This problem asks us to find the percentage change in the volume of a right circular cone when its dimensions, specifically the radius and height, are altered. We need to calculate how the volume changes when the radius increases by 20% and the height decreases by 25%.
The formula for the volume ($V$) of a right circular cone is:
$$ V = \frac{1}{3} \pi r^2 h $$
where '$r$' is the radius of the base and '$h$' is the height of the cone.
Let the original radius be '$r_1$' and the original height be '$h_1$'. The original volume is:
$$ V_1 = \frac{1}{3} \pi r_1^2 h_1 $$
Now, let's calculate the new dimensions:
The new volume, '$V_2$', using the new radius '$r_2$' and new height '$h_2$', is:
$$ V_2 = \frac{1}{3} \pi r_2^2 h_2 $$
Substitute the expressions for '$r_2$' and '$h_2$' into the formula:
$$ V_2 = \frac{1}{3} \pi (1.20 r_1)^2 (0.75 h_1) $$
Simplify the expression:
$$ V_2 = \frac{1}{3} \pi (1.44 r_1^2) (0.75 h_1) $$
Rearrange the terms to compare with the original volume '$V_1$':
$$ V_2 = \left( \frac{1}{3} \pi r_1^2 h_1 \right) \times (1.44 \times 0.75) $$
$$ V_2 = V_1 \times (1.44 \times 0.75) $$
Let's calculate the product of the factors for the radius squared and height:
$$ 1.44 \times 0.75 = 1.08 $$
So, the new volume is:
$$ V_2 = V_1 \times 1.08 $$
This means the new volume is 1.08 times the original volume.
To find the percentage increase, we calculate the difference between the new and original volumes and divide by the original volume, then multiply by 100%:
Percentage Increase $= \frac{V_2 - V_1}{V_1} \times 100\%$
Substitute '$V_2 = 1.08 V_1$':
Percentage Increase $= \frac{1.08 V_1 - V_1}{V_1} \times 100\%$
Percentage Increase $= \frac{0.08 V_1}{V_1} \times 100\%$
Percentage Increase $= 0.08 \times 100\%$
Percentage Increase $= 8\% $
Therefore, when the radius of the right circular cone is increased by 20% and its height is decreased by 25%, the volume of the cone increases by 8%.
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