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

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Patty has <math>20</math> coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have <math>70</math> cents more. How much are her coins worth?
 
Patty has <math>20</math> coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have <math>70</math> cents more. How much are her coins worth?
  
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<math> \mathrm{(A) \ } \</math> <math>1.15 \qquad \mathrm{(B) \ } \</math> <math>1.20 \qquad \mathrm{(C) \ } \</math> <math>1.25 \qquad \mathrm{(D) \ } \</math> <math>1.30 \qquad \mathrm{(E) \ } \</math> <math>1.35</math>
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<math> \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 </math>
  
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==Solution==
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== Solution 1 ==
  
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=== Solution 1 ===
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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>.
  
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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 = \</math> <math>1.15</math>.
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== Solution 2 ==
  
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=== Solution 2 ===
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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>.
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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.
 
  
 
== See also ==
 
== See also ==
  
 
{{AMC10 box|year=2004|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2004|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

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 (Problems • Answer Key • Resources)
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

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