# 1951 AHSME Problems/Problem 23

## Problem

The radius of a cylindrical box is $8$ inches and the height is $3$ inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is: $\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non-existent} \qquad\textbf{(E)}\ \text{none of these}$

## Solution

Let $x$ be the number of inches increased. We can set up an equation for $x$: $$8^2 \times (3+x)=(8+x)^2\times 3$$

Expanding gives $3x^2+48x+192=64x+192$.

Combining like terms gives the quadratic $3x^2-16x=0$

Factoring out an $x$ gives $x(3x-16)=0$.

So either $x=0$, or $3x-16=0$.

The first equation is not possible, because the problem states that the value has to be non-zero. The second equation gives the answer, $x = \boxed{5\frac{1}{3}}.$ Thus the answer is $\textbf{B}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 