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

(Solution)
(Solution)
Line 14: Line 14:
 
To solve the system of equations:
 
To solve the system of equations:
  
\begin{align}
+
 
p&=3c-4\\
+
<math>p=3c-4</math>
p&=4c-12\\
+
<math>p=4c-12</math>
3c-4&=4c-12\\
+
<math>3c-4=4c-12</math>
c&=8\\
+
<math>c=8</math>
p&=20.
+
<math>p=20.</math>
\end{align}
 
  
 
Let <math>x</math> be the number of years until Pete is twice as old as Claire.
 
Let <math>x</math> be the number of years until Pete is twice as old as Claire.

Revision as of 16:20, 19 December 2018

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 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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