Compound Interest: Practice

Each year multiplies the balance by the factor $1 + r$, and over $t$ years that factor is applied $t$ times.

$A = P(1+r)^t$

The form $P(1 + rt)$ is simple interest, which grows in a straight line instead.

One year multiplies the principal by $1 + 0.10 = 1.10$.

$A = 100\,(1.10)^1 = 110$

The balance is $110$ dollars.

Two years multiply by $1.10$ twice, that is, by $1.10^2 = 1.21$.

$A = 100\,(1.10)^2 = 100 \times 1.21 = 121$

Simple interest would give only $120$ dollars, so the extra dollar is interest earned on the first year's interest.

The letters stand for principal $P$, rate $r$, time $t$, and final amount $A$.

$A = P(1+r)^t$

Here $P$ is the principal, the amount you put in at the very start.

Compound interest pays interest on the interest already earned, while simple interest never does.

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

So compound interest pulls ahead and the gap widens over time.

Adding $r$ of a balance $B$ to itself gives $B + rB$, which factors as a single multiplication.

$B + rB = (1 + r)B$

So one year scales the balance by $1 + r$.

Use $I = Prt$ with $P = 250$, $r = 0.08$, and $t = 1$.

$I = 250 \times 0.08 \times 1 = 20$

The interest is $20$ dollars.

One year multiplies by $1.05$.

$A = 2000\,(1.05) = 2100$

The balance is $2{,}100$ dollars.

Continuous compounding uses the number $e$, the limit of $\left(1 + \frac{1}{m}\right)^m$ as $m$ grows.

$e \approx 2.718$

The value $3.14$ is $\pi$, a different constant.

Compound interest always applies the rate to the current balance, not the original deposit.

$110 \times 0.10 = 11$

So year 2 earns $11$ dollars, giving $121$ dollars. That is why it beats simple interest.

Divide the percent by $100$.

$12\% = \frac{12}{100} = 0.12$

Multiply the principal by the two-year growth factor.

$A = 300 \times 1.1025 = 330.75$

The balance is $330.75$ dollars.