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Difference between revisions of "1951 AHSME Problems/Problem 23"
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== 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>
+
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>
  
−
Expanding gives <math>8x^2+48x+72=9x+72</math>.
+
Expanding gives <math>3x^2+48x+192=64x+192</math>.
  
−
Combining like therms gives the quadratic <math>8x^2+39x=0</math>
+
Combining like terms gives the quadratic <math>3x^2-16x=0</math>
  
−
Factoring out an <math>x</math> gives <math>x(8x+39)=0</math>.
+
Factoring out an <math>x</math> gives <math>x(3x-16)=0</math>.
  
−
So either <math>x=0</math>, or <math>8x+39=0</math>.
+
So either <math>x=0</math>, or <math>3x-16=0</math>.
  
−
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>
+
The first solution is not possible, because the problem states that the value has to be non-zero. The second value gives the answer, <math>x = \boxed{\frac{16}{3}}.</math> Thus the answer is <math>\textbf{B}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 14:58, 14 June 2013

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 $\textbf{B}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 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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