Questions & Answers

Question

Answers

(a) 30 %

(b) 40 %

(c) 50 %

(d) 60 %

Answer
Verified

Let the length of the smaller cube be s.

Let A be the surface area of the smaller cube. It is given as 6 times the square of the length of the side.

$\Rightarrow A=6{{s}^{2}}$

Let ${{a}_{1}}$ be the cumulative surface area of 8 cubes. It will be 8 times surface area of single cube.

$\begin{align}

& \Rightarrow {{a}_{1}}=8\times 6{{s}^{2}} \\

& \Rightarrow {{a}_{1}}=48{{s}^{2}}......\left( 1 \right) \\

\end{align}$

Now, the volume of a cube is given as the cube of the length of the side.

Let v be the volume of one smaller cube.

$\Rightarrow v={{s}^{3}}$

Let ${{v}_{1}}$ be the cumulative volume of 8 smaller cubes. It will be 8 times the volume of one smaller cube.

$\Rightarrow {{v}_{1}}=8{{s}^{3}}$

Now, we know that the volume of the cubes will be conserved. Hence, cumulative volume of 8 smaller cubes will be equal to the volume of one bigger cube.

If V is the volume of the bigger cube, then $V={{v}_{1}}$.

$\Rightarrow V=8{{s}^{3}}$

Let a be the length of the side of the bigger cube.

The volume of the bigger cube is given as the cube of the length of the side of the bigger cube.

$\begin{align}

& \Rightarrow V={{a}^{3}} \\

& \Rightarrow {{a}^{3}}=8{{s}^{3}} \\

& \Rightarrow a=2s \\

\end{align}$

Thus, the length of the side of the bigger cube is 2s.

Let ${{a}_{2}}$ be the surface area of the bigger cube. It will be six times the square of the length of the side.

$\begin{align}

& \Rightarrow {{a}_{2}}=6{{\left( 2s \right)}^{2}} \\

& \Rightarrow {{a}_{2}}=6\times 4{{s}^{2}} \\

& \Rightarrow {{a}_{2}}=24{{s}^{2}}......\left( 2 \right) \\

\end{align}$

The percentage decrease in the surface area is given by the relation $pd=\dfrac{{{a}_{1}}-{{a}_{2}}}{{{a}_{1}}}\times 100%$ , where pd is percentage decrease, ${{a}_{1}}$ is cumulative surface area of 8 smaller squares and ${{a}_{2}}$ is the surface area of the bigger square.

We will substitute ${{a}_{1}}=48{{s}^{2}}$ and ${{a}_{2}}=24{{s}^{2}}$ in $pd=\dfrac{{{a}_{1}}-{{a}_{2}}}{{{a}_{1}}}\times 100\%$

$\begin{align}

& \Rightarrow pd=\dfrac{48{{s}^{2}}-24{{s}^{2}}}{48{{s}^{2}}}\times 100\% \\

& \Rightarrow pd=\dfrac{1}{2}\times 100\% \\

& \Rightarrow pd=50\% \\

\end{align}$