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> \mathrm{(A) \ } | + | <math> \mathrm{(A) \ } $ </math>1.15 \qquad \mathrm{(B) \ } $ <math>1.20 \qquad \mathrm{(C) \ } $ </math>1.25 \qquad \mathrm{(D) \ } $ <math>1.30 \qquad \mathrm{(E) \ } $ </math>1.35<math> |
==Solution== | ==Solution== | ||
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=== Solution 1 === | === Solution 1 === | ||
| − | She has <math>n< | + | 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>1.15</math>. |
=== Solution 2 === | === Solution 2 === | ||
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{{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}} | ||
Revision as of 11:12, 4 July 2013
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?
1.15 \qquad \mathrm{(B) \ } $ 
1.25 \qquad \mathrm{(D) \ } $ 
1.35$==Solution==
=== Solution 1 ===
She has$ (Error compiling LaTeX. Unknown error_msg)n
d=20-n
5n+10d=5n+10(20-n)=200-5n
10n+5d=10n+5(20-n)=100+5n
70
200-5n+70=100+5n
n=17
200-5n = $ 
.
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.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
