1989 AJHSME Problems/Problem 6

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 $\boxed{\text{C}}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1989 AJHSME Problems/Problem 6

Problem

Solution

See Also