Class 9: Volume of a Cuboid and a Cube

$\displaystyle \textbf{Question 1: } \text{A cuboid water tank is }6\text{ m long, }5\text{ m wide and }4.5\text{ m deep.}$
$\displaystyle \text{How many litres of water can it hold?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the cuboid tank: Length }(l)=6\text{ m, Breadth }(b)=5\text{ m, Height }(h)=4.5\text{ m.}$
$\displaystyle \text{Volume of the tank}=lbh$
$\displaystyle =6\times5\times4.5=135\text{ m}^3$
$\displaystyle 1\text{ m}^3=1000\text{ litres}$
$\displaystyle \therefore \text{Capacity of the tank}=135\times1000=135000\text{ litres}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{A cuboidal vessel is }10\text{ m long and }8\text{ m wide. How high must it be}$
$\displaystyle \text{made to hold }380\text{ cubic metres of a liquid?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=10\text{ m, Breadth }(b)=8\text{ m, Height}=h\text{ m.}$
$\displaystyle \text{Volume}=lbh$
$\displaystyle 380=10\times8\times h$
$\displaystyle h=\frac{380}{10\times8}=4.75\text{ m}$
$\displaystyle \therefore \text{The required height is }4.75\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{Find the cost of digging a cuboid pit }8\text{ m long, }6\text{ m broad}$
$\displaystyle \text{and }3\text{ m deep at the rate of Rs. }30\text{ per m}^3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=8\text{ m, Breadth }(b)=6\text{ m, Height }(h)=3\text{ m.}$
$\displaystyle \text{Volume of the pit}=lbh$
$\displaystyle =8\times6\times3=144\text{ m}^3$
$\displaystyle \text{Rate of digging}=\text{Rs. }30\text{ per m}^3$
$\displaystyle \therefore \text{Cost of digging}=144\times30=\text{Rs. }4320$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{If }V\text{ is the volume of a cuboid of dimensions }a,b,c\text{ and }S\text{ is its}$
$\displaystyle \text{surface area, then prove that }\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the cuboid}=V=abc\qquad\ldots\text{(i)}$
$\displaystyle \text{Surface area of the cuboid}=S=2(ab+bc+ca)\qquad\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i),}$
$\displaystyle \frac{S}{V}=\frac{2(ab+bc+ca)}{abc}$
$\displaystyle \frac{S}{V}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
$\displaystyle \therefore \frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: } \text{The areas of three adjacent faces of a cuboid are }x,y\text{ and }z\text{. If its volume}$
$\displaystyle \text{is }V\text{, prove that }V^2=xyz.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the dimensions of the cuboid be }a,b\text{ and }c.$
$\displaystyle \text{Then }V=abc$
$\displaystyle ab=x\qquad\ldots\text{(i)}$
$\displaystyle bc=y\qquad\ldots\text{(ii)}$
$\displaystyle ca=z\qquad\ldots\text{(iii)}$
$\displaystyle \text{Multiplying (i), (ii) and (iii),}$
$\displaystyle xyz=(ab)(bc)(ca)$
$\displaystyle xyz=a^2b^2c^2=(abc)^2$
$\displaystyle \therefore xyz=V^2$
$\displaystyle \therefore V^2=xyz$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: } \text{If the areas of three adjacent faces of a cuboid are }8\text{ cm}^2,\ 18\text{ cm}^2$
$\displaystyle \text{and }25\text{ cm}^2\text{, find the volume of the cuboid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the dimensions of the cuboid be }a,\ b\text{ and }c.$
$\displaystyle ab=8\qquad\ldots\text{(i)}$
$\displaystyle bc=18\qquad\ldots\text{(ii)}$
$\displaystyle ca=25\qquad\ldots\text{(iii)}$
$\displaystyle \text{Multiplying (i), (ii) and (iii),}$
$\displaystyle 8\times18\times25=(abc)^2$
$\displaystyle \therefore V=\sqrt{8\times18\times25}$
$\displaystyle =\sqrt{2^3\times2\times3^2\times5^2}=2^2\times3\times5$
$\displaystyle =60\text{ cm}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: } \text{The breadth of a room is twice its height and one-half of its length. If the}$
$\displaystyle \text{volume of the room is }512\text{ dm}^3\text{, find its dimensions.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height}=x\text{ dm.}$
$\displaystyle \text{Then breadth}=2x\text{ dm and length}=4x\text{ dm.}$
$\displaystyle \text{Volume}=lbh$
