MATHS M.C.Q

Let,
$l \rightarrow$ Length of the first cuboid
$\mathrm{b} \rightarrow$ Breadth of the first cuboid
$h \rightarrow$ Height of the first cuboid
Volume of the cuboid is $12 \mathrm{~cm}^3$
Dimensions of the new cuboid are,
Lenghth $(L)=2 l$
Breadth $(B)=2 b$
Height $(H)=2 h$
We are asked to find the volume of the new cuboid
We know that,
Volume of the new cuboid,
$\mathrm{V}^{\prime}=\mathrm{LBH}$
$=(2 \mathrm{l})(2 \mathrm{~b})(2 \mathrm{~h})$
$=8(\mathrm{lbh})$
$=8 \mathrm{~V}\{\text { Since, } \mathrm{V}=\mathrm{lbh}\}$
$=8 \times 12\left\{\text { Since, } \mathrm{V}=12 \mathrm{~cm}^3\right\}$
$=96 \mathrm{~cm}^3$
Thus, volume of the new cuboid is $96 \mathrm{~cm}^3$.
Hence, the correct option is $(d)$.

Dimensions of the wall are,
Length $(\mathrm{l})=8 \mathrm{~m}$
Breadth $(\mathrm{b})=20 \mathrm{~cm}$
$=0.2 \mathrm{~m}$
Height $(h)=4 \mathrm{~cm}$
Volume of the hall,
$V=l b h$
$=8 \times 4 \times 0.2$
$=6.4 \mathrm{~m}^3$
Cost of building the wall is $₹ 160$ .
Hence, the correct option is $(c)$.

Let, a $\rightarrow$ Initial edge of the cube
So, $V=a^3$
In the new cube, let,
$a^{\prime} \rightarrow$ Edge of new cube
Volume of the new cube,
$V^{\prime}=\left(a^{\prime}\right)^3$
$=(2 a)^3\left\{\text { Since, } a^{\prime}=2 a\right\}$
$=8 a^3$
$=8 \mathrm{~V}\left\{\text { Since, } a^3=V\right\}$
Volume of the new cube is $8\ V$ .
Hence, the correct choice is $(d)$.

Let,
$a \rightarrow$ Side of each cube
Length of the diagonal $=8 \sqrt{3} \mathrm{~cm}$
$\sqrt{3} \mathrm{a}=8 \sqrt{3}$
$a=8 \mathrm{~cm}$
We have to find the surface area of the cube
Surface area of the cube,
$=6 \mathrm{a}^2$
$=6 \times 8^2$
$=384 \mathrm{~cm}^2$
Thus, surface area of the cube is $384 \mathrm{~cm}^2$.
Hence, the correct choice is $(b)$.

Let,
$a \rightarrow $ Initial edge of the cube
$A \rightarrow $ Initial surface area of the cube
$a' \rightarrow $ Increased edge of the cube
$A' \rightarrow $ Increased surface area of the cube
We have to find the percentage increase in the surface in the surface area of the cube
Since it's given that
$\text{a}'=\text{a}+\text{a}\times\frac{50}{100}$
$=\frac{3}{2}\text{a}$
We have
$A^{\prime}=6\left(a^{\prime}\right)^2$
$=6\left(\frac{3}{2} a\right)^2\left\{\text { Since, } a^{\prime}=\frac{3}{2} a\right\}$
$=\frac{9}{4}\left(6 a^2\right)$
$=\frac{9}{4} \mathrm{~A}\left\{\text { Since, } A=6 a^2\right\}$
Percentage increase in surface area,
$=\frac{\text{A}'-\text{A}}{\text{A}}\times100$
$=\frac{\frac94\text{A}-\text{A}}{\text{A}}\times100$
$=\frac{\frac54\text{A}}{\text{A}}\times100$
$= 125$
Increase in surface area is $125\ %$.
Hence, the correct choice is $(d)$.

A cube has total $12$ edges.
Let, $a \rightarrow$ edge of the cube
Sum of all the edges of the cube $=12 a$
$36=12 a$
$\mathrm{a}=3 \mathrm{~cm}$
Volume of that cube,
$V=a^3$
$=3^3$
$=27 \mathrm{~cm}^3$
Volume of the cube is $27 \mathrm{~cm}^3$.
Hence, the correct option is $(b)$.

