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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 16:06, 29 April 2021

Problem

If the markings on the number line are equally spaced, what is the number $\text{y}$?

[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]

$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$

Solution

Five steps are taken to get from $0$ to $20$. Each step is of equal size, so each step is $4$. Three steps are taken from $0$ to $y$, so $y=3\times 4=12\rightarrow \boxed{\text{C}}$.


There are five steps, and y is step number three. We can get our answer by multiplying since all the steps are the same. \[20\cdot \frac{3}{5}=12\]. So, our answer is $\boxed{\text{C}}$

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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