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Difference between revisions of "2019 AMC 8 Problems/Problem 7"

(→‎Solution 3)
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==Solution 1==
 
==Solution 1==
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We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables <math>x</math> and <math>y</math> for the scores on the last two tests. <cmath>\frac{76+94+87+x+y}{5} = 81,</cmath> <cmath>\frac{257+x+y}{5} = 81.</cmath> We can now cross multiply to get rid of the denominator. <cmath>257+x+y = 405,</cmath> <cmath>x+y = 148.</cmath> Now that we have this equation, we will assign <math>y</math> as the lowest score of the two other tests, and so: <cmath>x = 100,</cmath> <cmath>y=48.</cmath> Now we know that the lowest score on the two other tests is <math>\boxed{48}</math>.
 +
 +
~ aopsav
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 +
==Solution 2==
 
Right now, she scored <math>76, 94,</math> and <math>87</math> points, with a total of <math>257</math> points. She wants her average to be <math>81</math> for her <math>5</math> tests so she needs to score <math>405</math> points in total. She needs to score a total of <math>(405-257)  
 
Right now, she scored <math>76, 94,</math> and <math>87</math> points, with a total of <math>257</math> points. She wants her average to be <math>81</math> for her <math>5</math> tests so she needs to score <math>405</math> points in total. She needs to score a total of <math>(405-257)  
 
148</math> points in her <math>2</math> tests. So the minimum score she can get is when one of her <math>2</math> scores is <math>100</math>. So the least possible score she can get is <math>\boxed{\textbf{(A)}\ 48}</math>.
 
148</math> points in her <math>2</math> tests. So the minimum score she can get is when one of her <math>2</math> scores is <math>100</math>. So the least possible score she can get is <math>\boxed{\textbf{(A)}\ 48}</math>.
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Note: You can verify that <math>\boxed{48}</math> is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.
 
Note: You can verify that <math>\boxed{48}</math> is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.
 
~~ gorefeebuddie
 
~~ gorefeebuddie
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==Solution 2==
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 +
==Solution 3==
 
We can compare each of the scores with the average of <math>81</math>:
 
We can compare each of the scores with the average of <math>81</math>:
 
<math>76</math> <math>\rightarrow</math> <math>-5</math>,
 
<math>76</math> <math>\rightarrow</math> <math>-5</math>,
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So the last one has to be <math>-33</math> (since all the differences have to sum to <math>0</math>), which corresponds to <math>81-33 = \boxed{48}</math>.
 
So the last one has to be <math>-33</math> (since all the differences have to sum to <math>0</math>), which corresponds to <math>81-33 = \boxed{48}</math>.
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−
==Solution 3 (The Easiest One :P)==
 
−
We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables <math>x</math> and <math>y</math> for the scores on the last two tests. <cmath>\frac{76+94+87+x+y}{5} = 81,</cmath> <cmath>\frac{257+x+y}{5} = 81.</cmath> We can now cross multiply to get rid of the denominator. <cmath>257+x+y = 405,</cmath> <cmath>x+y = 148.</cmath> Now that we have this equation, we will assign <math>y</math> as the lowest score of the two other tests, and so: <cmath>x = 100,</cmath> <cmath>y=48.</cmath> Now we know that the lowest score on the two other tests is <math>\boxed{48}</math>.
 
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−
~ aopsav
 
  
 
==See Also==
 
==See Also==

Revision as of 20:26, 4 January 2020

Problem 7

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$, $94$, and $87$. In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?

$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$

Solution 1

We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{48}$.

~ aopsav

Solution 2

Right now, she scored $76, 94,$ and $87$ points, with a total of $257$ points. She wants her average to be $81$ for her $5$ tests so she needs to score $405$ points in total. She needs to score a total of $(405-257) 148$ points in her $2$ tests. So the minimum score she can get is when one of her $2$ scores is $100$. So the least possible score she can get is $\boxed{\textbf{(A)}\ 48}$. ~heeeeeeeheeeeee


Note: You can verify that $\boxed{48}$ is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer. ~~ gorefeebuddie

Solution 3

We can compare each of the scores with the average of $81$: $76$ $\rightarrow$ $-5$, $94$ $\rightarrow$ $+13$, $87$ $\rightarrow$ $+6$, $100$ $\rightarrow$ $19$;

So the last one has to be $-33$ (since all the differences have to sum to $0$), which corresponds to $81-33 = \boxed{48}$.

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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