$\displaystyle 4x\times2x\times x=512$
$\displaystyle 8x^3=512$
$\displaystyle x^3=64$
$\displaystyle x=4$
$\displaystyle \therefore \text{Height}=4\text{ dm}$
$\displaystyle \text{Breadth}=8\text{ dm}$
$\displaystyle \text{Length}=16\text{ dm}$
$\displaystyle \therefore \text{The dimensions of the room are }16\text{ dm, }8\text{ dm and }4\text{ dm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: } \text{A river }3\text{ m deep and }40\text{ m wide is flowing at the rate of }2\text{ km per hour.}$
$\displaystyle \text{How much water will fall into the sea in one minute?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Depth of the river}=3\text{ m, Width}=40\text{ m.}$
$\displaystyle \text{Speed of flow}=2\text{ km/h}=\frac{2\times1000}{60}\text{ m/min}=\frac{100}{3}\text{ m/min}$
$\displaystyle \text{Volume of water flowing in one minute}$
$\displaystyle =3\times40\times\frac{100}{3}=4000\text{ m}^3$
$\displaystyle \therefore \text{Water falling into the sea in one minute}=4000\text{ m}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{Water in a canal }30\text{ dm wide and }12\text{ dm deep is flowing with a}$
$\displaystyle \text{velocity of }100\text{ km per hour. How much area will it irrigate in }30\text{ minutes if }8\text{ cm}$
$\displaystyle \text{of standing water is desired?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Width of the canal}=30\text{ dm}=3\text{ m}$
$\displaystyle \text{Depth of the canal}=12\text{ dm}=1.2\text{ m}$
$\displaystyle \text{Velocity of flow}=100\text{ km/h}=\frac{100\times1000}{60}\text{ m/min}=\frac{10000}{6}\text{ m/min}$
$\displaystyle \text{Volume of water flowing in }30\text{ minutes}$
$\displaystyle =1.2\times3\times\frac{10000}{6}\times30=180000\text{ m}^3$
$\displaystyle \text{Depth of standing water}=8\text{ cm}=0.08\text{ m}$
$\displaystyle \therefore \text{Area irrigated}=\frac{180000}{0.08}=2250000\text{ m}^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{Three metal cubes with edges }6\text{ cm, }8\text{ cm and }10\text{ cm respectively are}$
$\displaystyle \text{melted together and formed into a single cube. Find the volume, surface area and}$
$\displaystyle \text{diagonal of the new cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total volume of the three cubes}=6^3+8^3+10^3$
$\displaystyle =216+512+1000=1728\text{ cm}^3$
$\displaystyle \text{Let the edge of the new cube}=a\text{ cm.}$
$\displaystyle a^3=1728$
$\displaystyle \therefore a=\sqrt[3]{1728}=12\text{ cm}$
$\displaystyle \therefore \text{Volume of the new cube}=12^3=1728\text{ cm}^3$
$\displaystyle \text{Surface area of the new cube}=6a^2$
$\displaystyle =6(12)^2=864\text{ cm}^2$
$\displaystyle \text{Diagonal of the new cube}=a\sqrt{3}$
$\displaystyle =12\sqrt{3}=20.78\text{ cm (approx.).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{Two cubes, each of volume }512\text{ cm}^3\text{, are joined end to end. Find the}$
$\displaystyle \text{surface area of the resulting cuboid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of each cube}=a\text{ cm.}$
$\displaystyle a^3=512$
$\displaystyle \therefore a=8\text{ cm}$
$\displaystyle \text{Dimensions of the resulting cuboid are }16\text{ cm, }8\text{ cm and }8\text{ cm.}$
$\displaystyle \text{Total surface area}=2(lb+bh+hl)$
$\displaystyle =2(16\times8+8\times8+8\times16)$
$\displaystyle =2(128+64+128)=640\text{ cm}^2$
$\displaystyle \therefore \text{Surface area of the resulting cuboid}=640\text{ cm}^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: } \text{Half cubic metre of a gold sheet is extended by hammering so as to cover}$
$\displaystyle \text{an area of }1\text{ hectare. Find the thickness of the gold sheet.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the gold sheet}=0.5\text{ m}^3$
$\displaystyle 1\text{ hectare}=10000\text{ m}^2$
$\displaystyle \text{Let the thickness of the gold sheet}=x\text{ m.}$
$\displaystyle \text{Volume}=\text{Area}\times\text{Thickness}$
$\displaystyle 0.5=10000x$
$\displaystyle x=\frac{0.5}{10000}=0.00005\text{ m}$
