Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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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? | + | == 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 coins that Joe has. Therefore, he must have <math>(x+3)</math> | + | 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 \ | + | <cmath>\begin{align*} |
− | Joe has | + | 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ |
− | <math>8-6=\boxed{\textbf{( | + | 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 | + | 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 | ||
− | + | ==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) \\ | ||
+ | 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> | ||
− | + | Finally, the answer is <math>q-n=\boxed{\textbf{(C) } 2}.</math> | |
− | |||
− | + | ~MRENTHUSIASM | |
− | |||
− | + | ==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 == | == 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}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 14:14, 3 July 2023
Contents
Problem
Joe has a collection of coins, consisting of -cent coins, -cent coins, and -cent coins. He has more -cent coins than -cent coins, and the total value of his collection is cents. How many more -cent coins does Joe have than -cent coins?
Solution 1 (One Variable)
Let be the number of -cent coins that Joe has. Therefore, he must have -cent coins and -cent coins. Since the total value of his collection is cents, we can write Joe has six -cent coins, nine -cent coins, and eight -cent coins. Thus, our answer is
~Nivek
Solution 2 (Two Variables)
Let the number of -cent coins be the number of -cent coins be and the number of -cent coins be
Set up the following two equations with the information given in the problem:
From there, multiply the second equation by to get
Subtract the first equation from the multiplied second equation to get or
Substitute in for into one of the equations to get
Finally, the answer is
- mutinykids
Solution 3 (Three Variables)
Let and be the numbers of -cent coins, -cent coins, and -cent coins in Joe's collection, respectively. We are given that Substituting into each of and and then simplifying, we have Subtracting from gives from which Substituting this into either or produces
Finally, the answer is
~MRENTHUSIASM
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.