If each side of a cube is reduced by 50%, the surface area will reduced by.

75%

The correct answer is Option 1: 75%

Let the original side of the cube = a

1. Original surface area

$S_1 = 6a^2$

2. New side after 50% reduction

$a - 50\% \text{ of } a = \frac{a}{2}$

3. New surface area

$S_2 = 6\left(\frac{a}{2}\right)^2 = 6 \times \frac{a^2}{4} = \frac{6a^2}{4} = \frac{3}{2}a^2$

4. Reduction in surface area

$6a^2 - \frac{3}{2}a^2 = \frac{12a^2 - 3a^2}{2} = \frac{9a^2}{2}$

5. Percentage reduction

$\frac{\frac{9a^2}{2}}{6a^2} \times 100 = \frac{9}{12} \times 100 = 75\%$