Compound Interest & Algebra

Q 01 Easy Rate → Decimal

A savings account earns an annual interest rate of 8%, compounded annually. You deposit $2,000. Which expression correctly represents the account value after 5 years?

A$2000(1.8)^5$

B$2000(1.08)^5$

C$2000(0.08)^5$

D$2000 + 2000(0.08)(5)$

📖 Explanation

The compound interest formula is $A = P(1 + r/n)^{nt}$. Here $P=2000$, $r=0.08$, $n=1$ (annual), $t=5$. So $A = 2000(1 + 0.08/1)^{1 \times 5} = 2000(1.08)^5$.

❌ Trap (A): 8% as a decimal is 0.08, not 0.8. 1 + 0.8 = 1.8 is wrong.
❌ Trap (D): That's the simple interest formula — it doesn't compound.

Q 02 Easy Identify P, r, t

In the formula $A = P(1 + r/n)^{nt}$, if $P = 1000$, $r = 0.05$, $n = 1$, and $t = 3$, what is the value of $A$?

A$\$1,150$

B$\$1,157.63$

C$\$1,157.63$

D$\$1,500$

📖 Explanation

$A = 1000(1.05)^3 = 1000 \times 1.157625 = \$1{,}157.63$

❌ Trap (A): $\$1,150$ = $1000 + 1000(0.05)(3)$ — that's simple interest, not compound.

Q 03 Easy n = compounding frequency

A bank compounds interest monthly. What is the correct value of n to use in the compound interest formula?

A$n = 1$

B$n = 4$

C$n = 52$

D$n = 12$

📖 Explanation

$n$ = number of times interest is compounded per year. Monthly = 12 times per year → $n = 12$.

Memory: Annual=1, Semi-annual=2, Quarterly=4, Monthly=12, Daily=365.

Q 04 Easy nt = total periods

Interest is compounded semi-annually (twice per year) for 6 years. What is the value of the exponent $nt$ in the formula $A = P(1 + r/n)^{nt}$?

A$6$

B$12$

C$3$

D$2$

📖 Explanation

$n = 2$ (semi-annual), $t = 6$ years. So $nt = 2 \times 6 = 12$.

❌ Trap (A): Many students forget to multiply and just use $t = 6$. The exponent is always $n \times t$, not just $t$.

Q 05 Easy Build the expression

$5,000 is invested at an annual interest rate of 6%, compounded quarterly (4 times per year). Which expression gives the value after 3 years?

A$5000(1.06)^3$

B$5000(1.015)^3$

C$5000(1.015)^{12}$

D$5000(1.06)^{12}$

📖 Explanation

$P=5000,\ r=0.06,\ n=4,\ t=3$
$r/n = 0.06/4 = 0.015$ → base = $1.015$
$nt = 4 \times 3 = 12$ → exponent = $12$

Answer: $5000(1.015)^{12}$

❌ Trap (A) & (B): Exponent should be $nt = 12$, not just $t = 3$.
❌ Trap (D): Rate should be divided by $n=4$, giving 0.015, not kept at 0.06.

Q 06 Easy Simple vs Compound

Maria borrows $3,000 at 5% simple interest per year for 4 years. How much interest (not total amount) does she pay?

A$\$600$

B$\$3,646.52$

C$\$646.52$

D$\$150$

📖 Explanation

Simple interest: $I = Prt = 3000 \times 0.05 \times 4 = \$600$.

❌ Trap (C): $\$646.52$ is the compound interest (annual) — this question specifies simple interest.
❌ Trap (D): Only multiplied for 1 year, forgot $t = 4$.

Q 07 Medium Identify the formula piece

An investment of $10,000 earns 4.8% annual interest, compounded monthly. What is the periodic interest rate (i.e., $r/n$) used each compounding period?

A$0.048$

B$0.004$

C$0.4$

D$0.0048$

📖 Explanation

Periodic rate = $r/n = 0.048 / 12 = 0.004$. This means each month, 0.4% interest is earned.

❌ Trap (A): That's the annual rate — not divided by $n$.
❌ Trap (D): Off by a factor of 10 — common arithmetic slip.

Q 08 Medium Which formula fits?

A college fund starts with $8,000 and grows at 3% per year, compounded annually for 18 years. Which expression gives the interest earned (not the total)?

A$8000(1.03)^{18}$

B$8000(0.03)(18)$

C$8000 - 8000(1.03)^{18}$

D$8000(1.03)^{18} - 8000$

📖 Explanation

Interest earned = Final Amount − Principal = $A - P$
$= 8000(1.03)^{18} - 8000$

❌ Trap (A): That's the total value $A$, not the interest alone.
❌ Trap (B): Simple interest formula — wrong model here.
❌ Trap (C): Subtraction is backwards — this gives a negative number.

