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kryton1212

  • 2 years ago

Mr. Chan will deposit $25000 in a bank account at the beginning of each year from 2007 to 2012. If the interest is compounded yearly and the interest rate is 4% p.a., find the total amount that he can get at the beginning of 2013. (corr. to 3 sig. fig.)

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  1. kryton1212
    • 2 years ago
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    @hartnn

  2. ganeshie8
    • 2 years ago
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    looks like annuity setup

  3. ganeshie8
    • 2 years ago
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    P = 25,000 r = .04 n = 6

  4. ganeshie8
    • 2 years ago
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    \(A = P[ (1+r)^n + (1+r)^{n-1} + .... (1+r)]\) = \(P[\frac{1-(1+r)^n }{ 1-(1+r)}]\)

  5. ganeshie8
    • 2 years ago
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    plug and chug..

  6. kryton1212
    • 2 years ago
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    ahhh....wait

  7. ganeshie8
    • 2 years ago
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    ha take ur time not an easy problem :)

  8. primeralph
    • 2 years ago
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    |dw:1372927372461:dw|

  9. kryton1212
    • 2 years ago
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    errr...my calculator shows that it's 165824.3866

  10. primeralph
    • 2 years ago
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    @ganeshie8 Why the long stuff?

  11. ganeshie8
    • 2 years ago
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    every year he is adding 25 grand

  12. kryton1212
    • 2 years ago
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    @primeralph this's a question about sequence

  13. primeralph
    • 2 years ago
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    Well, both ways work. Mine is just shorter.

  14. kryton1212
    • 2 years ago
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    @primeralph but your equation does match the answer, neither @ganeshie8 ..

  15. primeralph
    • 2 years ago
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    Uh hold on.

  16. ganeshie8
    • 2 years ago
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    prime i used ur formula and took a sum to arrive at the final formula

  17. primeralph
    • 2 years ago
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    @kryton1212 What's the answer?

  18. kryton1212
    • 2 years ago
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    172000 after corr. to 3 sig fig

  19. kryton1212
    • 2 years ago
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    uhh?

  20. kryton1212
    • 2 years ago
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    i mean the correct answer..

  21. primeralph
    • 2 years ago
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    It is not possible to get that much money after just 6 years.

  22. ganeshie8
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=25000%281-%281%2B.04%29%5E6%29%2F%281-%281%2B.04%29%29

  23. ganeshie8
    • 2 years ago
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    im getting 165,824 @kryton1212 prime read the q again

  24. kryton1212
    • 2 years ago
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    i don't know how to do... @hartnn show yourself-_-

  25. ganeshie8
    • 2 years ago
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    which formula wont work ?

  26. primeralph
    • 2 years ago
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    ohhh..........beginning of each year. Hold on then.

  27. kryton1212
    • 2 years ago
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    i also think that @ganeshie8 's formula is correct...maybe i should ask my teacher about this question. Let's move on to the next question that you both will be tagged by me a moment later

  28. kryton1212
    • 2 years ago
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    ...

  29. primeralph
    • 2 years ago
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    Hold on.

  30. ganeshie8
    • 2 years ago
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    okie

  31. kryton1212
    • 2 years ago
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    maybe @primeralph goes on

  32. primeralph
    • 2 years ago
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    Uh well, tag away if you're satisfied.

  33. kryton1212
    • 2 years ago
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    uh..

  34. ganeshie8
    • 2 years ago
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    move on to new problem, im pretty sure what we worked is correct. if you do manually 6 times also, you would get the same answer.

  35. kryton1212
    • 2 years ago
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    at the beginning of 2013..................

  36. kryton1212
    • 2 years ago
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    @primeralph do you have any other ideas?

  37. primeralph
    • 2 years ago
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    To check, you can do each step. If you get the same thing, let it go.

  38. kryton1212
    • 2 years ago
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    i am lazy XD...i will ask my teacher about this question tonight. thank you both

  39. ganeshie8
    • 2 years ago
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    np :)

  40. hartnn
    • 2 years ago
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    i am getting 172457.362022

  41. kryton1212
    • 2 years ago
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    ohh, how?

  42. ganeshie8
    • 2 years ago
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    3 sig figs, thats the right answer also... how ?

  43. hartnn
    • 2 years ago
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    i calculated for each year...6 calculations :\

  44. hartnn
    • 2 years ago
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    2008---25k(1+0.04)=26k 2009---(26k+25k)(1+0.04)=53040 2010---(53040+25k)(1+0.04)=81161.1 2011---(81161.1+25k)(1+0.04)=110408.064 2012---(110408.064+25k)(1+0.04)=140824.38 2013---(140824.38+25k)(1+0.04)=172457....

  45. kryton1212
    • 2 years ago
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    wow

  46. ganeshie8
    • 2 years ago
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    oh every year rounding off to 3 sigfigs is it

  47. kryton1212
    • 2 years ago
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    that means we use the wrong method @ganeshie8 .....

  48. ganeshie8
    • 2 years ago
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    may be

  49. kryton1212
    • 2 years ago
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    @hartnn million thanks. but how to present it in a shorter form...?

  50. hartnn
    • 2 years ago
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    thats what i am thinking... if we have it for just 2 years, \(P(1+r)+ [P(1+r)+P](1+r) = P [(1+r)+(1+r)(2+r)]\) so, maybe for 6 years, the formula will become \(P (1+r)[1+(2+r)[1+(3+r)[1+(4+r)[1+(5+r)[1+(6+r)]]]]]\) maybe there's a simplification involved, maybe...

  51. kryton1212
    • 2 years ago
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    wow, i think this's more complicated..

  52. hartnn
    • 2 years ago
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    hmm.....maybe they wanted us to calculate each year's separately...because i don't see any explicit formula for this...

  53. kryton1212
    • 2 years ago
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    uhh, anyway, thank you @hartnn

  54. hartnn
    • 2 years ago
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    hmm....welcome ^_^

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