Word problem assistance needed

Hello and good day:

I'm a homeschool mom and need a bit of help. I am not a math wiz and wish I were. I have some difficulty remembering how to do some things in math and could use assistance. Its always fun to see how it should have been done compared to what I did. Anyway, I am assisting my daughters and the math wiz is now in college. Could you help me remember how to set up this word problem?

To make an A the student needs a 94 and has averaged 91 on the first three tests. What does the student need to average on the last three tests to achieve the 94 for an A?

Re: Word problem assistance needed

How many tests are there in total?

Re: Word problem assistance needed

Assuming all four test count the same, then the average of 4 tests would be (a+ b+ c+ d)/4 where a, b, c, and d are the test scores. If this persons test scores are 91, 91, 91, and we call the score on the fourth test "x", then we have (91+ 91+ 91+ x)/4= (273+ x)/4= 94. If you multiply both sides by 4, you get 273+ x= 376. Solve that for x.

(Looks like this guy has a problem!)

Re: Word problem assistance needed

First, though not necessary, the way this problem is phrased makes it possible to just "see" the answer. When after half the tests done (3 down, 3 to go), you're averaging 3 points below your desired average, then to hit your desired average, the other half of your tests must average 3 points above your desired average. One way to "see" this is to imagine if the scores were 91, 91, 91, x, x, x, and averaged to 94 (of course, the first 3 tests were not necessarily all 91, though they could've been). Clearly that must be 91, 91, 91, 97, 97, 97, which would average out to 94, since they so average pairwise. The average is the "balancing point".

Prompting that intuition further, suppose there were 9 total tests. Then the "6 to go" would have to "over shoot" the desired average by only half as much as the first 3 under-shot it, since the 6 remaining amount to double the weight of the first 3. So if there were 6 tests to go, the average on them would have to be 94 + (1/2)(3) = 95.5.

If that helps your intuitions about averages, then great. If not, then ignore it - the comments I made above are completley unnecessary to solving problems with averages.

The main point to know is that the *key* to solving average problems is to work with *totals*. You do that by multplying the average by the number of things being averaged. It's much easier to work with totals; you can always move back to averages when needed by dividing by the number of things being averaged.

"Move to totals" is the approach to remember.

This problem: 6 tests, 3 taken averaging 91, seek to avergae 94, want to know average of the remaining 3 to get there.

TotalOfFirst3Tests = 3(91) = 273.

RequiredTotalOfAll6Tests = 6(94) = 564.

Thus RequiredTotalOfLast3Tests = RequiredTotalOfAll6Tests - TotalOfFirst3Tests = 564 - 273 = 291.

So RequiredTotalOfLast3Tests = 291, and that's the total of 3 tests, so the average on those 3 tests must be 291 / 3 = 97.

Answer: "To make an A, to have an average of 94 for the year, the student must average 97 on the final three tests."