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

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Leah has <math> 13 </math> 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?
 
Leah has <math> 13 </math> 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?
  
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}}\ 39\qquad\textbf{(E)}\ 41 </math>
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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>
  
 
==Solution==
 
==Solution==

Revision as of 09:08, 3 March 2015

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?

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

Solution

She has $p$ pennies and $n$ nickels, where $n + p = 13$. If she had $n+1$ nickels then $n+1 = p$, so $2n+ 1 = 13$ and $n=6$. So she has 6 nickels and 7 pennies, which clearly have a value of $\boxed{\textbf{(C)}\ 37}$ cents.

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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