College Algebra Quiz: Annuities And Loan Payments Intro
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Annuities And Loan Payments IntroQuestion 1 of 8

A loan of $15,000 is to be repaid with equal monthly payments over $44 yearsatyears at 9%9\% $ annual interest compounded monthly. What is the monthly payment amount?

15000⋅0.0075(1.0075)48(1.0075)48−115000 \cdot \frac{0.0075(1.0075)^{48}}{(1.0075)^{48} - 1}
15000⋅0.09(1.09)4(1.09)4−115000 \cdot \frac{0.09(1.09)^{4}}{(1.09)^{4} - 1}
15000⋅(1.0075)48−10.0075⋅(1.0075)4815000 \cdot \frac{(1.0075)^{48} - 1}{0.0075 \cdot (1.0075)^{48}}
15000⋅0.0075(1.0075)48−115000 \cdot \frac{0.0075}{(1.0075)^{48} - 1}
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College Algebra Quiz

College Algebra Quiz: Annuities And Loan Payments Intro

Practice Annuities And Loan Payments Intro in College Algebra 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 Annuities And Loan Payments Intro, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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

A loan of $15,000 is to be repaid with equal monthly payments over $44 yearsatyears at 9%9\% $ annual interest compounded monthly. What is the monthly payment amount?

  1. 15000⋅0.0075(1.0075)48(1.0075)48−115000 \cdot \frac{0.0075(1.0075)^{48}}{(1.0075)^{48} - 1} (correct answer)
  2. 15000⋅0.09(1.09)4(1.09)4−115000 \cdot \frac{0.09(1.09)^{4}}{(1.09)^{4} - 1}
  3. 15000⋅(1.0075)48−10.0075⋅(1.0075)4815000 \cdot \frac{(1.0075)^{48} - 1}{0.0075 \cdot (1.0075)^{48}}
  4. 15000⋅0.0075(1.0075)48−115000 \cdot \frac{0.0075}{(1.0075)^{48} - 1}
Explanation: For loan payments, use PMT = PV × [r(1+r)^n]/[(1+r)^n - 1] where PV = $15,000, r = 0.09/12 = 0.0075, and n = 48 months. This gives the formula in choice A. Choice B uses annual periods instead of monthly. Choice C inverts the fraction (gives present value of annuity instead of payment). Choice D is missing the (1+r)^n term in the numerator.

Question 2

A company establishes a sinking fund to replace equipment in 88 years. They deposit $2,000 at the end of each quarter into an account earning $5%5\% annualinterestcompoundedquarterly.Afterannual interest compounded quarterly. After 55 yearsofdeposits,theinterestratechangestoyears of deposits, the interest rate changes to 6%6\% annualinterestcompoundedquarterly.Whatisthetotalfundvalueafterthefullannual interest compounded quarterly. What is the total fund value after the full 88 $ years?

  1. 2000⋅(1.0125)20−10.0125+2000⋅(1.015)12−10.015⋅(1.0125)202000 \cdot \frac{(1.0125)^{20} - 1}{0.0125} + 2000 \cdot \frac{(1.015)^{12} - 1}{0.015} \cdot (1.0125)^{20}
  2. 2000⋅(1.0125)32−10.0125+2000⋅(1.015)12−10.0152000 \cdot \frac{(1.0125)^{32} - 1}{0.0125} + 2000 \cdot \frac{(1.015)^{12} - 1}{0.015}
  3. 2000⋅(1.05)5−10.05⋅(1.06)3+2000⋅(1.06)3−10.062000 \cdot \frac{(1.05)^{5} - 1}{0.05} \cdot (1.06)^{3} + 2000 \cdot \frac{(1.06)^{3} - 1}{0.06}
  4. 2000⋅(1.0125)20−10.0125⋅(1.015)12+2000⋅(1.015)12−10.0152000 \cdot \frac{(1.0125)^{20} - 1}{0.0125} \cdot (1.015)^{12} + 2000 \cdot \frac{(1.015)^{12} - 1}{0.015} (correct answer)
Explanation: When you encounter a sinking fund problem with changing interest rates, you need to treat it as two separate annuity calculations that connect at the rate change point. First, identify the key information: quarterly deposits of $2,000, initial rate of 5% annually (1.25% quarterly), rate changes to 6% annually (1.5% quarterly) after 5 years, total duration is 8 years. The correct approach requires two calculations. For the first 5 years (20 quarters), calculate the future value of the annuity at 1.25% quarterly: $2000⋅(1.0125)20−10.01252000 \cdot \frac{(1.0125)^{20} - 1}{0.0125} .Thisamountthengrowsfortheremaining3years(12quarters)atthenew1.5. This amount then grows for the remaining 3 years (12 quarters) at the new 1.5% rate: multiply by (1.015)12(1.015)^{12} $. For the last 3 years, calculate the future value of those deposits separately: 2000 \cdot \frac{(1.015)^{12} - 1}{0.015} . The total is the sum of both parts, which matches answer D. Answer A incorrectly compounds the second part backwards at the old rate. Answer B adds the terms without properly accounting for the time value - it fails to grow the first period's accumulated value through the second period. Answer C uses annual compounding instead of quarterly, treating this as if deposits occur annually rather than quarterly. Study tip: In multi-period problems with rate changes, always break the timeline at the change point. Calculate what happens in each period separately, then ensure earlier accumulations grow through all remaining periods at the appropriate rates.