We have;
Here $A_1, A_2$ and $A_3$ are the area of three adjacent faces of a cuboid.
But the areas of three adjacent faces of a cuboid are $\mathrm{lb}, \mathrm{bh}$ and hl where,
$l \rightarrow$ Length of the cuboid
$\mathrm{b} \rightarrow$ Breadth of the cuboid
$h \rightarrow$ Height of the cuboid
We have to find the volume of the cuboid
Here,
$\mathrm{A}_1 \mathrm{~A}_2 \mathrm{~A}_3=(\mathrm{lb})(\mathrm{bh})(\mathrm{hl})$
$=(\mathrm{lbh})(\mathrm{lbh})$
$=\mathrm{V}^2\{\text { Since, } \mathrm{V}=\mathrm{Ibh}\}$
$\mathrm{V}=\sqrt{\mathrm{A}_1 \mathrm{~A}_2 \mathrm{~A}_3}$
Thus, volume of the cuboid is $\sqrt{\mathrm{A}_1 \mathrm{~A}_2 \mathrm{~A}_3}$.
Hence, the correct choice is $(c)$.

Let,
$\mathrm{V}_1, \mathrm{~V}_2, \mathrm{~V}_3 \rightarrow$ Volume of the three cubes
$a_1, a_2, a_3 \rightarrow$ Side of the three cubes
We know that,
$a^3=V$
So,
$\text{a}^3_1=\text{V}$
$\text{a}^3_1=\frac18$
$\text{a}_1=\frac12\text{cm}$
Similarly,
$\text{a}^3_2=1$
$\text{a}_2=1\text{cm}$
And;
$\text{a}^3_3=8$
$\text{a}_2=2\text{cm}$
So the height of the resulting structure,
$=\frac12+1+2$
$= 0.5 +1 + 2$
$= 3.5\ cm$
The height of the structure is $3.5\ cm$.
Hence, the correct choice is $(a)$.

We have;
$l \rightarrow $ Diagonal of the cube
$V \rightarrow $ Volume of the cube
$a \rightarrow $ Side of the cube
We know that,
$\text{l}=\text{a}\sqrt{3}$
$\text{l}^3=3\sqrt{3}\text{a}^3$
$=3\sqrt{3}\text{V}$ {Since, $V = a^3$}
$3\sqrt{3}\text{V}=\text{l}^2$
So, the correct choice is $(d)$.

The area of the floor $(A)=15 \mathrm{~m}^2$
Height of the room $(h)=4 \mathrm{~m}$
We have to find the volume of the air in the room
So, capacity of the room to contain air,
$(V)=A h$
$=15 \times 4$
$=60 \times(10 \mathrm{dm})^3\{\text { Since, } 1 \mathrm{~m}=10 \mathrm{dm}\}$
$=60 \times 1000 \mathrm{dm}^3$
$=60,000 \mathrm{dm}^3$
Volume of the air contained in the room is $60,000 \mathrm{dm}^3$. So, the correct choice is $(d)$.

Length of the terrace,
$l=6 \mathrm{~m}$
$=60 \mathrm{dm}$
Breadth of the terrace,
$b=5 \mathrm{~m}$
$=1.5 \mathrm{dm}$
Height of the water level
$\mathrm{h}=15 \mathrm{~cm}$
$=1.5 \mathrm{dm}$
We have to find the quantity of water Quantity of water,
$V=l b h$
$=(60 \times 50 \times 1.5) \mathrm{dm}^3$
$=4500 \mathrm{dm}^3$
$=4500 \text { litre }$
The quantity of water is $4500$ litre
The correct option is $(d)$.

Let,
$a$ $\rightarrow$ Side of the cube
$\mathrm{V} \rightarrow$ Volume of the cube
$A$ $\rightarrow$ Surface area of the cube
We have,
$A=96 \mathrm{~cm}^2$
$6 a^2=96\left\{\text { Since, } A=6 a^2\right\}$
$a=4 \mathrm{~cm}$
So,
$V=a^3$
$=4^3$
$=64 \mathrm{~cm}^3$
Thus, volume of the cube is $64 \mathrm{~cm}^3$.
Hence, the correct choice is $(c)$.