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

(Created page with "==Problem 4== 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; ...")
 
m (See Also)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
==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?
 
The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
Line 50: Line 50:
 
After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60.  Thus the answer is <math>60-45=15</math> <math>\boxed{\text{(A)}}</math>.
 
After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60.  Thus the answer is <math>60-45=15</math> <math>\boxed{\text{(A)}}</math>.
  
 +
==Solution 2==
 +
We see that each dot is <math>15</math> units away from the nearest one above it. So the answer is <math>\boxed{\text{A}}  </math>.
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=1999|num-b=3|num-a=5}}
 
{{AMC8 box|year=1999|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Latest revision as of 20:02, 14 April 2017

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 $\boxed{\text{A}}$.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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. AMC logo.png