Question 13 Marks
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.
AnswerVolume of the earth dug out $=50 \times 40 \times 7=14000\ m^3$
Let ' $h$ ' be the rise in the height of the field Therefore,
volume of the field (cuboidal) = Volume of the earth dug out
$\Rightarrow 200 \times 150 \times h=14000$
$\Rightarrow\text{h}=\frac{14000}{200\times150}=0.47\text{m}$
View full question & answer→Question 23 Marks
A field is in the form of a rectangular length $18\ m$ and width $15\ m$. A pit $7.5\ m$ long, 6m 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.
AnswerLet $'h'$ metres be the rise in the level of field Volume of earth taken out from the pit $= 7.5 \times 6 \times 0.8 = 36m^3$
Area of the field on which the earth taken out is to be spread $= 18 \times 15 - 7.5 \times 6 = 225m^2$
Now, Area of the field $* h$ = Volume of the earth taken out from the pit
$\Rightarrow 225 \times h = 7.5 \times 6 \times 0.8$
$\Rightarrow\text{h}=\frac{36}{225}=0.16\text{m}=16\text{cm}$
View full question & answer→Question 33 Marks
Find the lateral surface area and total surface area of a cube of edge $10\ cm.$
AnswerCube of edge $(a)=10 cm$ We know that,
Cube Lateral Surface Area $=4 a ^2=4(10 \times 10)=400 cm^2$
Total Surface Area $=6 a ^2=6 \times 10^2=600 cm^2$
View full question & answer→Question 43 Marks
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?
AnswerGiven that: The external dimensions of the wooden box are as follows:
Length $(I)=48 cm$,
Breadth $(b)=36 cm$,
Height $(h) =30 cm$
Now, the internal dimensions of the wooden box are as follows:
Length $(I)=48-(2 \times 1.5)=45 cm$
Breadth $(b)=36-(2 \times 1.5)=33 cm$
Height $(h)=30-(2 \times 1.5)=27 cm$
Internal volume of the wooden box
$=1 \times b \times$
$h cm { }^3=45 \times 33 \times 27 cm^3=40095\ cm^3$
Volume of the brick $=6 \times 3 \times 0.75=13.5\ cm^3$
$\therefore$ Number of bricks
$=\frac{40095}{13.5}=2970$ bricks
$\therefore 2970$ bricks can be kept inside the wooden box.
View full question & answer→Question 53 Marks
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.
AnswerVolume of the cube $= S^3 = 9^3 = 729cm^3$
Area of the base $= l \times b = 15 \times 12 = 180cm^2$
Rise in water level $=\frac{\text{Volume of the cube }}{\text{Area of base of rectangular vessel}}$
$=\frac{729}{180}$
$= 4.05\text{cm}.$
View full question & answer→Question 63 Marks
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.
AnswerVolume of the large cube $=v_1+v_2+v_3$
Let the edge of the third cube be ' $x$ ' $cm 12^3=6^3+8^3+a^3$
[Volume of cube $=$ side $^3$ ] $1728=216+512+x^3$
$\Rightarrow x^3=1728-728=1000$
$\Rightarrow x=10 cm$
Therefore, Side of third side $=10 cm$
View full question & answer→Question 73 Marks
If the areas of three adjacent face of a cuboid are $8cm^3, 18cm^3$ and $25\ cm^3$. Find the volume of the cuboid.
Answer$WKT$, if $x, y, z$ denote the areas of three adjacent faces of a cuboid.
$=x=1 \times b, y=b \times h, z=1 \times h$ Volume $(V)$ is given by
$V = l \times b \times h$
Now, $xyz = lb \times bh \times hl = V ^2$
Here $x =8 y =18 z =25 \therefore V^2=8 \times 18 \times 25=3600 \Rightarrow V=$ $60 cm^3$.
View full question & answer→Question 83 Marks
The areas of three adjacent faces of a cuboid are $x, y$ and $z$. If the volume is $V$, Prove that $V^2 = xyz.$
AnswerLet $a, b$ and $d$ be the length, breadth, and height of the cuboid.
Then, $x=a b y=b c z=c a$ and $V=a b c[V=I \times b \times h]$
$=x y z=a b \times b c \times c a=(a b c)^2$ and $V=a b c V^2=(a b c)^2$
Therefore, $V ^2=( xyz )$
View full question & answer→Question 93 Marks
Water in a rectangular reservoir having base $80\ m$ by $60\ m$ is $6.5\ m$ deep. In what time can the water be pumped 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 $15\ km/hr.$
AnswerFlow of water $= 15\ km/hr = 15000\ m/hr$
Volume of water coming out of the pipe in one hour,
$\Rightarrow\frac{20}{100}\times\frac{20}{100}\times15000=600\text{m}^3$
Volume of the tank $= 80 \times 60 \times 6.5 = 31200m^3$
Times taken to empty the tank $=\frac{\text{Volume of tank}}{\text{Volume of water coming out of pipe in one hour}}$
$=\frac{31200}{600}=52\text{ hours}$
View full question & answer→Question 103 Marks
Given that $1$ cubic cm 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.
