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 $\textbf{B}$.

See Also

 1951 AHSC (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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