Difference between revisions of "2004 AMC 12B Problems/Problem 5"
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== Solution 2 == | == Solution 2 == | ||
| − | Each time Isabella exchanges <math>7</math> U.S. dollars, she gets <math>7</math> Canadian dollars and <math>3</math> Canadian dollars extra. Isabella received a total of <math>60</math> Canadian dollars extra, therefore she exchanged <math>7</math> U.S. dollars <math>\frac{60}{3}=20</math> times. Thus <math>d=7\cdot20=\boxed{\mathrm{(A)}\ 5}</math>. | + | Each time Isabella exchanges <math>7</math> U.S. dollars, she gets <math>7</math> Canadian dollars and <math>3</math> Canadian dollars extra. Isabella received a total of <math>60</math> Canadian dollars extra, therefore she exchanged <math>7</math> U.S. dollars <math>\frac{60}{3}=20</math> times. Thus <math>d=7\cdot20=140</math>, and the sum of the digits is <math>\boxed{\mathrm{(A)}\ 5}</math>. |
== See Also == | == See Also == | ||
Revision as of 02:28, 26 November 2019
- The following problem is from both the 2004 AMC 12B #5 and 2004 AMC 10B #7, so both problems redirect to this page.
Contents
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 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 Isabella exchanges U.S. dollars, she gets Canadian dollars and 
Canadian dollars extra. Isabella received a total of Canadian dollars extra, therefore she exchanged U.S. dollars 
times. Thus 
, and the sum of the digits is .
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. 
