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Math Help - geometric progression word problem,

  1. #1
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    geometric progression word problem,

    A liquid is kept in a barrel. At the start of a year the barrel is filled with 160 litres of the liquid. Due to evaporation, at the end of every year the amount of liquid in the barrel is reduced by 15% of its volume at the start of the year.
    (a) Calculate the amount of liquid in the barrel at the end of the first year.
    Liquid at the start of the year = 160
    end of 1st year = 160x0.85 = 136ml


    (b) Show that the amount of liquid in the barrel at the end of ten years is approximately 31.5 litres.
    Liquid at end of year 10=160(0.85)^10=31.499 =31.5 litres


    At the start of each year a new barrel is filled with 160 litres of liquid so that, at the end of 20 years, there are 20 barrels containing liquid.

    (c) Calculate the total amount of liquid, to the nearest litre, in the barrels at the end of 20 years.
    need help with part 'c'

    thanks.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    The first barrel will have 85% of it's total liquid after the first year, 85% of what remains the second year, 85% of what's left after the second year, and etc... Do you recognize the pattern?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    The first barrel will have 85% of it's total liquid after the first year, 85% of what remains the second year, 85% of what's left after the second year, and etc... Do you recognize the pattern?
    How do I work out the first term, 'a' ?
    I know I have to use this formula  s_{n} = \frac{a(1- r^{n})}{1-r} to work out the total amount of liquid at the end of 20 years
    Last edited by Tweety; May 28th 2009 at 11:26 AM.
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  4. #4
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    Hello, Twetty!

    A liquid is kept in a barrel.
    At the start of a year the barrel is filled with 160 litres of the liquid.
    Due to evaporation, at the end of every year the amount of liquid in the barrel
    is reduced by 15% of its volume at the start of the year.

    At the start of each year a new barrel is filled with 160 litres of liquid so that,
    at the end of 20 years, there are 20 barrels containing liquid.

    (c) Calculate the total amount of liquid, to the nearest litre, in the barrels
    at the end of 20 years.
    At the end of 20 years . . .


    The first barrel has had 20 years of evaporation.
    . . It contains: . 160(0.85)^{20} liters.

    The second barrel has had 19 years of evaporation.
    . . It contains: . 160(0.85)^{19} liters.

    The third barrel has had 18 years of evaporation.
    . . It contains: . 160(0.85)^{18} liters.

    . . . . . . . . . \vdots

    The 18th barrel has had 3 years of evaporation.
    . . It contains: . 160(0.85)^3 liters.

    The 19th barrel has had 2 years of evaporation.
    . . It contains: . 160(0.85)^2 liters.

    The 20th barrel has had 1 year of evaporation.
    . . It contains: . 160(0.85) liters.


    Hence, the total amount of liquid at the end of 20 years is:

    . . S \;=\;160(0.85) + 160(0.85)^2 + 160(0.85)^3 + \hdots + 160(0.85)^{19} + 160(0.85)^{20}

    . . S \;=\;160(0.85)\underbrace{\bigg[1 + 0.85 + 0.85^2 + 0.85^3 + \hdots + 0.85^{19}\bigg]}_{\text{geometric sries}}

    The geometric series has a = 1,\;r = 0.85,\;n = 20
    . . Its sum is: . \frac{1-0.85^{20}}{1-0.85} \;=\;6.408269793


    Therefore: . S \;=\;160(0.85)(6.408269793) \;=\;871.5246918 \;\approx\;872 liters.

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