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

## 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

~IceMatrix

~savannahsolver

~pi_is_3.14