Question 3

An individual retirement account receives deposits of $300 at the beginning of each month for $2525 years,earningyears, earning 8%8\% annualinterestcompoundedmonthly.Whatistheaccountbalanceattheendoftheannual interest compounded monthly. What is the account balance at the end of the 2525 $ years?

  1. 300⋅(1.08)25−10.08⋅(1.08)300 \cdot \frac{(1.08)^{25} - 1}{0.08} \cdot (1.08)
  2. 300⋅(1+0.0812)300−10.0812300 \cdot \frac{(1 + \frac{0.08}{12})^{300} - 1}{\frac{0.08}{12}}
  3. 300⋅(1+0.0812)300−10.0812⋅(1+0.0812)300 \cdot \frac{(1 + \frac{0.08}{12})^{300} - 1}{\frac{0.08}{12}} \cdot (1 + \frac{0.08}{12}) (correct answer)
  4. 300⋅(1+0.0812)301−10.0812300 \cdot \frac{(1 + \frac{0.08}{12})^{301} - 1}{\frac{0.08}{12}}
Explanation: When you encounter a problem about regular deposits into an account with compound interest, you're dealing with an annuity. The key distinction is whether deposits occur at the beginning or end of each period, as this affects which formula to use. This problem involves deposits at the beginning of each month (an "annuity due"), so each payment earns interest for one additional period compared to end-of-period deposits. For monthly compounding at 8% annual interest, the monthly rate is 0.0812\frac{0.08}{12}. Over 25 years with monthly deposits, you have 25×12=30025 \times 12 = 300 payments. The standard annuity formula for end-of-period payments is: PMT⋅(1+r)n−1rPMT \cdot \frac{(1+r)^n - 1}{r} Since these are beginning-of-period payments, each deposit earns one extra month of interest, so we multiply by (1+r)(1+r): PMT⋅(1+r)n−1r⋅(1+r)PMT \cdot \frac{(1+r)^n - 1}{r} \cdot (1+r) Substituting our values: 300⋅(1+0.0812)300−10.0812⋅(1+0.0812)300 \cdot \frac{(1 + \frac{0.08}{12})^{300} - 1}{\frac{0.08}{12}} \cdot (1 + \frac{0.08}{12}) This matches choice C. Why the others are wrong:
  • A uses annual compounding (0.08 and 25 periods) instead of monthly
  • B uses the end-of-period annuity formula, missing the extra interest period for beginning-of-month deposits
  • D incorrectly uses 301 periods instead of the multiplication factor for annuity due
Study tip: Always identify whether deposits occur at the beginning or end of periods. Beginning-of-period deposits require multiplying the standard annuity formula by (1+r)(1+r) to account for the extra compounding period.

Question 4

Maria deposits $500 at the end of each quarter into an account that earns $6%6\% annualinterestcompoundedquarterly.Aftermakingdepositsforexactlyannual interest compounded quarterly. After making deposits for exactly 33 $ years, she immediately withdraws all the money. Which expression gives the total amount she receives?

  1. 500⋅(1.015)12−10.015500 \cdot \frac{(1.015)^{12} - 1}{0.015} (correct answer)
  2. 500⋅(1.06)12−10.06500 \cdot \frac{(1.06)^{12} - 1}{0.06}
  3. 500⋅(1.015)13−10.015500 \cdot \frac{(1.015)^{13} - 1}{0.015}
  4. 500⋅(1.06)3−10.06500 \cdot \frac{(1.06)^{3} - 1}{0.06}
Explanation: This is an ordinary annuity with quarterly payments of $500, quarterly interest rate of 6%/4 = 1.5% = 0.015, and n = 12 payments (4 per year × 3 years). The annuity formula is PMT × [(1+r)^n - 1]/r = 500 × [(1.015)^12 - 1]/0.015. Choice B uses annual rate instead of quarterly rate. Choice C incorrectly uses 13 periods instead of 12. Choice D uses annual compounding instead of quarterly.

