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? | ||
− | <math> \ | + | <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> |
− | ==Solution== | + | == Solution 1 == |
− | === | + | 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 == | |
− | + | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:53, 24 July 2014
Contents
Problem
Patty has coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have
cents more. How much are her coins worth?
Solution 1
She has nickels and
dimes. Their total cost is
cents. If the dimes were nickels and vice versa, she would have
cents. This value should be
cents more than the previous one. We get
, which solves to
. Her coins are worth
.
Solution 2
Changing a nickel into a dime increases the sum by cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by
cents, there are
more nickels than dimes. As the total count is
, this means that there are
nickels and
dimes, which is equal to
.
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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