Difference between revisions of "1951 AHSME Problems/Problem 23"

(Solution)
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 \minus{} existent} \qquad\textbf{(E)}\ \text{none of these}</math>
+
<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 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>.
+
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 $8$ 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 (ProblemsAnswer KeyResources)
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. AMC logo.png