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Math Help - Practice Paper Questions

  1. #1
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    Practice Paper Questions

    1. Sally bought identical red files. Each red file costs $1.40. When Sally bought 1 more blue file at $4.60, it increased the average cost of the red files to $1.80. How many files did Sally buy in total?

    2. There are a total of 200 blue, green and red balls. There are as many red balls as blue balls. There are twice as many red balls as blue balls. There are fewer green balls than red balls. The number of blue balls and red balls in each group is less than 100 and divisible by 3 and 4. How many green balls are there?

    3. Bella, Cody and Deborah had a collection of stickers. Cody and Deborah collected 7/10 of the stickers. Bella and Deborah collected 6/7 of the stickers. Bella and Cody collected 620 stickers. How many more stamps did Deborah collect than Cody?

    These are questions from a math practice paper I did today. I am going to continue on it tomorrow, so I need the solutions by today.
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  2. #2
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    1.

    t=total amount
    x=amount bought.

    \frac{t}{x}=1.40
    after we buy the $4.60 one we add 4.60 to the total and add 1 to the amount bought. this gives us a new average of 1.8
    \frac{t+4.60}{x+1}=1.80

    so we just solve for x;
    first equation gives t=1.4x plug this into second equation and solve for x. (x=7 not including the one she bought last. so 8 in total)
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  3. #3
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    Quote Originally Posted by AnnaRydell View Post
    1. Sally bought identical red files. Each red file costs $1.40. When Sally bought 1 more blue file at $4.60, it increased the average cost of the red files to $1.80. How many files did Sally buy in total?
    Think about how "average" is computed. Then: If she bought "f" red folders at $1.40 each how much did she spend on red folders? What then was (the expression for) her total cost? How many items did she buy, in total? What expression then stands for the average cost?

    Set this expression equal to the given average, and solve the resulting equation.

    Quote Originally Posted by AnnaRydell View Post
    2. There are a total of 200 blue, green and red balls. There are as many red balls as blue balls. There are twice as many red balls as blue balls. There are fewer green balls than red balls. The number of blue balls and red balls in each group is less than 100 and divisible by 3 and 4. How many green balls are there?
    Which is correct: "as many", or "twice as many"?

    Quote Originally Posted by AnnaRydell View Post
    3. Bella, Cody and Deborah had a collection of stickers. Cody and Deborah collected 7/10 of the stickers. Bella and Deborah collected 6/7 of the stickers. Bella and Cody collected 620 stickers. How many more stamps did Deborah collect than Cody?
    If Cody and Deborah collected 7/10 together, and Bella and Deborah collected 6/7 together, then (by subtracting), what fraction did Cody and Bella collect together? Use this information, along with the given total for Bella and Cody, to find the total number collected by all three. Then work backwards to find the individual amounts.

    If you get stuck, please reply showing your work and reasoning so far. Thank you!
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  4. #4
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    3.

    x is total stickers

    these are the equations

    (a)-----> B+D=\frac{6x}{7}

    (b)-----> B+C=620

    (c)-----> C+D=\frac{7x}{10}

    equation (a) is like that because its 6/7 of the total amount.

    from (a) we have C=\frac{x}{7} since all C,B,and D must add up to total stickers

    from (c) we have B=\frac{3x}{10}

    from (b) B+C=\frac{3x}{10}+\frac{x}{7}=\frac{31x}{70}=620

    so the this solves to x=1400

    we need (a)-(b)=D-C=\frac{6x}{7}-620=\frac{6(1400)}{7}-620=580 which is the difference between D and C.

    I know you dont like algebra anna but i hope you understand this. if i can think of a non-algebraic method i'll let you know
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  5. #5
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    Quote Originally Posted by stapel View Post

    Which is correct: "as many", or "twice as many"?

    question said something about 3 groups so im guessing thos are 3 different groups.

    still i get multiple working answers
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  6. #6
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    Quote Originally Posted by Krahl View Post
    question said something about 3 groups so im guessing thos are 3 different groups.
    Yes, there are three colors. And two of the colors are related in two conflicting ways: "as many" versus "twice as many".

