# Thread: Algebra Variable expression word problem finding averages

1. ## Algebra Variable expression word problem finding averages

Ok I managed to figure out the first part of this question on my own but the second part I just can't seem to wrap my brain around how to put the words into a solvable equation.

Here's the first part including the method I used to get the right answer:

Course Guide - To Get an A in a course, you must have an average of at least 90 points for four tests of 100 points each. For the first three tests, your scores are 87, 92, and 84. What must you score on the fourth exam to earn a 90% average for the course.

87 + 92 + 84 + x/4 = 90

263 + x/4 = 90

multiply both sides by 4

which will leave me with...

263 + x = 360

minus 263 from both sides and the it becomes

x = 97 (which is the answer in the book)

Now part two I just can't get. Please see the question below and if you are able to help that would be great.

Repeat excercise 75 if the fourth test is weighted so that it counts for twice as much as the first three tests.

2. Originally Posted by suddenlysmarter
Repeat excercise 75 if the fourth test is weighted so that it counts for twice as much as the first three tests.
If the fourth test is counted twice as much, simply recopy your previous equation but multiply the x by 2, and divide the sum by 5:
(87 + 92 + 84 + 2x)/5 = 90
(263 + 2x)/5 = 90

Now solve for x.

01

EDIT: This is assuming you mean that the fourth test is counted twice as much as any of the other 3 tests, as opposed to counting twice as much as the combined score of the three tests.

3. The answer in the book for the question is 93.5

I would need to see it worked out with that answer in order to fully understand how it was achieved. Thank you so much for your help so far.

(87 + 92 + 84 + 2x)/5 = 90
(263 + 2x)/5 = 90
263 + 2x = 450

(I think you can go on from here. )

01

5. That's the right way for sure. Thank you. Can you explain the logic of why you divide by 5 when there is only 4 numbers?

6. Perhaps because the one test counts twice as much, so it counts as two tests...?

7. Yes. It is exactly the same as (87 + 92 + 84 + x+ x)/5 = 90, counting the last test as if it were two tests.

Notice why it has to be that way: If you got 90 on every tests, counting them all equally, the average is (90+ 90+ 90+ 90)/4= 90, of course. If the last test counts twice as much as the rest and you only divided by 4, you would get (90+ 90+ 90+ 2*90)/4= 450/4= 112.5 which makes no sense! Dividing by 5 gives (90+ 90+ 90+ 2*90)= 450/5= 90.

More generally, you could average 4 numbers, $x_1, x_2, x_3, x_4$ weighting each test as you like by multiplying each number by something: $(nx_1+ mx_2+ px_3+ px_4)/(m+ n+ p+ q)$.