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Math Help - algebraic word problems

  1. #1
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    algebraic word problems

    1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?

    2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?

    3)two cars are traveling toward each other, starting at the same time 480km apart. if they meet after 4 hours, and one car is traveling 40kph faster than the other, how fast is each car moving?

    4) bill has a collection of 65 nickels and dimes. their total worth is $5.50. find how many nickels and dimes bill has.

    can someone please show me how to do these??????
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  2. #2
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    Quote Originally Posted by pspman354 View Post

    2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?
    \frac{x}{y}= \frac{6}{5} ...(1)

    x-y = 8 ...(2)

    Now solve.
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  3. #3
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    Quote Originally Posted by pspman354 View Post
    1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?

    20 Litres of 65% acid solution = 20 x 0.65 = 13L acid

    a Litres of 60% acid solution = 0.6a L acid

    b Litres of 80% acid solution = 0.8b L acid.

    You want
    0.6a + 0.8b = 13 Litres of acid
    a + b = 20 L solution

    Solve for a and b.
    Attempt the first one. Then apply the same method to the other questions.
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  4. #4
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    \frac{x}{y}= \frac{6}{5} ...(1)



    how do i solve for x or y using this?
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  5. #5
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    Quote Originally Posted by pspman354 View Post
    \frac{x}{y}= \frac{6}{5} ...(1)



    how do i solve for x or y using this?
    you can't you need both equations


    \frac{x}{y}= \frac{6}{5} ...(1)

    x-y = 8 ...(2)

    Using (2) x=8+y and back into (1) gives

    \frac{8+y}{y}= \frac{6}{5}

    now finish it.
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