If the slant height of a cone is decreased by 15 percent and the radius of its base is increased by 20 percent, then by what percent will its curved surface area change?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 percent increase
The curved surface area of a cone is \(\pi rl\), so it changes in proportion to the product of the radius and the slant height.
New radius \(=1.2r\) (20% more)
New slant height \(=0.85l\) (15% less)
New area \(=\pi(1.2r)(0.85l)=1.02\,\pi rl\)
The area becomes 102% of the old one: an increase of 2%
Shortcut: For a product of two quantities changed by \(x\%\) and \(y\%\), the net change is \(x+y+\frac{xy}{100}\).
\(\displaystyle 20+(-15)+\frac{20\times(-15)}{100}=5-3=2\%\) increase