Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AMC 8 Problems/Problem 10"

(LaTeX solution, added problem and see also)
Line 12: Line 12:
  
 
{{AMC8 box|year=2000|num-b=9|num-a=11}}
 
{{AMC8 box|year=2000|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Revision as of 23:36, 4 July 2013

Problem

Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grow half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?

$\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

Solution

Shea has grown $20\%$, so she was originally $\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase. Ara also started off at $50$ inches. Since Ara grew half as much as Shea, Ara grew $10 \div 2 = 5$ inches. Therefore, Ara is now $50+5=55$ inches tall which is choice $\boxed{E}.$

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_10&oldid=55821"