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Compound Interest: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The semiannual balance

An account receives a single deposit of $1400$ dollars and pays $6\%$ compounded semiannually for $3$ years, with no further deposits or withdrawals. Find the balance at the end of the $3$ years, to the nearest cent.
1. Hint 1
  Each period pays only a share of the yearly rate, and the exponent counts periods rather than years.
2. Hint 2
  Semiannual compounding means $n=2$ , so the rate per period is $0.06\div2$ and the number of periods is $2(3)$ .
Answer

$1671.67$ dollars.

Full solution

Semiannual compounding gives $n=2$ , so each period pays
$0.06\div2=0.03$
and over $3$ years there are
$\begin{aligned} nt&=2(3)\\ &=6 \end{aligned}$
periods.

Substituting into the formula gives
$A=1400(1.03)^6$
The sixth power is $1.03^6\approx1.194052$ , so the balance is
$A\approx1671.67$
dollars.

Using the exponent $3$ instead of $6$ would count years rather than periods and understate the balance.

Answer

$1671.67$ dollars.

Key idea

Compounding $n$ times a year divides the rate by $n$ and multiplies the exponent by $n$ , so the exponent is $nt$ .
Problem 1: Mark complete
Problem 2 The extra deposit

An account already contains $160$ dollars. A single extra deposit, a whole number of dollars, is made now, before any interest is earned. The account then pays $10\%$ compounded annually for $2$ years with no further changes, and the final balance must be at least $300$ dollars. What is the smallest extra deposit that works?
1. Hint 1
  The money already present and the new deposit grow together as one principal.
2. Hint 2
  Find the smallest principal whose two-year balance reaches $300$ dollars, then remove the amount already present.
3. Hint 3
  A principal below the required amount leaves the balance short, so round the principal up to the next whole dollar.
Answer

$88$ dollars, which makes the principal $248$ dollars.

Full solution

Two years multiply the principal by
$1.10^2=1.21$
so a principal of $P$ dollars must satisfy
$1.21P\ge300$

Dividing gives
$P\ge247.93\ldots$
Because the $160$ dollars present and the extra deposit are both whole numbers of dollars, $P$ is a whole number, so the smallest principal that works is $248$ dollars.

The extra deposit is then
$248-160=88$
dollars.
Checking,
$248(1.21)=300.08$
dollars clears the target, while a principal of $247$ dollars gives only
$247(1.21)=298.87$
dollars.

Answer

$88$ dollars, which makes the principal $248$ dollars.

Key idea

When a compounded balance must clear a target, divide back to the principal and round that principal up, since rounding down leaves the balance short.
Problem 2: Mark complete
Problem 3 The retained interest

An account earns $36$ dollars of interest during its first year at $8\%$ compounded annually. All interest stays in the account, and there are no other deposits or withdrawals. How many more dollars of interest does the account earn during its second year than during its first?
1. Hint 1
  The second interest payment is figured on the balance that already includes the first payment.
2. Hint 2
  Recover the principal from the first payment and the decimal rate, then add the retained interest before applying the rate again.
Answer

$2.88$ dollars more; the second payment is $38.88$ dollars.

Full solution

The first payment is $0.08P=36$ , so the principal is
$\begin{aligned} P&=36\div0.08\\ &=450 \end{aligned}$
dollars.
Retaining that interest makes the balance
$450+36=486$
dollars.

The second payment is
$0.08(486)=38.88$
dollars, which exceeds the first by
$38.88-36=2.88$
dollars.

That excess is exactly $8\%$ of the retained $36$ dollars, the interest the first payment earns for itself.

Answer

$2.88$ dollars more; the second payment is $38.88$ dollars.

Key idea

Retained interest joins the balance, so each payment exceeds the one before it by the rate applied to that earlier payment.
Problem 3: Mark complete
Problem 4 The two deposits

An account pays $5\%$ compounded annually. It receives $400$ dollars at the start, then $250$ dollars immediately after the first annual interest payment. No other changes occur. Find the balance immediately after the second annual interest payment, and state how many years each deposit earned interest.
1. Hint 1
  The deposits entered at different times, so they earn different numbers of interest payments.
2. Hint 2
  Either follow the year-end balances or grow each deposit over its own time in the account.
Answer

$703.50$ dollars; the first deposit earns for $2$ years and the second for $1$ year.

Full solution

After the first payment, the original deposit has grown to
$400(1.05)=420$
dollars.
Adding the new deposit makes
$420+250=670$
dollars.

The second payment gives
$670(1.05)=703.50$
dollars.

Checking by separate deposits gives
$400(1.05)^2=441$
dollars and
$250(1.05)=262.50$
dollars.
Their sum is $703.50$ dollars, with two years of interest on the first deposit and one on the second.