Question 5

A lottery winner chooses to receive $75,000 per year for $2020 years(firstpaymentinoneyear)insteadofalumpsumtoday.Ifmoneycanbeinvestedatyears (first payment in one year) instead of a lump sum today. If money can be invested at 5.5%5.5\% $ annual interest, what lump sum today would be equivalent to this annuity?

  1. 75000⋅1−(1.055)−210.055⋅(1.055)75000 \cdot \frac{1 - (1.055)^{-21}}{0.055} \cdot (1.055)
  2. 75000⋅(1.055)20−10.05575000 \cdot \frac{(1.055)^{20} - 1}{0.055}
  3. 75000⋅1−(1.055)−200.05575000 \cdot \frac{1 - (1.055)^{-20}}{0.055} (correct answer)
  4. 75000⋅20(1.055)20\frac{75000 \cdot 20}{(1.055)^{20}}
Explanation: When you encounter annuity problems, you're dealing with present value calculations—finding what a series of future payments is worth today. Since the first payment occurs in one year (not immediately), this is an ordinary annuity. The present value of an ordinary annuity uses the formula: PV=PMT⋅1−(1+r)−nrPV = PMT \cdot \frac{1 - (1 + r)^{-n}}{r}, where PMT is the payment amount, r is the interest rate, and n is the number of periods. Here, you have PMT=$75,000PMT = \$75,000, r=0.055r = 0.055, and n=20n = 20 payments. Substituting these values gives: 75000⋅1−(1.055)−200.05575000 \cdot \frac{1 - (1.055)^{-20}}{0.055}. This matches answer choice C perfectly. Answer A uses (1.055)−21(1.055)^{-21} instead of (1.055)−20(1.055)^{-20} and multiplies by an extra (1.055)(1.055) factor. The 21 periods would apply if there were 21 payments, and the extra multiplication suggests confusion between ordinary and annuity due formulas. Answer B shows (1.055)20−10.055\frac{(1.055)^{20} - 1}{0.055}, which is the future value formula for an ordinary annuity. This calculates what the payments would be worth after 20 years, not their present value today. Answer D simply divides the total payments by a discount factor, which isn't how present value works. You can't just discount the sum of all payments as if they were a single lump sum. Remember: for ordinary annuities (first payment in one year), use the present value formula with the negative exponent. Watch out for future value formulas—they look similar but have positive exponents.

Question 6

A company wants to accumulate $500,000 for equipment replacement in $66 years.Theyplantomakequarterlydepositsintoanaccountearningyears. They plan to make quarterly deposits into an account earning 6.8%6.8\% annualinterestcompoundedquarterly.Duetocashflowconstraints,theycanonlyafforddepositsof$15,000perquarterforthefirst$ annual interest compounded quarterly. Due to cash flow constraints, they can only afford deposits of $15,000 per quarter for the first $3years.Whatquarterlydepositamountisneededforthefinalyears. What quarterly deposit amount is needed for the final3$$ years to reach their goal?

  1. 500000−15000⋅(1.017)24−10.017(1.017)12−10.017\frac{500000 - 15000 \cdot \frac{(1.017)^{24} - 1}{0.017}}{\frac{(1.017)^{12} - 1}{0.017}}
  2. 500000−15000⋅(1.017)12−10.017⋅(1.017)12(1.017)12−10.017\frac{500000 - 15000 \cdot \frac{(1.017)^{12} - 1}{0.017} \cdot (1.017)^{12}}{\frac{(1.017)^{12} - 1}{0.017}} (correct answer)
  3. 500000(1.017)12−10.017−15000\frac{500000}{\frac{(1.017)^{12} - 1}{0.017}} - 15000
  4. 500000−15000⋅12⋅(1.017)12(1.017)12−10.017\frac{500000 - 15000 \cdot 12 \cdot (1.017)^{12}}{\frac{(1.017)^{12} - 1}{0.017}}
Explanation: When you encounter annuity problems with changing payment amounts, you need to treat each period separately and account for how earlier deposits continue to grow. This problem involves two distinct annuity periods. For the first 3 years (12 quarters), $15,000 deposits earn interest. For the final 3 years, unknown deposits of amount $Paremade.Thequarterlyinterestrateisare made. The quarterly interest rate is6.8% ÷ 4 = 1.7% = 0.017$. The correct approach requires calculating: (1) the future value of the first 12 payments at the end of 6 years, and (2) the future value of the final 12 payments. Their sum must equal $500,000. The first 12 payments accumulate to $15000 \cdot \frac{(1.017)^{12} - 1}{0.017} \cdot (1.017)^{12}$$. The (1.017)^{12}$ factor is crucial—it represents the additional 3 years of compound growth after these payments end. The final 12 payments accumulate to P⋅(1.017)12−10.017P \cdot \frac{(1.017)^{12} - 1}{0.017}. Setting their sum equal to $500,000 and solving for $P$ gives answer choice B. Choice A incorrectly uses 24 periods for the first annuity, treating all payments as if made throughout 6 years. Choice C ignores the time value relationship between the two payment periods entirely. Choice D uses simple rather than compound interest for the first period's growth (15000⋅12⋅(1.017)1215000 \cdot 12 \cdot (1.017)^{12}) and omits the annuity formula. Strategy tip: In multi-period annuity problems, always track when each payment stops and how long it continues to compound afterward. Draw a timeline to visualize the cash flows.

