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Difference between revisions of "2015 AMC 10A Problems/Problem 8"

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==Problem==
 
==Problem==
  
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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 ?
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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>
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The answer is <math>\boxed{\textbf{(B) }4}</math>.
 
The answer is <math>\boxed{\textbf{(B) }4}</math>.
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==See Also==
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{{AMC10 box|year=2015|ab=A|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 17:53, 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

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