Answer

$703.50$ dollars; the first deposit earns for $2$ years and the second for $1$ year.

Key idea

Deposits made at different times receive different numbers of compounding factors.
Problem 4: Mark complete
Problem 5 The withdrawal record

An account starts with $800$ dollars at $8\%$ compounded twice per year for $2$ years. Immediately after the second interest payment, $100$ dollars is withdrawn. No other changes occur. Find the balance immediately after the fourth interest payment, and the total interest earned over the $2$ years, each to the nearest cent.
1. Hint 1
  Apply each period rate to the balance actually present during that period.
2. Hint 2
  The rate per half-year is $0.08\div2$ , and $2$ years hold $nt=4$ periods, with the withdrawal between the second and the third.
Answer

Final balance: $827.73$ dollars. Total interest: $127.73$ dollars.

Full solution

The half-year rate is $0.04$ and the account runs for
$\begin{aligned} nt&=2(2)\\ &=4 \end{aligned}$
periods.
The first two periods give
$800(1.04)^2=865.28$
dollars, and the withdrawal leaves
$865.28-100=765.28$
dollars.

The last two periods produce
$765.28(1.04)^2\approx827.73$
dollars.

The deposit less the withdrawal accounts for
$800-100=700$
dollars of that balance, so the total interest is
$827.73-700=127.73$
dollars.
The four separate payments, $32$ , $33.28$ , $30.61$ and $31.84$ dollars, add to the same total.

Answer

Final balance: $827.73$ dollars. Total interest: $127.73$ dollars.

Key idea

An intervening withdrawal changes the principal for later periods, so a single unchanged-principal power does not describe the whole history.
Problem 5: Mark complete
Problem 6 The balance gap

Two accounts earn $8\%$ compounded annually for $3$ years. Their initial principals differ by $80$ dollars. There are no deposits, withdrawals, or fees. By how many dollars do their final balances differ, and by how many dollars has that gap increased? Give both amounts in dollars and cents.
1. Hint 1
  Both principals are multiplied by the same growth factor.
2. Hint 2
  Factor the common growth factor out of the difference of the two final balances.
Answer

Final gap: $100.78$ dollars. Increase in gap: $20.78$ dollars.

Full solution

If the smaller principal is $P$ dollars, the balance gap is
$\big((P+80)-P\big)(1.08)^3$
dollars.
Since $1.08^3=1.259712$ , the final gap is
$80(1.259712)\approx100.78$
dollars.

The gap has increased by
$100.78-80=20.78$
dollars.
The difference behaves exactly like an additional $80$ dollars earning the same interest.

Answer

Final gap: $100.78$ dollars. Increase in gap: $20.78$ dollars.

Key idea

Equal compounding factors multiply a difference in principals by that same factor.
Problem 6: Mark complete
Problem 7 The fee comparison

Two plans hold $500$ dollars for $1$ year at an annual rate of $8\%$ . Plan A compounds annually with no fee. Plan B compounds continuously and deducts a $2$ dollar fee at the end. Which plan leaves more money after the fee, and by how much? Use $e^{0.08}\approx1.083287$ and round the difference to the nearest cent.
1. Hint 1
  Compare the amounts actually left after each plan applies its growth and any fee.
2. Hint 2
  For the continuous plan, apply the fee after computing $Pe^{rt}$ .
Answer

Plan A leaves more; the difference is about $0.36$ dollars.

Full solution

Plan A ends with
$500(1.08)=540$
dollars.
Plan B grows to
$500e^{0.08}\approx541.64$
dollars before its fee.

Deducting the fee leaves
$541.64-2=539.64$
dollars, so Plan A finishes ahead by
$540-539.64=0.36$
dollars.

Continuous compounding is worth about $1.64$ dollars more than annual compounding here, and the $2$ dollar fee is larger than that advantage.

Answer

Plan A leaves more; the difference is about $0.36$ dollars.

Key idea

A fee larger than the advantage from more frequent compounding reverses which plan finishes ahead.
Problem 7: Mark complete
Problem 8 The two-year claim

A principal $P>0$ earns a positive annual decimal rate $r$ , with no money added or removed. Lena claims that after exactly $2$ years, annual compounding exceeds simple interest by $Pr^2$ . Decide whether she is correct, justifying algebraically, and then state, in terms of $P$ and $r$ , by how much annual compounding exceeds simple interest after exactly $3$ years.
1. Hint 1
  Write the two balances over the same period before comparing them.
2. Hint 2
  Expand the annual factor raised to the number of years, then subtract the simple-interest balance.
3. Hint 3
  Over three years every term of the expanded cube matters, not only the squared one.
Answer

Yes, the two-year excess is $Pr^2$ ; after $3$ years the excess is $3Pr^2+Pr^3$ .

