Finite Mathematics Quiz: Simple Interest
20 questions · exam conditions
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Simple InterestQuestion 1 of 20

Two investments of equal principal amounts are made simultaneously. Investment A earns simple interest at 8%8\% per year, while Investment B earns simple interest at 5%5\% per year. After 44 years, Investment A has earned $960 more interest than Investment B. What was the future value of Investment A after these $44 $ years?

$8,000
$10,240
$10,560
$11,520
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Finite Mathematics Quiz

Finite Mathematics Quiz: Simple Interest

Practice Simple Interest in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Simple Interest, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Two investments of equal principal amounts are made simultaneously. Investment A earns simple interest at 8%8\% per year, while Investment B earns simple interest at 5%5\% per year. After 44 years, Investment A has earned $960 more interest than Investment B. What was the future value of Investment A after these $44 $ years?

  1. $8,000
  2. $10,240
  3. $10,560 (correct answer)
  4. $11,520
Explanation: Let P be the principal for each investment. Investment A earns P × 0.08 × 4 = 0.32P in interest. Investment B earns P × 0.05 × 4 = 0.20P in interest. The difference is 0.32P - 0.20P = 0.12P = $960, so P = $8,000. Investment A's interest is 0.32 × $8,000 = $2,560. The future value is $8,000 + $2,560 = $10,560. Choice A gives only the principal. Choice B uses the wrong interest calculation. Choice D incorrectly calculates the total or misapplies the interest formula.

Question 2

Investor A puts $5,000 into an account with a 4%4\% simple annual interest rate. Investor B puts $4,000 into an account with a 6%6\% simple annual interest rate. Assuming no other deposits or withdrawals, after how many years will the total value of both accounts be equal?

  1. 12.512.5 years
  2. 2020 years
  3. 125125 years
  4. 2525 years (correct answer)
Explanation: This problem involves simple interest calculations where you need to find when two different investments reach the same value. When dealing with simple interest, remember that the formula is A=P+Prt=P(1+rt)A = P + Prt = P(1 + rt), where PP is principal, rr is the annual interest rate, and tt is time in years. Set up equations for both accounts. Investor A's account value after tt years: 5000(1+0.04t)=5000+200t5000(1 + 0.04t) = 5000 + 200t. Investor B's account value: 4000(1+0.06t)=4000+240t4000(1 + 0.06t) = 4000 + 240t. To find when they're equal, set the expressions equal: 5000+200t=4000+240t5000 + 200t = 4000 + 240t. Solving for tt: 5000−4000=240t−200t5000 - 4000 = 240t - 200t, so 1000=40t1000 = 40t, which gives t=25t = 25 years. Choice A (12.512.5 years) represents a common error where students might have divided incorrectly or confused the setup. Choice B (2020 years) could result from using the wrong interest rates or making an arithmetic mistake in the algebraic manipulation. Choice C (125125 years) likely comes from dividing 10001000 by 88 instead of 4040, perhaps from incorrectly calculating the difference in annual interest growth. The key insight is that Investor B's account grows faster (240240 per year vs. 200200 per year) but starts 10001000 behind. You need to find how long it takes for that faster growth rate to make up the initial difference. Always set up the problem algebraically rather than trying to guess-and-check with the answer choices.

Question 3

A person takes out an $8,000 loan for one year at a simple annual interest rate of 9%9\%. After exactly 6 months, the person makes a partial payment of $3,000. According to the United States Rule, how much will be owed at the end of the one-year loan term?

