and iii) we get

$xyz = ab \times bc \times ac$

$\Rightarrow xyz = (abc)^2 = V^2$. Hence proved.

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Question 6: If the areas of three adjacent faces of a cuboid are $8 \ cm^2$, $18 \ cm^3$ and $25 \ cm^3$. Find The volume of the cuboid

Area of adjacent faces $= 8 \ cm^2, 18 \ cm^2 \ and \ 25 \ cm^2$

Let the dimensions be Length $(l) = a$     Breadth $(b) = b$     Height $(h) = c$. Given

$ab = 8$ … … … … … i)

$bc = 18$ … … … … … ii)

$ac = 25$ … … … … … iiii)

Multiplying i), ii) and iii) we get

$8 \times 18 \times 25 = (abc)^2$

Therefore $V = \sqrt{8 \times 18 \times 25} = 4 \time 3 \times 5 = 60 \ cm^3$

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Question 7: The breadth of a room is twice its height; one half of its length and the volume of the room is $512 \ cu. \ dm$. Find its dimensions.

Let the dimensions be Length $(l) = 4x$     Breadth $(b) = 2x$     Height $(h) = x$

Volume $= 512 \ dm^3$

Therefore $4x \times 2x \times x = 512$

$\Rightarrow 8x^3 = 512$

$\Rightarrow x^3 = 64$

$\Rightarrow x = 4$

Therefore dimensions will be Length $(l) = 16 \ dm$     Breadth $(b) = 8 \ dm$     Height $(h) = 4 \ dm$

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Question 8: A river $3 \ m$ deep and $40 \ m$ wide is flowing at the rate of $2$ km per hour. How much water will fall into the sea in a minute?

Dimension of River: Depth $= 3 \ m$ and Width $= 40 \ m$

Velocity of rate of flow $= 2 \ km/hr =$ $\frac{2 \times 1000}{60}$ $m/min =$  $\frac{100}{3}$ $\ m/min$

Therefore volume of water flowing in a minute $= 3 \times 40 \times$ $\frac{100}{3}$ $= 4000 \ m^3$

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Question 9: Water in a canal $30 \ dm$ wide and $12 \ dm$ deep, is flowing with a velocity of $100$ km per hour. How much area will it irrigate in $30$ minutes if $8 \ cm$ of standing water is desired?

Dimension of Canal: Depth $= 12 \ dm = 1.2 \ m$ and Width $= 30 \ dm = 3 \ m$

Velocity $= 100 \ km/hr =$ $\frac{100 \times 1000}{60}$ $m/min =$ $\frac{10000}{6}$ $\ m/min$

Therefore volume of flow in 30 minutes $= 1.2 \times 3 \times$ $\frac{10000}{6}$ $\times 30 = 180000 \ m^3$

Therefore area irrigated $=$ $\frac{180000}{0.08}$ $= 2250000 \ m^2$

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Question 10: Three metal cubes with edges $6 \ cm, 8 \ cm$ and $10 \ cm$ respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.

Total volume of the three cubes $= 6^3 + 8^3 + 10^3 = 1728 \ cm^3$

Therefore dimension of new cube $= \sqrt[3]{1728} = 12 \ cm$

Therefore volume $= 12^2 = 1728 \ cm$

Surface area $= 6(12^2) = 864 \ cm^2$

Diagonal $= \sqrt{3} . 12 = 20.78 \ cm$

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Question 11: Two cubes, each of volume $512 \ cm^3$ are joined end to end. Find the surface area of the resulting cuboid.

Volume of cube $= 512 \ cm^3$

$\Rightarrow l^3 = 512 \Rightarrow l = 8 \ cm$

Therefore dimension of the cuboid

Length $(l) = 16 \ cm$     Breadth $(b) = 8 \ cm$     Height $(h) = 8 \ cm$

Therefore total surface are $= 2 (lb + bh + lh) = 2(16 \times 8 + 8 \times 8 + 8 \times 16) = 640 \ cm^3$

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Question 12: Half cubic meter of gold-sheet is extended by hammering so as to cover an area of $1$ hectare. Find the thickness of the gold-sheet.

