Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of <math>5.</math> To have exactly <math>17</math> different multiples of <math>5,</math> we will need to make up to <math>85</math> cents. If all twelve coins were 5-cent coins, we will have <math>60</math> cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain <math>5</math> cents, and as we need to gain <math>25</math> cents, the answer is | Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of <math>5.</math> To have exactly <math>17</math> different multiples of <math>5,</math> we will need to make up to <math>85</math> cents. If all twelve coins were 5-cent coins, we will have <math>60</math> cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain <math>5</math> cents, and as we need to gain <math>25</math> cents, the answer is | ||
| + | <math>\boxed{\textbf{(C) } 5}</math> | ||
| + | |||
| + | ==Solution 3== | ||
| + | Since the coins are 5-cent and 10-cent, there are 17 combinations, and there are 12 coins, we can set up the following system of equations: | ||
| + | a+b=12 | ||
| + | 5a+10b= 17 | ||
| + | |||
| + | Manipulating the equations gives b=5, so there are 5 10-cent coins and the answer is | ||
<math>\boxed{\textbf{(C) } 5}</math> | <math>\boxed{\textbf{(C) } 5}</math> | ||
Revision as of 23:54, 25 November 2016
Problem 13
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
Solution 1
Let Claudia have 
5-cent coins and 
10-cent coins. It is easily observed that any multiple of 
between and 
inclusive can be obtained by a combination of coins. Thus, 
combinations can be made, so 
. But the answer is not 
because we are asked for the number of 10-cent coins, which is 
Solution 2
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of 
To have exactly 
different multiples of 
we will need to make up to 
cents. If all twelve coins were 5-cent coins, we will have 
cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain cents, and as we need to gain 
cents, the answer is
Solution 3
Since the coins are 5-cent and 10-cent, there are 17 combinations, and there are 12 coins, we can set up the following system of equations: a+b=12 5a+10b= 17
Manipulating the equations gives b=5, so there are 5 10-cent coins and the answer is
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