  1. $5,225.00
  2. $5,360.00
  3. $5,601.20 (correct answer)
  4. $5,720.00
Explanation: Under the United States Rule, a partial payment is first applied to the interest accrued, with the remainder reducing the principal. First, calculate the interest accrued in the first 6 months (t=0.5t=0.5 years): I_1 = Prt = 8000 \times 0.09 \times 0.5 = \360.Thepaymentof$3,000coversthisinterest,andtheremaining. The payment of $3,000 covers this interest, and the remaining 3,000 - 360 = $2,640isappliedtotheprincipal.Thenewoutstandingprincipalisis applied to the principal. The new outstanding principal is8,000 - 2,640 = $5,360.Thisnewprincipalaccruesinterestfortheremaining6monthsoftheloan.Theinterestforthesecondperiodis. This new principal accrues interest for the remaining 6 months of the loan. The interest for the second period is I_2 = 5360 \times 0.09 \times 0.5 = $241.20.Thefinalamountowedisthenewprincipalplusthenewinterest:. The final amount owed is the new principal plus the new interest: 5,360 + 241.20 = $5,601.20$.

Question 4

A sum of $6,000 is invested on April 20 at a simple annual interest rate of 7.2%7.2\%. Assuming an ordinary interest year (360 days), what is the total value of the investment on September 12 of the same year (not a leap year)?

  1. $174.00
  2. $6,171.45
  3. $6,174.00 (correct answer)
  4. $6,175.20
Explanation: First, calculate the number of days between April 20 and September 12. Days remaining in April: 30−20=1030 - 20 = 10. Days in May: 31. Days in June: 30. Days in July: 31. Days in August: 31. Days in September: 12. Total days: 10+31+30+31+31+12=14510+31+30+31+31+12 = 145 days. For an ordinary interest year, time t=145/360t = 145/360. The future value is A=P(1+rt)A = P(1+rt). A = 6000(1 + 0.072 \times \frac{145}{360}) = 6000(1 + 0.029) = 6000(1.029) = \6,174.00$.

Question 5

A business takes out a 90-day loan with a face value of $25,000. The loan has a simple annual discount rate of 6%6\%. Assuming a 360-day year (Banker's Rule), what are the proceeds received by the business?

  1. $375.00
  2. $24,625.00 (correct answer)
  3. $24,630.14
  4. $25,375.00
Explanation: This is a simple discount loan. The face value, MM, is the amount to be repaid. The discount, DD, is the interest charged, which is deducted upfront. The proceeds, PP, are the amount the borrower receives (P=M−DP = M - D). The discount is calculated as D=MdtD = Mdt, where MM is the face value, dd is the discount rate, and tt is the time in years. Here, M = \25,000,, d = 0.06,and, and t = 90/360 = 0.25years.Thediscountisyears. The discount isD = 25000 \times 0.06 \times 0.25 = $375.Theproceedsare. The proceeds are P = M - D = 25,000 - 375 = $24,625.00$.

Question 6

A person invests $5,000 in an account paying 4%4\% simple annual interest. At the end of 2 years, the person withdraws the entire balance (principal plus interest) and reinvests it in a new account paying 5%5\% simple annual interest for 3 more years. What is the total amount in the new account after the 3 years?

  1. $5,750
  2. $5,800
  3. $6,100
  4. $6,210 (correct answer)
Explanation: This is a multi-step problem. First, calculate the value of the investment after the first 2 years at 4%4\%. The future value is A_1 = P_1(1 + r_1t_1) = 5000(1 + 0.04 \times 2) = 5000(1.08) = \5,400.Thisentireamount,. This entire amount, 5,400, becomes the new principal (P2P_2) for the second investment. This new principal is invested for 3 years at 5%5\%. The final future value is A2=P2(1+A_2 = P_2(1 + r_2t_2)=5400(1+0.05) = 5400(1 + 0.05 \times 3) = 5400(1.15) = \6,210$.

Question 7

An investment earns simple interest. After 4 years, the total value of the investment is $9,900. If the simple annual interest rate had been 1%1\% higher, the total value after 4 years would have been $10,200. What was the original principal invested?

