Question 15 Marks
A solid cube of side $12 cm$ is cut into $8$ identical cubes. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the total surface area of all the small cubes formed.
AnswerHere, the cube of side $12 \ cm$ is divided into $8$ cubes of side $9 \ cm.$
Given that,
Their volumes are equal.
The volume of big cube of $12 \ cm =$ Volume of $8$ cubes of side $a \ cm$
$($Side of the big cube$)^3=8 \times ($Side of the small cube$) ^3$
$(12)^3=8 \times a^3$
$ \Rightarrow a^3=\frac{12 \times 12 \times 12}{8}$
$ \Rightarrow a^3=6^3 \ cm ^3$
$ \Rightarrow a=6 \ cm$
Ratio of their surface $=\frac{\text { Surface area of the original cube }}{\text { Total surface area of the small cube }}$
$ =\frac{6(\text { side of big cube })^2}{8 \times 6(\text { side of small cube })^2}$
$ =\frac{6 \times 12 \times 12}{8 \times 6 \times 6 \times 6}=\frac{4}{8}=1: 2$
So, the ratio is $1: 2$ View full question & answer→Question 25 Marks
The length of a hall is double its breadth. Its height is $3 m$. The area of its four walls $($including doors and windows$)$ is $108\ m^2,$ find its volume.
AnswerLet the breadth be $x$
and the length be $2 x$
Height $=3\ m$
Area of four walls $=108\ m ^2$
$\Rightarrow 2( l + b ) h =108$
$\Rightarrow 2(2 x+x) 3=108$
$\Rightarrow 6 \times 3 x=108$
$\Rightarrow 3 x=\frac{108}{6}$
$\Rightarrow x=\frac{18}{3}=6\ m$
$\therefore$ Breadth $= x =6\ m$
and length $=2 x=12\ m$
Hence, Volume $=l \times b \times h$
$=12 \times 6 \times 3$
$=216\ m ^3$
View full question & answer→Question 35 Marks
The floor of a rectangular hall has a perimeter $250\ m$. If the cost of painting the four walls at the rate of $₹ 10\ per\ m^2$ is ₹. $15,000,$ find the height of the hall.
AnswerLet the length, breadth, and height of the rectangular hall be $l \ m, b\ m$, and $h\ m$ respectively.
The perimeter of the floor of hall $= 2(l + b)$
$250\ m = 2(l + b)$
$( l + b )=\frac{250}{2}=125\ cm$
Area of four walls $=$ Area of cuboid $–$ Area of floor $–$ Area of top
$= 2 (lb + bh + hl) – (l \times b) – (l \times b)$
$= 2(lb) + 2 (bh) + 2(hl) – 2lb = 2 lh + 2 bh$
$= 2h(l + b)$
$= 2h \times 125 [$From $(i)]$
$= 250\ h m^2$
Area of four walls $= 250\ h m^2$
Cost of painting $1$ m$^2$ area $=₹ 10$
Cost of painting $250\ h m ^2$ area $=₹ 10 \times 250\ h=2500\ h$
$15000=2500\ h$
$h=\frac{15000}{2500}$
The height of the hall is $6\ m.$ View full question & answer→Question 45 Marks
The curved surface area and the volume of a toy, cylindrical in shape, are $132 \ cm^2$ and $462\ cm^3$ respectively. Find, its diameter and its length.
AnswerLet the radius of a toy $=r$ and
height of the toy $= h$
The curved surface area of a toy $=132\ cm^2$
$ =>2 \pi r h=132\ cm ^2$
$\Rightarrow 2 \pi r h=132\ cm ^2$
$ \Rightarrow r=\frac{132}{2 \pi \times h}\ cm ^2 \ldots . . .(\text { i) }$
Also, volume of a toy $=462\ cm ^3$
$\Rightarrow \pi r^2 h=462\ cm ^3$
$\Rightarrow r^2=\frac{462}{\pi \times h}$
Now, substitute the volume of $r$, we get
$\frac{(132)^2}{(2)^2 \times(\pi)^2 \times h^2}=\frac{462}{\pi \times h}$
$ \Rightarrow \frac{132^2}{4 \times \pi \times h}=462$
$ \Rightarrow 4 \times \pi \times h=\frac{132 \times 132}{462}$
$\Rightarrow h=\frac{132 \times 132}{462 \times \pi \times h}$
$\Rightarrow h=\frac{132 \times 132 \times 7}{462 \times 22 \times 4}=3\ cm$
Now, put the value of $h$ in eq. $(i)$, we get
$r=\frac{132 \times 7}{2 \times 22 \times 3}=7\ cm$
$\therefore$ Diameter of the toy $=2 \times r$
$=2 \times 7\ cm =14\ cm$
View full question & answer→Question 55 Marks
The external dimensions of an open wooden box are $65 \ cm, 34 \ cm$, and $25 \ cm$. If the box is made up of wood $2 \ cm$ thick, find the capacity of the box and the volume of wood used to make it.
