Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1969 AHSME Problems/Problem 26"

(→‎Problem)
(Solution to Problem 26)
 
Line 17: Line 17:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
 
 +
Because the arch has a height of <math>16</math> inches, an equation that models the arch is <math>y = ax^2 + 16</math>, where <math>x</math> is the horizontal distance from the center and <math>y</math> is the height.  The arch has a span of <math>40</math> inches, so the arch meets the ground <math>20</math> inches from the center.  That means <math>0 = 400a + 16</math>, so <math>a = -\frac{1}{25}</math>.
 +
 
 +
Thus, the equation that models height based on distance from the center <math>y = -\frac{1}{25}x^2 + 16</math>, so the height of the arch <math>5</math> inches from the center is <math>-\frac{1}{25} \cdot 5^2 + 16 = \boxed{\textbf{(B) } 15}</math> inches.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:41, 9 June 2018

Problem

[asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]

A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:

$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

Solution

Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$, where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$, so $a = -\frac{1}{25}$.

Thus, the equation that models height based on distance from the center $y = -\frac{1}{25}x^2 + 16$, so the height of the arch $5$ inches from the center is $-\frac{1}{25} \cdot 5^2 + 16 = \boxed{\textbf{(B) } 15}$ inches.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1969_AHSME_Problems/Problem_26&oldid=95027"