Difference between revisions of "2014 AMC 12B Problems/Problem 1"
m (Added templates) |
m |
||
| Line 9: | Line 9: | ||
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. | 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. | ||
| + | == 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 12:22, 22 February 2014
Problem
Leah has 
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 
pennies and 
nickels, where 
. If she had 
nickels then 
, so 
and 
. So she has 6 nickels and 7 pennies, which clearly have a value of 
cents.
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