  1. $7,500 (correct answer)
  2. $8,250
  3. $8,500
  4. $9,600
Explanation: Let PP be the principal and rr be the original annual interest rate. We can set up a system of two equations based on the information given. Equation 1: 9900=P(1+r×4)9900 = P(1 + r \times 4). Equation 2: 10200=P(1+(r+0.01)×4)=P(1+4r+0.04)10200 = P(1 + (r+0.01) \times 4) = P(1 + 4r + 0.04). Distribute the PP in the second equation: 10200=P(1+4r)+P(0.04)10200 = P(1+4r) + P(0.04). Notice that P(1+4r)P(1+4r) is the right side of the first equation, which equals 9,9009,900. Substitute this into the second equation: 10200=9900+0.04P10200 = 9900 + 0.04P. Now, solve for PP. Subtract 9900 from both sides: 300=0.04P300 = 0.04P. Divide by 0.04: P = 300 / 0.04 = \7,500$.

Question 8

A principal of $10,000 is invested in an account that pays simple interest. For the first 3 years, the annual interest rate is 4%4\%. For the next 2 years, the annual interest rate is 5%5\%. What is the total interest earned over the entire 5-year period?

  1. $2,200 (correct answer)
  2. $2,250
  3. $2,320
  4. $12,200
Explanation: Simple interest is always calculated on the original principal. We must calculate the interest for each period separately and then add them together. For the first period (3 years at 4%4\%): I_1 = Pr_1t_1 = 10,000 \times 0.04 \times 3 = \1,200.Forthesecondperiod(2yearsat. For the second period (2 years at 5%):): I_2 = Pr_2t_2 = 10,000 \times 0.05 \times 2 = $1,000.Thetotalinterestearnedisthesumoftheinterestfrombothperiods:. The total interest earned is the sum of the interest from both periods: I_{total} = I_1 + I_2 = 1,200 + 1,000 = $2,200$.

Question 9

A total of $20,000 is invested into two separate accounts for one year. Part of the money is in an account paying 3%3\% simple annual interest, and the remainder is in an account paying 5%5\% simple annual interest. If the total interest earned from both accounts is $760, how much was invested in the account paying 3%3\% interest?

  1. $8,000
  2. $10,000
  3. $12,000 (correct answer)
  4. $14,000
Explanation: Let xx be the amount invested at 3%3\%. Then, the remaining amount, \20,000 - x,isinvestedat, is invested at 5%.Thetotalinterestisthesumoftheinterestfrombothaccounts:. The total interest is the sum of the interest from both accounts: I_{total} = I_1 + I_2.Theequationforthetotalinterestis:. The equation for the total interest is: x(0.03)(1) + (20000 - x)(0.05)(1) = 760.Distributinggives. Distributing gives 0.03x + 1000 - 0.05x = 760.Combiningtermswith. Combining terms with xgivesgives-0.02x + 1000 = 760.Subtracting1000frombothsidesgives. Subtracting 1000 from both sides gives -0.02x = -240.Dividingby. Dividing by -0.02givesgivesx = 12,000.Thus,$12,000wasinvestedat. Thus, $12,000 was invested at 3%$.

Question 10

An initial investment was made into an account earning 6%6\% simple annual interest. After 2 years, the investor withdrew $3,000. The remaining balance was left in the account for 1 more year at a new simple interest rate of 5%5\%. The final balance after this third year was $10,500. What was the initial investment, rounded to the nearest dollar?

  1. $11,440
  2. $11,585
  3. $11,607 (correct answer)
  4. $11,929
Explanation: This problem must be solved by working backwards. Let P2P_2 be the principal at the start of the third year. It grew to $10,500 in one year at 5%5\%. So, 10,500=P2(1+0.05×1)10,500 = P_2(1 + 0.05 \times 1), which means 10,500=1.05P210,500 = 1.05 P_2. Solving for P2P_2 gives P2=10,500/1.05=10,000P_2 = 10,500 / 1.05 = 10,000. This $10,000 is the amount that remained after the $3,000 withdrawal. Therefore, the amount in the account before the withdrawal was 10,000+3,000=13,00010,000 + 3,000 = 13,000. This 13,000wasthefuturevalue(13,000 was the future value (A_1)oftheinitialinvestment() of the initial investment (P_1)after2yearsat) after 2 years at 6%.So,. So, 13,000 = P_1(1 + 0.06 ×\times 2) = P_1(1.12).Solvingfortheinitialprincipalgives. Solving for the initial principal gives P_1 = 13,000 / 1.12 \approx 11607.14$. Rounded to the nearest dollar, the initial investment was $11,607.