$\displaystyle =0.005\text{ cm}=0.05\text{ mm}$
$\displaystyle \therefore \text{Thickness of the gold sheet}=0.05\text{ mm}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: } \text{A metal cube of edge }12\text{ cm is melted and formed into three smaller}$
$\displaystyle \text{cubes. If the edges of the two smaller cubes are }6\text{ cm and }8\text{ cm, find the}$
$\displaystyle \text{edge of the third smaller cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of the third cube}=x\text{ cm.}$
$\displaystyle \text{Volume of the large cube}=12^3=1728\text{ cm}^3$
$\displaystyle \therefore 6^3+8^3+x^3=1728$
$\displaystyle 216+512+x^3=1728$
$\displaystyle x^3=1000$
$\displaystyle x=\sqrt[3]{1000}=10\text{ cm}$
$\displaystyle \therefore \text{The edge of the third cube is }10\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: } \text{The dimensions of a cinema hall are }100\text{ m, }50\text{ m and }18\text{ m. How many}$
$\displaystyle \text{persons can sit in the hall, if each person requires }150\text{ m}^3\text{ of air?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=100\text{ m, Breadth }(b)=50\text{ m, Height }(h)=18\text{ m.}$
$\displaystyle \text{Volume of the hall}=lbh$
$\displaystyle =100\times50\times18=90000\text{ m}^3$
$\displaystyle \text{Air required by each person}=150\text{ m}^3$
$\displaystyle \therefore \text{Number of persons}=\frac{90000}{150}=600$
$\displaystyle \therefore 600\text{ persons can sit in the hall.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: } \text{Given that }1\text{ cm}^3\text{ of marble weighs }0.25\text{ kg, the weight of a marble block}$
$\displaystyle 28\text{ cm in width and }5\text{ cm thick is }112\text{ kg. Find the length of the block.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the marble block}=x\text{ cm.}$
$\displaystyle \text{Width}=28\text{ cm, Thickness}=5\text{ cm.}$
$\displaystyle \text{Volume of the marble block}=28\times5\times x=140x\text{ cm}^3$
$\displaystyle 1\text{ cm}^3\text{ of marble weighs }0.25\text{ kg.}$
$\displaystyle \therefore 140x\times0.25=112$
$\displaystyle x=\frac{112}{140\times0.25}=3.2\text{ cm}$
$\displaystyle \therefore \text{The length of the marble block is }3.2\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: } \text{A box with a lid is made of }2\text{ cm thick wood. Its external length, breadth}$
$\displaystyle \text{and height are }25\text{ cm, }18\text{ cm and }15\text{ cm respectively. How many cubic centimetres}$
$\displaystyle \text{of a liquid can be placed in it? Also, find the volume of the wood used in it.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{External length}=25\text{ cm, Breadth}=18\text{ cm, Height}=15\text{ cm.}$
$\displaystyle \text{Thickness of the wood}=2\text{ cm.}$
$\displaystyle \text{Internal length}=25-2(2)=21\text{ cm}$
$\displaystyle \text{Internal breadth}=18-2(2)=14\text{ cm}$
$\displaystyle \text{Internal height}=15-2(2)=11\text{ cm}$
$\displaystyle \text{Capacity of the box}=21\times14\times11=3234\text{ cm}^3$
$\displaystyle \text{External volume}=25\times18\times15=6750\text{ cm}^3$
$\displaystyle \therefore \text{Volume of wood used}=6750-3234=3516\text{ cm}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: } \text{The external dimensions of a closed wooden box are }48\text{ cm, }36\text{ cm and}$
$\displaystyle 30\text{ cm. The box is made of }1.5\text{ cm thick wood. How many bricks of size}$
$\displaystyle 6\text{ cm}\times3\text{ cm}\times0.75\text{ cm can be put in this box?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{External length}=48\text{ cm, Breadth}=36\text{ cm, Height}=30\text{ cm.}$
$\displaystyle \text{Thickness of the wood}=1.5\text{ cm.}$
$\displaystyle \text{Internal length}=48-2(1.5)=45\text{ cm}$
$\displaystyle \text{Internal breadth}=36-2(1.5)=33\text{ cm}$
$\displaystyle \text{Internal height}=30-2(1.5)=27\text{ cm}$
$\displaystyle \text{Internal volume of the box}=45\times33\times27=40095\text{ cm}^3$
$\displaystyle \text{Volume of one brick}=6\times3\times0.75=13.5\text{ cm}^3$
