Difference between revisions of "2004 AMC 12B Problems/Problem 5"
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== Solution == | == Solution == | ||
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| − | Isabella had <math>60+d</math> Canadian dollars. Setting up an equation we get <math>d = \ | + | Isabella had <math>60+d</math> Canadian dollars. Setting up an equation we get <math>d=\frac{7}{10}\cdot(60+d)</math>, which solves to <math>d=140</math>, and the sum of digits of <math>d</math> is <math>\boxed{\mathrm{(A)}\ 5}</math>. |
=== Solution 2 === | === Solution 2 === | ||
Revision as of 20:15, 22 July 2014
- The following problem is from both the 2004 AMC 12B #5 and 2004 AMC 10B #7, so both problems redirect to this page.
Problem
On a trip from the United States to Canada, Isabella took 
U.S. dollars. At the border she exchanged them all, receiving 
Canadian dollars for every 
U.S. dollars. After spending 
Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ?
Solution
Solution 1
Isabella had 
Canadian dollars. Setting up an equation we get 
, which solves to 
, and the sum of digits of is 
.
Solution 2
Each time Isabelle exchanges U.S. dollars, she gets Canadian dollars and 
Canadian dollars extra. Isabelle received a total of Canadian dollars extra, therefore she exchanged U.S. dollars 
times. Thus 
.
See Also
| 2004 AMC 12B (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 AMC 12 Problems and Solutions | |
| 2004 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
