Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (→‎Solution 3: LaTeX'ed the solution and uncentered the less important equations.)
Line 36: Line 36:
  
 
==Solution 3==
 
==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.
+
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:
 
Set up the following two equations with the information given in the problem:
Line 42: Line 42:
 
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath>
 
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath>
  
−
From there, multiply the second equation by 25 to get
+
From there, multiply the second equation by <math>25</math> to get <cmath>50x+25y=500.</cmath>
−
<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>
  
−
Subtract the first equation from the multiplied second equation to get
 
−
<cmath>35x=210 \Rightarrow x=6</cmath>
 
−
Substitute <math>6</math> in for <math>x</math> into one of the equations to get <cmath>y=8</cmath>
 
 
Finally, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>.
 
Finally, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>.
  

Revision as of 14:04, 25 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 $25$-cent coins 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, multiply the second equation by $25$ to get \[50x+25y=500.\]

Subtract the first equation from the multiplied second equation to get $35x=210,$ or $x=6.$

Substitute $6$ in for $x$ into one of the equations to get $y=8.$

Finally, 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&oldid=160921"