Difference between revisions of "1986 AHSME Problems/Problem 6"
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==Problem== | ==Problem== | ||
| + | Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length <math>r</math> is found to be <math>32</math> inches. After rearranging the blocks as in Figure 2, length <math>s</math> is found to be <math>28</math> inches. How high is the table? | ||
| − | + | <asy> | |
| + | size(300); | ||
| + | defaultpen(linewidth(0.8)+fontsize(13pt)); | ||
| + | path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; | ||
| + | path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; | ||
| + | path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; | ||
| + | draw(table^^shift((14,0))*table); | ||
| + | filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); | ||
| + | draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); | ||
| + | draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); | ||
| + | unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); | ||
| + | unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); | ||
| + | label("$r$",(8.5,5.75)); | ||
| + | label("$s$",(22,5.75)); | ||
| + | </asy> | ||
| − | + | <math>\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}</math> | |
| − | |||
| − | |||
| − | <math>\textbf{(A)}\ | ||
| − | \textbf{(B)}\ | ||
| − | \textbf{(C)}\ | ||
| − | \textbf{(D)}\ | ||
| − | \textbf{(E)}\ | ||
==Solution== | ==Solution== | ||
| − | + | Let <math>h</math>, <math>l</math>, and <math>w</math> represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have <math>l+h-w=32</math>, and from Figure 2, <math>w+h-l=28</math>. Adding the equations gives <math>2h=60 \implies h=30</math>, which is <math>\boxed{C}</math>. | |
== See also == | == See also == | ||
Latest revision as of 17:11, 1 April 2018
Problem
Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length 
is found to be 
inches. After rearranging the blocks as in Figure 2, length 
is found to be 
inches. How high is the table?
![[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]](http://latex.artofproblemsolving.com/5/1/0/51083ef4d298fc4ae8f25d1d08d41f8e76b5ada8.png)
Solution
Let 
, 
, and 
represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have 
, and from Figure 2, 
. Adding the equations gives 
, which is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
