Logan starts an IRA (Individual Retirement Account) at the age of 28 28 to save for retirement. He deposits $400 $ ⁢ 400 each month. Upon retirement at the age of 65 65 , his retirement savings is $1,161,278.01 $ ⁢ 1,161,278.01 . Determine the amount of money Logan deposited over the length of the investment and how much he made in interest upon retirement.

A) Total amount deposited over the length of investment = $177600

B) Interest accrued upon retirement = $1,043,678.01

Step-by-step explanation:

From the question, we can see that he started the savings at the age of 28 and ended when he retired at the age of 65. Thus, number of years = 65 - 28 = 37 years

We are told he deposited $400 each month and that upon retirement it had grown to $1,161,278.01.

Since he deposited $400 each month, in a year which has 12 months, he deposited;

Amount deposited each year = 12 × 400 = $4800

Now, it took 37 years before he retired which was the point at which he stopped the deposit.

Thus;

Total Amount deposited upon retirement = Amount deposited each year × number of years he deposited.

Thus;

Total amount deposited over the length of investment upon retirement = 4800 × 37 = $177600

Now,we have seen that his retirement savings reads $1,161,278.01.

Thus is way more than the deposits he made throughout his work years.

This means there was interest on the money.

Interest = $1,161,278.01 - $177600

Interest = $1,043,678.01

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

one set of data has more variation than the other. hope this !