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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 1==
 
==Solution 1==
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Let Claudia have <math>x</math> 5-cent coins and <math>\left( 12 - x \right)</math> 10-cent coins. It is easily observed that any multiple of <math>5</math> between <math>5</math> 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 <math>7,</math> because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{\textbf{(C) } 5}</math>
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Let Claudia have <math>x</math> 5-cent coins and <math>\left( 12 - x \right)</math> 10-cent coins. It is easily observed that any multiple of <math>5</math> between <math>5</math> 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 <math>7,</math> because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{\textbf{(C) } 5}</math> kurt
  
 
==Solution 2==
 
==Solution 2==
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<math>\boxed{\textbf{(C) } 5}</math>
 
<math>\boxed{\textbf{(C) } 5}</math>
  
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For small scale substitutions of 5 cent for 10 cent, like in this problem, it is quite fast, however, if the problem is a little bit more complex, knowing that you need 85 cents, explained above, it is also possible to use a system of equations. It would be
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==Solution 3 (Quick Insight)==
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5x+10y=85
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x+y=12
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Notice that for every <math>d</math> dimes, any multiple of <math>5</math> less than or equal to <math>10d + 5(12-d)</math> is a valid arrangement. Since there are <math>17</math> in our case, we have <math>10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } 5}</math>.
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Solving this x(5-cent coins)=7 and y(10-cent coins)=5, so again the answer is <math>\boxed{\textbf{(C) } 5}</math>
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~MrThinker
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== Solution 4 ==
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Dividing by 5cents to reduce clutter:
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<math>D </math> double coins and <math>S </math> single coins can reach any value between <math>1</math> and <math>2D + S</math>. Set <math>2D + S = 17 </math> and <math>D + S = 12</math>.
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Subtract to get <math>D = \boxed{\textbf{(C) } 5}</math>.
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~oinava
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==Video Solution==
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https://youtu.be/F2iyhLzmCB8
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 08:28, 31 October 2024

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 1

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 $12 - 7 = \boxed{\textbf{(C) } 5}$ kurt

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

Solution 3 (Quick Insight)

Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$. Therefore, the answer is $\boxed{\textbf{(C) } 5}$.

~MrThinker

Solution 4

Dividing by 5cents to reduce clutter:


$D$ double coins and $S$ single coins can reach any value between $1$ and $2D + S$. Set $2D + S = 17$ and $D + S = 12$.

Subtract to get $D = \boxed{\textbf{(C) } 5}$.

~oinava

Video Solution

https://youtu.be/F2iyhLzmCB8

~savannahsolver

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

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