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

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{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #5]] and [[2004 AMC 10B Problems|2004 AMC 10B #7]]}}
 
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #5]] and [[2004 AMC 10B Problems|2004 AMC 10B #7]]}}
 
== Problem ==
 
== Problem ==
On a trip from the United States to Canada, Isabella took <math>d</math> 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 <math>d</math> Canadian dollars left. What is the sum of the digits of <math>d</math>?  
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On a trip from the United States to Canada, Isabella took <math>d</math> U.S. dollars. At the border she exchanged them all, receiving <math>10</math> Canadian dollars for every <math>7</math> U.S. dollars. After spending <math>60</math> Canadian dollars, she had <math>d</math> Canadian dollars left. What is the sum of the digits of <math>d</math>?  
  
<math>(\mathrm {A}) 5\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 7 \qquad (\mathrm {D}) 8 \qquad (\mathrm {E}) 9</math>
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<math>\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9</math>
  
== Solution ==
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== Solution 1 ==
  
=== Solution 1 ===
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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 = \dfrac{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{5} \Longrightarrow \mathrm{(A)}.</math>
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== Solution 2 ==
  
=== Solution 2 ===
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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 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>.
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== Video Solution 1==
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https://youtu.be/WVK4SZAoD5E
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~Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2004|ab=B|num-b=4|num-a=6}}
 
{{AMC12 box|year=2004|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2004|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2004|ab=B|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 19:22, 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 $d$ 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 $d=7\cdot20=140$, and the sum of the digits is $\boxed{\mathrm{(A)}\ 5}$.

Video Solution 1

https://youtu.be/WVK4SZAoD5E

~Education, the Study of Everything

See Also

2004 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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