$\displaystyle \therefore \text{Number of bricks}=\frac{40095}{13.5}=2970$
$\displaystyle \therefore 2970\text{ bricks can be put in the box.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: } \text{How many cubic centimetres of iron are there in an open box whose external}$
$\displaystyle \text{dimensions are }36\text{ cm, }25\text{ cm and }16.5\text{ cm, the iron being }1.5\text{ cm thick}$
$\displaystyle \text{throughout? If }1\text{ cm}^3\text{ of iron weighs }15\text{ g, find the weight of the empty box in kg.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{External length}=36\text{ cm, Breadth}=25\text{ cm, Height}=16.5\text{ cm.}$
$\displaystyle \text{Thickness of the iron}=1.5\text{ cm.}$
$\displaystyle \text{Internal length}=36-2(1.5)=33\text{ cm}$
$\displaystyle \text{Internal breadth}=25-2(1.5)=22\text{ cm}$
$\displaystyle \text{Internal height}=16.5-1.5=15\text{ cm}$
$\displaystyle \text{External volume}=36\times25\times16.5=14850\text{ cm}^3$
$\displaystyle \text{Internal volume}=33\times22\times15=10890\text{ cm}^3$
$\displaystyle \therefore \text{Volume of iron}=14850-10890=3960\text{ cm}^3$
$\displaystyle \text{Weight of }1\text{ cm}^3\text{ of iron}=15\text{ g}$
$\displaystyle \therefore \text{Weight of the empty box}=3960\times15=59400\text{ g}$
$\displaystyle =59.4\text{ kg}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: } \text{A cube of edge }9\text{ cm is immersed completely in a rectangular vessel containing}$
$\displaystyle \text{water. If the dimensions of the base are }15\text{ cm and }12\text{ cm, find the rise in the}$
$\displaystyle \text{water level in the vessel.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the cube}=9^3=729\text{ cm}^3$
$\displaystyle \text{Base area of the vessel}=15\times12=180\text{ cm}^2$
$\displaystyle \text{Let the rise in water level}=h\text{ cm.}$
$\displaystyle 180h=729$
$\displaystyle \therefore h=\frac{729}{180}=4.05\text{ cm}$
$\displaystyle \therefore \text{The rise in water level is }4.05\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: } \text{A rectangular container, whose base is a square of side }5\text{ cm, stands on a}$
$\displaystyle \text{horizontal table and holds water up to }1\text{ cm from the top. When a cube is placed}$
$\displaystyle \text{in the water, it is completely submerged, the water rises to the top and }2\text{ cm}^3\text{ of}$
$\displaystyle \text{water overflows. Calculate the volume of the cube and also the length of its edge.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the base of the container}=5\times5=25\text{ cm}^2$
$\displaystyle \text{Height of the empty space above water}=1\text{ cm}$
$\displaystyle \text{Volume of the empty space}=25\times1=25\text{ cm}^3$
$\displaystyle \text{Volume of water overflowing}=2\text{ cm}^3$
$\displaystyle \therefore \text{Volume of water displaced by the cube}=25+2=27\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the cube}=27\text{ cm}^3$
$\displaystyle \text{Let the edge of the cube}=a\text{ cm.}$
$\displaystyle a^3=27$
$\displaystyle \therefore a=3\text{ cm}$
$\displaystyle \therefore \text{The volume of the cube is }27\text{ cm}^3\text{ and its edge is }3\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: } \text{A field is }200\text{ m long and }150\text{ m broad. There is a plot, }50\text{ m long and}$
$\displaystyle 40\text{ m broad, near the field. The plot is dug }7\text{ m deep and the earth taken out is}$
$\displaystyle \text{spread evenly on the field. By how many metres is the level of the field raised?}$
$\displaystyle \text{Give the answer to the second decimal place.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the field}=200\times150=30000\text{ m}^2$
$\displaystyle \text{Volume of earth dug out}=50\times40\times7=14000\text{ m}^3$
$\displaystyle \text{Let the rise in the level of the field}=h\text{ m.}$
$\displaystyle \text{Volume of earth spread over the field}=30000h$
$\displaystyle \therefore 30000h=14000$
$\displaystyle h=\frac{14000}{30000}=0.4666\ldots\text{ m}$
