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

Revision as of 15:05, 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 (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 Solution

https://youtu.be/ZiZVIMmo260

https://youtu.be/BLTrtkVOZGE

~savannahsolver

Video Solution

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

~ pi_is_3.14

@@ Line 36: / Line 36: @@
 ==Solution 3 (Three Variables)==
+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) \\
+n+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*}
+n+q&=20, \hspace{17.5mm} &(1\star) \\
+n+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> Finally, the answer is <math>q-n=\boxed{\textbf{(C) } 2}.</math>
+~MRENTHUSIASM
 ==Video Solution==

2018 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 10A Problems/Problem 8"

Revision as of 15:05, 25 August 2021

Contents

Problem

Solution 1 (One Variable)

Solution 2 (Two Variables)

Solution 3 (Three Variables)

Video Solution

Video Solution

See Also