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  1. #1
    777
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    algebra word problem

    Please can anyone solve the following questions and show working pls.?

    4. The radii of three concentric circles are in the ratio 1: 2: 3. Find the ratio of the area between the two inner circles to that between the two outer circles.

    5. A dealer allows a discount of 12% on the marked price. How much per cent above the cost price, must he mark his goods so as to make a profit of 10%?

    6. From a 5 litres of 20% solution of alcohol in water, 2 litres of solution is taken out and 2 litres of water is added to it. Find the strength of alcohol in the new solution.
    Last edited by mr fantastic; December 13th 2008 at 12:58 AM. Reason: 16 questions were originally posted. I've distributed them over six newly created threads.
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  2. #2
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    Quote Originally Posted by 777 View Post
    Please can anyone solve the following questions and show working pls.?

    4. The radii of three concentric circles are in the ratio 1: 2: 3. Find the ratio of the area between the two inner circles to that between the two outer circles.

    5. A dealer allows a discount of 12% on the marked price. How much per cent above the cost price, must he mark his goods so as to make a profit of 10%?

    6. From a 5 litres of 20% solution of alcohol in water, 2 litres of solution is taken out and 2 litres of water is added to it. Find the strength of alcohol in the new solution.
    4. Your ratio of the radii 1:2:3 means that

    r_2 = 2r_1 and r_3 = 3r_1

    The area of the circles are as follows...

    A_1 = \pi r_1^2

    A_2 = \pi r_2^2
     = \pi(2r_1)^2
     = 4\pi r_1^2
     = 4A_1

    The area between the two inner circles (which is known as a toros) is therefore

    T_1 = A_2 - A_1
     = 4A_1 - A_1
     = 3A_1.


    Similarly A_3 = \pi r_3^2
     = \pi(3r_1)^2
     = 9\pi r_1^2
     = 9A_1.

    So the area between the two outer circles is therefore

    T_2 = A_3 - A_2
     = 9A_1 - 4A_1
     = 5A_1
     = \frac{5}{3}T_1.

    So the ratio of the inner toros to the outer toros is therefore

    1:\frac{5}{3} or 3:5.
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    Hello, 777!

    We can talk our way through this one . . .


    6. From a 5 litres of 20% solution of alcohol in water,
    2 litres of solution is taken out and 2 litres of water is added to it.
    Find the strength of alcohol in the new solution.
    Since water is being removed and replaced,
    . . we will consider the amount of water at each stage.


    We start with 5 liters of solution which is 80% water.
    . . It contains: . 0.80 \times 5 \:=\:4.4 liters of water.

    Two liters of solution is removed.
    . . This contains: . 0.80 \times 2 \:=\:1.76 liters of water.
    So there are: . 4.4 - 1.76 \:=\:2.64 liters of water left.

    Then two liters of pure water is added.
    . . So there are: . 2.64 + 2 \:=\:4.64 liters of water.


    We have 5 liters of new solution of which 4.64 liters is water.
    . . It is: . \frac{4.64}{5} \:=\:0.928 \:=\:92.8\% water.

    Therefore, the new solution is: . 100\% - 92.8\% \:=\:7.2\% alcohol.

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  4. #4
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    Hello, 777!

    #5 is a tricky one . . .


    5. A dealer allows a discount of 12% on the selling price.
    What percent above the cost must he mark his goods so as to make a profit of 10%?

    We have: . \text{(Cost)} + \text{(Markup)} \:=\:\text{(Selling price)} \quad\Rightarrow\quad C + M \:=\:S .[1]


    Suppose Markup is p percent of Cost: . M \:=\:pC

    Then [1] becomes: . S \:=\:C + pC \:=\:(1+p)C

    He offers a 12% discount, so he sells it for: . 0.88(1+p)C .[2]

    This amount is to give him a profit which is 10% of cost.
    . . That is, the sale price should be: .  C + 0.10C \:=\:1.1C .[3]


    Equate [2] and [3]: . 0.88(1+p)C \;=\;1.1C

    Divide by 0.88C\!:\;\;1+p \:=\:1.25 \quad\Rightarrow\quad p \:=\:0.25


    He should mark it up 25% of cost
    . . for the sale price to bring a 10% profit.

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