2015 AMC 10A Problems/Problem 8

The following problem is from both the 2015 AMC 12A #6 and 2015 AMC 10A #8, so both problems redirect to this page.

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 Trahee's current age and $c$ be her brother'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 Trahee is twice as old as her brother.

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

$20+x=16+2x$

$x=4$

The answer is $\boxed{\textbf{ }4}$ .

Video Solution

https://youtu.be/g8lPXUg-K_I

~savannahsolver

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

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2015 AMC 10A Problems/Problem 8

Contents

Problem

Solution

Video Solution

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