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

Revision as of 13:11, 6 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

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 $8-6 = \boxed{\textbf{(C) } 2}$

~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 = $\boxed{\textbf{(C) } 2}$ .

Solution 3

Let the number of 5-cent coins be x, the number of 10-cent coins be x+3, and the number of quarters 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, you multiply the second equation by 25 to get $50x+25y=500$

You subtract the first equation from the multiplied second equation to get $35x=210 \Rightarrow x=6$ You can plug that value into one of the equations to get $y=8$ So, the answer is $8-6=\boxed{\textbf{(C) } 2}$ .

- 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

@@ Line 36: / Line 36: @@
 ==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.
+Let the number of 5-cent coins be x, the number of 10-cent coins be x+3, and the number of quarters be y.
-You make the two equations:
+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>

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

Revision as of 13:11, 6 August 2021

Contents

Problem

Solution 1

Solution 2

Solution 3

Video Solution

Video Solution

See Also