Difference between revisions of "2006 AMC 12A Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Mary is <math>20%</math> older than Sally, and Sally is <math>40%</math> younger than Danielle. The sum of their ages is <math>23.2</math> years. How old will Mary be on her next birthday? | + | Mary is <math>20\%</math> older than Sally, and Sally is <math>40\%</math> younger than Danielle. The sum of their ages is <math>23.2</math> years. How old will Mary be on her next birthday? |
<math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11</math> | <math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11</math> |
Revision as of 19:16, 5 January 2008
Problem
Mary is older than Sally, and Sally is younger than Danielle. The sum of their ages is years. How old will Mary be on her next birthday?
Solution
Let be Mary's age, let be Sally's age, and let be Danielle's age. We have , and . The sum of their ages is . Therefore, , and . Then . Mary will be on her next birthday. The answer is B.
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |