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

Revision as of 19:31, 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

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

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 $x$ 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

@@ Line 21: / Line 21: @@
 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>.
-==Solution 3==
+==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>.

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