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Math Help - simultaneous problem solving? year 10 extension work

  1. #1
    Newbie white's Avatar
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    simultaneous problem solving? year 10 extension work

    eeek these questions are confusing . year 10 extension work the teacher gave me.. need help. thank you

    q.1an investor received $1400 interest per annum from a sum of money , with part of it invested at 10% and the remainder at 7% simple interest. This investor found that if she interchanged the amounts she had invested she could increase her return by $90 per annum. calculate the total amount invested.

    q.2a tea wholesaler blends together three types of tea that normally sell for $10,$11,$12 per kilogram so as to obtain 100 kilograms f tea worth $11.20 per kilogram. If the same amounts of the two higher priced teas are used, calculate how much of each type must be used in the blend.
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  2. #2
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    Quote Originally Posted by white View Post
    [snip]
    q.2a tea wholesaler blends together three types of tea that normally sell for $10,$11,$12 per kilogram so as to obtain 100 kilograms f tea worth $11.20 per kilogram. If the same amounts of the two higher priced teas are used, calculate how much of each type must be used in the blend.
    I have a hunch Soroban is doing q1.

    I'll let you contemplate where the following two equations come from that you must solve simultaneously:

    100 = x + 2y .... (1)

    1120 = 10x + 23y .... (2)

    I get 40 kg each of the two higher priced teas and 20 kg of the lowest priced tea in the 100 kg of tea. Therefore .....
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  3. #3
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    Hello, white!

    I'll set these up for you . . .


    1. An investor received $1400 interest per annum from a sum of money
    with part of it invested at 10% and the remainder at 7% simple interest.
    This investor found that if she interchanged the amounts she had invested,
    she could increase her return by $90 per annum.
    Calculate the total amount invested.
    Let: . \begin{array}{ccc}x &=& \text{amount at 10\%} \\ y &=& \text{amount at 7\%} \end{array}

    x dollars at 10% produces 0.10x dollars in interest.
    y dollars at 7% produces 0.07y dollars in interest.

    The total interest is $1400: . 0.10x + 0.07y \:=\:1400\;\;{\color{blue}[1]}


    If the amounts are interchanged, the total interest is $1490.

    . . 0.07x + 0.10y \:=\:1490\;\;{\color{blue}[2]}


    Multiply [1] and [2] by 100: . \begin{array}{ccc}10x + 7y &=& 140,000 \\ 7x + 10y &=& 149,000 \end{array}




    2. A tea wholesaler blends together three types of tea that normally sell
    for $10, $11, $12 per kg, as to obtain 100 kilograms of tea worth $11.20 per kg.
    If the same amounts of the two higher priced teas are used,
    calculate how much of each type must be used in the blend.
    Let: . \begin{array}{ccc}x &=& \text{amount of \$10 tea} \\ y &=& \text{amount of \$11 tea} \\ y &=& \text{amount of \$12 tea}\end{array}

    The total amount of tea is 100 kg: . x + 2y \:=\:100\;\;{\color{blue}[1]}


    x kg of $10 tea has a value of: 10x dollars.
    y kg of $11 tea has a value of: 11y dollars.
    y kg of $12 tea has a value of: 12y dollrs.
    . . The total value is: . 10x +  23y dollars.

    But the mixture is 100 kg worth $11.20 per kg.
    . . Its total value is: . 100 \times 11.20 \:=\:1120

    There is our second equation: . 10x + 23y \:=\:1120\;\;{\color{blue}[2]}

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