Difference between revisions of "1951 AHSME Problems/Problem 23"
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So either <math>x=0</math>, or <math>3x-16=0</math>. | So either <math>x=0</math>, or <math>3x-16=0</math>. | ||
| − | The first | + | 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 == | ||
Revision as of 13:58, 14 June 2013
Problem
The radius of a cylindrical box is 
inches and the height is 
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 
be the number of inches increased. We can set up an equation for : ![\[8^2 \times (3+x)=(8+x)^2\times 3\]](http://latex.artofproblemsolving.com/8/5/2/85244f9a7205dd77fd62da4dedf132ea43206b5c.png)
Expanding gives 
.
Combining like terms gives the quadratic 
Factoring out an gives 
.
So either 
, or 
.
The first equation is not possible, because the problem states that the value has to be non-zero. The second equation gives the answer, 
Thus the answer is 
.
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 | ||
