Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 4 (Cheese))
 
(44 intermediate revisions by 17 users not shown)
Line 1: Line 1:
−
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?
+
== Problem ==
 +
 
 +
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==
+
==Solution 1 (One Variable)==
−
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
+
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
−
<cmath>5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10(x+3)+25(20-2x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6</cmath>  
+
<cmath>\begin{align*}
−
Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is  
+
5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\
−
<math>8-6=\boxed{2}</math>
+
5x + 10x + 30 + 500 - 50x &= 320 \\
 +
35x &= 210 \\
 +
x &= 6.
 +
\end{align*}</cmath>  
 +
Joe has six <math>5</math>-cent coins, nine <math>10</math>-cent coins, and eight <math>25</math>-cent coins. Thus, our answer is  
 +
<math>8-6 = \boxed{\textbf{(C) } 2}.</math>
  
 
~Nivek
 
~Nivek
  
−
==Solution 2==
+
==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.
+
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>
 +
 
 +
Set up the following two equations with the information given in the problem:
 +
<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath>
 +
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath>
 +
 
 +
From there, multiply the second equation by <math>25</math> to get <cmath>50x+25y=500.</cmath>
 +
 
 +
Subtract the first equation from the multiplied second equation to get <math>35x=210,</math> or <math>x=6.</math>
 +
 
 +
Substitute <math>6</math> in for <math>x</math> into one of the equations to get <math>y=8.</math>
 +
 
 +
Finally, the answer is <math>8-6=\boxed{\textbf{(C) } 2}.</math>
 +
 
 +
- mutinykids
  
−
We know that the value of the coins add up to 320 cents.
+
==Solution 3 (Three Variables)==
−
Thus, we have 5n + 10d + 25q = 320. Let this be (1).
+
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*}
 +
n+d+q&=23, &(1) \\
 +
5n+10d+25q&=320, &(2) \\
 +
d&=n+3. &(3)
 +
\end{align*}</cmath>
 +
Substituting <math>(3)</math> into each of <math>(1)</math> and <math>(2)</math> and then simplifying, we have
 +
<cmath>\begin{align*}
 +
2n+q&=20, \hspace{17.5mm} &(1\star) \\
 +
3n+5q&=58. &(2\star)
 +
\end{align*}</cmath>
 +
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>
  
−
We know that there are 23 coins.
+
Finally, the answer is <math>q-n=\boxed{\textbf{(C) } 2}.</math>
−
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.
+
~MRENTHUSIASM
−
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.
+
==Solution 4 (Cheese)==
 +
The problem states that there are 23 coins, and subtracting the 3 "extra" 10-cent coins will give 20 coins total. 20 is not a multiple of 3, and so there must be more 25-cent coins. So the "extra" 25-cent coins, when subtracted, should give a multiple of 3. The only answer choice that gives this is <math>\boxed{\textbf{(C) } 2}.</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.
+
~Yvz2900
  
−
Plugging d into d - 3 = n, n = 6.
+
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/zbcnOfDJmQI
  
−
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.
+
~Education, the Study of Everything
  
−
Thus, the answer is 8 - 6 = 2.
+
 
 +
 
 +
==Video Solutions==
 +
https://youtu.be/ZiZVIMmo260
 +
 
 +
 
 +
https://youtu.be/BLTrtkVOZGE
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/HISL2-N5NVg?t=1861
 +
 
 +
~pi_is_3.14
  
 
== See Also ==
 
== See Also ==
Line 37: Line 82:
 
{{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}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:19, 27 October 2025

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 six $5$-cent coins, nine $10$-cent coins, and eight $25$-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

Solution 4 (Cheese)

The problem states that there are 23 coins, and subtracting the 3 "extra" 10-cent coins will give 20 coins total. 20 is not a multiple of 3, and so there must be more 25-cent coins. So the "extra" 25-cent coins, when subtracted, should give a multiple of 3. The only answer choice that gives this is $\boxed{\textbf{(C) } 2}.$

~Yvz2900

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/zbcnOfDJmQI

~Education, the Study of Everything


Video Solutions

https://youtu.be/ZiZVIMmo260


https://youtu.be/BLTrtkVOZGE

~savannahsolver

Video Solution by OmegaLearn

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

~pi_is_3.14

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&oldid=264292"