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:

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non \minus{} existent} \qquad\textbf{(E)}\ \text{none of these}$ (Error compiling LaTeX. Unknown error_msg)

Solution

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

Expanding gives $8x^2+48x+72=9x+72$ .

Combining like therms gives the quadratic $8x^2+39x=0$

Factoring out an $x$ gives $x(8x+39)=0$ .

So either $x=0$ , or $8x+39=0$ .

The first solution is not possible, because the problem states that the value has to be non-zero. However, the second value also does not work because it gives a negative value. Therefore, the answer is $\boxed{\textbf{(D)}\ \text{non \minus{} existent} }$ (Error compiling LaTeX. Unknown error_msg)

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1951 AHSME Problems/Problem 23

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