Question

Let A and B be two solid spheres such that the surface area of B is 300\% higher than the surface area of A. The volume of A is found to be k\% lower than the volume of B. Determine the value of k:

• A. 85.5
• B. 92.5
• C. 87.5
• D. 88.5
• E. 90.5

Hint:

"300% higher" means add 300% to the original, so multiply by 4.

Answer 1

The Correct Option is C

Step 1: Surface Area Relationship:
Surface area of sphere = \(4\pi r^2\).
Let radius of A = \(r_A\), radius of B = \(r_B\).
Surface area of B is 300% higher than A means:
\(SA_B = SA_A + 300% \text{ of } SA_A = SA_A + 3SA_A = 4SA_A\)
So, \(4\pi r_B^2 = 4 \times 4\pi r_A^2 \Rightarrow r_B^2 = 4r_A^2 \Rightarrow r_B = 2r_A\).

Step 2: Volume Relationship:
Volume of sphere = \(\frac{4}{3}\pi r^3\).
\(V_B = \frac{4}{3}\pi (2r_A)^3 = \frac{4}{3}\pi \times 8r_A^3 = 8V_A\)
So, \(V_A = \frac{1}{8}V_B = 0.125V_B\).

Step 3: Finding k%:
Volume of A is lower than volume of B by:
\(V_B - V_A = V_B - 0.125V_B = 0.875V_B = 87.5% \text{ of } V_B\)
Thus, \(k = 87.5\).