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 "2015 AMC 10A Problems/Problem 8"

m (→‎Problem)
m (→‎See also: Added See Also section.)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be 2 : 1 ?
+
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be <math>2</math> : <math>1</math> ?
  
 
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8 </math>
 
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8 </math>
Line 32: Line 32:
  
 
The answer is <math>\boxed{\textbf{(B) }4}</math>.
 
The answer is <math>\boxed{\textbf{(B) }4}</math>.
 +
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=A|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Revision as of 16:53, 4 February 2015

Problem

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ : $1$ ?

$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$

Solution

This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.

The first statement can be written as $p-2=3(c-2)$. The second statement can be written as $p-4=4(c-4)$

To solve the system of equations:

$p=3c-4$

$p=4c-12$

$3c-4=4c-12$

$c=8$

$p=20$

Let $x$ be the number of years until Pete is twice as old as Claire.

$20+x=2(8+x)$

$20+x=16+2x$

$x=4$

The answer is $\boxed{\textbf{(B) }4}$.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 AMC 10 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=2015_AMC_10A_Problems/Problem_8&oldid=67355"