SOLUTION: Going into the final exam, which will count as two tests, Jack has test scores 80, 83, 71, 61, and 95. What score does Jack need on the final in order to have an average score of 8

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Question 1208697: Going into the final exam, which will count as two tests, Jack has test scores 80, 83, 71, 61, and 95. What score does Jack need on the final in order to have an average score of 80?
My setup, which leads to the wrong answer, is this:
(80 + 83 + 71 + 61 + 95 + 2x)/6 = 80
Here, 2x = 2 tests. This is my understanding. I am wrong.
A. Why is my setup wrong?
B. If you were taking a test and came across this word problem, how would you break it down leading to the right equation or setup?
Found 3 solutions by josgarithmetic, greenestamps, math_tutor2020:
Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!
The final examination may count as 2x. that is two tests at x score each; but this is doubled, so 2x.








Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


The response from the other tutor shows a straightforward solution using the standard method for determining the average.

When the numbers being averaged are "close together", it is sometimes faster and easier to work with the "overs" and "unders" instead of with the numbers themselves.

See if this solution method "works" for you.

Compare each test score to the desired average of 80:

   1  80   0
   2  83  +3
   3  71  -9
   4  61 -19
   5  95 +15
 ------------
  total  -10

With his 5 tests, he is a total of 10 points short of his desired average, so he must make up those 10 points on the final. Then, since the final counts twice as much as each test, his score on the final must be 10/2 = 5 points above the desired average.

So his score on the final must be 85 to get an overall average of 80.

ANSWER: 85
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Your setup equation is very close to being correct.
The 6 in the denominator should be 7.
This is because we basically have these test scores: {80, 83, 71, 61, 95, x, x}
x shows up twice since the final counts as two tests.
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