Difference between revisions of "2022 AMC 8 Problems/Problem 5"
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Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned <math>6</math> years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is <math>30</math> years. How many years older than Bella is Anna? | Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned <math>6</math> years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is <math>30</math> years. How many years older than Bella is Anna? | ||
| − | <math>\textbf{(A)} | + | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } ~5</math> |
==Solution== | ==Solution== | ||
| + | |||
| + | Five years ago, Bella was <math>6</math> years old, and the kitten was <math>0</math> years old. | ||
| + | |||
| + | Today, Bella is <math>11</math> years old, and the kitten is <math>5</math> years old. It follows that Anna is <math>30-11-5=14</math> years old. | ||
| + | |||
| + | Therefore, Anna is <math>14-11=\boxed{\textbf{(C) } 3}</math> years older than Bella. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 10:22, 28 January 2022
Problem
Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 
years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 
years. How many years older than Bella is Anna?
Solution
Five years ago, Bella was years old, and the kitten was 
years old.
Today, Bella is 
years old, and the kitten is 
years old. It follows that Anna is 
years old.
Therefore, Anna is 
years older than Bella.
~MRENTHUSIASM
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
