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| − | ==Problem==
| + | #REDIRECT [[2007 AMC 10B Problems/Problem 12]] |
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| − | Tom's age is <math>T</math> years, which is also the sum of the ages of his three children. His age <math>N</math> years ago was twice the sum of their ages then. What is <math>T/N</math> ?
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| − | <math>\mathrm {(A)} 2</math> <math>\mathrm {(B)} 3</math> <math>\mathrm {(C)} 4</math> <math>\mathrm {(D)} 5</math> <math>\mathrm {(E)} 6</math>
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| − | ==Solution==
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| − | <math>T=a+b+c</math>
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| − | <math>T-N=2(a-N+b-N+c-N)=2(a+b+c)-6N=2T-6N</math>
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| − | <math>2T-6N=T-N</math>
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| − | <math>T=5N</math>
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| − | <math>T/N=5 \Rightarrow \mathrm {(D)}</math>
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| − | ==See Also==
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| − | {{AMC12 box|year=2007|ab=B|num-b=7|num-a=9}}
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