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Math Help - My rice problem

  1. #1
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    My rice problem

    A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

    A.24
    B.32
    C.30
    D.33

    Please explain the solution clearly.With many thanks.
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  2. #2
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    Hello, nimanamjouyan!

    A rice company decides to combine a first-rate rice that costs 5000 per kg
    with a second-rate type costs 4000 per kg.
    The result is a mixture of rice which costs 4600 per kg.
    How many kilograms of the first rate rice are there in 40 kg of the mixture?

    . . (A)\;24 \qquad (B)\;32 \qquad (C)\;30 \qquad (D)\;33

    Let \,x = kg of first-rate rice.
    Then 40-x = kg of second-rate rice.


    We have \,x kg of first-rate rice at 5000/kg.
    . . Its value is: . 5000x

    We have 40-x kg of second-rate rice at 4000/kg.
    . . Its value is: . 4000(40-x)

    The total value of the mixture is:
    . . 5000x + 4000(40-x) \:=\:1000x + 160,\!000 .[1]


    But we know that mixture will be 40 kg of rice at 4600/kg.
    . . Its value will be: . 40 \times 4600 \:=\:184,\!000 .[2]


    We just described the value of the mixture in two ways.

    There is our equation! . \hdots\;\;1000x + 160,\!000 \:=\:184,\!000

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  3. #3
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    Quote Originally Posted by nimanamjouyan View Post
    A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

    A.24
    B.32
    C.30
    D.33

    Please explain the solution clearly.With many thanks.
    Let x= mass of the first rate rice in 1 Kg of mixture
    y= mass of second rate rice in 1 Kg of mixture.

    (x and y both measured in Kg)

    Form 2 equations connecting x and y and solve them.
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  4. #4
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    reply

    I have tried forming two equations like that but I couldn't get an answer.
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  5. #5
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    Well, you saw Soroban form the equations. Do you understand how he did that?
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  6. #6
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    a simplified solution

    mark a cost line in relative costs per kg

    40 46 50
    delta 46-40 = 6 parts of 50
    delta 50-46 = 4 parts of 40
    makes 10 parts of 46

    to make 40 kg of 46 cost requires .6 x 40 = 24 kg of 50 and .4 x40 = 16 kg of 40



    bjh
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  7. #7
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    Quote Originally Posted by nimanamjouyan View Post
    I have tried forming two equations like that but I couldn't get an answer.
    x+y=1
    5000x+4000y=4600

    (it is an alternate solution to the one provided above by Soroban, hence x and y have different meanings, leading to different equations)
    Last edited by Ithaka; January 31st 2011 at 06:53 AM. Reason: clarification
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  8. #8
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    Quote Originally Posted by nimanamjouyan View Post
    A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

    A.24
    B.32
    C.30
    D.33

    Please explain the solution clearly.With many thanks.
    You have 2 different weights for which you need to solve for only one.

    To find 2 unknowns, you need 2 clues.
    By formulating the clues as a pair of equations, you can solve the equations.
    Only one weight is needed, so we can eliminate the 2nd weight from our equations.

    The clues are

    (1) the sum of the weights is 40 kg
    (2) the sum of the prices is 4600(40)=184000

    Price is (weight in kg)(price per kg)

    Formulating these, let x=weight of first rate rice, y= weight of second rate rice.

    x+y=40

    x(5000)+y(4000)=184000

    Therefore, y can be eliminated by substituting it for 40-x.

    y=40-x\Rightarrow\ 5000x+4000(40-x)=4600

    allows us to find x.

    Alternatively, multiply both sides of x+y=40 by 4000

    and subtract the resulting pair of equations to find x.
    This also eliminates y.

    viz....

    4000x+4000y=160000

    5000x+4000y=184000

    Subtracting these

    5000x+4000y-(4000x+4000y)=184000-160000

    5000x-4000x+4000y-4000y=24,000

    1000x=24,000
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