Difference between revisions of "2007 AMC 10B Problems/Problem 12"
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| + | {{duplicate|[[2007 AMC 12B Problems|2007 AMC 12B #8]] and [[2007 AMC 10B Problems|2007 AMC 10B #12]]}} | ||
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==Problem == | ==Problem == | ||
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==See Also== | ==See Also== | ||
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| + | {{AMC12 box|year=2007|ab=B|num-b=7|num-a=9}} | ||
{{AMC10 box|year=2007|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2007|ab=B|num-b=11|num-a=13}} | ||
Revision as of 16:26, 5 June 2011
- 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 
years, which is also the sum of the ages of his three children. His age 
years ago was twice the sum of their ages then. What is 
Solution
Tom's age years ago was 
. The ages of his three children years ago was 
since there are three people. If his age years ago was twice the sum of the children's ages then,
See Also
| 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 | ||
