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

m
Line 14: Line 14:
 
==Solution 2==
 
==Solution 2==
 
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>-5</math>
+
<math>76</math> <math>\rightarrow</math> <math>-5</math>,
<math>94</math> --> <math>+13</math>
+
<math>94</math> <math>\rightarrow</math> <math>+13</math>,
<math>87</math> --> <math>+6</math>
+
<math>87</math> <math>\rightarrow</math> <math>+6</math>,
<math>100</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>.

Revision as of 15:19, 29 December 2019

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

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 2

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 (ProblemsAnswer KeyResources)
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