https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Try&feedformat=atomAoPS Wiki - User contributions [en]2021-02-27T10:44:20ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=909162018 AMC 10A Problems/Problem 82018-02-10T04:52:25Z<p>Try: /* Solution 3 */</p>
<hr />
<div>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?<br />
<br />
<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 </math><br />
<br />
==Solution 1==<br />
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<br />
<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> <br />
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br />
<math>8-6=\boxed{\textbf{(C) } 2}</math><br />
<br />
~Nivek<br />
<br />
==Solution 2==<br />
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.<br />
<br />
We know that the value of the coins add up to 320 cents.<br />
Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br />
<br />
We know that there are 23 coins.<br />
Thus, we have n + d + q = 23. Let this be (2).<br />
<br />
We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br />
Thus, we have d - 3 = n.<br />
<br />
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.<br />
<br />
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.<br />
<br />
Plugging d into d - 3 = n, n = 6.<br />
<br />
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br />
<br />
Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(C) } 2}</math>.<br />
<br />
==Solution 3==<br />
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.<br />
<br />
You make the two equations:<br />
<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath><br />
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath><br />
<br />
From there, you multiply the second equation by 25 to get<br />
<cmath>50x+25y=500</cmath><br />
<br />
You subtract the first equation from the multiplied second equation to get<br />
<cmath>35x=210 \Rightarrow x=6</cmath><br />
You can plug that value into one of the equations to get <cmath>y=8</cmath><br />
So, the answer is <math>8-6=\boxed{\textbf{(C) } 2}</math>.<br />
<br />
- mutinykids<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Tryhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=909152018 AMC 10A Problems/Problem 82018-02-10T04:52:06Z<p>Try: /* Solution 2 */</p>
<hr />
<div>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?<br />
<br />
<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 </math><br />
<br />
==Solution 1==<br />
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<br />
<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> <br />
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br />
<math>8-6=\boxed{\textbf{(C) } 2}</math><br />
<br />
~Nivek<br />
<br />
==Solution 2==<br />
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.<br />
<br />
We know that the value of the coins add up to 320 cents.<br />
Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br />
<br />
We know that there are 23 coins.<br />
Thus, we have n + d + q = 23. Let this be (2).<br />
<br />
We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br />
Thus, we have d - 3 = n.<br />
<br />
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.<br />
<br />
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.<br />
<br />
Plugging d into d - 3 = n, n = 6.<br />
<br />
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br />
<br />
Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(C) } 2}</math>.<br />
<br />
==Solution 3==<br />
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.<br />
<br />
You make the two equations:<br />
<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath><br />
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath><br />
<br />
From there, you multiply the second equation by 25 to get<br />
<cmath>50x+25y=500</cmath><br />
<br />
You subtract the first equation from the multiplied second equation to get<br />
<cmath>35x=210 \Rightarrow x=6</cmath><br />
You can plug that value into one of the equations to get <cmath>y=8</cmath><br />
So, the answer is <math>8-6=\boxed{\textbf{(B) } 2}</math>.<br />
<br />
- mutinykids<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Tryhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=909142018 AMC 10A Problems/Problem 82018-02-10T04:51:41Z<p>Try: /* Solution 1 */</p>
<hr />
<div>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?<br />
<br />
<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 </math><br />
<br />
==Solution 1==<br />
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<br />
<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> <br />
Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br />
<math>8-6=\boxed{\textbf{(C) } 2}</math><br />
<br />
~Nivek<br />
<br />
==Solution 2==<br />
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.<br />
<br />
We know that the value of the coins add up to 320 cents.<br />
Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br />
<br />
We know that there are 23 coins.<br />
Thus, we have n + d + q = 23. Let this be (2).<br />
<br />
We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br />
Thus, we have d - 3 = n.<br />
<br />
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.<br />
<br />
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.<br />
<br />
Plugging d into d - 3 = n, n = 6.<br />
<br />
Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br />
<br />
Thus, the answer is 8 - 6 = <math>\boxed{\textbf{(B) } 2}</math>.<br />
<br />
==Solution 3==<br />
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.<br />
<br />
You make the two equations:<br />
<cmath>5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290</cmath><br />
<cmath>x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20</cmath><br />
<br />
From there, you multiply the second equation by 25 to get<br />
<cmath>50x+25y=500</cmath><br />
<br />
You subtract the first equation from the multiplied second equation to get<br />
<cmath>35x=210 \Rightarrow x=6</cmath><br />
You can plug that value into one of the equations to get <cmath>y=8</cmath><br />
So, the answer is <math>8-6=\boxed{\textbf{(B) } 2}</math>.<br />
<br />
- mutinykids<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Try