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

(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
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Let <math>x</math> be the number of 5-cent stamps that Joe has. Therefore, he must have <math>(x+3)</math> 10-cent stamps and <math>(23-(x+3)-x)</math> 25-cent stamps. 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
 
<cmath>5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6</cmath>  
 
<cmath>5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6</cmath>  
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Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. 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{2}</math>
  

Revision as of 18:54, 8 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?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution 1

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 \[5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6\] Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is $8-6=\boxed{2}$

~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.

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

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