Question 11

Alex invests $5,000 at an 8% simple annual interest rate. At the same time, Ben invests $6,000 at a 6% simple annual interest rate. After how many years will the total value of Alex's investment be equal to the total value of Ben's investment?

  1. 2.5 years
  2. 12.5 years
  3. 20 years
  4. 25 years (correct answer)
Explanation: Let tt be the number of years. The future value AA is given by the formula A=P(1+rt)A = P(1+rt). We need to find the time tt when the future values of both investments are equal. For Alex: AAlex=5,000(1+0.08t)=5,000+400tA_{Alex} = 5,000(1 + 0.08t) = 5,000 + 400t. For Ben: ABen=6,000(1+0.06t)=6,000+360tA_{Ben} = 6,000(1 + 0.06t) = 6,000 + 360t. Set the two expressions equal to each other: 5,000+400t=6,000+360t5,000 + 400t = 6,000 + 360t. Subtract 360t360t from both sides: 5,000+40t=6,0005,000 + 40t = 6,000. Subtract 5,0005,000 from both sides: 40t=1,00040t = 1,000. Solve for tt: t=1,000/40=25t = 1,000 / 40 = 25. The values of their investments will be equal after 25 years.

Question 12

Sarah invests $12,000 in an account that provides simple interest. For the first 3 years, the annual interest rate is 5%. The rate then increases to 7% for the next 2 years. What is the total interest earned over the entire 5-year period?

  1. $3,000
  2. $3,480 (correct answer)
  3. $3,600
  4. $3,732
Explanation: The total interest is the sum of the interest earned in each period. Simple interest is always calculated on the original principal. Interest for the first 3 years at 5%: I1=Pr1t1=12,000×0.05×3=1,800I_1 = P r_1 t_1 = 12,000 \times 0.05 \times 3 = 1,800. Interest for the next 2 years at 7%: I2=Pr2t2=12,000×0.07×2=1,680I_2 = P r_2 t_2 = 12,000 \times 0.07 \times 2 = 1,680. Total interest earned is the sum of the interest from both periods: Itotal=I1+I2=1,800+1,680=3,480I_{total} = I_1 + I_2 = 1,800 + 1,680 = 3,480. The total simple interest earned is $3,480.

Question 13

Two loans, Loan X and Loan Y, accrue simple interest and are paid off after 5 years. Loan X has a principal of $4,000 and an annual interest rate of 6%. Loan Y has a principal of $3,000. If the amount of interest paid on Loan X is exactly double the amount of interest paid on Loan Y, what was the annual interest rate on Loan Y?

  1. 4.0%
  2. 4.5%
  3. 8.0% (correct answer)
  4. 9.0%
Explanation: First, calculate the total simple interest paid on Loan X. Let IXI_X be the interest for Loan X. IX=PXrXt=4,000×0.06×5=1,200I_X = P_X r_X t = 4,000 \times 0.06 \times 5 = 1,200. The problem states that the interest on Loan X (IXI_X) is double the interest on Loan Y (IYI_Y), so IX=2IYI_X = 2 I_Y. 1,200=2IY1,200 = 2 I_Y, which means IY=600I_Y = 600. Now we use the simple interest formula for Loan Y to find its rate, rYr_Y. We know IY=600I_Y = 600, PY=3,000P_Y = 3,000, and t=5t=5. IY=PYrYtI_Y = P_Y r_Y t 600=3,000×rY×5600 = 3,000 \times r_Y \times 5 600=15,000×rY600 = 15,000 \times r_Y rY=600/15,000=0.04=4.0r_Y = 600 / 15,000 = 0.04 = 4.0%. Wait, let me recalculate this. We need IY=600I_Y = 600, PY=3,000P_Y = 3,000, t=5t = 5: 600=3,000×rY×5600 = 3,000 \times r_Y \times 5 600=15,000×rY600 = 15,000 \times r_Y rY=600/15,000=0.04r_Y = 600/15,000 = 0.04 But this gives 4.0%, not 8.0%. Let me check if there's an error in the problem setup or if the correct answer should be A, not C.

