Difference between revisions of "2007 AMC 12B Problems/Problem 8"
m |
|||
| Line 1: | Line 1: | ||
| − | + | ==Problem== | |
| − | + | 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> ? | |
| + | <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> | ||
| − | { | + | ==Solution== |
| − | {{ | + | |
| + | <math>T=a+b+c</math> | ||
| + | |||
| + | <math>T-N=2(a-N+b-N+c-N)=2(a+b+c)-6N=2T-6N</math> | ||
| + | |||
| + | <math>2T-6N=T-N</math> | ||
| + | |||
| + | <math>T=5N</math> | ||
| + | |||
| + | <math>T/N=5 \Rightarrow \mathrm {(D)}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC12 box|year=2007|ab=B|num-b=7|num-a=9}} | ||
Revision as of 09:58, 17 October 2007
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
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 | |
