Results 1 to 7 of 7

Math Help - Algebra Variable expression word problem finding averages

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    15

    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.

    Answer is:

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by suddenlysmarter View Post
    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.

    EDIT #2: I misread your equation. You forgot the parentheses earlier.
    Last edited by yeongil; May 19th 2009 at 02:24 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    15
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    May 2009
    Posts
    527
    As I stated earlier, I misread your equation.

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

    (I think you can go on from here. )


    01
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    15
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Perhaps because the one test counts twice as much, so it counts as two tests...?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,537
    Thanks
    1392
    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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ration expression word problem please help!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 10th 2010, 04:43 PM
  2. Algebra 2 word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 2nd 2009, 02:37 PM
  3. Two-Variable Linear Systems, Word Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 25th 2009, 03:29 PM
  4. Word Problem for Algebra.
    Posted in the Algebra Forum
    Replies: 25
    Last Post: August 24th 2009, 09:31 PM
  5. Replies: 1
    Last Post: November 11th 2007, 02:12 PM

Search Tags


/mathhelpforum @mathhelpforum