    Unless there are "zero" of each (which conflicts with other parts of the exercise), then we need corrections of the original post.
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  7. #7
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    Edit: yeah your right stapel
    Last edited by Krahl; August 3rd 2009 at 05:50 AM.
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  8. #8
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    Quote Originally Posted by AnnaRydell View Post

    2. There are a total of 200 blue, green and red balls. There are twice as many red balls as blue balls. There are fewer green balls than red balls. The number of blue balls and red balls in each group is less than 100 and divisible by 3 and 4. How many green balls are there?
    Sorry! Muddle-headed me...Above is the edited question^^^
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  9. #9
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    Quote Originally Posted by AnnaRydell View Post
    2. There are a total of 200 blue, green and red balls. There are twice as many red balls as blue balls. There are fewer green balls than red balls. The number of blue balls and red balls in each group is less than 100 and divisible by 3 and 4. How many green balls are there?
    Well, the number of red balls is a multiple of 12 that is also less than 100, so I'll pick 96. There are twice as many red balls as blue balls, so there would be 48 blue balls. That means that there must be 56 green balls, which is less than 96.

    It looks like that this is not the only answer, though.


    01
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  10. #10
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    Quote Originally Posted by AnnaRydell View Post
    2. There are a total of 200 blue, green and red balls. There are twice as many red balls as blue balls. There are fewer green balls than red balls. The number of blue balls and red balls in each group is less than 100 and divisible by 3 and 4. How many green balls are there?
    The number of blues is a multiple of twelve, such that twice this multiple (the number of reds) is still less than 100. So list out multiples:

    blues: 12, 24, 36, 48, 60, 72, ...

    reds: 24, 48, 72, 96, 120, 144, ...

    Clearly, the only plausible options are:

    (blues, reds): (12, 24), (24, 48), (36, 72), (48, 96)

    The total is 200, and the number of greens is less than the number of reds. Then the number of greens is:

    (12, 24): 200 - 12 - 24 = ...?

    (24, 48): 200 - 24 - 48 = ...?

    (36, 72): 200 - 36 - 72 = ...?

    (48, 96): 200 - 48 - 96 = ...?

    Which option(s) work(s)?
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  11. #11
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    Hello, Anna!

    2. There is a total of 200 blue, green and red balls.
    There are twice as many red balls as blue balls.
    There are fewer green balls than red balls.
    The number of blue balls and red balls is less than 100 and divisible by 3 and 4.
    How many green balls are there?

    Let: . \begin{array}{ccc}R &=& \text{no. of red balls} \\ B &=& \text{no. of blue balls} \\ G &=& \text{no. of green balls}\end{array}

    There are fewer green balls than red balls: . G \:<\:R .[1]

    The number of blue balls is divisible by 12: . B \:=\:12k for some positive integer k.

    The number of red balls is twice the number of blue balls: . R \:=\:2B \:=\:24k

    The number of red balls is less than 100: . 24k \:<\:100 \quad\Rightarrow\quad k \:\leq \:4 .[2]

    There is a total of 200 balls: . R + B + G \:=\:200 \quad\Rightarrow\quad 24k + 12k + G \:=\:200

    . . We have: . 36k + G \:=\:200 \quad\Rightarrow\quad G \:=\:200 - 36k


    From [2], there are four cases to consider: . k \:=\:4,3,2,1

    . . If k = 4, then: . R = 96,\;B = 48,\;G \:=\:200 - 36(4) \quad\Rightarrow\quad G\:=\:56

    . . If k = 3, then: . R = 72,\;N =36,\;G \:=\:92 . . . which contradicts [1].
    . . The same happens for k \:=\:2\text{ or }1.


    Therefore, the only solution is: . \begin{Bmatrix} R &=& 96 \\ B  &=& 48 \\ G &=& 56 \end{Bmatrix}\quad\hdots \text{ There are 56 green balls.}

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