Volume of gold $= 0.5 \ m^3$

1 hectare $= 10000 \ m^2$

Let the thickness of the gold sheet $= x$

Therefore $0.5 = x \times 10000 \Rightarrow x =$ $\frac{0.5}{10000}$ $m$

$\Rightarrow x =$ $\frac{0.5 \times 100}{10000}$ $\ cm =$ $\frac{0.5 \times 10}{100}$ $\ mm = 0.05 \ mm$

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Question 13: A metal cube of edge $12 \ cm$ is melted and formed into three smaller cubes. If the edges of the two smaller cubes are $6 \ cm$ and $8 \ cm$, find the edge of the third smaller cube.

If $x$ is the dimension of the third cube

Volume of big cube $= 12^2 = 1728 \ cm^3$

Therefore $6^3 + 8^3 + x^3 = 1728$

$\Rightarrow x^3 = 1000$

$\Rightarrow x = 10 \ cm$

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Question 14: The dimensions of a cinema hall are $100 \ m, 50 \ m$ and $18 \ m$. How many persons can sit in the hall, if each Person requires $150 \ m^3$ of air.

Dimension of a the hall:

Length $(l) = 100 \ m$     Breadth $(b) = 50 \ m$     Height $(h) = 18 \ m$

Space required by each person is $150 \ m^3$

Therefore no of people that can sit in the hall $=$ $\frac{100 \times 50 \times 18}{150}$ $= 600$

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Question 15: Given that $1 \ cm^3$ of marble weighs $0.25 \ kg$, the weight of marble block $28 \ cm$ in width and $5 \ cm$ thick is $112 \ kg$. Find the length of the block.

Dimension of a the hall:

Length $(l) = 28 \ cm$     Breadth $(b) = 5 \ cm$     Height $(h) = x$

Volume of the marble block $= 28 \times 5 \times x = 140 x$

Therefore $112 = 140x \times 0.25$

$\Rightarrow x =$ $\frac{112}{140 \times 0.25}$ $= 3.2 \ cm$

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Question 16: A box with lid is made of $2 \ cm$ thick wood. Its external length, breadth and height are $25 \ cm, 18 cm$ and $15 \ cm$ respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.

External dimension of a the box:

Length $(l) = 25 \ cm$     Breadth $(b) = 18 \ cm$     Height $(h) = 15 \ cm$

Thickness of the wood $= 2 cm$

Internal dimension of a the box:

Length $(l) = 21 \ cm$     Breadth $(b) = 14 \ cm$     Height $(h) = 11 \ cm$

Internal Volume $= 21 \times 14 \times 11 = 3234 \ cm^3$

External Volume $= 25 \times 18 \times 15 = 6750 \ cm^3$

Therefor Volume of wood used $= 6756-3234 = 3516 \ cm^3$

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Question 17: The external dimensions of a closed wooden box are $48 \ cm, 36 \ cm, 30 \ cm$. The Box is made of $1.5 \ cm$ thick wood. How many bricks of size $6 cm \times 3 \ cm \times 0.75 \ cm$ can be put in this box?

External dimension of a the box:

Length $(l) = 48 \ cm$     Breadth $(b) = 36 \ cm$     Height $(h) = 30 \ cm$

Thickness of the wood $= 1.5 cm$

Internal dimension of a the box:

Length $(l) = 45 \ cm$     Breadth $(b) = 33 \ cm$     Height $(h) = 27 \ cm$

Internal Volume $= 45 \times 33 \times 27 = 38880 \ cm^3$

Volume of the brick $= 6 \times 3 \times 0.75 = 13.5 \ cm^3$

Hence the number of bricks $= \frac{38880}{13.5} = 2970$

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Question 18: How many cubic centimeters of iron are there in an open box whose external dimensions are $36 \ cm$, $25 \ cm$ and $16.5 \ cm$, the iron being $1.5 \ cm$ thick throughout? If $1 \ cm^3$ of iron weighs $15 \ g$, find the weight of the empty-box in kg.

External dimension of a the box:

Length $(l) = 36 \ cm$     Breadth $(b) = 25 \ cm$     Height $(h) = 16.5 \ cm$

Thickness of the wood $= 1.5 cm$

Internal dimension of a the box:

Length $(l) = 33 \ cm$     Breadth $(b) = 22 \ cm$     Height $(h) = 15 \ cm$

External Volume $= 36 \times 25 \times 16.5 = 14850 \ cm^3$

Internal Volume $= 33 \times 22 \times 15 = 10890 \ cm^3$

Volume of Iron $= 14850-10890 = 3960 \ cm^3$

Weight of Iron box $= 3960 \times 15 = 59400 \ gm \ or \ 59.4 \ kg$

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Question 19: A cube of $9 \ cm$ edge is immersed completely in a rectangular vessel containing Water. If the dimensions of the base are $15 \ cm$ and $12 \ cm$, find the rise in water level in the vessel.

