Difference between revisions of "1986 AHSME Problems/Problem 6"

(Created page with "==Problem== Solve <math>t</math> in this system of equation: <cmath>t-w+l = 32\quad \\ t-l+w = 28</cmath> <math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 3...")
 
(Added a solution with explanation)
 
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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?
  
Solve <math>t</math> in this system of equation:
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<asy>
 +
size(300);
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defaultpen(linewidth(0.8)+fontsize(13pt));
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path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle;
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path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle;
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path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle;
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draw(table^^shift((14,0))*table);
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filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray);
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draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5));
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draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5));
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unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle);
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unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle);
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label("$r$",(8.5,5.75));
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label("$s$",(22,5.75));
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</asy>
  
<cmath>t-w+l = 32\quad \\
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<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>
t-l+w = 28</cmath>
 
 
 
<math>\textbf{(A)}\ 28 \qquad
 
\textbf{(B)}\ 29 \qquad
 
\textbf{(C)}\ 30 \qquad
 
\textbf{(D)}\ 31 \qquad
 
\textbf{(E)}\ 32 </math>  
 
  
 
==Solution==
 
==Solution==
 
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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 ==
{{AHSME box|year=1986|num-b=4|num-a=6}}   
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{{AHSME box|year=1986|num-b=5|num-a=7}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:11, 1 April 2018

Problem

Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ 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]

$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$

Solution

Let $h$, $l$, and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$, and from Figure 2, $w+h-l=28$. Adding the equations gives $2h=60 \implies h=30$, which is $\boxed{C}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AHSME Problems and Solutions

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