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

(Solution 1)
(Solution 1)
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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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Another way to do this (to save on computation) is to compare each of the scores with the average of <math>81</math>:
 +
<math>76</math> --> <math>-5</math>
 +
<math>94</math> --> <math>+13</math>
 +
<math>87</math> --> <math>+6</math>
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<math>100</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>.
  
 
==See Also==
 
==See Also==

Revision as of 22:24, 24 November 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


Another way to do this (to save on computation) is to compare each of the scores with the average of $81$: $76$ --> $-5$ $94$ --> $+13$ $87$ --> $+6$ $100$ --> $19$

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

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