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anonymous

  • 5 years ago

You have £1,000 in your savings account. At the end of each month, the bank adds an extra 0.2% of interest to the outstanding balance and you make a withdrawal of £10. How much is in the savings account after this has happened for 36 months, rounded to the nearest pound? (I guess you need to solve it with a linear difference equation, but I cannot figure it out)

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  1. anonymous
    • 5 years ago
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    \[a ^{n}x _{0} +b (a ^{n}-1)/(a-1)\] I found this formula for it and it gives the good answer but I do not see why. (a^n=1.002^36 , xo=1000 , b=-10)

  2. anonymous
    • 5 years ago
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    yeah i have to write this down. certainly cannot do it on the fly. if at all!

  3. anonymous
    • 5 years ago
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    wait, why isn't it just \[1000(1.002)^{10}-10(1.002)^9\]?

  4. anonymous
    • 5 years ago
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    actually my exponents are ridiculous. it is 36 months not ten. so i think maybe \[1000(1.002)^{36}-10(1.002)^{35}\]

  5. anonymous
    • 5 years ago
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    i just wrote out the first three months and saw what i got.

  6. anonymous
    • 5 years ago
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    I am not sure here at all, I found that formula and it did give me the correct answer, but I can only see the answer unfortunately

  7. anonymous
    • 5 years ago
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    first month P(1.002) second month (P(1.002)-10)(1.002)=P(1.002)^2-10(1.002) third month (P(1.002)^2-10(1.002))-10)(1.002)=P(1.002)^3-10(1.002)^2-10(1.002)

  8. anonymous
    • 5 years ago
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    oh wait maybe you get a geometric series for the ten part.

  9. anonymous
    • 5 years ago
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    yeah that formula does look like a geometric series

  10. anonymous
    • 5 years ago
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    first part is definitely p(1.002)^36

  11. anonymous
    • 5 years ago
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    lets see what the last one is.

  12. anonymous
    • 5 years ago
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    I think I see now why it is so

  13. anonymous
    • 5 years ago
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    guess it depends on whether you take out ten pounds at the end. if so you will get 10+10(1.002)+10(1.002)^2+...+10(1.002)^35

  14. anonymous
    • 5 years ago
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    \[x _{n}=a ^{n} x _{0} +(a ^{n-1} +....+a+1)b\]

  15. anonymous
    • 5 years ago
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    and the last part is a series

  16. anonymous
    • 5 years ago
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    equal to a^n -1/a-1

  17. anonymous
    • 5 years ago
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    yeah i think so. i bet this simplifies to get essentially what i wrote at the beginning. if my algebra was good enuf i could do it

  18. anonymous
    • 5 years ago
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    did you try 1000(1.002)^36-10(1.002)^35?

  19. anonymous
    • 5 years ago
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    no, but I can

  20. anonymous
    • 5 years ago
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    that gives a positive amount, that cannot be right, the interest is around 2 and you take out 10

  21. anonymous
    • 5 years ago
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    we can use 1+r+r^2+...+r^35=\[\frac{r^{36}-1}{1-r}\] with r = 1.002

  22. anonymous
    • 5 years ago
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    3.7289 rounded

  23. anonymous
    • 5 years ago
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    yes that is what I thought too, but isnt the denominator r-1 ?

  24. anonymous
    • 5 years ago
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    multiply by ten to get 37.289

  25. anonymous
    • 5 years ago
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    yea typo

  26. anonymous
    • 5 years ago
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    denominator in this case is .002

  27. anonymous
    • 5 years ago
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    oh wait my calculation is wrong

  28. anonymous
    • 5 years ago
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    i divided by .02

  29. anonymous
    • 5 years ago
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    the answer was something around 700

  30. anonymous
    • 5 years ago
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    i am off by a decimal

  31. anonymous
    • 5 years ago
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    :) happens

  32. anonymous
    • 5 years ago
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    \[\frac{(1.002)^{36}-1}{.002}=37.289\]

  33. anonymous
    • 5 years ago
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    according to my calculator.

  34. anonymous
    • 5 years ago
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    multiply by ten to get 372.89

  35. anonymous
    • 5 years ago
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    according to my brain

  36. anonymous
    • 5 years ago
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    subtract from 1000(1.002)^36 = 1074.578 rounded

  37. anonymous
    • 5 years ago
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    gives the correct answer! sweet

  38. anonymous
    • 5 years ago
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    and i get as a final answer (as they say) 701.68

  39. anonymous
    • 5 years ago
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    whew.

  40. anonymous
    • 5 years ago
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    glad we got this worked out even if a couple of false starts

  41. anonymous
    • 5 years ago
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    maths is like this, try hard and fail a lot

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