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Several AMC 10 Problems

  1. Feb 12, 2008 #1
    AMC 10 year 2000

    here are several problems that i found while practicing for AMC 10, really wish someone can give a thorough explaination of how they are to be solved. The answers are at the end.

    22) One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Anegla drank a qarter of the total amount of milk and a sixth of the totalamount of coffee. How many people are in the family?
    a)3 b)4 c)5 d)6 e)7

    23) When the mean, median and moode of the list:
    10, 2, 5, 2, 4, 2, x
    are arranged in increasing order, they form a "non-constant arithmetic progressions". What is the sum of all possible real value of x?
    a)3 b)6 c)9 d)17 e)20
    *also would you please explain non-constant arithmetic progression? I think i have some confusion on that THANK YOU

    24) Lef f be a function ofr which f(x/3) = x^2 + x + 1. Find the sum of all values of z for which f(3z) = 7
    a) -1/3 b) -1/9 c)0 d) 5/9 e) 5/3
    this problem i tried to solve. since 3z is to be plugged into f(x/3) then it can be written as f(3z/3) = f(z) = 7 = x^2 + x + 1; i solved that and got -1/3 (a), but i don't think it is correct.

    The answers to the problems: 22:(c) 23:(e) 24:(b)
     
  2. jcsd
  3. Feb 12, 2008 #2

    arildno

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    I'll indulge you with the first one:
    Let x be the total amount of ounces milk, y the total amount of ounces coffee, and z the number of persons in Anfgela's family.

    What restrictions MUST lie on x, y and z:
    Clearly, all three numbers must be positive!
    Furthermore, z must be a whole number!

    Now, the total amount of milk+coffee clearly satisfies the equation:

    x+y=z*8 (1)

    For Angie, we have the following equation that also holds for x and y:

    x/4+y/6=8 (2)

    Let us multiply all terms in (2) with 12 and all terms in (1) with 2.

    This yields the system of equations:

    2x+2y=16z (1*)

    3x+2y=96 (2*)

    Subtracting (1*) from (2*) yields:

    3x-2x+2y-2y=96-16z

    or:

    x=16*(6-z) (**)

    Now, deduce your result by (**) and that (1) must hold, with the above-mentioned restrictions on x,y and z.
     
  4. Feb 12, 2008 #3
    THANK YOU so much for that one!
     
  5. Feb 13, 2008 #4
    24)

    If f( x/3 ) = x^2 + x + 1 then f(x) must have the form a x^2 + b x + c.
    So let f(x) = a x^2 + b x + c which gives f( x/3 ) = a x^2 / 9 + b x / 3 + c.
    Identifying coefficients gives a = 9, b = 3 and c = 1.
    Solving f( 3z ) = 7 yields z1 = -1/3 and z2 = 2/9 which sums to -1/9.
     
  6. Feb 14, 2008 #5
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