Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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<math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math> | <math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math> | ||
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==Solution== | ==Solution== | ||
Let Claudia have <math>x</math> 5-cent coins and <math>12-x</math> 10-cent coins. It is easily observed that any multiple of 5 between 5 and <math>5x + 10(12 - x) = 120 - 5x</math> inclusive can be obtained by a combination of coins. Thus, <math>24 - x = 17</math> combinations can be made, so <math>x = 7</math>. But the answer is not 7, because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{5}</math>. <math>\textbf{(C)}</math> | Let Claudia have <math>x</math> 5-cent coins and <math>12-x</math> 10-cent coins. It is easily observed that any multiple of 5 between 5 and <math>5x + 10(12 - x) = 120 - 5x</math> inclusive can be obtained by a combination of coins. Thus, <math>24 - x = 17</math> combinations can be made, so <math>x = 7</math>. But the answer is not 7, because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{5}</math>. <math>\textbf{(C)}</math> | ||
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| + | ==Alternate Solution== | ||
| + | Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of 5. To have exactly 17 different multiples of 5, we will need to make up to 85 cents. If all twelve coins were 5-cent coins, we will have 60 cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain 5 cents, and as we need to gain 25 cents, the answer is | ||
| + | <math>\boxed{5}</math>. <math>\textbf{(C)}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2015|ab=A|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:00, 5 February 2015
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
Let Claudia have 
5-cent coins and 
10-cent coins. It is easily observed that any multiple of 5 between 5 and 
inclusive can be obtained by a combination of coins. Thus, 
combinations can be made, so 
. But the answer is not 7, because we are asked for the number of 10-cent coins, which is 
. 
Alternate Solution
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of 5. To have exactly 17 different multiples of 5, we will need to make up to 85 cents. If all twelve coins were 5-cent coins, we will have 60 cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain 5 cents, and as we need to gain 25 cents, 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. 
