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

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== Problem ==
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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?
 
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?
  
 
<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==
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==Solution 1==
Let <math>x</math> be the number of 5-cent stamps that Joe has. Therefore, he must have <math>(x+3)</math> 10-cent stamps and <math>(23-(x+3)-x)</math> 25-cent stamps. 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 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
<math>\begin{align*}
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<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>  
5x+10(x+3)+25(23-(x+3)-x) &=320 \\
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Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is  
5x+10(x+3)+25(20-2x) &=320 \\
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<math>8-6 = \boxed{\textbf{(C) } 2}</math>
5x+10x+30+500-50x &=320 \\
 
35x &=210 \\
 
x &=6
 
\end{align*}</math>
 
Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is  
 
<math>8-6=\boxed{2}</math>
 
  
 
~Nivek
 
~Nivek
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==Solution 2==
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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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We know that the value of the coins add up to 320 cents.
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Thus, we have 5n + 10d + 25q = 320. Let this be (1).
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We know that there are 23 coins.
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Thus, we have n + d + q = 23. Let this be (2).
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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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Thus, we have d - 3 = n.
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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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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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Plugging d into d - 3 = n, n = 6.
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Plugging d and q into the (2) we had at the beginning of this problem, q = 8.
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Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(C) } 2}</math>.
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==Solution 3==
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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.
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You make the two equations:
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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>
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From there, you multiply the second equation by 25 to get
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<cmath>50x+25y=500</cmath>
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You subtract the first equation from the multiplied second equation to get
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<cmath>35x=210 \Rightarrow x=6</cmath>
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You can plug that value into one of the equations to get <cmath>y=8</cmath>
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So, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>.
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- mutinykids
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==Video Solution==
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https://youtu.be/ZiZVIMmo260
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https://youtu.be/BLTrtkVOZGE
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~savannahsolver
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== Video Solution ==
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https://youtu.be/HISL2-N5NVg?t=1861
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 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
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{{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 22:21, 27 January 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

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 $5x + 10(x+3) + 25(23-(x+3)-x) = 320 \Rightarrow 5x + 10x + 30 + 500 - 50x = 320 \Rightarrow 35x = 210 \Rightarrow x = 6$ 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

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.

We know that the value of the coins add up to 320 cents. Thus, we have 5n + 10d + 25q = 320. Let this be (1).

We know that there are 23 coins. 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. 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.

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.

Plugging d into d - 3 = n, n = 6.

Plugging d and q into the (2) we had at the beginning of this problem, q = 8.

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

Solution 3

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: \[5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290\] \[x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20\]

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

You subtract the first equation from the multiplied second equation to get \[35x=210 \Rightarrow x=6\] You can plug that value into one of the equations to get \[y=8\] So, the answer is $8-6=\boxed{\textbf{(C) } 2}$.

- mutinykids

Video Solution

https://youtu.be/ZiZVIMmo260

https://youtu.be/BLTrtkVOZGE

~savannahsolver

Video Solution

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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