Difference between revisions of "2001 AMC 8 Problems/Problem 3"

(Solution)
Line 11: Line 11:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=2|num-a=4}}
 
{{AMC8 box|year=2001|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Revision as of 23:37, 4 July 2013

Problem

Granny Smith has <dollar/>63. Elberta has <dollar/>2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?

$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$

Solution

Since Anjou has $\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $$63$, Anjou has $\frac{1}{3}\times63=21$ dollars. Elberta has $$2$ more than this, so she has $$23$, or $\boxed{\text{E}}$.

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

Invalid username
Login to AoPS