Difference between revisions of "2004 AMC 10B Problems/Problem 15"

(Solution)
m (Solution 2)
 
Line 9: Line 9:
 
She has <math>n</math> nickels and <math>d=20-n</math> dimes. Their total cost is <math>5n+10d=5n+10(20-n)=200-5n</math> cents. If the dimes were nickels and vice versa, she would have <math>10n+5d=10n+5(20-n)=100+5n</math> cents. This value should be <math>70</math> cents more than the previous one. We get <math>200-5n+70=100+5n</math>, which solves to <math>n=17</math>. Her coins are worth <math>200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.
 
She has <math>n</math> nickels and <math>d=20-n</math> dimes. Their total cost is <math>5n+10d=5n+10(20-n)=200-5n</math> cents. If the dimes were nickels and vice versa, she would have <math>10n+5d=10n+5(20-n)=100+5n</math> cents. This value should be <math>70</math> cents more than the previous one. We get <math>200-5n+70=100+5n</math>, which solves to <math>n=17</math>. Her coins are worth <math>200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.
  
=== Solution 2 ===
+
== Solution 2 ==
  
Changing a nickel into a dime increases the sum by <math>5</math> cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by <math>70</math> cents, there are <math>70/5=14</math> more nickels than dimes. As the total count is <math>20</math>, this means that there are <math>17</math> nickels and <math>3</math> dimes.
+
Changing a nickel into a dime increases the sum by <math>5</math> cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by <math>70</math> cents, there are <math>70/5=14</math> more nickels than dimes. As the total count is <math>20</math>, this means that there are <math>17</math> nickels and <math>3</math> dimes, which is equal to <math>\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 23:53, 23 July 2014

Problem

Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?

$\textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qquad\textbf{(D)}\ \textdollar 1.30\qquad\textbf{(E)}\ \textdollar 1.35$

Solution 1

She has $n$ nickels and $d=20-n$ dimes. Their total cost is $5n+10d=5n+10(20-n)=200-5n$ cents. If the dimes were nickels and vice versa, she would have $10n+5d=10n+5(20-n)=100+5n$ cents. This value should be $70$ cents more than the previous one. We get $200-5n+70=100+5n$, which solves to $n=17$. Her coins are worth $200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}$.

Solution 2

Changing a nickel into a dime increases the sum by $5$ cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by $70$ cents, there are $70/5=14$ more nickels than dimes. As the total count is $20$, this means that there are $17$ nickels and $3$ dimes, which is equal to $\boxed{\mathrm{(A)\ }\textdollar1.15}$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS