SharkeyBoy
Member
- Joined
- Nov 15, 2012
- Messages
- 180
- Gender
- Male
- HSC
- 2013
Hey guys, I've been looking at this question for like the past hour and I still don't get how my answer is wrong. (I've probably made a silly mistake somewhere :L) anyway, the question says:
A superannuation fund paid 6% p.a. for the first 10 years and then 10% p.a. after that. if Thanh put $5000 into this fund at the end of each year, how much would she have at the end of 25 years?
My final answer as will see below is $224766.38, when it apparently should be $240 652.62
If it helps, the textbook is Maths in Focus and the question is 16 from Challenge Exercise 8
My working:
A1 = 5000
A2 = 5000(1 + 6%) + 5000
...
A10 = 5000(1 + 6%)^9 + 5000(1 + 6%)^8 + ... + 5000
= 5000 (((1 + 6%)^10 - 1)/6%)
A11 = A10 + 5000
A12 = A10 + 5000(1.1) + 5000
A13 = A10 + 5000(1.1)^2 + 5000(1.1) + 5000
...
A25 = A10 + 5000(1.1)^14 + 5000(1.1)^13 + ... + 5000(1.1) + 5000
= A10 + 5000 ((1.1^15 - 1)/0.1)
= 224766. 38
Thanks
A superannuation fund paid 6% p.a. for the first 10 years and then 10% p.a. after that. if Thanh put $5000 into this fund at the end of each year, how much would she have at the end of 25 years?
My final answer as will see below is $224766.38, when it apparently should be $240 652.62
If it helps, the textbook is Maths in Focus and the question is 16 from Challenge Exercise 8
My working:
A1 = 5000
A2 = 5000(1 + 6%) + 5000
...
A10 = 5000(1 + 6%)^9 + 5000(1 + 6%)^8 + ... + 5000
= 5000 (((1 + 6%)^10 - 1)/6%)
A11 = A10 + 5000
A12 = A10 + 5000(1.1) + 5000
A13 = A10 + 5000(1.1)^2 + 5000(1.1) + 5000
...
A25 = A10 + 5000(1.1)^14 + 5000(1.1)^13 + ... + 5000(1.1) + 5000
= A10 + 5000 ((1.1^15 - 1)/0.1)
= 224766. 38
Thanks