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 $x$ : $8^2 \times (3+x)=(8+x)^2\times 3$

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 solution is not possible, because the problem states that the value has to be non-zero. The second value gives the answer, $x = \boxed{\frac{16}{3}}.$ Thus the answer is $\textbf{B}$ .

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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