English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Ok, so...
If someone begins working at $10 an hour. Each year he gets a raise of $1/hour. He's working 40 hours a week and 52 weeks a year. He deducts 10,000 a year for living expenses, and puts the rest in an investment that yields 4% compounded annually.

How many years will it take him to have saved $300,000 in his investment?

Thanks a bunch!

2006-11-23 16:19:23 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

Oh bother...
I'ts not a problem in my math or anything! I'm just trying to figure out my job earnings.

2006-11-23 16:39:39 · update #1

5 answers

earnings in the first year=52*40*10
=$20800
expenses=10000
savings=10800
savings in the second year=10800+2080
third year=10800+4160
=10800+2*2080
in n years the savings will be 10800+2080n
interest for the firs Principal of 10800
=10800(104)/100=10800(26/25)^n
for the second year=(10800+1*2080)(26/25)^n-1
for the last year={10800+(n-1)*2080}(26/25)
summation of this must be equal to 300,000
solve

2006-11-23 16:41:54 · answer #1 · answered by raj 7 · 0 0

Make 3columns. The first column is how much he makes (his hourly rate times 2080 hours/year worked) The second column is how much he puts in his savings account. The third column is how much he has in the acount at the end of the year.

Start on the 1'st row with $20800 $10900 and $10900. Then just keep adding rows for weach year until the total in the 3'rd column is $300,000 ☺


Doug

2006-11-24 00:27:34 · answer #2 · answered by doug_donaghue 7 · 0 1

1 $10,800
2 $24,112
3 $40,036
4 $58,678
5 $80,145
5 $104,551
6 $132,013
7 $162,653
8 $196,600
9 $233,984
10 $274,943
11 $319,621

2006-11-24 01:06:03 · answer #3 · answered by Helmut 7 · 0 0

It happens in his 27th year of work

Note 40*52 = 2080

Savings at end of 1st year = 10*2080 - 10000 = 10800
Savings at end of 2nd year = 11*2080 - 10000 + 1.04*(10*2080 - 10000) = 11*2080 + 10*2080*1.04 - 10000 - 10000*1.04
= 2080(11 + 10*1.04) - 10000(1 + 1.04)
Savings at end of 3rd year = 12*2080 - 10000 + 1.04 *[11*2080 - 10000 + 1.04*(10*2080 - 10000)]
= 2080(12 + 11*1.04 + 10*1.04²) - 10000(1 + 1.04 + 1.04²)

Savings at end of nth Year
= 2080 [(10 + n) + (10 + n-1)*1.04 + (10+ n-2)*1.04² + .... +11*1.04^(n-2) + 10*1.04^(n -1)] - 10000(1 + 1.04 + 1.04² + ... + 1.04^(n-2) + 1.02^(n-1)) = 300000

And solve for n

2006-11-24 00:47:21 · answer #4 · answered by Wal C 6 · 0 0

6pm, but I can't make the party.

2006-11-24 00:21:11 · answer #5 · answered by Anonymous · 1 1

fedest.com, questions and answers