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Difference between revisions of "2019 AMC 8 Problems/Problem 23"

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==Problem 23==
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==Problem==
 
After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points. What was the total number of points scored by the other <math>7</math> team members?
 
After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points. What was the total number of points scored by the other <math>7</math> team members?
  
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==Solution 1==
 
==Solution 1==
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Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members be <math>x</math>, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x=\boxed{\textbf{(B)} 11}</math>. - juliankuang (lol im smart)
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Given the information above, we start with the equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math>,where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then, it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate remaining, giving <math>x=\boxed{\textbf{(B)} 11}</math>.  
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-scrabbler94
  
 
==Solution 2==
 
==Solution 2==
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Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94
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We first start by setting the total number of points as <math>28</math>, since <math>\text{LCM}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30</math> <math>(> 28)</math> ). Next, we can see that the total number of points scored is <math>56</math> as, if it is more than or equal to <math>84</math>, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math>. (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point.)
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-aops5234 -Edited by [[User: Penguin_Spellcaster|Penguin_Spellcaster]]
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==Solution 3 — Modular Arithmetic ==
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Adding together Alexa's and Brittany's fractions, we get <math>\frac{15}{28}</math> as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: <math>\frac{15x}{28x}</math> where <math>x</math> is the common ratio. Let <math>y</math> and <math>z</math> and <math>w</math> be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have <math>\frac{13x}{28x} = 15 + y + 2z + 0w.</math> We want all of our variables to be integers. Thus, we want <math>15 + y + 2z = 0 \pmod {13}.</math> Simplifying, <math>y+2z = 11 \pmod {13}.</math> The only possible value, as this integer sum has to be less than <math>7 \cdot 2 + 1 = 15,</math> must be 11. Therefore, <math>y+2z = 11,</math> and the answer is <math>\boxed{ \textbf{(B) 11}}</math>.
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- ab2024
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==Solution 4: Answer choices==
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We can rewrite the question as an algebraic equation: <math>\frac{1}{4} x + \frac{2}{7} x + 15 + y = x</math>, where <math>x</math> represents the total amount of points and <math>y</math> the amount of points the <math>7</math> other players scored. From there, we add the two fractions to get <math>\frac{15}{28} x + 15 + y = x</math>. Subtracting <math>\frac{15}{28} x</math> from both sides, we get <math>\frac{13}{28} x = y + 15</math>. We multiply each side by <math>28</math> to get rid of the denominator, in which we get <math>13x = 420 + 28y</math>. Now let’s think of this logically. This equation is telling us that if you add <math>420</math> and <math>28</math> times the amount of points scored by the extra <math>7</math> players, you get <math>13</math> times the amount of points total. And since we have to have a whole number of points total, this means that <math>420 + 28y</math> must be divisible by <math>13</math>. Plugging in all the answer choices for <math>y</math>, we find that the only answer that makes <math>420 + 28y</math> divisible by <math>13</math> is <math>\boxed{ \textbf{(B) 11}}</math>.
  
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==Solution 3==
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~ilee0820
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Fakesolve: We first start by setting the total number of points as <math>28</math>, since <math>\text{lcm}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 > 28</math>). Next, we assume that the total number of points scored is <math>56</math>. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234
 
  
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==Video explaining solution==  
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==Video Solution by Math-X (Learn to do this under a minute!!!)==
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https://youtu.be/IgpayYB48C4?si=JjQHbrlBpeox9TFq&t=7063
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~Math-X
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https://www.youtube.com/watch?v=fKjmw_zzCUU
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- Happytwin
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https://www.youtube.com/watch?v=jE-7Se7ay1c
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 +
Associated video
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https://www.youtube.com/watch?v=3Mae_6qFxoU&t=204s
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~ hi_im_bob
  
 
https://youtu.be/wsYCn2FqZJE
 
https://youtu.be/wsYCn2FqZJE
  
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https://www.youtube.com/watch?v=fKjmw_zzCUU
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https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5
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~ MathEx
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https://youtu.be/HISL2-N5NVg?t=4115
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~ pi_is_3.14
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https://youtu.be/dI8RzUHLqZc
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~savannahsolver
  
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https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5 ~ MathEx
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==Video Solution by The Power of Logic(1 to 25 Full Solution)==
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https://youtu.be/Xm4ZGND9WoY
  
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https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
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~Hayabusa1
  
 
==See Also==
 
==See Also==

Latest revision as of 09:33, 9 November 2024

Problem

After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$

Solution 1

Given the information above, we start with the equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$,where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$, or $28x+28\cdot 15=13t$. Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$. Then, it is easy to see $u=2$ ($t=56$) is the only candidate remaining, giving $x=\boxed{\textbf{(B)} 11}$.

-scrabbler94

Solution 2

We first start by setting the total number of points as $28$, since $\text{LCM}(4,7) = 28$. However, we see that this does not work since we surpass the number of points just with the information given ($28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30$ $(> 28)$ ). Next, we can see that the total number of points scored is $56$ as, if it is more than or equal to $84$, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: $56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45$, and thus, the other seven players would have scored a total of $56-45 = \boxed{\textbf{(B)} 11}$. (We see that this works since we could have $4$ of them score $2$ points, and the other $3$ of them score $1$ point.)

-aops5234 -Edited by Penguin_Spellcaster

Solution 3 — Modular Arithmetic

Adding together Alexa's and Brittany's fractions, we get $\frac{15}{28}$ as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: $\frac{15x}{28x}$ where $x$ is the common ratio. Let $y$ and $z$ and $w$ be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have $\frac{13x}{28x} = 15 + y + 2z + 0w.$ We want all of our variables to be integers. Thus, we want $15 + y + 2z = 0 \pmod {13}.$ Simplifying, $y+2z = 11 \pmod {13}.$ The only possible value, as this integer sum has to be less than $7 \cdot 2 + 1 = 15,$ must be 11. Therefore, $y+2z = 11,$ and the answer is $\boxed{ \textbf{(B) 11}}$.

- ab2024

Solution 4: Answer choices

We can rewrite the question as an algebraic equation: $\frac{1}{4} x + \frac{2}{7} x + 15 + y = x$, where $x$ represents the total amount of points and $y$ the amount of points the $7$ other players scored. From there, we add the two fractions to get $\frac{15}{28} x + 15 + y = x$. Subtracting $\frac{15}{28} x$ from both sides, we get $\frac{13}{28} x = y + 15$. We multiply each side by $28$ to get rid of the denominator, in which we get $13x = 420 + 28y$. Now let’s think of this logically. This equation is telling us that if you add $420$ and $28$ times the amount of points scored by the extra $7$ players, you get $13$ times the amount of points total. And since we have to have a whole number of points total, this means that $420 + 28y$ must be divisible by $13$. Plugging in all the answer choices for $y$, we find that the only answer that makes $420 + 28y$ divisible by $13$ is $\boxed{ \textbf{(B) 11}}$.

~ilee0820

Video Solution by Math-X (Learn to do this under a minute!!!)

https://youtu.be/IgpayYB48C4?si=JjQHbrlBpeox9TFq&t=7063

~Math-X


https://www.youtube.com/watch?v=fKjmw_zzCUU

- Happytwin

https://www.youtube.com/watch?v=jE-7Se7ay1c

Associated video

https://www.youtube.com/watch?v=3Mae_6qFxoU&t=204s

~ hi_im_bob

https://youtu.be/wsYCn2FqZJE

https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5

~ MathEx

https://youtu.be/HISL2-N5NVg?t=4115

~ pi_is_3.14

https://youtu.be/dI8RzUHLqZc

~savannahsolver

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AJHSME/AMC 8 Problems and Solutions

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