A spherical balloon is inflated, causing its radius to grow by 10%. By what percentage does its surface area increase?

Correct — C. The surface area of a sphere is 4πr², so it scales with the square of the radius.

New radius: r × 1.1
New area factor: 1.1² = 1.21
Increase: 1.21 − 1 = 0.21 = 21% → option (c)

Shortcut for two successive 10% rises: 10 + 10 + (10 × 10)⁄100 = 21%.

Why the others are wrong

• (a)
  10% — 10% treats surface area as if it grew in step with the radius. Area depends on r², so the 10% rise acts twice.
• (b)
  20% — 20% adds the two 10% rises but drops the cross term: 1.1² = 1 + 0.2 + 0.01, and that 0.01 is the missing 1%.
• (d)
  25% — 25% would need the radius to grow by about 11.8%, since √1.25 ≈ 1.118. The stated growth is 10%.

When every length of a solid is scaled by a factor k, every area is scaled by k² and every volume by k³. Constants such as 4 and π are unchanged, so they drop out of a percentage question.

Here k = 1.1, so the surface area becomes 1.21 times, a 21% rise. The volume would become 1.1³ = 1.331 times, a 33.1% rise.

Key facts

• Surface area of a sphere = 4πr².
• Scale every length by k and areas scale by k², volumes by k³.
• Two successive changes of x% and y% give a net change of x + y + xy⁄100 percent.
• A 10% longer radius raises surface area by 21% and volume by 33.1%.

Common traps

• Adding 10% + 10% = 20% and forgetting the 1% cross term.
• Using the volume exponent for a surface-area question: 1.1³ gives 33.1%, the answer to a different question.

No directly related past PYQ was found.