# 1988 AHSME Problems/Problem 3

## 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, $4\cdot 9=36 \implies \boxed{\text{A}}$.

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