Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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So, the answer is <math>8-6=\boxed{2}</math>, which is (C) | So, the answer is <math>8-6=\boxed{2}</math>, which is (C) | ||
| − | + | - mutinykids | |
== See Also == | == See Also == | ||
Revision as of 17:42, 9 February 2018
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?
Contents
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
![\[5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6\]](http://latex.artofproblemsolving.com/6/5/5/655aea321f9ff1a74e6b299f7f773096e7179750.png)
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 = 2.
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:
![\[5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290\]](http://latex.artofproblemsolving.com/5/3/e/53e8fc111ff08109cc62dddd09353e565375b7a3.png)
![\[x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20\]](http://latex.artofproblemsolving.com/1/8/c/18c3eb72b0da6b239961248a7db708ac07fad5b1.png)
From there, you multiply the second equation by 25 to get
![\[50x+25y=500\]](http://latex.artofproblemsolving.com/2/2/9/2299c4e5a787f326f10d7aad886147377955f992.png)
You subtract the first equation from the multiplied second equation to get
![\[35x=210 \Rightarrow x=6\]](http://latex.artofproblemsolving.com/a/d/0/ad02337b168a528f51c1073d517ca6bc6fc39322.png)
You can plug that value into one of the equations to get ![\[y=8\]](http://latex.artofproblemsolving.com/e/f/5/ef5d917e84293c831e265957b9abd8066b00eba1.png)
So, the answer is , which is (C)
- mutinykids
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. 
