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Math Help - help 2 algebra questions

  1. #1
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    help 2 algebra questions

    1. 2 pumps of different sizes, working together, can empty a fuel tank in 5 hours. The larger pump can emty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?

    2. How much water must be evaporated from 32 ounces of a 4% salt solution to make a 6% salt solution?
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  2. #2
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    Red face Sorry not a full solution

    Quote Originally Posted by u317dwang
    1. 2 pumps of different sizes, working together, can empty a fuel tank in 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
    Pump A flowrate = A
    Pump B flowrate = B
    Volume of fuel tank = T
    Number of hours it takes smaller pump(B) to empty tank = x

    eq1) 5A + 5B = T
    eq2) (x)B =T
    eq3) (x-4)A = T

    Deleted my bad work (thanks rgep and ticbol)
    Last edited by MathGuru; September 20th 2005 at 08:35 AM.
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  3. #3
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    If (4x-4)/(x+4) = -5 then multiplying up 4x-4 = -5x-20 so 9x = -16 which doesn't look right. The original equations were 5A + 5B = T, xB = T, (x-4)A = T which give B = T/x, A = T/(x-4) so T = 5A + 5B = 5T/(x-4) + 5T/x. Dividing through, 1 = 5/(x-4) + 5/x. Multiplying up, x(x-4) = 5x + 5(x-4), that is, x^2 - 14x + 20 = 0, x = 7 +- sqrt(49-20) = 1.6 or 12.4 approximately. Since x must be greater than 4 for the problem to make sense, the answer is 7+sqrt(29) ~ 12.4.
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  4. #4
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    1.) About some pumps...

    Let
    L = rate of larger pump
    S = rate of smaller pump
    F = total volume of fuel tank
    t = time the smaller pump can do the job alone.

    Volume = (rate)*(time) --------***

    "2 pumps of different sizes, working together, can empty a fuel tank in 5 hours."
    F = (L+S)*5 ---(1)

    " The larger pump can emty this tank in 4 hours less than the smaller one."
    If F = S*t ---------(2)
    then, F = L(t-4) ---(3)

    F from (1) = F from (2),
    5(L+S) = t(S)
    5L +5S = t(S)
    5L = t(S) -5S = S(t-5)
    L = S(t-5)/5 ---------------(4)

    F from (2) = F from (3),
    t(S) = (t-4)L
    L = t(S)/(t-4) -------------(5)

    L from (4) = L from (5),
    S(t-5)/5 = t(S)/(t-4)
    Divide both sides by S,
    (t-5)/5 = t/(t-4)
    Cross multiply,
    (t-5)(t-4) = 5t
    t^2 -9t +20 = 5t
    t^2 -9t -5t +20 = 0
    t^2 -14t +20 = 0
    Using the Quadratic Formula,
    t = {-(-14) +.-sqrt[(-14)^2 -4(1)(20)]} / (2*1)
    t = {14 +,-10.77}/2
    t = 12.39 or 1.62

    Since 1.62 cannot be in (t-4), then t = 12.39 hrs.
    That means, if left alone, the smaller pump can do the job in 12.39 hours. ---answer.

    ===============
    2.) About some salty solutions ...

    "How much water must be evaporated from 32 ounces of a 4% salt solution ..."
    Salt content = 4% of 32 oz = (0.04)(32) = 1.28 oz. salt.
    So, water content = 32 -1.28 = 30.72 oz. water.

    "...to make a 6% salt solution?"
    Salt content = 1.28 oz. salt also.
    Total weight of new solution = (1.28 / 0.06) = 21.33 oz. 6% solution.
    Water content = (21.33 -1.28) = 20.05 oz. water

    Therefore, (30.72 -20.05) = 10.67 oz. of water must be evaporated. ---answer.
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