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    Amy plans to join a savings scheme in which she will pay in £500 at the start of each year.

    One scheme that she is considering pays 6% interest on the amount in the account at the end of each year.

    For this scheme,

    a) find the amount of interest paid into the account at the end of the second year,

    b) show that after interest is piad at the end of the 8th year, the amount in the account will be £5246 to the nearest pound.

    Another scheme that she us considering pays 0.5% interest on the amount at the end of the each month.

    c) find, to the nearest pound, how much more or less will be in the account at the end of the 8th year under this scheme.



    Why isnt a) 500x1.06=530

    (530+500)1.06=1091.8

    Then subtract 1000 to get 91.8?


    For b) Why isnt it: (500x1.06)+(500+
    500x1.06)1.06 etc..
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    (Original post by Tynos)
    Amy plans to join a savings scheme in which she will pay in £500 at the start of each year.

    One scheme that she is considering pays 6% interest on the amount in the account at the end of each year.

    For this scheme,

    a) find the amount of interest paid into the account at the end of the second year,

    b) show that after interest is piad at the end of the 8th year, the amount in the account will be £5246 to the nearest pound.

    Another scheme that she us considering pays 0.5% interest on the amount at the end of the each month.

    [FONT=Arial]c) find, to the nearest pound, how much more or less will be in the account at the end of the 8th year under this scheme.



    Why isnt a) 500x1.06=530

    (530+500)1.06=1091.8

    Then subtract 1000 to get 91.8?

    because you would subtract 530 and 500
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    (Original post by TenOfThem)
    because you would subtract 530 and 500
    Ahh ok, what about part b?
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    For part (b), try looking at things in this manner:

    At the end of the first year, her account will have £ 500 (1.06) (do not attempt to simplify)

    At the end of the second year, her account will have [ 500(1.06) + 500 ] (1.06) = £ 500 (1.06)^2 + 500 (1.06)

    At the end of the third year, her account will have [ 500 (1.06)^2 + 500 (1.06) + 500 ] (1.06)

    = £ 500 (1.06)^3 + 500 (1.06)^2 + 500 (1.06)

    At the end of the 8th year, her account will therefore have

    = £ 500 (1.06)^8 + 500 (1.06)^7 + 500 (1.06)^ 6 +..................+500(1.06)

    = £ 500 [ (1.06)^8 +(1.06)^ 7 + (1.06)^ 6 +.............+1.06]

    You should be able to recognize the existence of a geometric progression (GP) at this stage, and it shouldn't be much of an issue to continue solving the problem.

    Hope it helps. Peace.
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    (Original post by WhiteGroupMaths)
    For part (b), try looking at things in this manner:

    At the end of the first year, her account will have £ 500 (1.06) (do not attempt to simplify)

    At the end of the second year, her account will have [ 500(1.06) + 500 ] (1.06) = £ 500 (1.06)^2 + 500 (1.06)

    At the end of the third year, her account will have [ 500 (1.06)^2 + 500 (1.06) + 500 ] (1.06)

    = £ 500 (1.06)^3 + 500 (1.06)^2 + 500 (1.06)

    At the end of the 8th year, her account will therefore have

    = £ 500 (1.06)^8 + 500 (1.06)^7 + 500 (1.06)^ 6 +..................+500(1.06)

    = £ 500 [ (1.06)^8 +(1.06)^ 7 + (1.06)^ 6 +.............+1.06]

    You should be able to recognize the existence of a geometric progression (GP) at this stage, and it shouldn't be much of an issue to continue solving the problem.

    Hope it helps. Peace.

    Ahh i understand now thanks.
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    No problem. Peace.
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    (Original post by Tynos)
    a) find the amount of interest paid into the account at the end of the second year,
    (Original post by TenOfThem)
    because you would subtract 530 and 500
    It's asking for the amount paid in at the end of the second year (using that strategy, you would need to subtract 1060!)

    But the answer is simply 530 \times 0.06
 
 
 
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