If the side of a cube is increased by 5%, then the surface area of a cube is increased by

Question

If the side of a cube is increased by 5%, then the surface area of a cube is increased by (A) 10% (B) 60% (C) 6% (D) 20%

Tardigrade Admin · Accepted Answer

Let one side of the cube be x and surface area be A So, dx =5 %=5 x / 100 Then, A=(6 x)2 ⇒ dA / dx =12 x ⇒ d A =(12 x ) d x ⇒ dA =(12 x )(5 x / 100) ⇒ d A =10 A / 100 ⇒ dA =10 %

Q. If the side of a cube is increased by 5%, then the surface area of a cube is increased by

10%

60%

6%

20%

Let one side of the cube be $x$ and surface area be $A$
So, $d x = 5% = 5 x /100$
Then, $A = (6 x)^{2}$
$\Rightarrow d A / d x = 12 x$
$\Rightarrow d A = (12 x) d x$
$\Rightarrow d A = (12 x) (5 x /100)$
$\Rightarrow d A = 10 A /100$
$\Rightarrow d A = 10%$

Q. If the side of a cube is increased by 5%, then the surface area of a cube is increased by

10%

60%

6%

20%

Let one side of the cube be x and surface area be A So, dx=5%=5x/100 Then, A=(6x)2 ⇒dA/dx=12x ⇒dA=(12x)dx ⇒dA=(12x)(5x/100) ⇒dA=10A/100 ⇒dA=10%

Let one side of the cube be $x$ and surface area be $A$
So, $d x = 5% = 5 x /100$
Then, $A = (6 x)^{2}$
$\Rightarrow d A / d x = 12 x$
$\Rightarrow d A = (12 x) d x$
$\Rightarrow d A = (12 x) (5 x /100)$
$\Rightarrow d A = 10 A /100$
$\Rightarrow d A = 10%$