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Math Help - simultaneous problem solving (2

  1. #1
    Newbie white's Avatar
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    simultaneous problem solving (2

    hey need help here

    1- a tent manufacturer produces two models, the Outback and the Bushwalker. From earlier sales records it is known that 20 per cent more of the Outback model is sold than the Bushwalker. A profit of $200 is made on each Outback sold, but $350 is made on each Bush walker. If during the next year a profit of $177 000 is planned, how many of each model must be sold.

    attempt
    outback- 20%more and $200p.each
    bushwalker- $350p.each

    so outback is 20/100 + 200x
    so bush walker is 350y

    where to form here ?

    2- OZ jeans has factories in Mydney and Selbourne. At the Mydney factory fixed costs are $28 000 per month and the cost of producing each pair of jeans is $30. At the selbourne factory, fixed costs are $35200 per month and each pair of jeans costs $24 to produce. during the next month OZ jeans must manufacture 6000 pairs of jeans. Calculate the production order for each factory, if th total manufacturing cost are to be the same.

    attempt
    Mydeny= 28,000 p.month +30x
    Selbourne 35200p.month +24x
    ...?

    Thanks for anyone who can help .
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by white View Post
    hey need help here

    1- a tent manufacturer produces two models, the Outback and the Bushwalker. From earlier sales records it is known that 20 per cent more of the Outback model is sold than the Bushwalker. A profit of $200 is made on each Outback sold, but $350 is made on each Bush walker. If during the next year a profit of $177 000 is planned, how many of each model must be sold.

    [snip]
    Let X be number of bushwalker and Y be number of outback.

    Then X = Y + 20% of Y => X = Y + Y/5 (since 20% is the same as 1/5) => X = 6Y/5.

    X = 6Y/5 .... (1)

    Profit: 350 X + 200 Y = 177 000 => 35X + 20Y = 17700 => 7X + 4Y = 3540

    7X + 4Y = 3540 .... (2)

    Solve equations (1) and (2) simultaneously (I suggest using the substitution method).
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  3. #3
    Flow Master
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    Quote Originally Posted by white View Post
    [snip]
    2- OZ jeans has factories in Mydney and Selbourne. At the Mydney factory fixed costs are $28 000 per month and the cost of producing each pair of jeans is $30. At the selbourne factory, fixed costs are $35200 per month and each pair of jeans costs $24 to produce. during the next month OZ jeans must manufacture 6000 pairs of jeans. Calculate the production order for each factory, if th total manufacturing cost are to be the same.

    attempt
    Mydeny= 28,000 p.month +30x
    Selbourne 35200p.month +24x
    ...?

    Thanks for anyone who can help .
    Let number of jeans at Mydney be M and number of jeans at Selbourne be S.

    Total number of jeans: M + S = 6000 ..... (1)

    Production costs: (28,000 + 30M) = (35,200 + 24S) => 30M - 24S = 7200 => 5M - 4S = 1200.

    5M - 4S = 1200 .... (2)

    Solve equations (1) and (2) simultaneously. (I suggest the elimination method).
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