AnswerExternal length $= 65\ cm$
External breath$ = 34 \ cm$
External height $= 25 \ cm$
Volume $= l \times b \times h$
$= 65 \times 34 \times 25$
$= 55,250 \ cm^3$
Thickness $= 2 \ cm$
Internal length $= 65 - 4 = 61 \ cm$
Internal breath $= 34 - 4 = 30 \ cm$
Internal height $= 25 - 2 = 23 \ cm$
Capacity of the box$ (V) = 61 \times 30 \times 23$
$= 42,090 \ cm^3$
The volume of wood used to make the box $= V_{ext} - V_{int}$
$= 65 \times 34 \times 25 \ cm^3 - 61 \times 30 \times 23 \ cm^3$
$= 55,250 - 42,090$
$= 13,160 \ cm^3$
View full question & answer→Question 65 Marks
The length, breadth, and height of a cuboid are in the ratio $6: 5 : 3.$ If its total surface area is $504 \ cm^2$, find its volume.
AnswerLet the length of the given cuboid $=6 x$
The breadth of the given cuboid $=5 x$
Height of the given cuboid $=3 x$
The total surface area of the given cuboid
$=2(l b+b h+h l) $
$=2(6 x \times 5 x+5 x \times 3 x+3 x \times 6 x) $
$=\left(30 x^2+15 x^2+18 x^2\right) $
$=2 \times 63 x^2=126 x^2$
But, we are given total surface area $=504 \ cm ^2$
$\therefore 126 x^2=504$
$\Rightarrow x^2=\frac{504}{126} $
$\Rightarrow x^2=4 $
$\Rightarrow x^2=(2)^2$
$\Rightarrow x =2 \ cm$
$\therefore$ Length of the given cuboid $=6 x$
$=6 \times 2 \ cm =12 \ cm$
Breadth of the given cuboid $=5 x$
$=5 \times 2 \ cm =10 \ cm$
Height of the given cuboid $=3 x$
$=3 \times 2 \ cm =6 \ cm$
Now, volume of the cuboid $=l \times b \times h$
$=12 \times 10 \times 6=720 \ cm ^3$
View full question & answer→Question 75 Marks
The sum of the radius and the height of a cylinder is $37\ cm$ and the total surface area of the cylinder is $1628 \ cm^2$. Find the height and the volume of the cylinder.
AnswerLet $r$ and $h$ be the radius and height of the solid cylinder respectively.
Given, $r + h =37 \ cm$
The total surface area of the cylinder $=1628 \ cm ^2 ($Given$)$
$\therefore 2 \pi r(r+h)=1628 \ cm ^2$
$\Rightarrow 2 \pi r \times 37=1628 \ cm ^2$
$\Rightarrow 2 \times \frac{22}{7} \times r \times 37=1628 \ cm ^2$
$\Rightarrow r=\frac{1628 \times 7}{2 \times 22 \times 37}=7 \ cm$
$rth =37 \ cm $
$\Rightarrow 7+ h =37 \ cm $
$\Rightarrow h =30 \ cm$
The volume of the cylinder $=\pi r^2 h$
$=\frac{22}{7} \times 7 \times 7 \times 30=4620 \ cm ^3$
View full question & answer→Question 85 Marks
The length, breadth, and height of a cuboid $($rectangular solid$)$ are $4 : 3: 2. \ (i)$ If its surface area is $2548\ cm^2$, find its volume.$(ii)$ If its volume is $3000\ m^3$, find its surface area.
