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 17:07, 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 $\boxed{\text{C}}$----stjwyl

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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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