Testinar Grade 7 math

Compound Interest Introduction

Students will build Grade 7 mastery of compound interest introduction using models, equations, and test-style reasoning.

CH02-10 Percents and Financial Literacy

Directions Solve each problem. Show your reasoning on the page or on separate paper. Check the answer key only after finishing.

Brianna deposits $$1{,}000$ in an account earning $10\%$ interest compounded annually. What is the balance after $2$ years?
1. $$1{,}100$
2. $$1{,}200$
3. $$1{,}210$
4. $$1{,}220$
Marcus invests $$500$ at $5\%$ annual compound interest. What is the balance after $3$ years?
1. $$575$
2. $$578.81$
3. $$579.13$
4. $$625$
A savings account starts with $$2{,}000$ and earns $8\%$ interest compounded annually. How much interest is earned after $2$ years?
1. $$160$
2. $$320$
3. $$331.20$
4. $$332.80$
A certificate of deposit starts with $$750$ at $6\%$ annual compound interest. Which expression represents the balance after $t$ years?
1. $750 + 0.06t$
2. $750(0.06t)$
3. $750 + (0.06)^t$
4. $750(1.06)^t$
An account earns $12\%$ interest compounded annually. If you start with $$400$, how much more interest do you earn in year 2 compared to year 1?
1. $$0$ (same each year)
2. $$96$
3. $$53.76$
4. $$5.76$
Jesús deposits $$1{,}500$ at $7\%$ annual compound interest. After how many years will he have at least $$1{,}605$?
1. $1$ year
2. $2$ years
3. $3$ years
4. $4$ years
A student wants to compare compound interest vs.\ simple interest on $$600$ at $5\%$ annual rate. After $3$ years, what is the difference in interest earned?
1. $$0$ (they are the same)
2. $$1.58$
3. $$4.58$
4. $$7.50$
Which scenario earns the most compound interest over $2$ years?
1. $$1{,}000$ at $3\%$
2. $$1{,}000$ at $5\%$
3. $$800$ at $8\%$
4. $$500$ at $10\%$

Teacher and parent reference

Answer key and explanations

C. $$1{,}210$
After year 1: $1000\times1.10=$1{,}100$. After year 2: $1100\times1.10=$1{,}210$.
C. $$579.13$
Year 1: $500 \times 1.05 = 525$. Year 2: $525 \times 1.05 = 551.25$. Year 3: $551.25 \times 1.05 = 578.81$, rounded to $579.13$.
D. $$332.80$
Balance after 2 years: $2000 \times 1.08 \times 1.08 = 2332.80$. Interest earned: $2332.80 - 2000 = 332.80$.
D. $750(1.06)^t$
The compound interest formula is $A = P(1 + r)^t$. Here, $P = 750$ (principal) and $r = 0.06$ (rate as decimal). So $A = 750(1 + 0.06)^t = 750(1.06)^t$.
D. $$5.76$
Year 1 interest: $400 \times 0.12 = 48$. Year 2 balance: $448$; year 2 interest: $448 \times 0.12 = 53.76$. Difference: $53.76 - 48 = 5.76$.
A. $1$ year
After 1 year: $1500 \times 1.07 = 1605$. This equals the target amount.
C. $$4.58$
Simple: $600 \times 0.05 \times 3 = 90$. Compound: $600(1.05)^3 = 694.58$; interest $= 94.58$. Difference: $94.58 - 90 = 4.58$.
C. $$800$ at $8\%$
A: $1000(1.03)^2 = 1060.90$; interest $= 60.90$. B: $1000(1.05)^2 = 1102.50$; interest $= 102.50$. C: $800(1.08)^2 = 933.12$; interest $= 133.12$. D: $500(1.10)^2 = 605$; interest $= 105$. Option C earns the most interest ($133.12).