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kryton1212

Bobby and Emily each got $100 red packet money from their grandmother when they were born. Then, their grandmother gave them red packet money on each of their birthdays. The amount of money given to Bobby was $50 more than the previous year, while the amount given to Emily was 20% more than the previous year. (a) How much did each of them get when they were 8 years old? --->Bobby=$500; Emily=$???? (b) Find the total amount of red packet money that each of them got until they were 8 years old. --->Bobby=$2700; Emily=$????

  • 9 months ago
  • 9 months ago

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  1. kryton1212
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    (c) Emily claims that she will get more packet money than Bobby when they are both 10 years old. (i) is Emily's claim correct? Why? (ii) Find the difference between their total red packet money when they are 10 years old.

    • 9 months ago
  2. kryton1212
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    @hartnn @ganeshie8 @primeralph

    • 9 months ago
  3. kryton1212
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    @Callisto

    • 9 months ago
  4. kryton1212
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    @mayankdevnani

    • 9 months ago
  5. xlegendx
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    i suppose not being in the tags means i still lack in notoriety huh

    • 9 months ago
  6. kryton1212
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    what?

    • 9 months ago
  7. kryton1212
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    nope, they help me a lot

    • 9 months ago
  8. kryton1212
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    @satellite73

    • 9 months ago
  9. hartnn
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    ok, the first part is just an arithmetic sequence, for bobby, (100,150,200,250....) so, first term = a1 = 100 and common difference = 50 9th term (after 8 years) = a1 + (n-1)d = 100 + 8*50 = $500 (correct) but for emily : (100 + 100*0.2 +100*0.2*0.2+...) = 100 (1+0.2 +0.2^2+0.2^3+...) now this is geometric sequence a1 = 1, r = 0.2 so, for 9th term (after 8 years) = a1 (r)^(n-1) = 1 * (0.2)^8 =.... ? multiply the result by 100 to get the answer...

    • 9 months ago
  10. hartnn
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    for b) part you will apply the sum formula instead of n'th term formula, with all other values being same

    • 9 months ago
  11. hartnn
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    you have those sum formulas ?

    • 9 months ago
  12. hartnn
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    and did you get that long explanation ? :P

    • 9 months ago
  13. kryton1212
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    why a1=1?

    • 9 months ago
  14. hartnn
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    for bobby, Arithmetic sequence n'th term formula \(T_n = a_1+(n-1)d\) sum formula \(S_n = (n/2)[2a_1+(n-1)d]\) for emily , geometric sequence : n'th term formula \(T_n = a_1r^{n-1}\) sum formula \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)

    • 9 months ago
  15. hartnn
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    for emily, the sequence will be 100 , 100*0.2 , 100*0.2*0.2 , .... any doubts in this ? so, taking 100 out, 100 ( 1, 0.2, 0.2*0.2,....) in the above sequence first term = a1 = 1

    • 9 months ago
  16. hartnn
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    if you want you can consider the sequence as 100, 100*0.2 , 100*0.2*0.2,.... then a1 = 1st term = 100 r= 0.2 (same) this will be easier :)

    • 9 months ago
  17. hartnn
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    but for emily : (100 + 100*0.2 +100*0.2*0.2+...) now this is geometric sequence a1 = 100, r = 0.2 so, for 9th term (after 8 years) = a1 (r)^(n-1) = 100 * (0.2)^8 =.... ?

    • 9 months ago
  18. kryton1212
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    oh, it seems easy. Let me try

    • 9 months ago
  19. kryton1212
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    but for part a. 100*(1* 0.2 ^(8) ) ???

    • 9 months ago
  20. hartnn
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    a1= 100, r = 0.2 , n= 9 \(\large T_9 = a_1r^{n-1} = 100(0.2)^8 =...?\) use calculator.....what exactly wast he doubt ?

    • 9 months ago
  21. kryton1212
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    uhh.

    • 9 months ago
  22. kryton1212
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    0.000256???!

    • 9 months ago
  23. hartnn
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    OMG, there must be some mistake in my method....it can't be 0.0256 let me review it...

