gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?
(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
DS24931.01
If you haven't already, you should know the interest formula \(A = P (1 + r)^t\) (you can discard n because it's only compounded annually, unless stated otherwise)
R= 20%
\(A = P (1.20)^t\)
We are trying to find the Value of A, in order to find A, we need to know P and t.
1.) Value increased by 44%. By using logic, we know that the value increased by 44%, which means we can determine how much time has passed if an investment is earning 20% a year, but we don't know the value because we don't know the starting (P) or current value.
INSUFFICENT 2.) Convert Statement 2 into math terms:
(P-600)(1.20)^(t-1) = 0.88[P*(1.20)^t]
Divide the whole equation by 0.88, and then divide (1.20)^(t-1).
(P-600)/0.88 = P(1.20)^t / (1.20)^(t-1)
The t cancels out, and now you can solve for P.
SUFFICIENT!Answer is B