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

## 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

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