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Galileo deposits 15,000 in a bank. During the first year, the bank credits an annual effective rate of interest i. For the second year, the bank credits an annual effective rate of interest (i-5%). At the end of two years, he has 18,093.75 in the bank. What would Galileo have in the bank at the end of three years, if the annual effective rate of interest were (i+9%) for each of the three years?

I'm a first-year student of Actuarial Science, and I tried to look for example question from my university depository. Out of 5 questions, this is the only one I can't think off on how to solve for the annual effective rate of interest.

Can somebody give me ideas on how to start or to solve for this problem?

Mohd Shahril
  • 131

2 Answers2

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$(15000(i+1))(i+.95)=18093.75$ solve for $i$.

The principle after the first year is $15000(i+1)$. This is used now to compute the principle after the second year the same way except we get $.05$ less interest.

You basically get a quadratic equation which will have 2 solutions but we need only one.

$15000(i^2+1.95i+.95)=18093.75$

$i^2+1.95i+.95=1.20625$

$i^2+1.95i-.25625=0$.

You can use the quadratic equation to solve this, which should be readily apparent hopefully,

$i=\frac{-1.95 \pm \sqrt{1.95^2+4(.25625)}}{2}$

$i=\frac{-1.95 \pm 2.197}{2}=- .975\pm 1.0985$

Possible values

$i={.1235,-2.02735}$

But you need $0 \le i \le 1$ so the interest rate was $.1235$.

note should double check answer.

marshal craft
  • 700
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The amount in the bank after one year is $15000(1+i)$ since $i$ is the effective annual rate for the first year. Since $i-0.05$ is the effective rate for the second year, after two years the account will be at $$15000(1+i)(1+i-0.05) = 18093.75.$$ You can solve this for $i.$

spaceisdarkgreen
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