Answer$(i)$ Surface area of cuboid $=2548 \ cm ^2$
Ratio in length, breadth and height of a cuboid $=4: 3: 2$
Let length $=4 x$, Breadth $=3 x$ and height $=2 x$
$\therefore$ Surface area $=2(4 x \times 3 x+3 x \times 2 x+2 x \times 4 x)$
$=2\left(12 x^2+6 x^2+8 x^2\right)$
$=2 \times 26 x^2=52 x^2$
$\therefore 52 x^2=2548$
$x^2=\frac{2548}{52}=49=(7)^2$
$\therefore x=7$
$\therefore$ Length $=4 x=4 \times 7=28 \ cm$
$\therefore$ Breadth $=3 x=3 \times 7=21 \ cm$
and height $=2 x=2 \times 7=14 \ cm$
$\therefore$ Volume $=l b h$
$=28 \times 21 \times 14 \ cm ^3=8232 \ cm ^2$
$(ii)$ If volume $=3000\ m ^3$
$\Rightarrow 4 x \times 3 x \times 2 x=3000$
$\Rightarrow 24 x^3=3000$
$\Rightarrow x^3=\frac{3000}{24}=125=(5)^3$
$\therefore x=5\ m$
Length $=5 \times 4=20$, breadth $=5 \times 3=15\ m$
and height $=5 \times 2=10\ m$
$\therefore$ Surface area $=2[l b+b h+h l]$
$=2[20 \times 15+15 \times 10+10 \times 20]\ m ^2$
$=2[300+150+200]\ m ^2$
$=2 \times 650=1300\ m ^2$
View full question & answer→Question 95 Marks
A cube of edge $6 \ cm$ and a cuboid with dimensions $4 \ cm \times x \ cm \times 15 \ cm$ are equal in volume. Find : $(i)$ the value of $x.\ (ii)$ the total surface area of the cuboid.$(iii)$ the total surface area of the cube.$(iv)$ which of these two has a greater surface and by how much?
AnswerEdge of a cube $=6 \ cm$
volume $=a^3=(6)^3=216 \ cm ^3$
Dimensions of a cuboid $=4 \ cm \times \ cm \times 15 \ cm$
volume $=60 \times \ cm ^3$
The volume of both is equal
$(i) \therefore 60 x=216 $
$ \Rightarrow x=\frac{216}{60}=\frac{36}{10}$
$\therefore x =3.6 \ cm$
$(ii)$ Total surface area of cuboid
$=2[l b+b h+h l]$
$=2[4 \times 3.6+3.6 \times 15+15 \times 4] \ cm ^2$
$=2[14.4+54.0+60] \ cm ^2$
$=128.4 \times 2=256.8 \ cm ^2$
$(iii)$ Total surface area of cube
$=6 a^2=6(6)^2=6 \times 36=216 \ cm ^2$
$(iv)$ Difference of surface areas $=256.8-216$
$=40.8 \ cm ^2$
$\therefore$ Surface area of cuboid is greater
View full question & answer→Question 105 Marks
The internal length, breadth, and height of a closed box are $1\ m, 80 \ cm,$ and $25 \ cm.$ respectively. If its sides are made of $2.5 \ cm$ thick wood; find : $(i)$ the capacity of the box $(ii)$ the volume of wood used to make the box.
AnswerInternal length of the closed box $=1 m =100 \ cm$
breadth $=80 \ cm$
height $=25 \ cm$
volume $=100 \times 80 \times 25$
$=200000 \ cm ^3$
External length of the box $=(100+2 \times 2.5)$
$=100+5=105 \ cm$
External breadth $=(80+2 \times 2.5)$
$=80+5=85 \ cm$
External height $=(25+2 \times 2.5)$
$=(25+5)=30 \ cm$
External volume $=105 \times 85 \times 30 \ cm ^3$
$=267750 \ cm ^3$
$(i)$ The capacity of the box $=100 \times 80 \times 25 \ cm ^3$
$=200000 \ cm ^3$
$=\frac{200000}{100 \times 100 \times 100}\ m ^3$
$=0.2\ m ^3$
$(ii)$ The volume of wood used to make the box
$=$ External volume $-$ internal volume
$ =267750$-$200000$
$ =67750 \ cm ^3$
$ =\frac{67750}{100 \times 100 \times 100}\ m ^3$
$ =0.06775\ m ^3$
View full question & answer→Question 115 Marks
The dining$-$hall of a hotel is $75\ m$ long;$ 60\ m$ broad and $16\ m$ high. It has five $-$ doors $4\ m$ by $3\ m$ each and four windows $3\ m$ by $1.6 m$ each. Find the cost of : $(i)$ papering its walls at the rate of $Rs. 12$ per $m ^2; (ii)$ carpetting its floor at the rate of $Rs. 25$ per $m ^2$.