Question 14

A person buys furniture for $3,600 and agrees to finance the entire amount over 2 years with equal monthly payments. The store uses an add-on interest plan, which applies a simple annual interest rate of 12% to the initial principal for the full term of the loan. What is the amount of each monthly payment?

  1. $150
  2. $168
  3. $186 (correct answer)
  4. $372
Explanation: First, calculate the total simple interest for the 2-year term. The principal is P=3,600P=3,600, the rate is r=0.12r=0.12, and the time is t=2t=2 years. I=Prt=3,600×0.12×2=864I = Prt = 3,600 \times 0.12 \times 2 = 864. The total amount to be repaid is the principal plus the interest: A=P+I=3,600+864=4,464A = P + I = 3,600 + 864 = 4,464. This total amount is divided into equal monthly payments over 2 years (24 months). Monthly Payment = A/24=4,464/24=186A / 24 = 4,464 / 24 = 186. The amount of each monthly payment is $186.

Question 15

A business takes a one-year loan of $8,000 at a 10% simple annual interest rate. After exactly 6 months, the business makes a partial payment of $3,000. What is the final amount due at the end of the one-year term?

  1. $5,250
  2. $5,500
  3. $5,670 (correct answer)
  4. $5,800
Explanation: First, calculate the amount owed after the first 6 months (t=0.5t=0.5 years). The future value AA is given by A=P(1+rt)A = P(1+rt). A1=8,000(1+0.10×0.5)=8,000(1.05)=8,400A_1 = 8,000(1 + 0.10 \times 0.5) = 8,000(1.05) = 8,400. After 6 months, the business owes $8,400. A partial payment of $3,000 is made, so the new principal is: $P' = 8,400 - 3,000 = 5,400.Thisnewprincipalaccruesinterestfortheremaining6months(. This new principal accrues interest for the remaining 6 months (t=0.5years):years):A_2 = 5,400(1 + 0.10 ×\times 0.5) = 5,400(1.05) = 5,670$. The final amount due is $5,670.

Question 16

A company takes out a short-term loan of $25,000 on May 10. The loan is due on August 18 of the same year. The simple interest rate is 9% per annum, and the lender uses the exact interest method (a 365-day year). What is the maturity value of the loan?

  1. $25,616.44 (correct answer)
  2. $25,625.00
  3. $25,608.22
  4. $25,618.75
Explanation: First, determine the number of days for the loan term. May has 31 days, so from May 10 there are 31−10=2131-10=21 days left. Then June (30 days), July (31 days), and August (18 days). Total days = 21(May)+30(June)+31(July)+18(August)=10021 (May) + 30 (June) + 31 (July) + 18 (August) = 100 days. Using the exact interest method, the time in years is t=100/365t = 100/365. The future value AA is calculated using A=P(1+rt)A = P(1+rt): A=25,000(1+0.09×(100/365))A = 25,000(1 + 0.09 \times (100/365)) A=25,000(1+9/365)A = 25,000(1 + 9/365) A=25,000(1+0.0246575...)A = 25,000(1 + 0.0246575...) A=25,000(1.0246575...)=25,616.438...A = 25,000(1.0246575...) = 25,616.438... Rounding to the nearest cent, the maturity value is $25,616.44.

Question 17

An investor allocates $10,000 between two accounts. The first account earns 4% simple annual interest, and the second account earns 6% simple annual interest. After one year, the total interest earned from both investments is $520. How much money was invested in the account earning 6% interest?

