Difference between revisions of "2018 AMC 10A Problems/Problem 8"

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== Problem ==
 
== Problem ==
  
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?
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Joe has a collection of <math>23</math> coins, consisting of <math>5</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-cent coins. He has <math>3</math> more <math>10</math>-cent coins than <math>5</math>-cent coins, and the total value of his collection is <math>320</math> cents. How many more <math>25</math>-cent coins does Joe have than <math>5</math>-cent coins?
  
 
<math>\textbf{(A) }  0  \qquad        \textbf{(B) }  1  \qquad    \textbf{(C) }  2  \qquad  \textbf{(D) }  3  \qquad  \textbf{(E) }  4 </math>
 
<math>\textbf{(A) }  0  \qquad        \textbf{(B) }  1  \qquad    \textbf{(C) }  2  \qquad  \textbf{(D) }  3  \qquad  \textbf{(E) }  4 </math>
  
==Solution 1==
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==Solution 1 (One Variable)==
Let <math>x</math> be the number of 5-cent coins that Joe has. Therefore, he must have <math>(x+3)</math> 10-cent coins and <math>(23-(x+3)-x)</math> 25-cent coins. Since the total value of his collection is 320 cents, we can write
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Let <math>x</math> be the number of <math>5</math>-cent coins that Joe has. Therefore, he must have <math>(x+3) \ 10</math>-cent coins and <math>(23-(x+3)-x) \ 25</math>-cent coins. Since the total value of his collection is <math>320</math> cents, we can write
<math>5x + 10(x+3) + 25(23-(x+3)-x) = 320 \Rightarrow 5x + 10x + 30 + 500 - 50x = 320 \Rightarrow 35x = 210 \Rightarrow x = 6</math>  
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<cmath>\begin{align*}
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is  
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5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\
<math>8-6 = \boxed{\textbf{(C) } 2}</math>
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5x + 10x + 30 + 500 - 50x &= 320 \\
 +
35x &= 210 \\
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x &= 6.
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\end{align*}</cmath>  
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Joe has <math>6 \ 5</math>-cent coins, <math>9 \ 10</math>-cent coins, and <math>8 \ 25</math>-cent coins. Thus, our answer is  
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<math>8-6 = \boxed{\textbf{(C) } 2}.</math>
  
 
~Nivek
 
~Nivek
  
==Solution 2==
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==Solution 2 (Two Variables)==
Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.
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Let the number of <math>5</math>-cent coins be <math>x,</math> the number of <math>10</math>-cent coins be <math>x+3,</math> and the number of <math>25</math>-cent coins be <math>y.</math>
  
We know that the value of the coins add up to 320 cents.
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Set up the following two equations with the information given in the problem:
Thus, we have 5n + 10d + 25q = 320. Let this be (1).
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<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath>
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<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath>
  
We know that there are 23 coins.
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From there, multiply the second equation by <math>25</math> to get <cmath>50x+25y=500.</cmath>
Thus, we have n + d + q = 23. Let this be (2).
 
  
We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.
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Subtract the first equation from the multiplied second equation to get <math>35x=210,</math> or <math>x=6.</math>
Thus, we have d - 3 = n.
 
  
Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.
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Substitute <math>6</math> in for <math>x</math> into one of the equations to get <math>y=8.</math>
  
Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.
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Finally, the answer is <math>8-6=\boxed{\textbf{(C) } 2}.</math>
  
Plugging d into d - 3 = n, n = 6.
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- mutinykids
  
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.
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==Solution 3 (Three Variables)==
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Let <math>n,d,</math> and <math>q</math> be the numbers of <math>5</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-cent coins in Joe's collection, respectively. We are given that
 +
<cmath>\begin{align*}
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n+d+q&=23, &(1) \\
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5n+10d+25q&=320, &(2) \\
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d&=n+3. &(3)
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\end{align*}</cmath>
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Substituting <math>(3)</math> into each of <math>(1)</math> and <math>(2)</math> and then simplifying, we have
 +
<cmath>\begin{align*}
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2n+q&=20, \hspace{17.5mm} &(1\star) \\
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3n+5q&=58. &(2\star)
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\end{align*}</cmath>
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Subtracting <math>(2\star)</math> from <math>5\cdot(1\star)</math> gives <math>7n=42,</math> from which <math>n=6.</math> Substituting this into either <math>(1\star)</math> or <math>(2\star)</math> produces <math>q=8.</math>
  
Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(C) } 2}</math>.
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Finally, the answer is <math>q-n=\boxed{\textbf{(C) } 2}.</math>
  
==Solution 3==
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~MRENTHUSIASM
So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.
 
