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 "2019 AMC 8 Problems/Problem 7"

(→‎Solution 2)
 
(24 intermediate revisions by 17 users not shown)
Line 1: Line 1:
−
==Problem 7==
+
==Problem==
−
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?
+
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>. <math>\frac{3}{4}</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.
Line 21: Line 20:
 
<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(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 09:30, 9 November 2024

Problem

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(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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_7&oldid=232579"