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Math Help - Grade 8 Algebra 1, Course 2 - Acid Solutions

  1. #1
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    Grade 8 Algebra 1, Course 2 - Acid Solutions

    THIS IS VERY URGENT

    This isn't actually an acid solution, just something like it. I would GREATLY appreciate it if you can help me understand this

    A man invested $1200 for one year, part at 4% and part at 6%. He earned 58$ from his investments. How much did he invest at each rate?

    The way I learned it was doing something like this

    1...|2.........|New
    .04 |.06......|.1
    x....|1200-x.|1200
    .04x|72-.06x|120

    -.02x = 48

    I know I'm doing something wrong there - The chart may look screwed up, but it was multiply down and add across.
    EDIT:
    ----------------
    I'd also like some help on understanding this:

    Jake can run downhill at a rate 2 times faster than he can run uphill. If the distance each way is 5km, and the total time for the trip is 30 minutes, what is his rate downhill?
    Last edited by Harryhit4; May 24th 2007 at 05:11 PM.
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  2. #2
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    Lexington, MA (USA)
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    Hello, Harryhit4!

    A man invested $1200 for one year, part at 4% and part at 6%.
    He earned $58 from his investments.
    How much did he invest at each rate?
    Your chart is not quite correct . . .

    He invested x dollars at 4%.
    . . His interest is: . 0.04x dollars.

    He invested the other 1200-x dollars at 6%.
    . . His interest is: . 0.06(1200-x) dollars.

    His total interest is $58.

    There is our equation: . 0.04x + 0.06(1200 - x) \:=\:58



    Jake can run downhill at a rate 2 times faster than he can run uphill.
    If the distance each way is 5km, and the total time for the trip is 30 minutes,
    what is his rate downhill?
    We will use: . \text{Distance }= \text{ Speed } \times \text{ Time}\quad\Rightarrow\quad T \:=\:\frac{D}{S}

    Let U = Jake's uphill speed (in km per hour).
    Then 2U = Jake's downhill speed.

    Running 5km uphill at U kph, his time is: . \frac{5}{U} hours.

    Running 5km downhill at 2U kph, his time is . \frac{5}{2U} hours.

    His total time is \frac{1}{2} hour (30 minutes).

    There is our equation: . \frac{5}{U} + \frac{5}{2U} \:=\:\frac{1}{2}

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  3. #3
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    Thanks

    Sadly, I didn't get enough time to check your answer before going to bed. But, for the first question, I just figured out during the test [A different problem, though] that I can use substitution as an alternative

    x + y = 1200
    .04x + .06y = 58

    The problem was that I kept thinking .04x + .06y was 1200, not 58.

    As for the second one, that type of problem wasn't on the test, so I was saved. I would have been completely wrong [Although I know how to do regular rate/time/distance problems]

    That helped me understand a lot, thanks!
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