2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 7 Easy

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?

  • A.

    48

  • B.

    52

  • C.

    66

  • D.

    70

  • E.

    74

Answer:A

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}.

 

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}.