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

Line 7: Line 7:
 
==Solution==
 
==Solution==
  
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 37 cents <math>\boxed</math>
+
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 37 cents <math>\boxedB</math>

Revision as of 15:49, 20 February 2014

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?

$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}}\ 39\qquad\textbf{(E)}\ 41$ (Error compiling LaTeX. Unknown error_msg)

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 37 cents $\boxedB$ (Error compiling LaTeX. Unknown error_msg)