BTGmoderatorDC 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has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested.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(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
OA B
Source: Official Guide
It is given that all interest has been reinvested; thus, the sum is invested at a compound rate of interest.
Say the sum invested is $P at the rate of 20% p.a. (given) for n years. Thus, the amount after n years would be, say $A.
Thus, A = P(1 + 20%)^n = 1.2^n*P
Let's take each statement one by one.
(1) The value of the investment has increased by 44% since it was first made.
=> A = P + 44% of P = 1.44P
Thus, 1.44P = 1.2^n*P => n = 2. Can't get the value of n. Insufficient.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
Since $600 is withdrawn by the end of (n - 1) years, the amount at the end of (n - 1) years = 1.2^(n-1)P - 600
Thus, the amount [1.2^(n-1)P - 600] was invested for the nth year (last one year).
=> [1.2^(n-1)P - 600]*1.2 = 88% of 1.2^n * P
1.2^n * P - 600 = 22/25 * 1.2^n * P
1.2^n *P = $6,000
Or, A = $6,000. Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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