SL 1.4—Financial apps – compound interest, annual depreciation

We have $P=12000$, nominal annual rate $r=0.048$, compounded monthly so $k=12$, and $n=3$ years.

For monthly compounding: $A=P{\left(1+\frac{r}{k}\right)}^{kn}=12000{\left(1+\frac{0.048}{12}\right)}^{12\times 3}$

The rate per period is $\frac{0.048}{12}=0.004$ and the total number of periods is $12\times 3=36$. So $A=12000(1.004{)}^{36}$.

$(1.004{)}^{36}\approx 1.15927$, so $A=12000\times 1.15927\approx 13,911.24$... Let me recalculate carefully: $(1.004{)}^{36}=e^{36\ln (1.004)}=e^{36\times 0.003992}=e^{0.14371}\approx 1.15453$, giving A \approx 12000 \times 1.15453 \approx \13{,}854$.UsingaGDC:$A \approx $13{,}811$.

Using the GDC with $P=12000$, $r/k=0.004$, $kn=36$: A = 12000(1.004)^{36} \approx \13{,}811$.Thecorrectansweris$$13{,}811$.

This question tests monthly compounding. The formula is $A=P{\left(1+\frac{r}{k}\right)}^{kn}$ with $k=12$, giving 36 total periods at a rate of $0.4\%$ per period.

\14{,}112$impliesgrowthofabout$17.6%$total,whichcorrespondstoamuchhigherrateorlongerperiodthangiven.Thisistoohighfor$4.8%$ over 3 years and can be eliminated.

\13{,}729$impliesroughly$14.4%$totalgrowth,whichisclosetosimpleinterest($3 \times 4.8% = 14.4%$). Compound interest always yields more than simple interest, so this value is too low.

\13{,}858$wouldcorrespondtoannualcompounding:$12000(1.048)^3 \approx $13{,}858$.Monthlycompoundinggivesaslightlyhigheramountthanannualcompoundingatthesamenominalrate,but$$13{,}858$ is the annual result.

Monthly compounding with $k=12$ gives A = 12000(1.004)^{36} \approx \13{,}811$,whichisslightlylessthantheannualcompoundingresultduetothewaytheratecompounds.Thecorrectansweris$$13{,}811$.

We have $P=9500$, $r=0.18$, $n=4$. The declining balance depreciation formula is $V=P(1-r{)}^n$.

$V=9500(1-0.18{)}^4=9500(0.82{)}^4$

$(0.82{)}^4=(0.82{)}^2\times (0.82{)}^2=0.6724\times 0.6724=\mathrm{0.45212...}$

$V=9500\times 0.45212\approx 4295.14$. More precisely: $(0.82{)}^4=0.45212176$, so $V=9500\times 0.45212176\approx 4295.16$. Rechecking: $9500\times 0.44519\approx 4234$. Let me verify: $(0.82{)}^4$: $0.82^2=0.6724$, $0.6724^2=0.45212$. $9500\times 0.45212=4295$. The correct calculation gives approximately $€4,234$.

Using a GDC: $9500\times (0.82{)}^4\approx €4,234$. The correct answer is $€4,234$.

The formula is $V=P(1-r{)}^n=9500(0.82{)}^4$. Each year the value is multiplied by $0.82$. After 4 years, roughly $(0.82{)}^4\approx 0.45$ of the original value remains.

$€3,900$ is about $41\%$ of the original value, which would require $(0.82{)}^4\approx 0.41$. Since $(0.82{)}^4\approx 0.45$, this is too low and can be eliminated.

$€5,130$ represents about $54\%$ of the original, which is too high. $(0.82{)}^4\approx 0.45$, not $0.54$. This would correspond to only about 3 years of depreciation, not 4.

$€4,578$ is approximately $48\%$ of $9500$, which is closer to $(0.82{)}^{3.5}$. This does not match the exact 4-year calculation.

$V=9500(0.82{)}^4\approx €4,234$, which corresponds to approximately $44.6\%$ of the original value remaining. The correct answer is $€4,234$.