Question

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=$2700; Emily=$????

Answer 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.

Answer 2

@hartnn @ganeshie8 @primeralph

Answer 3

@Callisto

Answer 4

@mayankdevnani

Answer 5

i suppose not being in the tags means i still lack in notoriety huh

Answer 6

what?

Answer 7

nope, they help me a lot

Answer 8

@satellite73

Answer 9

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...

Answer 10

for b) part you will apply the sum formula instead of n'th term formula, with all other values being same

Answer 11

you have those sum formulas ?

Answer 12

and did you get that long explanation ? :P

Answer 13

why a1=1?

Answer 14

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 \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)

Answer 15

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

Answer 16

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 :)

Answer 17

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 =.... ?

Answer 18

oh, it seems easy. Let me try

Answer 19

but for part a.
100*(1* 0.2 ^(8) ) ???

Answer 20

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 ?

Answer 21

uhh.

Answer 22

0.000256???!

Answer 23

OMG, there must be some mistake in my method....it can't be 0.0256
let me review it...

Answer 24

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 =.... ?

Answer 25

430?

Answer 26

yes, i get 430 too :)

Answer 27

thanks. and part b Emily?

Answer 28

revised formula for sum
\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

Answer 29

a1 = 100, r =0.2, n =9
you'll get the answer :)

Answer 30

i know why i cannot get the answer. i think n=8...
thankd

Answer 31

thanks*

Answer 32

after 1 year means n=2, (100*0.2 is the 2nd term)
so, after 8 years, means n=9

Answer 33

ah ha, i got it.

Answer 34

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?

Answer 35

not total amount, IN THAT YEAR

Answer 36

okay..

Answer 37

so, you will use n'th term formula or sum formula ?

Answer 38

n'th term

Answer 39

correct!
and since its \(\large 10th\) year, \(\large n=11\)

Answer 40

Emily=$619.174
Bobby=$600
then Emily's claim is correct, right?

Answer 41

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 :)

Answer 42

and part c ii ?
Bobby=$7700?
Emily=$??

Answer 43

ohh...for part 2 we need sum formula.....

Answer 44

yes

Answer 45

Emily : 100 * [(1.2)^11 -1] / 0.2 =... ?

i plugged in, r=0.2,a1 = 100, n=11

Answer 46

3215...

Answer 47

is that the answer in your book ?
difference = 7700-3215

Answer 48

the answer is 635

Answer 49

i am getting 3850 for bobby

Answer 50

3850-3215=635

Answer 51

Sum for bobby
\(\large (11/2)[2*100+10*50]=3850\)

Answer 52

why 2 times 100?

Answer 53

the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)

Answer 54

uh? why? it's fixed

Answer 55

?

Answer 56

yeah, thats the general formula for sum...

Answer 57

uh, OMG .....
okay, let me see. how about Emily's?

Answer 58

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

Answer 59

\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

Answer 60

ah ha, thanks

Answer 61

3215...
thank you so much

Answer 62

welcome ^_^