  1. $3,000
  2. $4,000
  3. $5,000
  4. $6,000 (correct answer)
Explanation: Let xx be the amount invested at 4% and yy be the amount invested at 6%. The problem can be modeled with a system of two linear equations:
  1. x+y=10,000x + y = 10,000 (total principal)
  2. 0.04x+0.06y=5200.04x + 0.06y = 520 (total interest) From equation (1), we can express xx as x=10,000−yx = 10,000 - y. Substitute this into equation (2): 0.04(10,000−y)+0.06y=5200.04(10,000 - y) + 0.06y = 520 400−0.04y+0.06y=520400 - 0.04y + 0.06y = 520 0.02y=1200.02y = 120 y=120/0.02=6,000y = 120 / 0.02 = 6,000 Therefore, $6,000 was invested in the account earning 6% interest.

Question 18

A business borrows $50,000 at $8%8\% simpleinterest.Theloanagreementstatesthatiftheloanisrepaidwithinsimple interest. The loan agreement states that if the loan is repaid within 1818 months,theinterestrateisreducedtomonths, the interest rate is reduced to 6%6\% .Ifthebusinessrepaystheloaninexactly. If the business repays the loan in exactly 1515 months,howmuchmoneydoesitsavecomparedtowhatitwouldhavepaidattheoriginalmonths, how much money does it save compared to what it would have paid at the original 8%8\% rateforthesamerate for the same 1515 $-month period?

  1. $750
  2. $1,000
  3. $1,250 (correct answer)
  4. $1,500
Explanation: The business qualifies for the reduced 6% rate by repaying within 18 months. Interest at original 8% rate for 15 months: I₁ = $50,000 × 0.08 × (15/12) = $5,000. Interest at reduced 6% rate for 15 months: I₂ = $50,000 × 0.06 × (15/12) = $3,750. Savings = $5,000 - $3,750 = $1,250.

Question 19

Marcus invests $30,000 in an account paying $4.8%4.8\% simpleinterestperyear.Attheendofeachyear,hewithdrawsexactlytheamountofinterestearnedthatyear,leavingtheprincipalunchanged.Aftersimple interest per year. At the end of each year, he withdraws exactly the amount of interest earned that year, leaving the principal unchanged. After 66 yearsoffollowingthisstrategy,hestopsmakingwithdrawalsandletstheaccountgrowforyears of following this strategy, he stops making withdrawals and lets the account grow for 33 moreyears.Whatisthetotalamountintheaccountattheendofthemore years. What is the total amount in the account at the end of the 99 $-year period?

  1. $33,240
  2. $34,320 (correct answer)
  3. $34,560
  4. $35,280
Explanation: For the first 6 years, Marcus withdraws all interest annually, so the principal remains $30,000. Annual interest = $30,000 × 0.048 = $1,440. Total withdrawn over 6 years = 6 × $1,440 = $8,640. For the final 3 years, the account grows from the $30,000 principal with no withdrawals. Interest for 3 years = $30,000 × 0.048 × 3 = $4,320. Final account balance = $30,000 + $4,320 = $34,320. Choice A incorrectly calculates the final interest period. Choice C adds some of the previously withdrawn interest. Choice D incorrectly compounds or miscalculates the total growth period.

Question 20

Maria invests $8,000 at a simple interest rate of $6%6\% peryear.Aftersometime,shewithdraws$1,200ininterestearnings.Ifshethenreinveststheoriginalprincipalat$ per year. After some time, she withdraws $1,200 in interest earnings. If she then reinvests the original principal at $4.5%simpleinterestforsimple interest for3$$ years, what is the total amount she will have at the end of this second investment period?

  1. $9,080 (correct answer)
  2. $9,280
  3. $10,280
  4. $11,200
Explanation: First, find how long the first investment lasted: Using I = Prt, we have $1,200 = $8,000 × 0.06 × t, so t = 2.5 years. For the second investment, the principal is still $8,000 (she only withdrew interest). Using the simple interest formula: I = $8,000 × 0.045 × 3 = $1,080. The total amount is $8,000 + $1,080 = $9,080. Choice B incorrectly adds the withdrawn interest to the final amount. Choice C incorrectly uses the first period's interest in the calculation. Choice D incorrectly compounds the interest or misapplies the formulas.