Difference between revisions of "2007 AMC 10B Problems/Problem 12"

m (Solution)
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T&=5N\
 
T&=5N\
 
T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*}</cmath>
 
T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*}</cmath>
 +
Not that actual values were not found.
  
 
==See Also==
 
==See Also==

Revision as of 21:27, 28 November 2018

The following problem is from both the 2007 AMC 12B #8 and 2007 AMC 10B #12, so both problems redirect to this page.

Problem

Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$?

$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$

Solution

Tom's age $N$ years ago was $T-N$. The ages of his three children $N$ years ago was $T-3N,$ since there are three people. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*} Not that actual values were not found.

See Also

2007 AMC 12B (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 12 Problems and Solutions
2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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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