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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=$2700; Emily=$????
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=$2700; Emily=$????

This Question is Closed

kryton1212
 2 years ago
Best 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.

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1@hartnn @ganeshie8 @primeralph

xlegendx
 2 years ago
Best ResponseYou've already chosen the best response.0i suppose not being in the tags means i still lack in notoriety huh

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1nope, they help me a lot

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1ok, 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...

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1you have those sum formulas ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1and did you get that long explanation ? :P

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1for 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 \(S_n = a_1 [\dfrac{r^n1}{r1}]\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1for 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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1if 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 :)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1but 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 =.... ?

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1oh, it seems easy. Let me try

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1but for part a. 100*(1* 0.2 ^(8) ) ???

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

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1ok, 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 =.... ?

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1thanks. and part b Emily?

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

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

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1i know why i cannot get the answer. i think n=8... thankd

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

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1i 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?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1not total amount, IN THAT YEAR

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1so, you will use n'th term formula or sum formula ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1correct! and since its \(\large 10th\) year, \(\large n=11\)

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1Emily=$619.174 Bobby=$600 then Emily's claim is correct, right?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1for 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 :)

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1and part c ii ? Bobby=$7700? Emily=$??

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1ohh...for part 2 we need sum formula.....

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1is that the answer in your book ? difference = 77003215

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1i am getting 3850 for bobby

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

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1yeah, thats the general formula for sum...

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.1uh, OMG ..... okay, let me see. how about Emily's?

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(S_n = a_1 [\dfrac{(r+1)^n1}{r}]\)

kryton1212
 2 years ago
Best ResponseYou've already chosen the best response.13215... thank you so much
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