Difference between revisions of "2004 AMC 12B Problems/Problem 5"
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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>. | 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>. | ||
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+ | == Video Solution 1== | ||
+ | https://youtu.be/WVK4SZAoD5E | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== See Also == | == See Also == |
Latest revision as of 18:22, 22 October 2022
- 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 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 .
Video Solution 1
~Education, the Study of Everything
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.