amount of interest earned

Question

bob places $1000 a year in his IRA for ten years and then invests $2000 a year for the next ten years.  Mary places $2000 a year in her IRA for ten years and then invests $1000 a year for the next ten years.  They both have invested $30000.  If they earn 8 percent annually, how much more will Mary have earned than Bob at the end of 20 years?

Answer 1

Hi Amy -- using rough "mid-point" estimates, Bob cooks 10k for 15yrs and 20k for 5yrs ... 

 

Mary cooks 20k for 15yrs and 10k for 5yrs ... 8% for 15yrs triples the mid-point outlays ...

 

BOB: 30k plus 21.6 ... 23.4 ... 25.4 ... 27.5 ... 30k ~ Bob 60k

 

MARY: 60k [2x Bob's 1st leg] plus 15k [half Bob's last leg] => Mary gets ~15k more :)

Answer 2

Bob, B = 1000( (1.08)^20 + (1.08)^19 + (1.08)^18 + (1.08)^17 + (1.08)^16 + (1.08)^15 + (1.08)^14 + (1.08)^13 + (1.08)^12 + (1.08)^11 )

+ 2000( (1.08)^10 + (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + (1.08) )

 

B = 1000 ((1.08)^11) ( (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1 )
+ 2000 (1.08) ( (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1 )

 

Let A = (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08 + 1

 

(1.08) A = (1.08)^10 + (1.08)^9 + (1.08)^8 + (1.08)^7 + (1.08)^6 + (1.08)^5 + (1.08)^4 + (1.08)^3 + (1.08)^2 + 1.08

 

(1.08) A - A = (1.08)^10 - 1

 

A = ( (1.08)^10 - 1 ) / 0.08

[N.B.: A is a geometric series. The formula for the first n terms is sum{1,n}(a*(r^n - 1)/(r - 1)). For A, a =1, r = 1.08, and n = 10. But every student should know how to derive that formula; which is essentially what I did.]

 

B = 1000 ((1.08)^11) A + 2000 (1.08) A

B = 1000 (1.08) A ( 1*(1.08)^10 + 2 )

 

Mary, M = 1000 (1.08) A ( 2*(1.08)^10 + 1 )

 

M - B = 1000 (1.08) A ( 2*(1.08)^10 + 1 ) - 1000 (1.08) A ( 1*(1.08)^10 + 2 )

M - B = 1000 (1.08) A ( 2*(1.08)^10 + 1 - 1*(1.08)^10 - 2 )

M - B = 1000 (1.08) A ( (1.08)^10 - 1 )

M - B = 1000 (1.08) ( (1.08)^10 - 1 ) ( (1.08)^10 - 1 ) / 0.08

M - B = 1000 (1.08) ( (1.08)^10) - 1 ) ^2 / 0.08

 

M - B ≈ $18,131.95

[N.B.: Every student on a precalculus - calculus track should make sure they understand this solution completely; they should be able to replicate it for similar problems without notes.]

Answer 3

Amy,

 

Since you did not specify how often the interest is compounded, I might not be able to give you an exact number, but let's try:

 

I'll assume (for simplicity's sake) that the interest is compounded annually.

 

Then A = P(1 + r/n)^nt = ($30,000)(1 + r/n)^nt         

 

Since I have assumed the number of times the interest is compounded is only once per year, n = 1 and our equation becomes:

 

A = ($30,000)(1 + 0.08)^t     

 

At the end of the first year

 

A₁ = ($30,000)(1.08) = 32,400 for both Bob and Mary

 

Here's where things get tricky.  At the beginning of year 2 Bob adds another $2,000 and Mary adds $1000 to the principal.

 

Bob's numbers in green, Mary's numbers in orange:

 

 

A₂ = ($32,400 + $2000)(1.08) = $37,152; A₃ = ($37,152 + $2000)  = $42,284.16   (Bob)

 

A₄ = ($42,284.16 + $2,000)(1.08) = $47,826.89

 

And for Mary

 

A₂ ($32,400 + $1,000)(1.08) = $36,072  A₃; = (36,072 + $1,000) = $40,037.76   (Mary)

 

A₄ = ($40,037.76 + $1,000)(1.08) = $44,320.78

 

At the end of 10 years, Bob will have $91,740.87 and Mary will have $78,25431.

 

Beginning at year 11, the situation reverses itself and Bob starts adding $1,000 per year while Mary increases her contributions to $2,000.

 

Here's where things get a bit embarrassing.  At the end of 20 years, I have Bob with $211,375.52 and Mary with only $200,236.17, which I don't think is the way things were supposed to work out.  I'll continue working on this to see if I can find a flaw in my logic.

 

 

Answer 4

This is an important lesson of the "time value of money." It's VERY important to start investing as early as possible. Of course timing is also important, but you don't have any control over that. There is a bit of luck involved with your initial entry to the marketplace. The best case is that Stocks are undervalued early and overvalued later in your career. This example presumes equivalent entry level circumstances and will therefore underscore how important it is to SAVE AS MUCH AS YOU CAN AS EARLY AS YOU CAN.
1)Bob puts in $1000 for 10 years and THEN transitions to $2000/yr later
Bob_Total = $1000*(1.08)^20 + $1000*(1.08)^19 + ... $2000*(1.08)^10 + $2000*(1.08)^9 + ...

2)Mary puts in $2000 for 10 years and THEN transitions to $1000/yr later
Mary_Total = $2000*(1.08)^20 + $2000*(1.08)^19 + ... $1000*(1.08)^10 + $1000*(1.08)^9 + ...

The difference (Diff) between the two investment strategies can be computed (Diff=Mary_Total-Bob_Total)

Diff= $1000[ (1.08)^20)+(1.08)^19+...(1.08)^11] - $1000[ (1.08)^10)+(1.08)^9+...(1.08)^1]

Note that the 2nd 10 terms can be subtracted from the 1st 10 terms of each 20 term total

Diff1= $1000[ 1.08^20 + 1.08^19 +... 1.08^11 ]=$1000*(33.777) (note: used geometric series for 33.777 factor)
Diff2= $1000[ 1.08^10 + 1.08^9 + ... 1.08^1 ] =$1000*(15.6455) (note: used geometric series for 15.6455 factor)
Diff= $1000*(33.777-15.6455)= $18,132 difference despite investing the same total amount.

Moral: Save as much as you can as early as possible.