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Math Help - concentration problem

  1. #1
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    Unhappy concentration problem

    Freezing weather is coming!
    The radiator in my 450 John Deere track loader
    is full with 18 quarts of fluid that is 15% antifreeze.
    Question: How much must I drain out into a disposable container,
    Then retighten drain plug, and pour back 100% pure
    antifreeze into radiator to make 18 quarts of 40%
    antifreeze.
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  2. #2
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    galactus's Avatar
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    Let x=amount to be drained.

    Original mixture less amount drained: (18-x)

    Amount of pure antifreeze=.15(18-x)

    plus

    pure antifreeze is x quarts.

    The new mixture is 40%= .4(18)=7.2

    So, we have the equation to solve:

    .15(18-x)+x=7.2
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  3. #3
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    i think i get it... so would the answer be -5.3 qts... that doesnt make sense
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Amelia View Post
    i think i get it... so would the answer be -5.3 qts... that doesnt make sense
    Check again. I get +5.29
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  5. #5
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    thank you so much

    is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

    I have a collection of nickels, dimes, and quarters
    in a money bag worth $7.95. If the number of
    quarters is one more than the number of nickels
    and the number of dimes is five more than the number
    of nickels, find the number of each type of coin.
    # Nickels ________
    # Dimes_________
    # Quarters________
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Amelia View Post
    is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

    I have a collection of nickels, dimes, and quarters
    in a money bag worth $7.95. If the number of
    quarters is one more than the number of nickels
    and the number of dimes is five more than the number
    of nickels, find the number of each type of coin.
    # Nickels ________
    # Dimes_________
    # Quarters________
    In future, please make a new thread for each new question.

    Okay we have 3 different types of coins.

    So:

    x + y + z = 7,95

    Nickels = x
    Quarters = y = x + 1
    Dimes = z = x + 5

    So you'll just need to substitute the values of these coins. And solve.

    a(x) + b(x + 1) + c(x + 5) = 7,95

    Where a;b;c are the monetary values of nickels, dimes, and quarters (not respectively!). I don't know the American system.
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  7. #7
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    would that make x 8.4 ? im confused a little: in america dime=.10 nickels=.5 and quarters=.25

    i'm confused
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Amelia View Post
    would that make x 8.4 ? im confused a little: in america dime=.10 nickels=.5 and quarters=.25

    i'm confused
    Can you double check the question?
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  9. #9
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    yea :
    I have a collection of nickels, dimes, and quarters
    in a money bag worth $7.95. If the number of
    quarters is one more than the number of nickels
    and the number of dimes is five more than the number
    of nickels, find the number of each type of coin.
    # Nickels ________
    # Dimes_________
    # Quarters________

    nickels are worth .5
    dimes are worth .10
    quarters are worth .25
    a dollare is 1.00
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Amelia View Post
    yea :
    I have a collection of nickels, dimes, and quarters
    in a money bag worth $7.95. If the number of
    quarters is one more than the number of nickels
    and the number of dimes is five more than the number
    of nickels, find the number of each type of coin.
    # Nickels ________
    # Dimes_________
    # Quarters________

    nickels are worth .5
    dimes are worth .10
    quarters are worth .25
    a dollare is 1.00
    Either the question has a fault, or I made a very stupid mistake in my calculations. The latter being very likely.

    You have the idea though, that should carry you through.
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  11. #11
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    Quote Originally Posted by Amelia View Post
    is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

    I have a collection of nickels, dimes, and quarters
    in a money bag worth $7.95. If the number of
    quarters is one more than the number of nickels
    and the number of dimes is five more than the number
    of nickels, find the number of each type of coin.
    # Nickels ________
    # Dimes_________
    # Quarters________
    Probably this comes too late but I'll post it nevertheless:

    Let x denote the number of nickel coins, then the value of these coins is 0.05x
    Let x+1 denote the number of quarter coins, then the value of these coins is 0.25(x+1)
    Let x+5 denote the number of dime coins, then the value of these coins is 0.1(x+5)

    Sum up the values

    0.05x+0.25(x+1)+0.1(x+5)=7.95

    Expand the brackets, collect like terms and solve for x.

    I've got x =18.
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  12. #12
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    oh so then what do i do with the x value?
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  13. #13
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Amelia View Post
    oh so then what do i do with the x value?
    Wait I get it. A nickel is 5 cents, and not 50 cents like you said. That's why Earboth's solution works out so nicely.
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  14. #14
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    oh yea.. lol stupid me..

    but i took x=18
    then that means their are 18 nickels
    then 18+1 which is 19 quarters
    then 18+5 which is 23..
    omg your right!!!!!! THANKK YOU SO MUCH!!! YOU GUYS ARE AWESOME!!!!!
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