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Math Help - [SOLVED] PSAT Practice Question with Inequalities

  1. #1
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    [SOLVED] PSAT Practice Question with Inequalities

    Hello

    While working on some practice PSAT questions, I came across this question:

    The various averages (arithmetic means) of three of the four numbers c, d, e, and f are calculated, and are arranged from greatest to least as follows

    The average of c, d, and e
    The average of d, e, and f
    The average of e, f, and c
    The average of f, c, and d

    Which of the following correctly orders c, d, e, and f from greatest to least?
    (A) c > d > f > e
    (B) d > f > e > c
    (C) d > e > c > f
    (D) e > c > f > d
    (E) e > d > c > f


    My answer was (A), but the correct answer was (E). I do not see how my work was wrong, other than I might have perhaps used the wrong method. I set it up like you do when you solve equations with two variables by subtracting one whole equation from another. Of course, I'm not totally positive if you can do this with inequalities, but I honestly don't see why not. Here's my work for clarity's sake:

    c + d + e > d + e + f
    -(d+ e + f > f + c + d)
    -----------------------
    f > e

    d + e + f > e + f + c
    -(e + f + c > f + c + d)
    ------------------------
    c > e

    So, I found that both c and f are greater than e and picked the only choice that had this written, (A).

    Anyone want to tell me if I'm doing this wrong, and, if so, how it is actually supposed to be done? Thanks.

    By the way, I didn't realize that I had posted this under elementary/middle school help at first. It is a high school question, but I simply saw the inequalities category so I posted there. Sorry about that.
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  2. #2
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    Okay, after going through it carefully I completely understand the way you're supposed to think it out and am surprised I didn't see it. I was just too busy thinking in terms of equations and knowing algebra, which isn't really what the PSAT is about.

    Of course, I still don't know how I arrived at the wrong conclusion with my original work. It appeared to be giving me the opposite answers of what it really was. What is the method for solving an inequality like this? Are you supposed to flip the sign when you're done?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seuzy13 View Post
    Hello

    While working on some practice PSAT questions, I came across this question:

    The various averages (arithmetic means) of three of the four numbers c, d, e, and f are calculated, and are arranged from greatest to least as follows

    The average of c, d, and e
    The average of d, e, and f
    The average of e, f, and c
    The average of f, c, and d

    Which of the following correctly orders c, d, e, and f from greatest to least?
    (A) c > d > f > e
    (B) d > f > e > c
    (C) d > e > c > f
    (D) e > c > f > d
    (E) e > d > c > f
    begin with the basics. what does it mean to take the average? it is the sum of the 3 numbers divided by 3. then we put them in descending order. so then

    (c + d + e)/3 > (d + e + f)/3 > (e + f + c)/3 > (f + c + d)/3

    => (c + d + e) > (d + e + f) > (e + f + c) > (f + c + d) ..........................we multiplied the above system by 3

    now, consider the first inequality: subtracting (d + e) from both sides, we get c > f

    consider the second inequality: subtracting (e + f) from both sides, we get d > c

    consider the third inequality: subtracting (f + c) from both sides, we get e > d

    choice (E) has all these inequalities intact, so it is the right choice
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  4. #4
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    Quote Originally Posted by seuzy13 View Post
    Hello

    While working on some practice PSAT questions, I came across this question:

    The various averages (arithmetic means) of three of the four numbers c, d, e, and f are calculated, and are arranged from greatest to least as follows

    The average of c, d, and e
    The average of d, e, and f
    The average of e, f, and c
    The average of f, c, and d

    Which of the following correctly orders c, d, e, and f from greatest to least?
    (A) c > d > f > e
    (B) d > f > e > c
    (C) d > e > c > f
    (D) e > c > f > d
    (E) e > d > c > f
    Hello

    Taking them in pairs from the top, in descending order.

    The average of c, d, and e
    The average of d, e, and f

    So c > f

    The average of d, e, and f
    The average of e, f, and c

    So d > c

    The average of e, f, and c
    The average of f, c, and d

    So e > d

    And then back up e > d > c > f

    Hope that helps a little.
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  5. #5
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    Yeah, I made everything way more complicated than it is, as usual.
    I know how to get the answer, and I understand the problem now, but I'm just wondering what was wrong with my original problems (besides being way too complicated).

    Thanks!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seuzy13 View Post
    Yeah, I made everything way more complicated than it is, as usual.
    I know how to get the answer, and I understand the problem now, but I'm just wondering what was wrong with my original problems (besides being way too complicated).

    Thanks!
    you just cannot do that with inequalities, i don't think, more over, if you could, the negative sign would flip the last inequality, as well as you would get negative quantities. that is, for example, you would no longer have a +f in your first system, it would become -f. so one side of your inequality would be negative, which introduces a whole bunch of problem. furthermore, if we can keep the inequality sign turned the right way, (which i don't think we can, this whole system makes no sense anyway ) you would end up with something like c - f > e - c, so you get 2c > e + f if we want to make everything positive. which tells us nothing.

    take my advice and just forget what you did. it was good that you tried though, but this is a bad thing to try. the PSAT is not the kind of exam that would ask you to pull such tricks with inequalities that would have even me baffled. i'm stupid, but not that stupid
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  7. #7
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    Right. I agree with you. Like I said, I have a tendency to complicate things, and when I first see some questions I latch onto this idea that it must be done this way, and don't look at the very obvious method to get the answer that doesn't even require working it out.

    And I realize that, yeah, it would be impossible to do this with inequalities. When you do it with equalities, you are pretty much subtracting the same value from both sides, because those two items in the second equation are equal to each other. Not the case with inequalities.

    Okay, that pretty much finishes this thread. Thanks, people.
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