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kryton1212

  • one year ago

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=$2700; Emily=$????

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  1. kryton1212
    • one year ago
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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.

  2. kryton1212
    • one year ago
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    @hartnn @ganeshie8 @primeralph

  3. kryton1212
    • one year ago
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    @Callisto

  4. kryton1212
    • one year ago
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    @mayankdevnani

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

  6. kryton1212
    • one year ago
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    what?

  7. kryton1212
    • one year ago
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    nope, they help me a lot

  8. kryton1212
    • one year ago
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    @satellite73

  9. hartnn
    • one year ago
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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...

  10. hartnn
    • one year ago
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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

  11. hartnn
    • one year ago
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    you have those sum formulas ?

  12. hartnn
    • one year ago
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    and did you get that long explanation ? :P

  13. kryton1212
    • one year ago
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    why a1=1?

  14. hartnn
    • one year ago
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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 \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)

  15. hartnn
    • one year ago
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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

  16. hartnn
    • one year ago
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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 :)

  17. hartnn
    • one year ago
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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 =.... ?

  18. kryton1212
    • one year ago
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    oh, it seems easy. Let me try

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

  20. hartnn
    • one year ago
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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 ?

  21. kryton1212
    • one year ago
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    uhh.

  22. kryton1212
    • one year ago
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    0.000256???!

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

  24. hartnn
    • one year ago
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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 =.... ?

  25. kryton1212
    • one year ago
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    430?

  26. hartnn
    • one year ago
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    yes, i get 430 too :)

  27. kryton1212
    • one year ago
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    thanks. and part b Emily?

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

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

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

  31. kryton1212
    • one year ago
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    thanks*

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

  33. kryton1212
    • one year ago
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    ah ha, i got it.

  34. kryton1212
    • one year ago
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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?

  35. hartnn
    • one year ago
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    not total amount, IN THAT YEAR

  36. kryton1212
    • one year ago
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    okay..

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

  38. kryton1212
    • one year ago
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    n'th term

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

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

  41. hartnn
    • one year ago
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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 :)

  42. kryton1212
    • one year ago
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    and part c ii ? Bobby=$7700? Emily=$??

  43. hartnn
    • one year ago
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    ohh...for part 2 we need sum formula.....

  44. kryton1212
    • one year ago
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    yes

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

  46. kryton1212
    • one year ago
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    3215...

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

  48. kryton1212
    • one year ago
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    the answer is 635

  49. hartnn
    • one year ago
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    i am getting 3850 for bobby

  50. hartnn
    • one year ago
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    3850-3215=635

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

  52. kryton1212
    • one year ago
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    why 2 times 100?

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

  54. kryton1212
    • one year ago
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    uh? why? it's fixed

  55. kryton1212
    • one year ago
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    ?

  56. hartnn
    • one year ago
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    yeah, thats the general formula for sum...

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

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

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

  60. kryton1212
    • one year ago
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    ah ha, thanks

  61. kryton1212
    • one year ago
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    3215... thank you so much

  62. hartnn
    • one year ago
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    welcome ^_^

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