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"

m (→‎Solution 3: LaTeX'ed the solution and uncentered the less important equations.)
(→‎Solution 1: LaTeX'ed the solution and used the align command.)
Line 6: Line 6:
  
 
==Solution 1==
 
==Solution 1==
−
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
+
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>  
+
<cmath>\begin{align*}
−
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is  
+
5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\
−
<math>8-6 = \boxed{\textbf{(C) } 2}</math>
+
5x + 10x + 30 + 500 - 50x &= 320 \\
 +
35x &= 210 \\
 +
x &= 6.
 +
\end{align*}</cmath>  
 +
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  
 +
<math>8-6 = \boxed{\textbf{(C) } 2}.</math>
  
 
~Nivek
 
~Nivek

Revision as of 14:08, 25 August 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 \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

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

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

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&oldid=160922"