Difference between revisions of "1989 AJHSME Problems/Problem 6"
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Five steps are taken to get from <math>0</math> to <math>20</math>. Each step is of equal size, so each step is <math>4</math>. Three steps are taken from <math>0</math> to <math>y</math>, so <math>y=3\times 4=12\rightarrow \boxed{\text{C}}</math>. | Five steps are taken to get from <math>0</math> to <math>20</math>. Each step is of equal size, so each step is <math>4</math>. Three steps are taken from <math>0</math> to <math>y</math>, so <math>y=3\times 4=12\rightarrow \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> | ||
==See Also== | ==See Also== |
Revision as of 17:06, 29 April 2021
Problem
If the markings on the number line are equally spaced, what is the number ?
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. . So, our answer is
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 |
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