Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 12B Problems/Problem 5"

(→‎Solution)
m (→‎Solution 2)
Line 9: Line 9:
 
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>.
 
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 ==
  
−
Each time Isabelle exchanges <math>7</math> U.S. dollars, she gets <math>7</math> Canadian dollars and <math>3</math> Canadian dollars extra. Isabelle received a total of <math>60</math> Canadian dollars extra, therefore she exchanged <math>7</math> U.S. dollars <math>60/3=20</math> times. Thus <math>d=7\cdot 20 = 140</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=\boxed{\mathrm{(A)}\ 5}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 20:18, 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 $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$?

$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$

Solution 1

Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$, which solves to $d=140$, and the sum of digits of $d$ is $\boxed{\mathrm{(A)}\ 5}$.

Solution 2

Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $\frac{60}{3}=20$ times. Thus $d=7\cdot20=\boxed{\mathrm{(A)}\ 5}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_5&oldid=62641"