Question

If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone shall: 
 

• A. increase by 25%
• B. decrease by 25%
• C. increase by 100%
• D. increase by 75%
• E. decrease by 20%

Hint:

When dimensions change by percentages, multiply the factors: \( (1 \pm \text{change}) \) for each dimension.

Answer 1

The Correct Option is B

Step 1: Volume Formula:
Volume of cone = \(V = \frac{1}{3} \pi r^2 h\)

Step 2: Changes in Dimensions:
Height increased by 200% means new height = \(h + 200\% \text{ of } h = h + 2h = 3h\)
Radius reduced by 50% means new radius = \(r - 50\% \text{ of } r = r - 0.5r = 0.5r\)

Step 3: New Volume:
\[ V' = \frac{1}{3} \pi (0.5r)^2 (3h) = \frac{1}{3} \pi (0.25r^2)(3h) \]
\[ V' = \frac{1}{3} \pi \times 0.25r^2 \times 3h = \frac{1}{3} \times 3 \times 0.25 \times \frac{1}{3} \pi r^2 h \]
\[ V' = 0.25 \times \frac{1}{3} \pi r^2 h = 0.25 V \]
So, new volume is 25% of original volume, i.e., decreased by 75%. Wait, check:
\[ V' = 0.25V \] means new volume is 25% of original, so decrease = \(V - 0.25V = 0.75V = 75\%\) decrease.
But option (B) says decrease by 25%. Let's recalc carefully:
\[ V' = \frac{1}{3} \pi (0.5r)^2 (3h) = \frac{1}{3} \pi (0.25r^2)(3h) = \frac{1}{3} \pi r^2 h \times 0.25 \times 3 = V \times 0.75 \]
So \(V' = 0.75V\), which means volume decreases by 25%.