Difference between revisions of "1986 AJHSME Problems/Problem 10"

(New page: ==Problem== A picture <math>3</math> feet across is hung in the center of a wall that is <math>19</math> feet wall. How many feet from the end of the wall is the nearest edge of the pict...)
 
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==Problem==
 
==Problem==
  
A picture <math>3</math> feet across is hung in the center of a wall that is <math>19</math> feet wall.  How many feet from the end of the wall is the nearest edge of the picture?
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A picture <math>3</math> feet across is hung in the center of a wall that is <math>19</math> feet wide.  How many feet from the end of the wall is the nearest edge of the picture?
  
 
<math>\text{(A)}\ 1\frac{1}{2} \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9\frac{1}{2} \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22</math>
 
<math>\text{(A)}\ 1\frac{1}{2} \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9\frac{1}{2} \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22</math>
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==Solution==
 
==Solution==
  
{{Solution}}
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Let's say that the distance from the picture to the wall is <math>x</math>. Since that distance will be on both sides of the picture (it's in the exact middle), we can say that
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<cmath>\begin{align*}
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x + 3 + x = 19 &\Rightarrow 2x+3=19 \\
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&\Rightarrow 2x=16 \\
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&\Rightarrow x=8 \\
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\end{align*}</cmath>
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<math>\boxed{\text{B}}</math>
  
 
==See Also==
 
==See Also==
  
[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=9|num-a=11}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:47, 7 August 2013

Problem

A picture $3$ feet across is hung in the center of a wall that is $19$ feet wide. How many feet from the end of the wall is the nearest edge of the picture?

$\text{(A)}\ 1\frac{1}{2} \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9\frac{1}{2} \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22$

Solution

Let's say that the distance from the picture to the wall is $x$. Since that distance will be on both sides of the picture (it's in the exact middle), we can say that \begin{align*} x + 3 + x = 19 &\Rightarrow 2x+3=19 \\ &\Rightarrow 2x=16 \\ &\Rightarrow x=8 \\ \end{align*}

$\boxed{\text{B}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
All AJHSME/AMC 8 Problems and Solutions

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