Hints and solutions to Guess and Check #6

In a geometry course the grade is based on six tests, each worth 100 points.
W. Orrier has an average of 88.5 on his first four tests. What is the lowest average he could obtain on his next two tests and still receive an overall average of 90 or better?

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• We're back to "guess and check" (or algebra).
• The tricky thing here is to figure out how to calculate the overall average given the "averaged" data in the problem.
• Since we know the average of the first four tests was 88.5, the grades on the four tests must have added up to 4 x 88.5 = 354. Because then W. Orrier would have divided 354 by 4 to get the average of 88.5.
• So to get the average for all six tests, we have to add the last two grades to 354.
• Now, let's think about solving our problem by "guess and check". For example, we could guess that W. Orrier needs an average of 95 on the last two tests. Then, to calculate the overall average, we would calculate
  (4 x 88.5 + 2 x 95)/6 = (354 + 190)/6 = 544/6 = 90.67

  So 95 is more than enough. We can then revise our guess downward and go again.

• A more systematic approach would realize that his total grade on all six tests needs to be at least 90 x 6 = 540. And he has 354 points so far, so he needs 186 more. That means he must average 186/2 = 93 on the last two tests to get a 90 average.
• The algebraic approach: Let a be his average on the last two tests. Then we must solve
  ((4)(88.5) + 2a)/6 ≥ 90.

  Multiply both sides by 6, subtract 354, and finally divide by 2 to get a ≥ 93.

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