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Difference between revisions of "1951 AHSME Problems/Problem 23"
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The radius of a cylindrical box is <math> 8</math> inches and the height is <math> 3</math> 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:
 
The radius of a cylindrical box is <math> 8</math> inches and the height is <math> 3</math> 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:
  
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<math> \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}</math>
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<math> \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}</math>
  
 
== Solution ==  
 
== Solution ==  
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Let <math>x</math> be the number of inches increased. We can set up an equation for <math>x</math>: <cmath>8\times (3+x)^2=(8+x)\times 3^2</cmath>
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Let <math>x</math> be the number of inches increased. We can set up an equation for <math>x</math>: <cmath>8^2 \times (3+x)=(8+x)^2\times 3</cmath>
  
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Expanding gives <math>8x^2+48x+72=9x+72</math>.
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Expanding gives <math>3x^2+48x+192=64x+192</math>.
  
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Combining like therms gives the quadratic <math>8x^2+39x=0</math>
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Combining like terms gives the quadratic <math>3x^2-16x=0</math>
  
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Factoring out an <math>x</math> gives <math>x(8x+39)=0</math>.
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Factoring out an <math>x</math> gives <math>x(3x-16)=0</math>.
  
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So either <math>x=0</math>, or <math>8x+39=0</math>.
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So either <math>x=0</math>, or <math>3x-16=0</math>.
  
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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 <math>\boxed{\textbf{(D)}\ \text{non \minus{} existent} }</math>
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The first equation is not possible, because the problem states that the value has to be non-zero. The second equation gives the answer, <math>x = \boxed{5\frac{1}{3}}.</math> Thus the answer is <math>\textbf{B}</math>.
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 22:01, 13 March 2015

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 by
Problem 22 		Followed by
Problem 24
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All AHSME Problems and Solutions

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