On January 1, 2005, a person’s savings account was worth $200,000. Every month thereafter, this person makes a cash contribution of $676 to the account. If the fund was expected to be worth $400,000 on January 1, 2010, what annual rate of interest was being earned on this fund? (4.15)

Question

On January 1, 2005, a person's savings account was worth [latex]\$ 200,000[/latex]. Every month thereafter, this person makes a cash contribution of [latex]\$ 676[/latex] to the account. If the fund was expected to be worth [latex]\$ 400,000[/latex] on January 1, 2010, what annual rate of interest was being earned on this fund? (4.15)

Accepted Answer

Number of monthly deposits [latex]=(5[/latex] years [latex])(12[/latex] months [latex]/ \mathrm{yr})=60[/latex]

[latex]\$ 400,000=\$ 200,000\left(F / P, i^{\prime} / ight.[/latex] month, 60[latex])+\$ 676\left(F / A, i^{\prime} / ight.[/latex] month, 60[latex])[/latex]

Try [latex]i^{\prime} /[/latex] month [latex]=0.75 \%: \$ 400,000>\$ 364,126.69, herefore i^{\prime} /[/latex] month [latex]>0.75 \%[/latex]

Try [latex]i^{\prime} /[/latex] month [latex]=1 \%: \quad \$ 400,000<\$ 418,548.72, herefore i^{\prime} /[/latex] month [latex]<1 \%[/latex]

Using linear interpolation:

[latex]\frac{i^{\prime} / ext { month }-0.75 \%}{\$ 400,000-\$ 364,126.69}=\frac{1 \%-0.75 \%}{\$ 418,548.72-\$ 364,126.69} ; i^{\prime} / ext { month }=0.9148 \%[/latex]

Therefore, [latex]i^{\prime} /[/latex] year [latex]=(1.009148)^{12}-1=0.1155[/latex] or [latex]11.55 \%[/latex] per year

Question 4.T-Y-S.X: On January 1, 2005, a person’s savings account was worth $20...

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