If three solid spheres of radii 3 cm, 4 cm and 5 cm are melted and made into one solid sphere, then by how much percent will the surface area be reduced?

Correct option is B

Given:
Three spheres of radius 3 cm, 4 cm, and 5 cm
Formula Used:
Volume of the sphere = $$\frac 43$$​ × π × R³,
Where R is the radius of the sphere
Solution:
Let radius of the newly formed sphere =  R
​$$\frac{4}{3} \times \pi \times R^3 = \frac{4}{3} \times \pi \times (3^3 + 4^3 + 5^3)$$​​
 R = 6 cm
Total surface of sphere = 4 × π × R²
Total surface area of all three-sphere =$$4 \pi (3^2 + 4^2 + 5^2) = 200\pi \\$$​
Total surface area of recast sphere = $$= 4 \pi R^2 = 4 \pi \times 6^2 = 144\pi \\$$​
Percentage decrease in surface area = $$= \frac{200\pi - 144\pi}{200\pi} \times 100 = 28\%$$​