# Each edge of a cube is increased by $50\% $. The percent of increase in the surface area of the cube is:

$(a){\text{ 50}}$

$(b){\text{ 125}}$

$(c){\text{ 150}}$

$(d){\text{ 300}}$

$(e){\text{ 750}}$

Answer

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Hint: Find the original surface area and new surface area of the cube by using given data. Percent increase in the surface area of the cube is found by $x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$.

Let us assume the original side to be $a$

Then, the original surface area $S = 6{a^2}$

It is given in the question that the side of the cube is increased by $50\% $

Then, the new side becomes

$a' = a + \dfrac{1}{2}a$

$ \Rightarrow a = \dfrac{{2a + a}}{2}$

$ \Rightarrow a = \dfrac{{3a}}{2}$

Therefore, the new surface area comes out to be

$s = 6{\left( {\dfrac{{3a}}{2}} \right)^2}$

$ \Rightarrow s = 6\left( {\dfrac{{9{a^2}}}{4}} \right)$

$ \Rightarrow s = \dfrac{{(3 \times 9){a^2}}}{2}$

$ \Rightarrow s = \dfrac{{27{a^2}}}{2}$

Hence, let be the $x$ increase of surface area in percentage, which is given as

$x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$

$ \Rightarrow x = \dfrac{{\dfrac{{27{a^2}}}{2}{\text{ - }}6{a^2}}}{{6{a^2}}} \times 100$

\[ \Rightarrow x = \dfrac{{\dfrac{{27{a^2} - 12{a^2}}}{2}}}{{6{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{{27{a^2} - 12{a^2}}}{{6{a^2} \times 2}} \times 100\]

\[ \Rightarrow x = \dfrac{{15{a^2}}}{{12{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{5}{4} \times 100\]

$\therefore x = 125\% $

So, the required solution is $(b){\text{ 125}}$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the various formulas of surface area, volume, etc. of solids. Here, in this question, we have calculated both the surface areas before and after the increase in the side of the cube. On comparing the two, the required solution can be reached.

Let us assume the original side to be $a$

Then, the original surface area $S = 6{a^2}$

It is given in the question that the side of the cube is increased by $50\% $

Then, the new side becomes

$a' = a + \dfrac{1}{2}a$

$ \Rightarrow a = \dfrac{{2a + a}}{2}$

$ \Rightarrow a = \dfrac{{3a}}{2}$

Therefore, the new surface area comes out to be

$s = 6{\left( {\dfrac{{3a}}{2}} \right)^2}$

$ \Rightarrow s = 6\left( {\dfrac{{9{a^2}}}{4}} \right)$

$ \Rightarrow s = \dfrac{{(3 \times 9){a^2}}}{2}$

$ \Rightarrow s = \dfrac{{27{a^2}}}{2}$

Hence, let be the $x$ increase of surface area in percentage, which is given as

$x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$

$ \Rightarrow x = \dfrac{{\dfrac{{27{a^2}}}{2}{\text{ - }}6{a^2}}}{{6{a^2}}} \times 100$

\[ \Rightarrow x = \dfrac{{\dfrac{{27{a^2} - 12{a^2}}}{2}}}{{6{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{{27{a^2} - 12{a^2}}}{{6{a^2} \times 2}} \times 100\]

\[ \Rightarrow x = \dfrac{{15{a^2}}}{{12{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{5}{4} \times 100\]

$\therefore x = 125\% $

So, the required solution is $(b){\text{ 125}}$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the various formulas of surface area, volume, etc. of solids. Here, in this question, we have calculated both the surface areas before and after the increase in the side of the cube. On comparing the two, the required solution can be reached.

Last updated date: 22nd Sep 2023

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