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very true

oh ok

That was the FIRST answer I gave:  99.95 gm

School year is ending for a lot of folks...PLUS it is the weekend....less to do or question about.

how many grams remaining tho

Just substitute t = 4 into the equation to get  (I think your equation is REALLY 100(.99988)^t

100 (.99988)^4 = 99.952 gm after 4 years  =99.95  gm (rounded to the nearest hundredth)

99.95/100 x 100% = 99.95 % remains  

- (1/9)^3 =   - (1/9 x 1/9 x 1/9) =   -1/729

ALSO        (-1/9) ^3  = (-1/9) x (-1/9) x (-1/9) =   - 1/729

lol hes right but heres the answer anyway: 862.0689655172413793

14/100

abc is a button on a calculator

its the button for a fraction

thx pavlov

(1/9)^3 = 1/9 x 1/9 x 1/9 =  1 / 729

THIS IS WHAT HAPPENS WHEN QUESTIONS ARE WRITTEN UNCLEARLY OR WITH MISTAKES

OK....if you want them BOTH to receive the SAME amont when they EACH turn 21

Amount available when 17 y/o turns 21  = 10,000(1.02)^16 = 13727.86

    x = amount they BOTH will get

    Money after 17y/o gets first share = 13727.86-x

Money when the 12 y/o turns 21 is the same 'x ' and this equals : 

x = (13727.86-x)(1.02)^20    Solve for 'x'

x = (13727.86-x)(1.4859473)      then x = 8205.68          THey BOTH will get this amount when they EACH turn 21

OK.....done BOTH ways ....THIS is the end of the 'saga'

~THE END   hahahahahaha

Solve for t:
t^6 = 8-t^3

Subtract 8-t^3 from both sides:
t^6+t^3-8 = 0

Substitute x = t^3:
x^2+x-8 = 0

Add 8 to both sides:
x^2+x = 8

Add 1/4 to both sides:
x^2+x+1/4 = 33/4

Write the left hand side as a square:
(x+1/2)^2 = 33/4

Take the square root of both sides:
x+1/2 = sqrt(33)/2 or x+1/2 = -sqrt(33)/2

Subtract 1/2 from both sides:
x = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2

Substitute back for x = t^3:
t^3 = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2

Taking cube roots gives (sqrt(33)/2-1/2)^(1/3) times the third roots of unity:
t = -(1/2 (1-sqrt(33)))^(1/3) or t = (1/2 (sqrt(33)-1))^(1/3) or t = (-1)^(2/3) (1/2 (sqrt(33)-1))^(1/3) or x+1/2 = -sqrt(33)/2

Subtract 1/2 from both sides:
t = -(1-sqrt(33))^(1/3)/2^(1/3) or t = (sqrt(33)-1)^(1/3)/2^(1/3) or t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3) or x = -1/2-sqrt(33)/2

Substitute back for x = t^3:
t = -(1-sqrt(33))^(1/3)/2^(1/3) or t = (sqrt(33)-1)^(1/3)/2^(1/3) or t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3) or t^3 = -1/2-sqrt(33)/2

Taking cube roots gives (-1/2-sqrt(33)/2)^(1/3) times the third roots of unity:
Answer: |  t = -(1-sqrt(33))^(1/3)/2^(1/3)    or    t = (sqrt(33)-1)^(1/3)/2^(1/3)    or    t = ((-1)^(2/3) (sqrt(33)-1)^(1/3))/2^(1/3)    or    t = (1/2 (-1-sqrt(33)))^(1/3)    or    t = -(1/2 (1+sqrt(33)))^(1/3)    or    t = -((-1)^(2/3) (1/2 (1+sqrt(33)))^(1/3))

The fairest way of distributing the money would be as follows:

When the mother died, she should have left instuctions in her Will that the $10,000 be split evenly between the two children, or $5,000 per child. At the same time, she should have put a provision in her Will that the money is to be distributed ONLY when each child turns 21.

In that case, the 17-year old would have received $6,863.93 when he/she turned 21.

And the 12-year old would have received $10,199.44, when he/she turned 21.

Even though there is a difference of $3,335.51 between the two amounts, in reality they are equivalent, because their PV was equivalent when she died. The difference of $3,335,51 in FV is because of the extra 5 years that the money grew @ 8% comp. quarterly.

Sorry , Guest....this WAS addressed in the Question AND the naswer EP gave.

"no vision that they are to get equal mounts when they reach age 21"

and

"..and the remaining will be distributed in 36 quarters and the amounts can be different?????     Well there is a whole bunch of possibilites depending on how it is split when the first kid turns 21 !  "

IF the question said they were to receive the same amounts when they turn 21 , that would be different...BUT IT DOESN'T SAY THAT !