  
You make the two equations:
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==Video Solutions==
<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath>
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https://youtu.be/ZiZVIMmo260
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath>
 
  
From there, you multiply the second equation by 25 to get
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~IceMatrix
<cmath>50x+25y=500</cmath>
 
  
You subtract the first equation from the multiplied second equation to get
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https://youtu.be/BLTrtkVOZGE
<cmath>35x=210 \Rightarrow x=6</cmath>
 
You can plug that value into one of the equations to get <cmath>y=8</cmath>
 
So, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>.
 
  
- mutinykids
+
~savannahsolver
 
 
==Video Solution==
 
https://youtu.be/ZiZVIMmo260
 
  
https://youtu.be/BLTrtkVOZGE
+
https://youtu.be/HISL2-N5NVg?t=1861
  
~savannahsolver
+
~pi_is_3.14
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:03, 29 October 2021

Problem

Joe has a collection of $23$ coins, consisting of $5$-cent coins, $10$-cent coins, and $25$-cent coins. He has $3$ more $10$-cent coins than $5$-cent coins, and the total value of his collection is $320$ cents. How many more $25$-cent coins does Joe have than $5$-cent coins?

$\textbf{(A) }   0   \qquad        \textbf{(B) }   1   \qquad    \textbf{(C) }   2   \qquad   \textbf{(D) }  3  \qquad  \textbf{(E) }   4$

Solution 1 (One Variable)

Let $x$ be the number of $5$-cent coins that Joe has. Therefore, he must have $(x+3) \ 10$-cent coins and $(23-(x+3)-x) \ 25$-cent coins. Since the total value of his collection is $320$ cents, we can write \begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. \end{align*} Joe has $6 \ 5$-cent coins, $9 \ 10$-cent coins, and $8 \ 25$-cent coins. Thus, our answer is $8-6 = \boxed{\textbf{(C) } 2}.$

~Nivek

Solution 2 (Two Variables)

Let the number of $5$-cent coins be $x,$ the number of $10$-cent coins be $x+3,$ and the number of $25$-cent coins be $y.$

Set up the following two equations with the information given in the problem: \[5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290\] \[x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20\]

From there, multiply the second equation by $25$ to get \[50x+25y=500.\]

Subtract the first equation from the multiplied second equation to get $35x=210,$ or $x=6.$

Substitute $6$ in for $x$ into one of the equations to get $y=8.$

Finally, the answer is $8-6=\boxed{\textbf{(C) } 2}.$

- mutinykids

Solution 3 (Three Variables)

Let $n,d,$ and $q$ be the numbers of $5$-cent coins, $10$-cent coins, and $25$-cent coins in Joe's collection, respectively. We are given that \begin{align*} n+d+q&=23, &(1) \\ 5n+10d+25q&=320, &(2) \\ d&=n+3. &(3) \end{align*} Substituting $(3)$ into each of $(1)$ and $(2)$ and then simplifying, we have \begin{align*} 2n+q&=20, \hspace{17.5mm} &(1\star) \\ 3n+5q&=58. &(2\star) \end{align*} Subtracting $(2\star)$ from $5\cdot(1\star)$ gives $7n=42,$ from which $n=6.$ Substituting this into either $(1\star)$ or $(2\star)$ produces $q=8.$

Finally, the answer is $q-n=\boxed{\textbf{(C) } 2}.$

~MRENTHUSIASM

Video Solutions

https://youtu.be/ZiZVIMmo260

~IceMatrix

https://youtu.be/BLTrtkVOZGE

~savannahsolver

https://youtu.be/HISL2-N5NVg?t=1861

~pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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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