$\displaystyle \therefore h=0.47\text{ m, correct to two decimal places.}$
$\displaystyle \therefore \text{The level of the field is raised by }0.47\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: } \text{A field is in the form of a rectangle of length }18\text{ m and width }15\text{ m. A pit,}$
$\displaystyle 7.5\text{ m long, }6\text{ m broad and }0.8\text{ m deep, is dug in a corner of the field and the}$
$\displaystyle \text{earth taken out is spread over the remaining area of the field. Find the extent to}$
$\displaystyle \text{which the level of the field has been raised.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the field}=18\times15=270\text{ m}^2$
$\displaystyle \text{Area of the pit}=7.5\times6=45\text{ m}^2$
$\displaystyle \therefore \text{Area over which the earth is spread}=270-45=225\text{ m}^2$
$\displaystyle \text{Volume of earth dug out}=7.5\times6\times0.8=36\text{ m}^3$
$\displaystyle \text{Let the rise in the level of the field}=h\text{ m.}$
$\displaystyle 225h=36$
$\displaystyle \therefore h=\frac{36}{225}=0.16\text{ m}$
$\displaystyle \therefore \text{The level of the field is raised by }0.16\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: } \text{A rectangular tank is }80\text{ m long and }25\text{ m broad. Water flows into it through}$
$\displaystyle \text{a pipe whose cross-sectional area is }25\text{ cm}^2\text{, at the rate of }16\text{ km/hour.}$
$\displaystyle \text{How much does the level of the water rise in the tank in }45\text{ minutes?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the base of the tank}=80\times25=2000\text{ m}^2$
$\displaystyle \text{Cross-sectional area of the pipe}=25\text{ cm}^2=\frac{25}{10000}\text{ m}^2$
$\displaystyle \text{Speed of flow}=16\text{ km/h}=\frac{16\times1000}{60}\text{ m/min}=\frac{800}{3}\text{ m/min}$
$\displaystyle \text{Volume of water flowing per minute}=\frac{25}{10000}\times\frac{800}{3}=\frac{2}{3}\text{ m}^3$
$\displaystyle \text{Volume of water flowing in }45\text{ minutes}=\frac{2}{3}\times45=30\text{ m}^3$
$\displaystyle \text{Let the rise in water level}=h\text{ m.}$
$\displaystyle 80\times25\times h=30$
$\displaystyle \therefore h=\frac{30}{2000}=\frac{3}{200}\text{ m}=0.015\text{ m}$
$\displaystyle =1.5\text{ cm}$
$\displaystyle \therefore \text{The water level rises by }1.5\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: } \text{Water in a rectangular reservoir having base }80\text{ m}\times60\text{ m is }6.5\text{ m}$
$\displaystyle \text{deep. In what time can the water be emptied by a pipe whose cross-section is a square}$
$\displaystyle \text{of side }20\text{ cm, if the water runs through the pipe at the rate of }15\text{ km/hr?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of water in the reservoir}=80\times60\times6.5=31200\text{ m}^3$
$\displaystyle \text{Side of the square pipe}=20\text{ cm}=0.2\text{ m}$
$\displaystyle \text{Cross-sectional area of the pipe}=0.2\times0.2=0.04\text{ m}^2$
$\displaystyle \text{Speed of flow}=15\text{ km/hr}=\frac{15\times1000}{60}=250\text{ m/min}$
$\displaystyle \text{Volume of water flowing per minute}=0.04\times250=10\text{ m}^3$
$\displaystyle \therefore \text{Time required}=\frac{31200}{10}=3120\text{ minutes}$
$\displaystyle =\frac{3120}{60}=52\text{ hours}$
$\displaystyle \therefore \text{The reservoir can be emptied in }52\text{ hours.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: } \text{A village having a population of }4000\text{ requires }150\text{ litres of water per head}$
$\displaystyle \text{per day. It has a tank measuring }20\text{ m}\times15\text{ m}\times6\text{ m. For how many days}$
$\displaystyle \text{will the water of this tank last?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the tank}=20\times15\times6=1800\text{ m}^3$
$\displaystyle \text{Daily water requirement}=4000\times150=600000\text{ litres}$
$\displaystyle =\frac{600000}{1000}=600\text{ m}^3$
$\displaystyle \therefore \text{Number of days}=\frac{1800}{600}=3$
