2006 AMC 8 Problems/Problem 4

Problem

Initially, a spinner points west. Chenille moves it clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions. In what direction does the spinner point after the two moves?

$[asy]size(96); draw(circle((0,0),1),linewidth(1)); draw((0,0.75)--(0,1.25),linewidth(1)); draw((0,-0.75)--(0,-1.25),linewidth(1)); draw((0.75,0)--(1.25,0),linewidth(1)); draw((-0.75,0)--(-1.25,0),linewidth(1)); label("$N$",(0,1.25), N); label("$W$",(-1.25,0), W); label("$E$",(1.25,0), E); label("$S$",(0,-1.25), S); draw((0,0)--(-0.5,0),EndArrow);[/asy]$

$\textbf{(A)}\ \text{north} \qquad \textbf{(B)}\ \text{east} \qquad \textbf{(C)}\ \text{south} \qquad \textbf{(D)}\ \text{west} \qquad \textbf{(E)}\ \text{northwest}$

Solution 1

If the spinner goes clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions, it ultimately goes counterclockwise $1 \dfrac{1}{2}$ which brings the spinner pointing $\boxed{\textbf{(B)}\ \text{east}}$ .

Solution 2 (Minor improvement)

Note that full revolutions do not matter, so this is equivalent to going clockwise $\dfrac{1}{4}$ revolutions and then counterclockwise $\dfrac{3}{4}$ revolutions, making it ultimately go counterclockwise $\dfrac{1}{2}$ , having the spinner point $\boxed{\textbf{(B)}\ \text{east}}$ .

~JeffersonJ

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

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2006 AMC 8 Problems/Problem 4

Contents

Problem

Solution 1

Solution 2 (Minor improvement)

See Also