Question

A deposit of X is made into a fund which pays an rate of 12 for 8 years. At the same time, 2X is deposited into another fund
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Answer #1

The future value of the deposits at the end of the fourth year are:

X(1 0.12) for the X-deposit

and 2Xe^{4\delta } for the 2X-deposit

The future value of the deposits at the end of the eight year are :

X(1 0.12) for the X-deposit

and 2Xe^{8\delta } for the 2X-deposit

Therefore the interest earned from the end ot the 4th year to the end of the 8th year is

X(1 0.12) X(1-0.12) for the X-deposit

and  2Xe2Xe   for the 2X-deposit

Since both fund earn the same amount of interest we have

X (1 0.12) X(1 0.12)=2Xe-2Xe

(1 0.12)-(1 - 0.12) 2e2e 8

( (1-0.12)(1 0.12)-4) e-e5 8

We need to solve for \delta, so let's use the following substitution

40 y= e

Then we have

y2-y((1-0.12)8( 0.12)-4)

y^2-y-0.55654367=0

So that

y=\frac{1\pm\sqrt{(-1)^2-4\cdot(- 0.55654367)}}{2}=\frac{1\pm1.796155}{2}

Since y can't be negative then

y=\frac{1+1.796155}{2}=1.398077

therefore

1.398077

\delta=\frac{1}{4}\text{ln}(1.398077)= 0.08377

So the answer is

(E)8.4%

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