Maintaining an Average

Maintaining an Average

Problem

Samatha needs to maintain a B average in courses in her major. This term she needs an 80% average in IT. In the first three exams, Sam averaged 75%. What is the lowest score she can get on the fourth exam and still have any possibility at all of getting an 80% average?

There are eight exams through the semester, Sam’s average is based on exams only, and all exams are weighted equally.

Solution

To get your arms around the question, think about what an average is: the sum of all the numbers in a set divided by the number of numbers.

Sam needs an 80% average, so the sum of all her scores divided by the number of exams adds up to 80. What do you know so far?

  • There are eight exams, so the denominator will be 8.
  • Sam’s average in the first three exams is 75. That means the sum of her first three scores divided by 3 is 75 — so the sum of her first three scores is 3 times 75, or 225.
  • You are looking for the lowest score that can possibly work.

Sam can get away with a lower score on this test if she does better on later ones. The lowest score she can afford this time is the one that will get her just an 80% average if she gets 100 on each one of the last four exams. That’s the score you’re looking for.

You need a name for the number you’re looking for. Call it x.

Now you can write an equation for the average.

80=\frac{225+x+100+100+100+100}{8}

80=\frac{625+x}{8}

And solve for x.

x=15

Sam can get a 15% on the fourth test and still get her 80% average for the course. All she has to do is get 100% on every test after that.

To check, add up all the scores, including the three 75s for the first three exams; 15 for the fourth exam, and 100 for each of the last four exams, and divide by 8:

\frac{75+75+75+15+100+100+100+100}{8}=80

Check.

It’s a silly question, but I was trying to replicate an SAT question that was, well, equally silly in the same way. I don’t recommend that you make a habit of counting on your future self to get perfect exam scores.

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