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kryton1212

  • 2 years 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
    • 2 years 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
    • 2 years ago
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    @hartnn @ganeshie8 @primeralph

  3. kryton1212
    • 2 years ago
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    @Callisto

  4. kryton1212
    • 2 years ago
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    @mayankdevnani

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

  6. kryton1212
    • 2 years ago
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    what?

  7. kryton1212
    • 2 years ago
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    nope, they help me a lot

  8. kryton1212
    • 2 years ago
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    @satellite73

  9. hartnn
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    you have those sum formulas ?

  12. hartnn
    • 2 years ago
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    and did you get that long explanation ? :P

  13. kryton1212
    • 2 years ago
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    why a1=1?

  14. hartnn
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    oh, it seems easy. Let me try

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

  20. hartnn
    • 2 years 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
    • 2 years ago
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    uhh.

  22. kryton1212
    • 2 years ago
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    0.000256???!

  23. hartnn
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    430?

  26. hartnn
    • 2 years ago
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    yes, i get 430 too :)

  27. kryton1212
    • 2 years ago
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    thanks. and part b Emily?

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

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

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

  31. kryton1212
    • 2 years ago
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    thanks*

  32. hartnn
    • 2 years 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
    • 2 years ago
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    ah ha, i got it.

  34. kryton1212
    • 2 years 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
    • 2 years ago
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    not total amount, IN THAT YEAR

  36. kryton1212
    • 2 years ago
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    okay..

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

  38. kryton1212
    • 2 years ago
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    n'th term

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

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

  41. hartnn
    • 2 years 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
    • 2 years ago
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    and part c ii ? Bobby=$7700? Emily=$??

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

  44. kryton1212
    • 2 years ago
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    yes

  45. hartnn
    • 2 years 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
    • 2 years ago
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    3215...

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

  48. kryton1212
    • 2 years ago
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    the answer is 635

  49. hartnn
    • 2 years ago
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    i am getting 3850 for bobby

  50. hartnn
    • 2 years ago
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    3850-3215=635

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

  52. kryton1212
    • 2 years ago
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    why 2 times 100?

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

  54. kryton1212
    • 2 years ago
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    uh? why? it's fixed

  55. kryton1212
    • 2 years ago
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    ?

  56. hartnn
    • 2 years ago
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    yeah, thats the general formula for sum...

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

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

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

  60. kryton1212
    • 2 years ago
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    ah ha, thanks

  61. kryton1212
    • 2 years ago
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    3215... thank you so much

  62. hartnn
    • 2 years ago
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    welcome ^_^

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