AnswerLength of the dining hall of a hotel $=75\ m$
The breadth of the dining hall of a hotel $=60\ m$
Height of the dining hall of a hotel $=16\ m$
$(i)$ Area of four walls of the dining hall $=2[L+B) \times H=2(75+60) \times 16$
$=2(135) \times 16$
$=270 \times 16$
$=4320\ m ^2$
Area of one door $=4 \times 3\ m ^2$
$=12\ m ^2$
Area of 5 doors $=12 \times 5=60\ m ^2$
Area of one window $=3 \times 1.6=4.8\ m ^2$
Area of 4 window $=4.8 \times 4=19.2\ m ^2$
Area of the walls to be papered
$=4320-(60+19.2)$
$=4320-79.2$
$=4240.8\ m ^2$
$ =4240.8 \times 12$
$ =\text { Rs. } 50889.60$
$(ii)$ Area of floor $=L \times B$
$ =75 \times 60$
$=4500\ m ^2$
Cost of carpetting the floor $@Rs. 25$ per $m ^2$
$=4500 \times 25$
$=\text { Rs. } 112500$
View full question & answer→Question 125 Marks
A room $5\ m$ long, $4.5\ m$ wide, and $3.6\ m$ high have one door $1.5\ m$ by $2.4\ m$ and two windows, each $1\ m$ by $0.75\ m$ . Find: $(i)$ the area of its walls, excluding door and windows ; $(ii)$ the cost of distempering its walls at the rate of $Rs. 4.50 per m ^2.(iii)$ the cost of painting its roof at the rate of $Rs. 9\ $ per $m ^2$.
AnswerLength of the room $=5\ m$
The breadth of the room $=4.5\ m$
Height of the room $=3.6\ m$
Area of the roof $= L \times B$
$=5 \times 4.5\ m ^2$
$=22.5\ m ^2$
Area of four walls $=2[L+B] \times H$
$=2[5+4.5] \times 3.6$
$=2(9.5) \times 3.6$
$=19 \times 3.6$
$=68.4\ m ^2$
Area of one door $=1.5 \times 2.4\ m ^2$
$=3.60\ m ^2$
$=3.6\ m ^2$
Area of one window $=1 \times 0.75\ m ^2$
$=0.75\ m ^2$
Area of 2 window $=0.75 \times 2\ m ^2$
$=1.5\ m ^2$
$(i)$ Area of four walls excluding door and windows $=68.4-(3.6+1.5)$
$= 68.4 - 5.1$
$= 63.3\ m^2$
$(ii)$ Cost of distempering four walls $@ Rs. 4.50$ per $m^2$
$= 63.3 \times 4.50$
$= Rs. 284.85$
$(iii)$ Cost of painting the roof $@ Rs.9$ per $m^2$
$= 22.5 \times 9$
$= Rs. 202.50$
View full question & answer→Question 135 Marks
Find the length of each edge of a cube, if its volume is : $(i)\ 216\ cm^3\ (ii)\ 1.728\ m^3$
Answer$(i)\ ($Edge$)^3=$ Volume of a cube
$($Edge$)^3=216 \ cm ^3$
$\Rightarrow$ Edge $=(216)^{\frac{1}{3}}$
$\Rightarrow$ Edge $=(3 \times 3 \times 3 \times 2 \times 2 \times 2)^{\frac{1}{3}}$
$\Rightarrow$ Edge $=3 \times 2$
$\Rightarrow$ Edge $=6 \ cm$.
