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If each edge of a cube is doubled,
$\left( i \right)$how many times will its surface area increase?
$\left( {ii} \right)$how many times will its volume increase?

Answer
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Hint: Surface area of a cube is $A = 6{l^2}$ and volume is $V = {l^3}$ where $l$ is the side of a cube.

According to the question,
First let’s assume that the side of a cube is $l$
Then the surface area of cube will be $ = 6 \times l \times l$

$\left( i \right)$Initial surface area$ = 6 \times l \times l$
Now as per the question if each edge is doubled then,
New surface area$ = 6 \times 2l \times 2l$$ = 4 \times \left( {6 \times l \times l} \right)$ that means New surface area is equal to four times of initial surface area.

So, the new surface area will be increased by $4$ times.

$\left( {ii} \right)$Initial volume$ = l \times l \times l$
Now as per the question if each edge is doubled then,
 Final volume$ = 2l \times 2l \times 2l = 8 \times l \times l \times l$ that means New final volume is equal to eight times of initial volume.

So, volume will be increased by $8$ times.

Note- This type of question is very easy to do. Always approach by assuming the side of the cube as in our solution we assumed the side of the cube to be $l$. After this just follow the steps and find the surface area and volume of the cube. Remember all the formulas related to cube