# Difference between revisions of "1999 AMC 8 Problems/Problem 4"

## Problem

The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?

$[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW); label("0",(0,0),S); label("1",(4,0),S); label("2",(8,0),S); label("3",(12,0),S); label("4",(16,0),S); label("5",(20,0),S); label("0",(0,0),W); label("15",(0,3),W); label("30",(0,6),W); label("45",(0,9),W); label("60",(0,12),W); label("75",(0,15),W); label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S); label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy]$

$\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

## Solution

After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\boxed{\text{(A)}}$.

## Solution 2

We see that each dot is $15$ units away from the nearest one above it. So the answer is $\boxed{\text{A}}$.