1

Determine the amount of interest earned from time $t=2$ to $t=4$ if $240$ is invested at $t=1$ and an additional $300$ is invested at $t=3$. Given, $a(1)=1.2$, $a(2)=1.5$, $a(3)=2.0$, $a(4)=3.0$

I tried finding $A(0)$ using the equation $A(1)=A(0).a(1)$. Hence $A(0)=200$

Hence I found $A(2)=300$, $A(3)=400$, $A(4)=600$

$I_{[2,3]}=A(0)[\text{new } A(3)-A(2)]=80000$

$I_{[3,4]}=A(0)[A(4)-\text{new }A(3)]$

From here I am stuck.I am messed up with the different times.

Tosh
  • 1,614
  • Is the interest rate supposed to be $20%$ per time period? – Henry Jan 23 '16 at 14:07
  • Do you have cumulative interest? An interest payout at t = 2, 3? – Pieter21 Jan 23 '16 at 14:31
  • @Henry, The amount of interest is 450. That is the answer. No idea about interest rate. – Tosh Jan 23 '16 at 15:48
  • 1
    If you don't know the interest rate, how did you get A(0)=200? –  Jan 23 '16 at 16:14
  • Wasn't some information missing from the problem statement? http://www.math.fsu.edu/~paris/MAP4170/M1S1.pdf (Accumulation function values at beginning of problem statement). – Moo Jan 23 '16 at 21:16
  • @Moo, thank you. The question is complete now. Any idea of the workings? – Tosh Jan 24 '16 at 01:25

1 Answers1

2

Answer:

$$a_{(2-4)} = \frac{a(4)}{a(2)}$$

$$A(2) = A(1).\frac{a(2)}{a(1)}$$

$$A(2) =240\times\frac{1.5}{1.2} = 300$$

$$A(4) = 300\times\frac{3}{1.5} = 600$$

$$A'(4) = A'(3)\frac{a(4)}{a(3)}$$

$$A'(4) = 300\times\frac{3}{2}$$

$$A'(4) = 450$$

$$I_1 = A(4)-A(2) = 600-300 = 300$$

$$I_2 = A'(4) - A'(3) = 450-300 = 150$$

$$I = I_1 +I_2$$

The required Interest:

$$I = 300+150 = 450$$

Regards

Satish

Satish Ramanathan
  • 11,984