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Math Help - PreCalculus Challenge puzzler

  1. #1
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    PreCalculus Challenge puzzler

    i received this problem and worked through it, but my solution is flawed! it seems to be a pretty backwards solution.

    The Problem
    While watching a basketball game one day on television, I saw the following situation:

    A player was fouled as he drove for the basket. At the bottom of the screen, it was stated that "Player X is making 78% of his free throws". The player missed the first free throw and made the second.

    Later in the game, the player was again fouled. This time, at the bottom of the screen, it said "Player X is making 76% of his free throws". The statistic had been updated after his first two attempts.

    Recognizing that the percentages presented on the screen were rouded to the nearest 1%, how many free throws had player X attempted and how many had he made? Be sure to find all possible solutions (if any).

    My Bad Solution

    First, I named things.

    # Made at first statistic=f
    % Made at first statistic=0.78f

    # Attempted at 2nd stat=f+2
    % Made at 2nd stat=0.76(f+2)
    # Made at 2nd stat=0.78f+1

    Then I made a plan.

    # Made @ 1st stat + # made in game = 76% of attempts @ 2nd stat

    0.78f+1=0.76(f+2)
    SOLVE FOR F (# attempted @ 1st stat)
    f=26

    SO...
    26 attempted at first stat
    20 made
    28 attempted at second stat
    21 made

    Right?? well...not really...

    The Issue

    78% of 26=20.28
    20/26 rounds to 0.77 (wrong percentage)
    76% of 28=21.28
    21/28 rounds to 0.75 (wrong again)

    the solution works when rounded from percentages to shots, but not from shots to percentages.

    I need to find a solution in which the set considered for shots is whole numbers and I think I need a solution with graphs and zeros so I can be sure I can solve for all possible solutions (if there is more than one).

    I'm stumped so any help is appreciated!!

    -Mike
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  2. #2
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    by the way I posted this in urgent help but I think it was moved...anyway this is sort of urgent, I need a fully worked out solution soon

    so anyone who can help would be appreciated!!
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  3. #3
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    Let x be the total number of shots he made out of n shots before he made the two shots.

    thus, \frac xn = 0.78 ..................(1)

    also, after he made the two shots, he made n + 2 shots in total. since he missed one, the number of shots he made becomes x + 1. at the end, he made 76% of his shots, so that

    \frac {x + 1}{n + 2} = 0.76 ...................(2)

    thus you have two equations and two unknowns. you can solve for x and n
    Last edited by Jhevon; September 28th 2008 at 09:27 AM. Reason: changed denominator
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  4. #4
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    wow I made that so complicated - thanks a lot for the help.

    -Mike
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  5. #5
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    when solved, I got

    n=38
    x=29.64

    same rounding problem occurs

    30/38=0.789
    29/38=0.763

    31/40=0.775
    30/40=0.75

    ???

    solving this backwards creates problems, I need a new method! any help is appreciated as I am very desperate!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mikedwd View Post
    when solved, I got

    n=38
    x=29.64

    same rounding problem occurs

    30/38=0.789
    29/38=0.763

    31/40=0.775
    30/40=0.75

    ???

    solving this backwards creates problems, I need a new method! any help is appreciated as I am very desperate!
    oh yes, sorry, it should be n + 2 in the denominator of my last fraction, not n + 1

    now those equations give you the right number
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    oh no. you made a mistake. the solution will be the same as the one you got. you got the right answer. the slight discrepancy comes from the "rounding up" they did, we should be off by about 1% remember?
    Yes but how is this rounding problem solved? the problem states that it will be rounded to the nearest 1%, which is a terrible coincidence for the numbers given (actually, I think the teacher did this on purpose to force us to use a different method to solve the problem)

    Quote Originally Posted by Jhevon View Post
    oh yes, sorry, it should be n + 2 in the denominator of my last fraction, not n + 1

    now those equations give you the right number
    oh yeah I thought that but I solved through anyway to see if it would be right, let me try this again
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mikedwd View Post
    Yes but how is this rounding problem solved? the problem states that it will be rounded to the nearest 1%, which is a terrible coincidence for the numbers given (actually, I think the teacher did this on purpose to force us to use a different method to solve the problem)
    see my previous post. i made a correction
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  9. #9
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    Yes, and I evaluated it.

    x=59.28
    n=76

    59/76=0.776 (good)
    60/78=0.769

    The methodology is still backwards! i thought i had arrived at a correct answer

    thanks for your help though
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mikedwd View Post
    Yes, and I evaluated it.

    x=59.28
    n=76

    59/76=0.776 (good)
    60/78=0.769

    The methodology is still backwards! i thought i had arrived at a correct answer

    thanks for your help though
    how did you get n = 76?
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  11. #11
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    Quote Originally Posted by mikedwd View Post
    i received this problem and worked through it, but my solution is flawed! it seems to be a pretty backwards solution.

    The Problem
    While watching a basketball game one day on television, I saw the following situation:

    A player was fouled as he drove for the basket. At the bottom of the screen, it was stated that "Player X is making 78% of his free throws". The player missed the first free throw and made the second.

    Later in the game, the player was again fouled. This time, at the bottom of the screen, it said "Player X is making 76% of his free throws". The statistic had been updated after his first two attempts.

    Recognizing that the percentages presented on the screen were rouded to the nearest 1%, how many free throws had player X attempted and how many had he made? Be sure to find all possible solutions (if any).
    You have a pair of inequalities defined by this problem:

    0.775 \le \frac{m}{a} < 0.785

    and:

    0.755 \le \frac{m+1}{a+2} < 0.765

    Each of these defines a region between a pair of lines in the first quadrant, and all integer grid points in the region where these regions overlap are feasible solutions. This is a closed region in the form of a parallelogram.

    RonL
    Last edited by CaptainBlack; September 28th 2008 at 11:14 AM.
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  12. #12
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    This makes good sense, and solves my rounding problem

    I'll try out some math on it (and not make any silly mistakes )

    oh and should those "less than" and "less than or equal to" signs be switched around?
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  13. #13
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    I need a little help solving this however.

    I put all the inequalities in terms of m (so 4 inequalities)

    so this is what I had:

    m<0.785a
    m(>or=)0.775a

    m<0.765a+0.53
    m(>or=)0.755a+0.51

    (i switched around the < and the (<or=) symbols because i thought you made a typo)

    i used the inequality application on my TI-84 Plus and graphed where the shades intersect, but I'm not sure how to find integer points within the defined region
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  14. #14
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    Quote Originally Posted by mikedwd View Post
    This makes good sense, and solves my rounding problem

    I'll try out some math on it (and not make any silly mistakes )

    oh and should those "less than" and "less than or equal to" signs be switched around?
    Yes, that is the usual convention.

    RonL
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  15. #15
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    Quote Originally Posted by mikedwd View Post
    I need a little help solving this however.

    I put all the inequalities in terms of m (so 4 inequalities)

    so this is what I had:

    m<0.785a
    m(>or=)0.775a

    m<0.765a+0.53
    m(>or=)0.755a+0.51

    (i switched around the < and the (<or=) symbols because i thought you made a typo)

    i used the inequality application on my TI-84 Plus and graphed where the shades intersect, but I'm not sure how to find integer points within the defined region
    I would do it by search, the largest value that a can take is less than 54, and the smallest more than 16 (find the points of intersection of the lines defining the feasible region to show this).

    So for each a between 17 and 53 I would check which values of m give points in the feasible region (they will all be close to 0.77a).

    RonL
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