Full solution

The compound balance is $P(1+r)^2$ and the simple balance is $P(1+2r)$ .
The squared factor expands to
$P(1+r)^2=P+2Pr+Pr^2$

Subtracting $P+2Pr$ leaves
$P(1+r)^2-P(1+2r)=Pr^2$
so Lena is correct.

Over $3$ years the compound balance expands to the four terms $P$ , $3Pr$ , $3Pr^2$ and $Pr^3$ , while the simple balance is only $P+3Pr$ .

Subtracting that simple balance leaves an excess of
$3Pr^2+Pr^3$
Since $P>0$ and $r>0$ , both excesses are positive, matching interest earned on earlier interest.

Answer

Yes, the two-year excess is $Pr^2$ ; after $3$ years the excess is $3Pr^2+Pr^3$ .

Key idea

Compounding beats simple interest by the interest earned on earlier interest, $Pr^2$ after two years and $3Pr^2+Pr^3$ after three.
Problem 8: Mark complete
Problem 9 Nadia's balance

Nadia invests $1200$ dollars at $3\%$ compounded semiannually for $2$ years, with no deposits or withdrawals. Her work reads $A=1200(1.015)^2$ , which she evaluates as $1236.27$ dollars.

Identify her error and give the correct balance to the nearest cent. Then decide whether any finite compounding schedule, on the same principal, rate, and time, could reach $1300$ dollars, using $e^{0.06}\approx1.061837$ to justify your decision.
1. Hint 1
  Her rate per period is right, so check how many periods $2$ years of semiannual compounding actually holds.
2. Hint 2
  The exponent is the total number of periods $nt$ , not the number of years.
3. Hint 3
  However often a finite schedule compounds, its balance stays below the continuous value $Pe^{rt}$ .
Answer

She used the exponent $2$ instead of $nt=4$ ; the correct balance is $1273.64$ dollars, and no finite schedule reaches $1300$ dollars: the continuous ceiling is about $1274.20$ dollars.

Full solution

Each half-year pays $0.015$ , which Nadia used correctly, but $2$ years hold
$\begin{aligned} nt&=2(2)\\ &=4 \end{aligned}$
periods, so her exponent counts years rather than periods.

The correct balance is
$1200(1.015)^4\approx1273.64$
dollars, about $37$ dollars above her figure.

Compounding more often raises the balance only toward the continuous value
$1200e^{0.06}\approx1274.20$
dollars.

Every finite schedule stays below that ceiling, and the ceiling is already below $1300$ dollars, so no finite schedule reaches the target.

Answer

She used the exponent $2$ instead of $nt=4$ ; the correct balance is $1273.64$ dollars, and no finite schedule reaches $1300$ dollars: the continuous ceiling is about $1274.20$ dollars.

Key idea

The exponent counts compounding periods, and however many periods there are the balance stays under the continuous ceiling $Pe^{rt}$ .
Problem 9: Mark complete
Problem 10 Omar's comparison

Account A holds $800$ dollars at an annual rate of $24\%$ for $1$ year. Account B holds $800$ dollars at an annual rate of $12\%$ for $2$ years. Both accounts compound $n$ times per year, with no deposits, withdrawals, or fees. Omar says that as $n$ grows without bound, the difference between their balances approaches zero, even though their balances at $n=1$ differ. Is he correct? Give both balances at $n=1$ and both limiting balances, using $e^{0.24}\approx1.271249$ and rounding to the nearest cent.
1. Hint 1
  The continuous growth factor depends on the product of the decimal rate and the time.
2. Hint 2
  Compare the two products $rt$ , then separately use $n=1$ for the two balances.
Answer

Yes. At $n=1$ : A $992$ dollars, B $1003.52$ dollars. Both limits: $800e^{0.24}\approx1017.00$ dollars.

Full solution

At $n=1$ , Account A gives
$800(1.24)=992$
dollars and Account B gives
$800(1.12)^2=1003.52$
dollars.

As the periods grow without bound, each balance approaches $Pe^{rt}$ .
Account A approaches
$800e^{0.24(1)}$
and Account B approaches
$800e^{0.12(2)}$
in dollars.

Both products $rt$ equal $0.24$ , so both limits are
$800e^{0.24}\approx1017.00$
dollars and the difference approaches zero, making Omar correct.
The unequal balances at $n=1$ do not prevent this shared limit.

Answer

Yes. At $n=1$ : A $992$ dollars, B $1003.52$ dollars. Both limits: $800e^{0.24}\approx1017.00$ dollars.

Key idea

Different rates and durations can have the same continuous-compounding limit when their products $rt$ agree.
Problem 10: Mark complete

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