Diameter of a jar cylindrical in shape is increased by 25%. By what percent must the height be decreased so that there is no change in its volume?

### Concept & Volume Preservation When the volume of a cylinder is kept constant, a change in its radius (or diameter) inversely affects its height. Because volume depends on the square of the radius ($V = \pi r^2 h$), the height must decrease by the square of the expansion factor of the radius. ### Step-by-Step Solution 1. **Understand the diameter/radius relationship:** - A $25\%$ increase in diameter means a $25\%$ increase in radius. - Let original radius $= r$ and original height $= h$. - New radius $r_2 = r + 0.25r = 1.25r = \frac{5}{4} r$. 2. **Equate the volumes:** - $V_1 = V_2$ - $\pi r^2 h_1 = \pi r_2^2 h_2$ - $\pi r^2 h_1 = \pi \left(\frac{5}{4} r\right)^2 h_2$ 3. **Solve for the new height ($h_2$):** - $r^2 h_1 = \frac{25}{16} r^2 h_2$ - $h_2 = \frac{16}{25} h_1 = 0.64 h_1$ 4. **Calculate the percentage decrease:** - Decrease $= h_1 - 0.64h_1 = 0.36h_1$ - Percentage decrease $= 0.36 \times 100 = 36\%$ ### Exam Strategy & Shortcut If the radius increases to $\frac{5}{4}$ of its original size, the $r^2$ term increases by $\left(\frac{5}{4}\right)^2 = \frac{25}{16}$. To maintain constant volume, the height must be multiplied by the reciprocal, $\frac{16}{25}$. The required reduction is $1 - \frac{16}{25} = \frac{9}{25}$, which is $36\%$. ### Common Pitfall A common pitfall is calculating the linear reciprocal decrease (e.g., assuming $25\% \Rightarrow \frac{1}{4}$ increase means a $\frac{1}{5}$ or $20\%$ decrease) instead of accounting for the squaring effect of the radius. ### Final Answer Therefore, the correct answer is **36**.