Algebra question

Question

Gail needs to earn a b in her algebra class. Her current test scores are 84, 76, 83, and 83. Her final exam is worth 6 test scores . In order to get a b Gail average must lie between 80 and 83 inclusive. What range of scores can Gail receive on the final exam to earn a b in the course.

Answer 1

If the final exam is worth 6 test scores, that is like having a total of 10 tests, including the ones listed. For the average to be the minimum average of 80, the total must be 800 (in order to divide by 10 to get 80), and 830 to get the maximum average of 83. So to get the final exam grades that make these averages happen, we need the two equations:

84 + 76 + 83 + 83 + 6x = 800

and

84 + 76 + 83 + 83 + 6x = 830

So, in each equation, we can add like terms and get

326 + 6x = 800

and

326 + 6x = 830

6x = 474

and

6x = 504

x = 79

and

x = 84.

Answer 2

Why anyone would not just try to get the highest score they possibly could escapes me, but here we go:

Average is defined as the sum of the numbers divided by the number of numbers. For example, the average of 3, 4, and 5 is (3 + 4 + 5)/3 = 12/3 = 4.

In this case, her scores are 84, 76, 83, and 83 and we'll let her final have a score of "x". But, because the final is worth 6 test scores, we can say the sum of her scores will be (84 + 76 + 83 + 83 + x + x + x + x + x + x) or (326 + 6x) and the number of scores will be 4 + the 6 equivalent test scores for the final or 10. So the Average = (326 + 6x)/10

So, using the low extreme we get:

80 = (326 + 6x)/10

800 = 326 + 6x

474 = 6x

x = 79

Using the high extreme we get:

83 = (326 + 6x)/10

830 = 326 + 6x

504 = 6x

x = 84

So to get a B in the class, she must score between 79 and 84 on the final.