Difference between revisions of "1951 AHSME Problems/Problem 23"
(→Solution) |
Mathgeek2006 (talk | contribs) m (→Problem) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 2: | Line 2: | ||
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: | ||
− | <math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non | + | <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 == | ||
Line 15: | Line 15: | ||
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 == | ||
Line 22: | Line 22: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:01, 13 March 2015
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:
Solution
Let be the number of inches increased. We can set up an equation for
:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.