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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
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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kryton1212Best ResponseYou've already chosen the best response.1
(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

kryton1212Best ResponseYou've already chosen the best response.1
@hartnn @ganeshie8 @primeralph
 9 months ago

xlegendxBest ResponseYou've already chosen the best response.0
i suppose not being in the tags means i still lack in notoriety huh
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
nope, they help me a lot
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
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 + (n1)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)^(n1) = 1 * (0.2)^8 =.... ? multiply the result by 100 to get the answer...
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
for b) part you will apply the sum formula instead of n'th term formula, with all other values being same
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
you have those sum formulas ?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
and did you get that long explanation ? :P
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
for bobby, Arithmetic sequence n'th term formula \(T_n = a_1+(n1)d\) sum formula \(S_n = (n/2)[2a_1+(n1)d]\) for emily , geometric sequence : n'th term formula \(T_n = a_1r^{n1}\) sum formula \(S_n = a_1 [\dfrac{r^n1}{r1}]\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
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

hartnnBest ResponseYou've already chosen the best response.1
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

hartnnBest ResponseYou've already chosen the best response.1
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)^(n1) = 100 * (0.2)^8 =.... ?
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
oh, it seems easy. Let me try
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
but for part a. 100*(1* 0.2 ^(8) ) ???
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
a1= 100, r = 0.2 , n= 9 \(\large T_9 = a_1r^{n1} = 100(0.2)^8 =...?\) use calculator.....what exactly wast he doubt ?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
OMG, there must be some mistake in my method....it can't be 0.0256 let me review it...
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
ok, its like compound interest formula! let me revise the formula n'th term \(T_n = a_1 (1+r)^{n1}\) here, a1 = 100, r=0.2, n = 9 gives T9 = 100 (1.2)^8 =.... ?
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
thanks. and part b Emily?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
revised formula for sum \(S_n = a_1 [\dfrac{(r+1)^n1}{r}]\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
a1 = 100, r =0.2, n =9 you'll get the answer :)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
i know why i cannot get the answer. i think n=8... thankd
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
after 1 year means n=2, (100*0.2 is the 2nd term) so, after 8 years, means n=9
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
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

hartnnBest ResponseYou've already chosen the best response.1
not total amount, IN THAT YEAR
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
so, you will use n'th term formula or sum formula ?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
correct! and since its \(\large 10th\) year, \(\large n=11\)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
Emily=$619.174 Bobby=$600 then Emily's claim is correct, right?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
for bobby first term = a1 = 100 and common difference = 50 11th term (after 10 years) = a1 + (n1)d = 100 + 10*50 =600 (correct ) T11 = 100 * (1.2)^10 = 619.174 (correct) and YES , emily's claim was correct :)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
and part c ii ? Bobby=$7700? Emily=$??
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
ohh...for part 2 we need sum formula.....
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
Emily : 100 * [(1.2)^11 1] / 0.2 =... ? i plugged in, r=0.2,a1 = 100, n=11
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
is that the answer in your book ? difference = 77003215
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
i am getting 3850 for bobby
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
Sum for bobby \(\large (11/2)[2*100+10*50]=3850\)
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n1)d]\)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
uh? why? it's fixed
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
yeah, thats the general formula for sum...
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
uh, OMG ..... okay, let me see. how about Emily's?
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
sum formula i posted above, the "revised" one Emily : 100 * [(1.2)^11 1] / 0.2
 9 months ago

hartnnBest ResponseYou've already chosen the best response.1
\(S_n = a_1 [\dfrac{(r+1)^n1}{r}]\)
 9 months ago

kryton1212Best ResponseYou've already chosen the best response.1
3215... thank you so much
 9 months ago
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