$\displaystyle \therefore \text{The water in the tank will last for }3\text{ days.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: } \text{A child playing with building blocks, which are in the shape of cubes, has built}$
$\displaystyle \text{a structure as shown in the adjoining figure. If the edge of each cube is }3\text{ cm,}$
$\displaystyle \text{find the volume of the structure built by the child.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of cubes}=15$
$\displaystyle \text{Volume of one cube}=3^3=27\text{ cm}^3$
$\displaystyle \therefore \text{Volume of the structure}=15\times27=405\text{ cm}^3$
$\displaystyle \therefore \text{The volume of the structure is }405\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: } \text{A godown measures }40\text{ m}\times25\text{ m}\times10\text{ m. Find the maximum}$
$\displaystyle \text{number of wooden crates, each measuring }1.5\text{ m}\times1.25\text{ m}\times0.5\text{ m, that can}$
$\displaystyle \text{be stored in the godown.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the godown}=40\text{ m}\times25\text{ m}\times10\text{ m}$
$\displaystyle \text{Dimensions of each crate}=1.5\text{ m}\times1.25\text{ m}\times0.5\text{ m}$
$\displaystyle \text{Arrange the crates so that }1.25\text{ m lies along }40\text{ m, }0.5\text{ m along }25\text{ m}$
$\displaystyle \text{and }1.5\text{ m along }10\text{ m.}$
$\displaystyle \text{Number of crates along the length}=\frac{40}{1.25}=32$
$\displaystyle \text{Number of crates along the breadth}=\frac{25}{0.5}=50$
$\displaystyle \text{Number of crates along the height}=\left\lfloor\frac{10}{1.5}\right\rfloor=6$
$\displaystyle \therefore \text{Maximum number of crates}=32\times50\times6=9600$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: } \text{A wall of length }10\text{ m was to be built across an open ground. The height of}$
$\displaystyle \text{the wall is }4\text{ m and its thickness is }24\text{ cm. If this wall is to be built using bricks}$
$\displaystyle \text{of dimensions }24\text{ cm}\times12\text{ cm}\times8\text{ cm, how many bricks are required?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the wall}=10\text{ m}\times0.24\text{ m}\times4\text{ m}$
$\displaystyle \text{Volume of the wall}=10\times0.24\times4=9.6\text{ m}^3$
$\displaystyle \text{Dimensions of one brick}=0.24\text{ m}\times0.12\text{ m}\times0.08\text{ m}$
$\displaystyle \text{Volume of one brick}=0.24\times0.12\times0.08=0.002304\text{ m}^3$
$\displaystyle \text{Number of bricks}=\frac{9.6}{0.002304}=4166.67$
$\displaystyle \therefore \text{Bricks required}=4167$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: } \text{If the volume of a cube is }V\text{ m}^3\text{, its surface area is }S\text{ m}^2\text{ and the}$
$\displaystyle \text{length of a diagonal is }d\text{ m, prove that }6\sqrt{3}V=Sd.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the cube be }a\text{ m.}$
$\displaystyle \text{Volume of the cube}=a^3=V$
$\displaystyle \text{Surface area of the cube}=6a^2=S$
$\displaystyle \text{Diagonal of the cube}=\sqrt{3}\,a=d$
$\displaystyle \therefore Sd=(6a^2)(\sqrt{3}\,a)$
$\displaystyle =6\sqrt{3}\,a^3$
$\displaystyle =6\sqrt{3}\,V$
$\displaystyle \therefore 6\sqrt{3}V=Sd$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: } \text{Water is being filled in an aquarium at the rate of }12.5\text{ litres per minute.}$
$\displaystyle \text{If the aquarium is }2\text{ m long and }80\text{ cm wide and it is filled in }96\text{ minutes,}$
$\displaystyle \text{find the height of the aquarium.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=2\text{ m, Breadth }(b)=80\text{ cm}=0.8\text{ m, Height}=h\text{ m.}$
$\displaystyle \text{Water filled in }96\text{ minutes}=12.5\times96=1200\text{ litres}$
$\displaystyle =\frac{1200}{1000}=1.2\text{ m}^3$
$\displaystyle \text{Volume of the aquarium}=lbh$
$\displaystyle 2\times0.8\times h=1.2$
$\displaystyle h=\frac{1.2}{2\times0.8}=0.75\text{ m}$
$\displaystyle =75\text{ cm}$
$\displaystyle \therefore \text{The height of the aquarium is }75\text{ cm.}$
$\displaystyle \\$