Q 09 Medium Compare compounding types

Jake invests $4,000 at 6% annual rate for 2 years. Bank A compounds annually; Bank B compounds quarterly. Which statement is true?

ABank A gives more because annual compounding uses the full rate

BBank B gives more because more frequent compounding earns more

CBoth banks give the same amount

DIt depends on the principal amount

📖 Explanation

Bank A: $4000(1.06)^2 = \$4{,}494.40$
Bank B: $4000(1.015)^8 = \$4{,}509.15$

More frequent compounding → interest earns interest more often → higher total. Bank B wins.

Rule: Same annual rate, more compounding periods = more money.

Q 10 Medium Solve for t (inverse)

You invest $1,000 at 10% annual interest, compounded annually. Approximately how many years will it take to double your money?

Hint: Use the Rule of 72: years ≈ 72 ÷ interest rate (%)

A5 years

B10 years

C7 years

D20 years

📖 Explanation

Rule of 72: $72 \div 10 = 7.2 \approx 7$ years.
Exact check: $1000(1.10)^7 = \$1{,}948.72$ (close to double); $(1.10)^{7.27} \approx 2$.

❌ Trap (B): 10 years is the simple interest doubling time ($I = Prt → 1 = rt$). Compound interest doubles faster!

Q 11 Medium Verbal → Formula

A car loan of $15,000 charges 9% annual interest, compounded monthly for 5 years. Which expression gives the total amount owed at the end?

A$15000\!\left(1 + \tfrac{0.09}{12}\right)^{60}$

B$15000(1.09)^{60}$

C$15000\!\left(1 + \tfrac{0.09}{12}\right)^{5}$

D$15000(1.09)^{5}$

📖 Explanation

$n=12$ (monthly), $t=5$ → $nt = 60$. Periodic rate = $0.09/12 = 0.0075$.
Answer: $15000(1.0075)^{60}$ = $15000\!\left(1+\tfrac{0.09}{12}\right)^{60}$

❌ Trap (B): Uses annual rate in base without dividing — WRONG.
❌ Trap (C): Correct base, but forgets to multiply exponent: uses $t=5$ not $nt=60$.

Q 12 Medium Rearrange for P

You want to have $20,000 in 10 years. The account earns 5% annual interest, compounded annually. Which expression gives the principal P you need to deposit today?

A$P = 20000 \times (1.05)^{10}$

B$P = 20000 - (1.05)^{10}$

C$P = \dfrac{20000}{(1.05)^{10}}$

D$P = 20000 \times (0.05)^{10}$

📖 Explanation

Rearrange $A = P(1 + r)^t$ for $P$:
$P = \dfrac{A}{(1+r)^t} = \dfrac{20000}{(1.05)^{10}} \approx \$12{,}278.27$

This is called finding the Present Value (PV) — what money is worth today.
❌ Trap (A): Multiplies instead of divides — that would give the future value of a future value.

Q 13 Hard % Increase vs. Compound

A population of 50,000 grows at 2% per year. Which expression gives the population after t years?

A$50000 + 50000(0.02)t$

B$50000(1.02)^t$

C$50000(0.02)^t$

D$50000(1.2)^t$

📖 Explanation

Exponential growth: $A = P(1 + r)^t = 50000(1.02)^t$. This is the same structure as annual compound interest with $n=1$.

❌ Trap (A): Linear growth — ignores compounding.
❌ Trap (D): 20% ≠ 2%. Decimal error: $1 + 0.2 = 1.2$ corresponds to 20% growth.

Q 14 Hard Depreciation = decay

A car worth $25,000 depreciates (loses value) at 15% per year. What is its value after 4 years?

A$25000(1.15)^4$

B$25000 - 25000(0.15)(4)$

C$25000(0.15)^4$

D$25000(0.85)^4$

📖 Explanation

Depreciation = exponential decay: $A = P(1 - r)^t = 25000(1 - 0.15)^4 = 25000(0.85)^4 \approx \$13{,}050$.

Key: Decay uses $(1 - r)$, growth uses $(1 + r)$.
❌ Trap (A): $(1.15)^4$ represents 15% growth — not depreciation.
❌ Trap (B): Simple depreciation — ignores compounding effect on the declining base.

Q 15 Hard Two-step: multiple investments

Account X has $6,000 at 4% compounded annually for 3 years. Account Y has $5,500 at 5% compounded annually for 3 years. After 3 years, which account has more money?

