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

Latest revision as of 22:04, 26 June 2023

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 his cousin.

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

$20+x=16+2x$

$x=4$

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

Video Solution (CREATIVE THINKING)

https://youtu.be/PL_0kjg62lU

~Education, the Study of Everything

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.

@@ Line 34: / Line 34: @@
 The answer is <math>\boxed{\textbf{(B) }4}</math>.
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/PL_0kjg62lU
+~Education, the Study of Everything
 ==Video Solution==

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