AnswerLet the length of the marble block be $'l'$ cm Volume of the marble block
$= l \times b \times h cm^3 = l \times 28 \times 5cm^3 $
Therefore, weight of the marble square $= 140l \times 0.25\ kg$
As mentioned in the question, weight of the marble $= 112\ kgs$
Therefore, $= 112 = 140l × 0.25$
$\Rightarrow\text{l}=\frac{112}{140\times0.25}=3.2\text{cm}.$
View full question & answer→Question 113 Marks
Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
AnswerLength of the new cuboid
$=3\ a$ Breadth of the cuboid
$= a$ Height of the new cuboid
$=a$ The Total surface area of the new cuboid
$(T S A)=2(l b+b h+h l)(T S A)_1=2(3 a \times a+a \times a+a \times 3 a)(T S A)_1=14 a^2$
The Total Surface area of three cubes $(\text { TSA })_2=3 \times 6 a^2(T S A)_2=18 a^2$
$\therefore\ \frac{\text{TSA}_1}{\text{TSA}_2}=\frac{14\text{a}^2}{18\text{a}^2}$
$\therefore$ Ratio is $7: 9$.
View full question & answer→Question 123 Marks
Find the cost of digging a cuboidal pit $8m$ long, $6m$ broad and $3m$ deep at the rate of $₹ 30$ per $\mathrm{m}^3$
AnswerGiven that: Length of the cuboidal pit $(\mathrm{l})=8 \mathrm{~m}$
Breadth of the cuboidal pit $(\mathrm{b})=6 \mathrm{~m}$
Depth of the cuboidal pit $(\mathrm{h})=$ $3\ m$
Volume of the Cuboidal pit $=1 \times \mathrm{b} \times \mathrm{h}=8 \times 6 \times 3=144 \mathrm{~m}^3$
Cost of digging $1 \mathrm{~m}^3=₹ 30$
Cost of digging $144 \mathrm{~m}^3$
$=144 \times 30=₹ 4320$
View full question & answer→Question 133 Marks
The dimensions of a cinema hall are $100 \mathrm{~m}, 50 \mathrm{~m}, 18 \mathrm{~m}$. How many persons can sit in the hall, if each person requires $150 \mathrm{~m}^3$ of air?
AnswerGiven that Volume of cinema hall $=100 \times 50 \times 18 \mathrm{~m}^3$
Volume of air required by each person $=150 \mathrm{~m}^3$
Number of persons who sit in the hall
$=\frac{\text { Volume of cinema hall }}{\text { Volume of air required by each person }}=\frac{100 \times 50 \times 18}{150}=600$
[Since, $\mathrm{V}=\mathrm{l} \times \mathrm{b} \times \mathrm{h}$ ]
Therefore, number of persons who can sit in the hall $=600$ members.
View full question & answer→Question 143 Marks
Three metal cubes with edges $6\ cm, 8\ cm, 10\ cm$ respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.
AnswerLet ' $a$ ' be the length of each edge of the new cube.
Then $a^3=\left(6^3+8^3+10^3\right) \mathrm{cm}^3 \Rightarrow a^3=1728 \Rightarrow a=12$
Therefore, Volume of the new cube $=a^3=1728 \mathrm{~cm}^3$
Surface area of the new cube $=6 a^2=6 \times(12)^2=864 \mathrm{~cm}^2$
Diagonal of the newly formed cube $=\sqrt{3} \mathrm{a}=12 \sqrt{3} \mathrm{~cm}$
View full question & answer→Question 153 Marks
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 solid 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.
AnswerLet the length of each edge of the cube be ' $x$ ' $cm$ Then,
volume of the cube $=$ Volume of water inside the tank + Volume of water that overflowed
$x^3=(5 \times 5 \times 1)+2 x^3=27 x=3 \mathrm{~cm}$
Hence, volume of the cube $=27 \mathrm{~cm}^3$ And edge of the cube $=3 \mathrm{~cm}$
View full question & answer→Question 163 Marks
A $4\ cm$ cube is cut into $1\ cm$ cubes. Calculate the total surface area of all the small cubes.
AnswerEdge of the cube $(a)=4 \mathrm{~cm}$
Volume of the cube $=a^3 a=4^3=64 \mathrm{~cm}^3$
Edge of the cube $=1 \mathrm{~cm}^3 $
$\therefore$ Total number of small cubes $=\frac{64 \mathrm{~cm}^3}{1 \mathrm{~cm}^3}=64 $
$\therefore$ Total Surface area of all the cubes $=64 \times 6 \times 1=384 \mathrm{~cm}^2$
View full question & answer→Question 173 Marks
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 $1$ cubic cm of iron weighs $15\ gms$. Find the weight of the empty box in kg.
AnswerGiven:
Outer dimensions of iron:
Length $(l)=36 \mathrm{~cm}$
Breadth $(b)=25 \mathrm{~cm}$
Height $(h)=16.5 \mathrm{~cm}$
Inner dimensions of iron:
Length $(l)=36-(2 \times 1.5)=33 \mathrm{~cm}$
Breadth (b) $=25-(2 \times 1.5)=22 \mathrm{~cm}$
Height $(h)=16.5-1.5=15 \mathrm{~cm}$
Volume of Iron $=$ Outer volume - Inner volume
$=(36 \times 25 \times 16.5)-(33 \times 22 \times 15)$
$=3960 \mathrm{~cm}^3$
Weight of Iron $=3960 \times 15=59400$ grams $=59.4 \mathrm{kgs}$
View full question & answer→