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Thread: [MATH] Money word problems( simple )

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    Default [MATH] Money word problems( simple )

    We learned how to do these in world problems

    Here are the formulas

    And the 3 problems(theres more, but i'm hoping id get it what to do by then)

    http://www.youtube.com/watch?v=q8L3DPmvijQ

    Since most people dislike the video idea

    Heres the first problem, number three

    Id like to know HOW to do it.


    Using only 32 - cent and 20 cent stamps, aaron put 3.36 postage on a package he sent to his sister. He used twice as many 32-cent stamps as 20 - cent stamps. Determine how many of each type of stamp he used .
    May someone post just how to start it?

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    If you wish, add me on MSN and i can help.
    Too lazy to decipher pic and video ;[. Lol

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    Lol, not many people like the video, second person

    Ill add ya

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    Quote Originally Posted by YoHoJo View Post
    If you wish, add me on MSN and i can help.
    Too lazy to decipher pic and video ;[. Lol
    Hehe. Me too. I offered to help if he typed it up for me

    Scripts: Edgeville Chop & Bank, GE Merchanting Aid
    Tutorials: How to Dominate the Grand Exchange

    Quote Originally Posted by YoHoJo View Post
    I like hentai.

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    Hes showering right now, after I finish drying him off I'll tell him to edit his first post and make it a lot neater.

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    you found that a video was easier to upload than just taking a picture?
    Proud owner of "Efferator" my totally boted main account!
    "You see, sometimes, science is not a guess" -Xiaobing Zhou (my past physics professor, with heavy Chinese accent)

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    He did pic, and video, both failed . Typing FTW!

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    Quote Originally Posted by YoHoJo View Post
    He did pic, and video, both failed . Typing FTW!
    Yohojo, thanks for drying me off!

    @Garrett: im on msn.
    @Brain: I couldn't fit one whole problem on a pic.
    @ Hojo: Edited first post, posted number 3

    Brain, and i finished off this math packet!

    Done, Solved, Resolved, what ever

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    Ok, this one isn't too hard.
    State the amount of one kind of stamps X, in this case it's the 20 stamps, so that amount is called X.
    For every stamp of 20 cent, we need two stamps of 32 cent, so for every X of 20, we need two X's of 32, so for every 20*X, we need 32*2X.
    Work that one out and you get the solution

    32(2x)+20x = 336.
    64x + 20x = 336
    84x = 336
    x = 336/84
    x = 4

    So you need 4 20 cent stamps, and 8 32 cent stamps, if we recalculate that it perfectly fits

    Edit: ninja'd, oh well.

    Edit2: bored so I figured I'd do 2 as well.

    Danielle spent $17 at an amusement park for admnission and rides. If she paid $5 for admission, and the rides cost $3 each, what is the total number of rides that she went on?

    highlight me

    Ok, so we gonna state the amount of rides, again, X. The total number of dollars she spent must be 17.

    So that makes:

    ****** = 17.

    She ALWAYS pays $5 entrance, no matter the amount of rides, we enter that in our formula:

    ***** + 5 = 17.

    She pays 3 dollars for every ride, so every X she pays 3, that makes X*3

    X*3 + 5 = 17

    As I'm lazy 3x + 5 = 17
    3x = 12
    x = 12/3 = 4

    So she did a total number of 4 rides.


    love'er


    Mohammed keeps quarters, nickles and dimes in his change jar. He has a total of 52 coins. He has three more quarters than dimes, and five fewer nickles than dimes. How many dimes does Mohammed have?


    Ok, again, outcome = 52

    **** = 52

    We state the amount of dimes = X, so the amount of quarters = x+3, amount of dimes = x, and amount of nickles = (x-5)

    Makes

    (X+3) + X + (X-5) = 52

    X + 3 + X + X - 5 = 52
    3X + 3 - 5 = 52
    3X -2 = 52
    3X = 54 Boreas: add 2 to both sides, you get 3X = 52+2
    X = 54/3 = 18


    First two are correct, I'm doubting about the 3rd one but it should be correct Boreas: 3rd is correct too

    Have fun!
    Last edited by Boreas; 10-20-2009 at 04:17 PM.
    I made a new script, check it out!.

