Difference between revisions of "1988 AHSME Problems/Problem 3"

(Created page with "==Problem== <asy> draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6...")
 
(Solution)
 
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==Solution==
 
==Solution==
 
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We first notice that the paper strips cover up part of the others. Since the width of the overlap is <math>1</math> and the length of the overlap is <math>1</math>, the area of
 
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each of the strips with the overlap is <math>(10\cdot 1)-1=9</math>. Since there are 4 strips, <math>4\cdot 9=36 \implies \boxed{\text{A}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 05:26, 31 August 2015

Problem

[asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(4,5)--(5,5)--(5,4)--(4,4)); [/asy]

Four rectangular paper strips of length $10$ and width $1$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?

$\text{(A)}\ 36 \qquad  \text{(B)}\ 40 \qquad  \text{(C)}\ 44 \qquad  \text{(D)}\ 98 \qquad  \text{(E)}\ 100$


Solution

We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$, the area of each of the strips with the overlap is $(10\cdot 1)-1=9$. Since there are 4 strips, $4\cdot 9=36 \implies \boxed{\text{A}}$.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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