Volume of cube $= 9^3 = 729 \ cm^3$

Therefore rise in water $=$ $\frac{729}{15 \times 12}$ $= 4.05 \ cm$

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Question 20: A rectangular container, whose base is a square of side $5 \ cm$, stands on a horizontal table, and holds water up to, $1 \ cm$ from the top. when a cube is placed in the water it is completely submerged, the water rises to the top and $2 \ cubic \ cm$ of water overflows. Calculate the volume of the cube and also the length of its edge.

Dimension of a the box:

Length $(l) = 5 \ cm$     Breadth $(b) = 5 \ cm$     Height $(h) = x$

Let the dimension of the cube $= a$

Volume of water over flowing $= 5 \times 5 \times 1 + 2 = 27 \ cm^3$

Therefore $a^3 = 27 \Rightarrow a = 3 \ cm$

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Question 21: A field is $200 \ m$ long and $150 \ m$ broad. There is a plot, $50 \ m$ long and $40 \ m$ broad, near the field. The plot is dug $7 \ m$ deep and the earth taken out is spread evenly on the field. By how many meters is the level of the field raised? Give the answer to the second place of decimal.

Dimension of field: Length $(l) = 200 \ m$     Breadth $(b) = 150 \ m$

Dimension of plot: Length $(l) = 50 \ m$     Breadth $(b) = 40 \ m$

Depth of dig $= 7 \ cm$

If $h$ is the rise in the level of the ground

Therefore $200 \times 150 \times h = 50 \times 40 \times 7$

$\Rightarrow h = 0.467 \ m$

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Question 22: A field is in the form of a rectangle of length $18 \ m$ and width $15 \ m$. A pit, $7.5 \ m$ long, $6 \ m$ broad and $0.8 \ m$ deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.

Dimension of field: Length $(l) = 18 \ m$     Breadth $(b) = 15 \ m$

Dimension of pit: Length $(l) = 7.5 \ m$     Breadth $(b) = 6.0 \ m$   Depth $(h) = 0.8 \ m$

Volume of mud $= 7.5 \times 6.5 \times 0.8 = 36 \ m^3$

Area of field $= 18 \times 15 = 270 \ m^2$

Area of pit $= 7.5 \times 6 = 45 \ m^2$

Therefore the area of spread $= 270 - 45 = 225 \ m^2$

Let $h$ be the raise in the eight

Therefore $h =$ $\frac{36}{225}$ $= 0.16 \ m$

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Question 23: A rectangular tank is $80 \ m$ long and $25 \ m$ broad. Water flows into it through a pipe-whose cross-section is $25 \ cm^2$, at the rate of $16 \ km/hour$. How much the level of the water rises in the tank in $45 \ minutes$.

Dimension of tank base: Length $(l) = 80 \ m$     Breadth $(b) = 25 \ m$

Rate of flow of water $= 16 \ km/hr =$ $\frac{16 \times 1000}{60}$ $m/min =$ $\frac{800}{3}$ $m/min.$

Volume of water flow per minute $=$ $\frac{25}{10000}$ $\times \frac{800}{3}$ $= \frac{2}{3}$ $m^3/min$

Therefore in $45$ minutes, let $h$ be the rise in water level.

Therefore $80\times 25 \times h =$ $\frac{2}{3}$ $\times 45$

$\Rightarrow h =$ $\frac{2 \times 4.5}{3 \times 80 \times 25}$ $=$ $\frac{3}{200}$ $m =$ $\frac{3}{2}$ $cm$

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Question 24: Water in a rectangular reservoir having base $80 \ m \times 60 \ m$ is $6.5 \ m$ deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side $20 \ cm$, if the water runs through the pipe at the rate of $15 \ km/hr$.

Dimension of the reservoir:

Length $(l) = 80 \ m$     Breadth $(b) = 60 \ m$     Height $(h) = 6.5 \ m$

Rate of flow of water $= 15 \ km/hr =$ $\frac{15 \times 1000}{60}$ $m/min = 250 \ m/min$

Therefore time to empty the tank $=$ $\frac{80 \times 60 \times 6.5}{0.2 \times 0.2 \times 50}$ $= 3120 \ minutes \ or \ 52 hours$

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Question 25: A village having a population of $4000$ requires $150$ liters of water per head per day. It has a tank measuring $20 \ m \times 15 \ m \times 6 \ m$. For how many days will the water of this tank last?

