2006 AMC 8 Problems/Problem 4: Difference between revisions
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\textbf{(E)}\ \text{northwest}</math> | \textbf{(E)}\ \text{northwest}</math> | ||
== Solution == | == Solution 1 == | ||
If the spinner goes clockwise <math> 2 \dfrac{1}{4}</math> revolutions and then counterclockwise <math> 3 \dfrac{3}{4}</math> revolutions, it ultimately goes counterclockwise <math> 1 \dfrac{1}{2} </math> which brings the spinner pointing <math> \boxed{\textbf{(B)}\ \text{east}} </math>. | If the spinner goes clockwise <math> 2 \dfrac{1}{4}</math> revolutions and then counterclockwise <math> 3 \dfrac{3}{4}</math> revolutions, it ultimately goes counterclockwise <math> 1 \dfrac{1}{2} </math> which brings the spinner pointing <math> \boxed{\textbf{(B)}\ \text{east}} </math>. | ||
== Solution 2 (Minor improvement) == | |||
Note that full revolutions do not matter, so this is equivalent to going clockwise <math> \dfrac{1}{4}</math> revolutions and then counterclockwise <math> \dfrac{3}{4}</math> revolutions, making it ultimately go counterclockwise <math> \dfrac{1}{2} </math>, having the spinner point <math> \boxed{\textbf{(B)}\ \text{east}} </math>. | |||
~JeffersonJ | |||
==Video Solution by WhyMath== | |||
https://youtu.be/vhRtbl9iV30 | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=3|num-a=5}} | {{AMC8 box|year=2006|num-b=3|num-a=5}} | ||
{{MAA Notice}} | |||
Latest revision as of 13:20, 29 October 2024
Problem
Initially, a spinner points west. Chenille moves it clockwise 
revolutions and then counterclockwise 
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]](http://latex.artofproblemsolving.com/3/9/f/39f6c3905c23037677847312e6fe6254d0dd9923.png)
Solution 1
If the spinner goes clockwise revolutions and then counterclockwise revolutions, it ultimately goes counterclockwise 
which brings the spinner pointing 
.
Solution 2 (Minor improvement)
Note that full revolutions do not matter, so this is equivalent to going clockwise 
revolutions and then counterclockwise 
revolutions, making it ultimately go counterclockwise 
, having the spinner point .
~JeffersonJ
Video Solution by WhyMath
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
