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 (→‎See also: Added See Also section.)
(→‎Video Solution)
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #6]] and [[2015 AMC 10A Problems|2015 AMC 10A #8]]}}
 
==Problem==
 
==Problem==
  
Line 12: Line 13:
  
 
To solve the system of equations:
 
To solve the system of equations:
 +
  
 
<math>p=3c-4</math>
 
<math>p=3c-4</math>
Line 21: Line 23:
 
<math>c=8</math>
 
<math>c=8</math>
  
−
<math>p=20</math>
+
<math>p=20.</math>
  
−
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 his cousin.
  
 
<math>20+x=2(8+x)</math>
 
<math>20+x=2(8+x)</math>
Line 32: Line 34:
  
 
The answer is <math>\boxed{\textbf{(B) }4}</math>.
 
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==
 +
https://youtu.be/g8lPXUg-K_I
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2015|ab=A|num-b=7|num-a=9}}
 +
{{AMC12 box|year=2015|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_8&oldid=194822"