Dimension of the tank:

Length $(l) = 20 \ m$     Breadth $(b) = 15 \ m$     Height $(h) = 6 \ m$

Volume of water required daily $= 4000 \times$ $\frac{150}{1000}$ $= 600 \ m^3$

Therefore the number of days the water will last $=$ $\frac{20 \times 15 \times 6}{600}$ $= 3 \ days.$

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Question 26: A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the adjoining figure. If the edge of each cube is $3 \ cm$, find the volume of the structure built by the child.

Number of blocks $= 15$

Volume of one block $= 3 \times 3 \times 3 = 27 \ cm^3$

Volume of the structure $= 15 \times 27 = 405 \ cm^3$

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Question 27: A godown measures $40 \ m \times 25 \ m \times 10 \ m$. Find the maximum number of wooden crates each measuring $1.5 \ m \times 1.25 \ m \times 0.5 \ m$ that can be stored in the godown.

Dimension of the godown:

Length $(l) = 40 \ m$     Breadth $(b) = 25 \ m$     Height $(h) = 10 \ m$

Dimension of the crate:

Length $(l) = 1.5 \ m$     Breadth $(b) = 1.25 \ m$     Height $(h) = 0.5 \ m$

Therefore the number of wooden crates that can be stored $=$ $\frac{40 \times 25 \times 10}{1.5 \times 1.25 \times 0.5}$ $= 10666 \ crates$

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Question 28: A wall of length $10 \ m$ was to be built across an open ground. The height of the wall is $4 \ m$ and thickness of the wall is $24 \ cm$. If this wall is to be built up with bricks whose dimensions are $24 cm \times 12 cm \times 8 cm$, how many bricks would be required?

Dimension of the wall:

Length $(l) = 10 \ m$     Breadth $(b) = 0.24 \ m$     Height $(h) = 4 \ m$

Dimension of the brick:

Length $(l) = 0.24 \ m$     Breadth $(b) = 0.12 \ m$     Height $(h) = 0.08 \ m$

Therefore the number of bricks needed $=$ $\frac{10 \times 0.24 \times 4}{0.24 \times 0.12 \times 0.08}$ $= 4167 \ bricks$

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Question 29: If the volume of a cube is $V \ m^3$, its surface are is $S \ m^2$ and the length of a diagonal is $d$ meters, prove that $6\sqrt{3}V = Sd$

Volume of the cube $= V$

Surface are of the cuber $= S$

Diagonal $= d$

Let the side of the cube $= a$

Therefore $a^3 = V$

$6a^2 = S$

$\sqrt{3} a = d$

Therefore $Sd = 6\sqrt{3} a^3$

$Sd = 6\sqrt{3} V$. Hence proved.

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Question 30: The adjoining figure shows a victory stand, each face of which is rectangular. All measurements are in centimeters. Find its volume and surface area.

Volume $= 12 \times 40 \times 50 + 40 \times 50 \times 40+24 \times 40 \times 50$

$= 2400 + 80000+48000 = 152000 \ cm^3$

Total surface area $=$ Front area $+$ Back area $+$ top area $+$ side area

$= 12 \times 40 + 40 \times 150+40 \times 24 + 28 \times 40+16 \times 40+ 2(12 \times 50+40 \times 50+50 \times 24)$

$= 480 + 6000+960+1120+640+2(600+2000+1200)$

$= 9200 + 7600 = 16800 \ cm^2$

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Question 31: Water is being filled in an aquarium at the rate of $12.5$ liters per minute. If the aquarium is $2 \ m$ long and $80 \ cm$ wide and it is filled in $96$ minutes, find the height of the aquarium.

Dimension of the aquarium:

Length $(l) = 2 \ m$     Breadth $(b) = 80 \ cm$     Height $(h) = h$

Therefore flow in 90 minutes $=$ $\frac{12.5}{1000}$ $\times 96 = 1.2 \ m^3$

Therefore $2 \times 0.80 \times h = 1.2$

$\Rightarrow$ $\frac{1.2}{2 \times 0.8}$ $= 0.75 \ m$

$\Rightarrow h = 75 \ cm$

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