    • 9 months ago
  24. hartnn
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    ok, its like compound interest formula! let me revise the formula n'th term \(T_n = a_1 (1+r)^{n-1}\) here, a1 = 100, r=0.2, n = 9 gives T9 = 100 (1.2)^8 =.... ?

    • 9 months ago
  25. kryton1212
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    430?

    • 9 months ago
  26. hartnn
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    yes, i get 430 too :)

    • 9 months ago
  27. kryton1212
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    thanks. and part b Emily?

    • 9 months ago
  28. hartnn
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    revised formula for sum \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

    • 9 months ago
  29. hartnn
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    a1 = 100, r =0.2, n =9 you'll get the answer :)

    • 9 months ago
  30. kryton1212
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    i know why i cannot get the answer. i think n=8... thankd

    • 9 months ago
  31. kryton1212
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    thanks*

    • 9 months ago
  32. hartnn
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    after 1 year means n=2, (100*0.2 is the 2nd term) so, after 8 years, means n=9

    • 9 months ago
  33. kryton1212
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    ah ha, i got it.

    • 9 months ago
  34. kryton1212
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    i want to know , in part c, the statement said "Emily claims that she will get more red packet money than Bobby when they are both 10 years old", total amount or the money in that year?

    • 9 months ago
  35. hartnn
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    not total amount, IN THAT YEAR

    • 9 months ago
  36. kryton1212
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    okay..

    • 9 months ago
  37. hartnn
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    so, you will use n'th term formula or sum formula ?

    • 9 months ago
  38. kryton1212
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    n'th term

    • 9 months ago
  39. hartnn
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    correct! and since its \(\large 10th\) year, \(\large n=11\)

    • 9 months ago
  40. kryton1212
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    Emily=$619.174 Bobby=$600 then Emily's claim is correct, right?

    • 9 months ago
  41. hartnn
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    for bobby first term = a1 = 100 and common difference = 50 11th term (after 10 years) = a1 + (n-1)d = 100 + 10*50 =600 (correct ) T11 = 100 * (1.2)^10 = 619.174 (correct) and YES , emily's claim was correct :)

    • 9 months ago
  42. kryton1212
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    and part c ii ? Bobby=$7700? Emily=$??

    • 9 months ago
  43. hartnn
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    ohh...for part 2 we need sum formula.....

    • 9 months ago
  44. kryton1212
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    yes

    • 9 months ago
  45. hartnn
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    Emily : 100 * [(1.2)^11 -1] / 0.2 =... ? i plugged in, r=0.2,a1 = 100, n=11

    • 9 months ago
  46. kryton1212
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    3215...

    • 9 months ago
  47. hartnn
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    is that the answer in your book ? difference = 7700-3215

    • 9 months ago
  48. kryton1212
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    the answer is 635

    • 9 months ago
  49. hartnn
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    i am getting 3850 for bobby

    • 9 months ago
  50. hartnn
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    3850-3215=635

    • 9 months ago
  51. hartnn
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    Sum for bobby \(\large (11/2)[2*100+10*50]=3850\)

    • 9 months ago
  52. kryton1212
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    why 2 times 100?

    • 9 months ago
  53. hartnn
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    the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)

    • 9 months ago
  54. kryton1212
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    uh? why? it's fixed

    • 9 months ago
  55. kryton1212
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    ?

    • 9 months ago
  56. hartnn
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    yeah, thats the general formula for sum...

    • 9 months ago
  57. kryton1212
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    uh, OMG ..... okay, let me see. how about Emily's?

    • 9 months ago
  58. hartnn
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    sum formula i posted above, the "revised" one Emily : 100 * [(1.2)^11 -1] / 0.2

    • 9 months ago
  59. hartnn
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    \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

    • 9 months ago
  60. kryton1212
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    ah ha, thanks

    • 9 months ago
  61. kryton1212
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    3215... thank you so much

    • 9 months ago
  62. hartnn
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    welcome ^_^

    • 9 months ago
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