Difference between revisions of "2014 AMC 12B Problems/Problem 1"

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<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41 </math>
 
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41 </math>
  
==Solution==
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==Video Solution 1 (Quick and Easy)==
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https://youtu.be/rNhg3QxwEO0
  
She has <math>p</math> pennies and <math>n</math> nickels, where <math>n + p = 13</math>. If she had <math>n+1</math> nickels then <math>n+1 = p</math>, so <math>2n+ 1 = 13 </math> and <math>n=6</math>. So she has 6 nickels and 7 pennies, which clearly have a value of <math>\boxed{\textbf{(C)}\ 37}</math> cents.
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~Education, the Study of Everything
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2014|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:40, 14 November 2022

Problem

Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?

$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41$

Video Solution 1 (Quick and Easy)

https://youtu.be/rNhg3QxwEO0

~Education, the Study of Everything

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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

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