Difference between revisions of "2015 AMC 10A Problems/Problem 13"

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==Alternate Solution==
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==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>
 
<math>\boxed{\textbf{(C) } 5}</math>
 
  
 
==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 11:50, 11 April 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?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$


Solution

Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$. But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{\textbf{(C) } 5}$


Solution #2

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 $\boxed{\textbf{(C) } 5}$

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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

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