AAccount X, because it started with more money

BAccount Y, because the higher interest rate overcomes the lower principal

CThey end up equal

DAccount X, because 4% over 3 years adds up to more

📖 Explanation

X: $6000(1.04)^3 = 6000 \times 1.12486 \approx \$6{,}749$
Y: $5500(1.05)^3 = 5500 \times 1.15763 \approx \$6{,}367$

Wait — Account X wins! Let's re-examine...
X ≈ \$6,749 > Y ≈ \$6,367. The higher principal in X outweighs the rate advantage of Y here.

⚠️ Lesson: Always calculate — don't assume rate or principal alone determines the winner. The correct answer is A — Account X has more.

Q 16 Hard Exponential vs Linear

A bacteria culture starts at 200 cells and doubles every hour. Which equation models the number of cells after h hours?

A$N = 200 + 200h$

B$N = 200h^2$

C$N = 200(2)^h$

D$N = 200(1.02)^h$

📖 Explanation

"Doubles every hour" means multiply by 2 each hour: $N = 200 \cdot 2^h$. This is exponential growth with base 2.

❌ Trap (A): Linear — adds 200 each hour rather than multiplying.
❌ Trap (D): $1.02^h$ = 2% growth per hour — not doubling.

Q 17 Hard Read carefully: what is asked?

An account has $2,500 at 6% compounded monthly. After 2 years, you withdraw $500. Which expression gives the balance immediately after the withdrawal?

A$2500\!\left(1+\tfrac{0.06}{12}\right)^{24} - 500$

B$(2500 - 500)\!\left(1+\tfrac{0.06}{12}\right)^{24}$

C$2500\!\left(1+\tfrac{0.06}{12}\right)^{24}$

D$2000\!\left(1+\tfrac{0.06}{12}\right)^{24}$

📖 Explanation

The $2,500 grows for all 2 years first, then $500 is withdrawn after that growth.
Balance = $2500(1.005)^{24} - 500$

❌ Trap (B): Subtracts \$500 before growth — wrong! The withdrawal happens after 2 years.
❌ Trap (D): Same error — reducing principal first is incorrect here.

Q 18 Hard Effective annual rate

Bank X offers 5% compounded annually. Bank Y offers 4.9% compounded monthly. Which bank offers the better effective annual rate?

ABank X, because 5% compounded annually is already the effective rate

BBank Y, because monthly compounding amplifies returns

CThey are equal

DCannot be compared without knowing the principal

📖 Explanation

Effective Annual Rate (EAR) = $\left(1 + r/n\right)^n - 1$
Bank X EAR: $(1.05)^1 - 1 = 5\%$
Bank Y EAR: $(1 + 0.049/12)^{12} - 1 \approx 5.012\% - 1 = 5.012\%$

Wait — Bank Y's EAR ≈ 5.012% > Bank X's 5%. So Bank Y is slightly better.

⚠️ This is a precision trap. Monthly compounding of 4.9% beats annual 5% slightly. Answer is B.

Q 19 Hard Graph interpretation

The graph of an investment account shows a curve that increases faster and faster over time (concave up). Which type of interest does this account use?

ASimple interest, because it adds a fixed amount each year

BCompound interest, because each year earns interest on prior interest

CSimple interest, because the rate is constant

DNeither — no interest formula produces a curve

📖 Explanation

Simple interest → straight line (linear: $A = P + Prt$)
Compound interest → curve, accelerating upward (exponential: $A = P(1+r)^t$)

The curve "gets steeper over time" because interest is earned on a growing base — classic compound growth shape.

Q 20 Hard Multi-step reasoning

Sarah invests $P at rate r compounded n times per year. After t years, she wants her money to triple. Which equation correctly models this goal?

A$3 = \left(1 + r\right)^t$

B$P = 3P\left(1 + r/n\right)^{nt}$

C$3P = P + P(r)(t)$

D$3 = \left(1 + r/n\right)^{nt}$

📖 Explanation

Triple means $A = 3P$. Substituting into $A = P(1 + r/n)^{nt}$:
$3P = P(1 + r/n)^{nt}$
Divide both sides by $P$: $\boxed{3 = (1 + r/n)^{nt}}$

❌ Trap (A): Omits $n$ — only works when compounding is annual ($n=1$).
❌ Trap (C): Simple interest formula — wrong model.
❌ Trap (B): The sides are flipped and setup is backward.

Compound Interest& Algebra

Compound Interest
& Algebra