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    Quote Originally Posted by Markus View Post
    Ok, this one isn't too hard.
    State the amount of one kind of stamps X, in this case it's the 20 stamps, so that amount is called X.
    For every stamp of 20 cent, we need two stamps of 32 cent, so for every X of 20, we need two X's of 32, so for every 20*X, we need 32*2X.
    Work that one out and you get the solution

    32(2x)+20x = 336.
    64x + 20x = 336
    84x = 336
    x = 336/84
    x = 4

    So you need 4 20 cent stamps, and 8 32 cent stamps, if we recalculate that it perfectly fits

    Edit: ninja'd, oh well.

    Edit2: bored so I figured I'd do 2 as well.

    Danielle spent $17 at an amusement park for admnission and rides. If she paid $5 for admission, and the rides cost $3 each, what is the total number of rides that she went on?

    highlight me

    Ok, so we gonna state the amount of rides, again, X. The total number of dollars she spent must be 17.

    So that makes:

    ****** = 17.

    She ALWAYS pays $5 entrance, no matter the amount of rides, we enter that in our formula:

    ***** + 5 = 17.

    She pays 3 dollars for every ride, so every X she pays 3, that makes X*3

    X*3 + 5 = 17

    As I'm lazy 3x + 5 = 17
    3x = 12
    x = 12/3 = 4

    So she did a total number of 4 rides.


    love'er


    Mohammed keeps quarters, nickles and dimes in his change jar. He has a total of 52 coins. He has three more quarters than dimes, and five fewer nickles than dimes. How many dimes does Mohammed have?


    Ok, again, outcome = 52

    **** = 52

    We state the amount of dimes = X, so the amount of quarters = x+3, amount of dimes = x, and amount of nickles = (x-5)

    Makes

    (X+3) + X + (X-5) = 52

    X + 3 + X + X - 5 = 52
    3X + 3 - 5 = 52
    3X -2 = 52
    3X = 54 Boreas: add 2 to both sides, you get 3X = 52+2
    X = 54/3 = 18


    First two are correct, I'm doubting about the 3rd one but it should be correct Boreas: 3rd is correct too

    Have fun!
    for the 3.36 one, I was thinking like that also, but I thought of a, in my opinion, fast and easier way.

    you have twice as many 32 cent stamps as 20 cent ones. So you have 32+32+20. That is equal to 84. 336/84 = 4. You have four 20 cent stamps, and four sets of two 32 cent stamps (Eight 32 cent stamps).



    Then what makes math so awesome, checking to make sure that was right, 32*8+20*4 = 336, we are correct.

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    Quote Originally Posted by Baked0420 View Post
    for the 3.36 one, I was thinking like that also, but I thought of a, in my opinion, fast and easier way.

    you have twice as many 32 cent stamps as 20 cent ones. So you have 32+32+20. That is equal to 84. 336/84 = 4. You have four 20 cent stamps, and four sets of two 32 cent stamps (Eight 32 cent stamps).



    Then what makes math so awesome, checking to make sure that was right, 32*8+20*4 = 336, we are correct.
    woops

    I ment to edit this message but instead i clicked new reply, didn't notice. Sorry about double post

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    Quote Originally Posted by Baked0420 View Post
    for the 3.36 one, I was thinking like that also, but I thought of a, in my opinion, fast and easier way.

    you have twice as many 32 cent stamps as 20 cent ones. So you have 32+32+20. That is equal to 84. 336/84 = 4. You have four 20 cent stamps, and four sets of two 32 cent stamps (Eight 32 cent stamps).



    Then what makes math so awesome, checking to make sure that was right, 32*8+20*4 = 336, we are correct.
    Thanks everyone

    blahblahblahblahblah

    <3 no post count in this section, but seriously, no one else post on this thread(unless your heart desires you too )

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    Quote Originally Posted by Baked0420 View Post
    for the 3.36 one, I was thinking like that also, but I thought of a, in my opinion, fast and easier way.

    you have twice as many 32 cent stamps as 20 cent ones. So you have 32+32+20. That is equal to 84. 336/84 = 4. You have four 20 cent stamps, and four sets of two 32 cent stamps (Eight 32 cent stamps).



    Then what makes math so awesome, checking to make sure that was right, 32*8+20*4 = 336, we are correct.
    That was the first way that came to mind for me also. However the algebraic method is better practice because you will end up having to use a similar method on harder problems that will come along eventually. And yes, using multiple methods to check answers in math is awesome. In language/history etc it's called arguing .

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