$(ii)\ ($Edge$)^3=$ Volume of a cube
$($Edge$)^3=1.728 \ cm ^3$
$\Rightarrow(iii)\ ($Edge$)^3=\frac{1.728}{1.000}=\frac{1728}{1000}$
$\Rightarrow$ Edge $=\left(\frac{1728}{1000}\right)^{\frac{1}{3}}$
$\Rightarrow$ Edge $=\left(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}{10 \times 10 \times 10}\right)^{\frac{1}{3}}$
$\Rightarrow$ Edge $=\frac{2 \times 2 \times 3}{10}$
$\Rightarrow$ Edge $=\frac{12}{10}\ m$
$\Rightarrow$ Edge $=1.2\ m$
View full question & answer→Question 145 Marks
The length, breadth, and height of a cuboid are in the ratio $6: 5: 3$. If its total surface area is $504\ cm^2$; find its dimensions. Also, find the volume of the cuboid.
AnswerLet length of the cuboid $=6 x$
Breadth of the cuboid $=5 x$
Height of the cuboid $=3 x$
Total surface area of the given cuboid $=2(\mid \times b+b \times h+h \times l)$
$=2(6 \times \times 5 x+5 \times \times 3 x+3 \times \times 6 x)=2(30 \times 2+15 \times 2+18 \times 2)$
$ =2 \times 63 \times 2=126 x^2$
But we are given total surface area of the given cuboid $=504 \ cm ^2$
$126 x^2=504 \ cm ^2$
$ \Rightarrow x^2=\frac{504}{126}$
$ \Rightarrow x^2=4$
$ \Rightarrow x=\sqrt{4}$
$ \Rightarrow x=2 \ cm .$
Length of the cuboid $=6 x=6 \times 2=12 \ cm$
Breadth of the cuboid $=5 x =5 \times 2=10 \ cm$
Height of the cuboid $=3 x =3 \times 2=6 \ cm$
Volume of the cuboid $=1 \times b \times h =12 \times 10 \times 6=720 \ cm ^3$
View full question & answer→Question 155 Marks
The length, breadth, and height of a cuboid are in the ratio $5 : 3: 2.$ If its volume is $240 \ cm^3;$ find its dimensions. Also, find the total surface area of the cuboid.
AnswerLet length of the given cuboid $=5 x$
Breadth of the given cuboid $=3 x$
Height of the given cuboid $=2 x$
Volume of the given cuboid $=$ Length $x$ Breadth $x$ Height $=5 x \times 3 x \times 2 x =30 x ^3$
But we are given volume $=240 \ cm ^3$
$30 x ^3=240 \ cm ^3$
$\Rightarrow x ^3=\frac{240}{30}$
$\Rightarrow x ^3=8$
$\Rightarrow x =8^{\frac{1}{3}}$
$\Rightarrow x=(2 \times 2 \times 2)^{\frac{1}{3}}$
$\Rightarrow x =2 \ cm$
Length of the given cube $=5 x=5 \times 2=10 \ cm$
Breadth of the given cube $=3 x=3 \times 2=6 \ cm$
Height of the given cube $=2 x =2 \times 2=4 \ cm$
Total surface area of the given cuboid $=2(\mid \times b+b \times h+h \times l)$
$=2(10 \times 6+6 \times 4+4 \times 10)=2(60+24+40)=2 \times 124$
$=248 \ cm ^2$
View full question & answer→Question 165 Marks
A closed box is cuboid in shape with length $=40 \ cm$, breadth $=30 \ cm$ and height $=50 \ cm$. It is made of a thin metal sheet. Find the cost of metal sheet required to make 20 such boxes, if $1 m^2$ of metal sheet costs $Rs. 45 .$
Answer
Length of closed box $(l) = 40 \ cm$
Breadth$ (b) = 30 \ cm$
and height $(h) = 50 \ cm$
Total surface area $= 2 (lb + bh + hl)$
$= 2 (40 \times 30 + 30 \times 50 + 50 \times 40) \ cm^2$
$= 2 (1200 + 1500 + 2000) \ cm^2$
$= 2 \times 4700$
$= 9400 \ cm^2$
Surface area of sheet used for $20$ such boxes $= 9400 \times 20 = 188000 \ cm^2$
$\therefore 188000 \ cm ^2=\frac{188000}{100 \times 100}=18.8\ m ^2$
Cost of $1\ m^2$ sheet $= Rs. 45$
Total cost $=₹ 18.8 \times 45=₹ 846$ View full question & answer→