Difference between revisions of "1989 AJHSME Problems/Problem 6"
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| − | There are five steps, and y is step number three. We can get our answer by multiplying since all the steps are the same. <cmath>20\cdot \frac{3}{5}=12</cmath>. So, our answer is <math>\boxed{\text{C}}</math> | + | |
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| + | There are five steps, and y is step number three. We can get our answer by multiplying since all the steps are the same. <cmath>20\cdot \frac{3}{5}=12</cmath>. So, our answer is <math>\boxed{\text{C}}</math>----stjwyl | ||
==See Also== | ==See Also== | ||
Latest revision as of 16:07, 29 April 2021
Problem
If the markings on the number line are equally spaced, what is the number 
?
![[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a) { draw((4a,-1)--(4a,1)); } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]](http://latex.artofproblemsolving.com/7/5/1/7517fe933dee763f38fde43a7c3ab4b690d1f175.png)
Solution
Five steps are taken to get from 
to 
. Each step is of equal size, so each step is 
. Three steps are taken from to 
, so 
.
There are five steps, and y is step number three. We can get our answer by multiplying since all the steps are the same. ![\[20\cdot \frac{3}{5}=12\]](http://latex.artofproblemsolving.com/7/4/e/74e62e16d01402a32062fed641c8b462b2f910b2.png)
. So, our answer is 
----stjwyl
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
