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

Revision as of 12:20, 4 July 2013

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, $\begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*}$

2007 AMC 12B (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 12 Problems and Solutions

2007 AMC 10B (Problems • Answer Key • Resources)
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

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

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	{{AMC10 box\|year=2007\|ab=B\|num-b=11\|num-a=13}}		{{AMC10 box\|year=2007\|ab=B\|num-b=11\|num-a=13}}
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Difference between revisions of "2007 AMC 10B Problems/Problem 12"

Revision as of 12:20, 4 July 2013

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