Compound Interest: Free Response

Convert the rate to a decimal, $r = 0.04$, and read off $P = 1200$ and $t = 3$. Apply $A = P(1+r)^t$ directly.

$A = 1200\,(1.04)^3$

Squaring gives $1.04^2 = 1.0816$, and one more factor gives $1.04^3 = 1.124864$, so

$A = 1200 \times 1.124864 = 1349.8368.$

Rounded to the nearest cent, the certificate matures at $1{,}349.84$ dollars.

Solving $A = P(1+r)^t$ for the principal means dividing by the growth factor instead of multiplying by it.

$P = \frac{A}{(1+r)^t} = \frac{2000}{1.124864}$

Carrying out the division,

$P = \frac{2000}{1.124864} = 1777.9927\ldots$

Rounded to the nearest cent, the customer must deposit about $1{,}777.99$ dollars today.

Compare the two formulas directly: simple interest is $A_{\text{simple}} = P(1+rt)$ and compound interest is $A_{\text{compound}} = P(1+r)^t$.

At $t = 1$ both reduce to the same expression,

$A_{\text{simple}} = P(1+r), \qquad A_{\text{compound}} = P(1+r)^1 = P(1+r),$

because only a single year's interest has been added under either rule; there has been no earlier interest yet for compounding to reinvest.

From year $2$ on, compound interest applies the rate to a balance that already includes the first year's interest, while simple interest keeps applying the rate to the original principal only. That extra reinvested interest earns interest of its own in every later year, so

$P(1+r)^t > P(1 + rt) \quad \text{for every whole number } t > 1.$

So simple interest matches compound interest exactly once, at the end of year $1$, and falls behind in every year after that. It never leads.

Check each piece of the line against what it is supposed to represent. The per-period rate $\frac{0.08}{4} = 0.02$ is correct: dividing the annual rate by the number of compounding periods per year is exactly right.

The exponent is where the line goes wrong. Compounding quarterly for $2$ years does not mean the factor $1.02$ is applied $2$ times; it means the factor is applied once every quarter, and there are $4$ quarters in each of the $2$ years, for a total of

$nt = 4 \times 2 = 8$

quarterly periods. The friend used the number of years where the number of periods belongs.

Keep the correct per-period rate and replace only the exponent.

$A = 3000\,(1.02)^{8}$

Raising $1.02$ to the eighth power gives $1.02^{8} = 1.171659$, so

$A = 3000 \times 1.171659 = 3514.977.$

Rounded to the nearest cent, the balance owed is $3{,}514.98$ dollars, well above the friend's $3{,}121.20$ dollars.

The letter $n$ counts how many times a year the balance is multiplied by a period's growth factor, and $t$ counts how many years pass, so their product $nt$ is the total number of times that multiplication happens. Treating the exponent as $t$ alone assumes the balance is multiplied only once a year, when it is really multiplied $n$ times a year.

Over $2$ years at $4$ periods a year, the true count is

$nt = 4 \times 2 = 8$

quarterly periods. Each of those eight applications is a separate chance for the balance to earn interest on its own already-earned interest, which is why the corrected total in part B is noticeably larger than a calculation that only multiplies twice.

Quarterly compounding at an annual rate of $6\%$ means $n = 4$ periods a year, each paying $\frac{0.06}{4} = 0.015$, over $nt = 4 \times 2 = 8$ periods total.

$A = 5000\left(1 + \frac{0.06}{4}\right)^{8} = 5000 \times 1.12649258$

$A = 5632.4629.$

Rounded to the nearest cent, Account X holds $5{,}632.46$ dollars after $2$ years.

Continuous compounding uses $A = Pe^{rt}$ with $rt = 0.06 \times 2 = 0.12$.

$A = 5000\,e^{0.12} = 5000 \times 1.12749685$

$A = 5637.48425.$

Rounded to the nearest cent, Account Y holds $5{,}637.48$ dollars after $2$ years.

Comparing the two balances,

$5637.48 - 5632.46 = 5.02,$

Account Y is ahead by about $5.02$ dollars.

The reason is not particular to quarterly compounding: it holds against monthly, daily, or any other finite schedule at this same nominal rate. A finite schedule, no matter how frequent, still credits interest at separated instants, and between two credits the balance sits still, earning nothing on itself until the next credit arrives. Continuous compounding removes every such gap, crediting interest at every instant, so it is always crediting interest on interest sooner than any finite schedule can. That is why the continuously compounded balance is the highest a fixed nominal rate can produce, a ceiling every finite schedule climbs toward but never passes.

Losing $15\%$ of the current value each year is the same as keeping $100\% - 15\% = 85\%$ of it, so one year multiplies the value by the fixed factor $0.85$, the same way one year of compound interest multiplies a balance by $1 + r$. Repeating that multiplication for $t$ years gives

$V = 40000\,(1 - 0.15)^{t} = 40000\,(0.85)^{t}.$

Because the $15\%$ comes off whatever the value happens to be that year, not off the original $40{,}000$ dollars, the value gets multiplied by $0.85$ again and again, exactly the mechanism that made the interest formula an exponent instead of a product.

Substitute $t = 4$ into the expression from part A and use the given power.

$V = 40000\,(0.85)^{4} = 40000 \times 0.52200625$

$V = 20880.25.$

The equipment is worth $20{,}880.25$ dollars after $4$ years.

The estimate $4 \times 15\% = 60\%$ of the original price is exactly the simple-interest pattern used in reverse: it subtracts the same fixed amount, $15\%$ of $40{,}000$ dollars, in every one of the four years, regardless of how much value is left. That is a straight-line decline, the decay analogue of $A = P(1 + rt)$.

The actual depreciation is compound: every year's $15\%$ is taken from whatever the CURRENT value is, which is smaller than $40{,}000$ dollars in every year after the first. Comparing the two after $4$ years,

$20880.25 > 16000,$

the compound value is higher, because the second year loses $15\%$ of a smaller number than the first year did, the third year loses $15\%$ of a smaller number still, and so on. The two methods agree only on how much is lost in year $1$; after that, compounding leaves consistently more value behind.

Apply $A = P(1+r)^t$ with $P = 180$ and the given growth factor.

$A = 180\,(1.06)^{5} = 180 \times 1.338226$

$A = 240.88.$

The predicted price after $5$ years is $240.88$ dollars.

Evaluate the price at both candidate years and compare each to the budget.

Year $5$:

$180\,(1.06)^{5} = 180 \times 1.338226 = 240.88,$

still under $250$ dollars.

Year $6$:

$180\,(1.06)^{6} = 180 \times 1.418519 = 255.33,$

which is over $250$ dollars. Since year $5$ has not yet crossed the budget and year $6$ has, the price first exceeds $250$ dollars in year $6$.

In the equation

$180(1.06)^t = 250,$

the unknown $t$ is stuck in the exponent, not sitting as a plain factor. Every quantity solved for so far in this lesson, whether a balance or a principal, has been found by running the formula forward or by dividing, and neither of those operations can pull an exponent down on its own.

Without a tool for that, the honest way to answer a question like "when does this cross a target" is to evaluate the formula at whole-number candidates and watch for the value to cross the target, exactly what part B does with years $5$ and $6$. That bracket is not a shortcut standing in for a real solution; it is the correct method available at this point, and it pins down the answer for a whole-year question just as precisely as an exact method would. A tool for isolating the exponent directly is introduced in the next lesson.