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%

Answer 1

The Correct Option is A

If the side of a cube is increased by 5%, then we need to find the percentage increase in the surface area of the cube.

Let the original side of the cube be 'a'. Then the original surface area is \(6a^2\).

If the side is increased by 5%, the new side is \(a + 0.05a = 1.05a\).

The new surface area is \(6(1.05a)^2 = 6(1.1025a^2) = 6.615a^2\).

The increase in surface area is \(6.615a^2 - 6a^2 = 0.615a^2\).

The percentage increase in surface area is \(\frac{0.615a^2}{6a^2} \times 100 = \frac{0.615}{6} \times 100 = 0.1025 \times 100 = 10.25\%\).

Since 10.25% is closest to 10%, the correct option is (A) 10%.

Answer 2

Let the side of the cube be $ a $.

Then the surface area of the cube is:

$$ A = 6a^2. $$

If the side is increased by 5%, the new side is:

$$ a' = a + 0.05a = 1.05a. $$

The new surface area is:

$$ A' = 6(a')^2 = 6(1.05a)^2 = 6(1.05^2)a^2 = 6(1.1025)a^2 = 1.1025(6a^2) = 1.1025A. $$

The increase in surface area is:

$$ A' - A = 1.1025A - A = 0.1025A. $$

The percentage increase is:

$$ \frac{A' - A}{A} \times 100 = \frac{0.1025A}{A} \times 100 = 0.1025 \times 100 = 10.25\%. $$