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

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==Problem 7==
 
==Problem 7==
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Shauna takes five tests, each worth a maximum of <math>100</math> points. Her scores on the first three tests are <math>76</math>, <math>94</math>, and <math>87</math>. In order to average <math>81</math> for all five tests, what is the lowest score she could earn on one of the other two tests?
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Shauna takes five tests, each worth a maximum of <math>100</math> points. Her scores on the first three tests are <math>76</math> , <math>94</math> , and <math>87</math> . In order to average <math>81</math> for all five tests, what is the lowest score she could earn on one of the other two tests?
  
 
<math>\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74</math>
 
<math>\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74</math>
  
 
==Solution 1==
 
==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 <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>.\frac{3}{4}
+
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
 
~ aopsav
  
 
==Solution 2==
 
==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, for 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. This means 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 next <math>2</math> tests. Since the maximum score she can get on one of her <math>2</math> tests is <math>100</math>, the least possible score she can get is <math>\boxed{\textbf{(A)}\ 48}</math>.
−
 
 
  
 
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.
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<math>94</math> <math>\rightarrow</math> <math>+13</math>,
 
<math>94</math> <math>\rightarrow</math> <math>+13</math>,
 
<math>87</math> <math>\rightarrow</math> <math>+6</math>,
 
<math>87</math> <math>\rightarrow</math> <math>+6</math>,
−
<math>100</math> <math>\rightarrow</math> <math>19</math>;
+
<math>100</math> <math>\rightarrow</math> <math>+19</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>.
 
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>.
  
−
==See Also==
+
==Solution 4==
 +
We know that she scored <math>76, 94,</math> and <math>87</math> points on her first <math>3</math> tests for a total of <math>257</math> points and that she wants her average to be <math>81</math> for her <math>5</math> total tests. Therefore, she needs to score a total of <math>405</math> points. In addition, one of the final <math>2</math> tests needs to be the maximum of <math>100</math> points, to make the final test score—the one that we are looking for—the lowest score possible for her to earn. We can now see here that the sum of <math>76+94+100</math> has a units digit of <math>0</math> and that the final test score must have a units digit ending with a <math>5</math>. Now, <math>87</math> needs to be added to a number that makes the sum divisible by <math>5</math>. Among the answer choices of <math>48, 52, 66, 70,</math> and <math>74</math>, only <math>\boxed{\textbf{(A)}\ 48}</math> has a units digit that works. (<math>48+87=135</math>, giving us a units digit of <math>5</math>.)
 +
 
 +
~ saxstreak
 +
 
 +
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=ESh8VTHfpwe7idKH&t=2030
 +
 
 +
~Math-X
 +
 
 +
== Video Solution 2 ==
 +
The Learning Royal : https://youtu.be/8njQzoztDGc
 +
== Video Solution 3 ==
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=8
 +
 
 +
==Video Solution 4==
 +
https://youtu.be/CEUuZiHL8c8
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution 5==
 +
https://youtu.be/3t6MQ2MK1fs
 +
 
 +
~ saxstreak
 +
 
 +
==Video Solution by OmegaLearn ==
 +
https://youtu.be/rQUwNC0gqdg?t=1399
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/s0O7YM2W3Bs
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 +
==See also==
 
{{AMC8 box|year=2019|num-b=6|num-a=8}}
 
{{AMC8 box|year=2019|num-b=6|num-a=8}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:41, 30 December 2023

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, for 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. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ tests is $100$, the least possible score she can get is $\boxed{\textbf{(A)}\ 48}$.

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.

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 $81-33 = \boxed{48}$.

Solution 4

We know that she scored $76, 94,$ and $87$ points on her first $3$ tests for a total of $257$ points and that she wants her average to be $81$ for her $5$ total tests. Therefore, she needs to score a total of $405$ points. In addition, one of the final $2$ tests needs to be the maximum of $100$ points, to make the final test score—the one that we are looking for—the lowest score possible for her to earn. We can now see here that the sum of $76+94+100$ has a units digit of $0$ and that the final test score must have a units digit ending with a $5$. Now, $87$ needs to be added to a number that makes the sum divisible by $5$. Among the answer choices of $48, 52, 66, 70,$ and $74$, only $\boxed{\textbf{(A)}\ 48}$ has a units digit that works. ($48+87=135$, giving us a units digit of $5$.)

~ saxstreak

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=ESh8VTHfpwe7idKH&t=2030

~Math-X

Video Solution 2

The Learning Royal : https://youtu.be/8njQzoztDGc

Video Solution 3

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=8

Video Solution 4

https://youtu.be/CEUuZiHL8c8

~savannahsolver

Video Solution 5

https://youtu.be/3t6MQ2MK1fs

~ saxstreak

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=1399

~ pi_is_3.14

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/s0O7YM2W3Bs

~Education, the Study of Everything

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

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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