algebra problem

Question

17.After the first 4 quizzes, Ron’s quiz average is an 87. There are two quizzes remaining for the

quarter and the maximum score for any individual quiz is 100. What is the minimum grade Ron

can score on one of these two quizzes and still score high enough on the other so that the final

quiz average of the six quizzes is at least a 90?

Answer 1

(87+87+87+87+x+x)/6=90

348+2x=540

2x=540-348

2x=192

x=96, so he can get 96 and 96 but you want the minimum on one of them

96-4=92 and 96+4=100 (you can't subtract 6 because 96+6>100)

so, 92 is the minimum

check: 87+87+87+87+92+100=540

540/6=90

try 91 as the minimum

4*87+91+100=539

539/6=89.8333...=90 rounded up

if the teacher rounds off (up) then 91 is okay for the min

also, if the teacher rounds (up), then 90 works too

4*87+90+100=538

538/6=89.666...=90 rounded up

Answer 2

SUM = average * count

The total = 87*4 = (90-3)*4 = 360-12 = 348

To achieve an A, the total over 6 quizzes must be 6*90=540

So he needs 540-348 = 192 points

192/2 = 96

Therefore he must score 96 or higher on the next two quizzes.

Answer 3

Ron has 348 points (4 x 87)

Ron needs 540 points (6 x 90)

Ron must score at least 192 points on the two tests. (540 - 348)

Ron cannot score less than 92 on one of the tests.