Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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Pi is 3.14 (talk | contribs) (→Video Solution) |
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+ | == 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? | 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? | ||
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==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 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 | ||
− | < | + | <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> |
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is | Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is | ||
− | <math>8-6=\boxed{2}</math> | + | <math>8-6 = \boxed{\textbf{(C) } 2}</math> |
~Nivek | ~Nivek | ||
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Plugging d and q into the (2) we had at the beginning of this problem, q = 8. | Plugging d and q into the (2) we had at the beginning of this problem, q = 8. | ||
− | Thus, the answer is 8 - 6 = 2. | + | Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(C) } 2}</math>. |
+ | |||
==Solution 3== | ==Solution 3== | ||
+ | So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y. | ||
+ | |||
+ | You make the two equations: | ||
+ | <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, you multiply the second equation by 25 to get | ||
+ | <cmath>50x+25y=500</cmath> | ||
+ | |||
+ | You subtract the first equation from the multiplied second equation to get | ||
+ | <cmath>35x=210 \Rightarrow x=6</cmath> | ||
+ | You can plug that value into one of the equations to get <cmath>y=8</cmath> | ||
+ | So, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>. | ||
+ | |||
+ | - 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 == | == 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 23:21, 27 January 2021
Contents
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?
Solution 1
Let be the number of 5-cent coins that Joe has. Therefore, he must have 10-cent coins and 25-cent coins. Since the total value of his collection is 320 cents, we can write Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is
~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 = .
Solution 3
So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.
You make the two equations:
From there, you multiply the second equation by 25 to get
You subtract the first equation from the multiplied second equation to get You can plug that value into one of the equations to get So, the answer is .
- mutinykids
Video Solution
~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.