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

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

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

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 $5x+10y=85$ $x+y=12$ Solving this $x$ (5-cent coins) $= 7$ and $y$ (10-cent coins) $= 5$ , so again the answer is $\boxed{\textbf{(C) } 5}$

Video Solution

https://youtu.be/F2iyhLzmCB8

~savannahsolver

@@ Line 1: / Line 1: @@
 ==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?
+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 his coins. How many 10-cent coins does Claudia have?
 <math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math>
+==Solution 1==
+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>
-==Solution==
+==Solution 2==
-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>
+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>
+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
+<cmath>5x+10y=85</cmath>
+<cmath>x+y=12</cmath>
+Solving this <math>x</math> (5-cent coins) <math>= 7</math> and <math>y</math> (10-cent coins) <math>= 5</math>, so again the answer is <math>\boxed{\textbf{(C) } 5}</math>
-==Alternate Solution==
+==Video 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
+https://youtu.be/F2iyhLzmCB8
-<math>\boxed{5}</math>.  <math>\textbf{(C)}</math>
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2015|ab=A|num-b=12|num-a=14}}
 {{MAA Notice}}
+[[Category:Introductory Number Theory Problems]]

2015 AMC 10A (Problems • Answer Key • Resources)
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