Question 7

A retiree has $450,000 in a retirement account earning $4%4\% annualinterest.Shewantstowithdrawequalamountsattheendofeachyearforannual interest. She wants to withdraw equal amounts at the end of each year for 2525 $ years, leaving a balance of $100,000 after the final withdrawal. What should be the annual withdrawal amount?

  1. 450000⋅0.041−(1.04)−25−10000025\frac{450000 \cdot 0.04}{1 - (1.04)^{-25}} - \frac{100000}{25}
  2. 350000⋅0.041−(1.04)−25\frac{350000 \cdot 0.04}{1 - (1.04)^{-25}} (correct answer)
  3. (450000−100000)⋅0.04⋅(1.04)25(1.04)25−1\frac{(450000 - 100000) \cdot 0.04 \cdot (1.04)^{25}}{(1.04)^{25} - 1}
  4. 450000⋅0.041−(1.04)−25−100000⋅0.04\frac{450000 \cdot 0.04}{1 - (1.04)^{-25}} - 100000 \cdot 0.04
Explanation: When you encounter a retirement withdrawal problem with a remaining balance, you're dealing with an annuity combined with a present value calculation. The key insight is recognizing that you need to find the withdrawal amount for a smaller principal—the original amount minus the present value of the desired final balance. The correct approach starts with the $450,000 initial balance, but you must account for the $100,000 that needs to remain after 25 years. The present value of that $100,000 (discounted back 25 years at 4%) must be subtracted from the initial amount. This gives you $\450,000 - $100,000(1.04)^{-25} = $350,000 as your effective principal for the annuity calculation. Using the standard annuity payment formula \frac{PV \cdot r}{1-(1+r)^{-n}} with this $350,000 gives you choice B. Choice A incorrectly tries to subtract a simple division of the final balance by years, ignoring time value of money. Choice C uses the annuity due formula (payments at the beginning of periods) and incorrectly calculates with the full $350,000 difference. Choice D attempts to subtract only the annual interest on the final balance, which doesn't properly account for the present value requirement. Study tip: In retirement problems with remaining balances, always convert that final amount to present value first, then subtract it from the initial principal before applying the annuity formula. Don't let the multiple components confuse you—break it into steps.

Question 8

Jennifer wants to accumulate $25,000 in $55 yearsbymakingequaldepositsattheendofeachyearintoanaccountearningyears by making equal deposits at the end of each year into an account earning 7%7\% annualinterest.However,shemissesherpaymentinyearannual interest. However, she misses her payment in year 33 .Ifshemakestheplannedpaymentinyears. If she makes the planned payment in years 1,2,4,1, 2, 4, andand 55 $, what equal annual payment amount will still allow her to reach her goal?

  1. 250004⋅0.07(1.07)4−1\frac{25000}{4} \cdot \frac{0.07}{(1.07)^4 - 1}
  2. 25000(1.07)5+(1.07)4+(1.07)2+(1.07)1\frac{25000}{(1.07)^5 + (1.07)^4 + (1.07)^2 + (1.07)^1}
  3. 25000⋅0.07(1.07)5−1⋅54\frac{25000 \cdot 0.07}{(1.07)^5 - 1} \cdot \frac{5}{4}
  4. 25000(1.07)4+(1.07)3+(1.07)1+1\frac{25000}{(1.07)^4 + (1.07)^3 + (1.07)^1 + 1} (correct answer)
Explanation: When you encounter annuity problems with missing payments, you need to think about the future value of each individual payment rather than using standard annuity formulas. Since Jennifer misses year 3, you can't apply the regular annuity formula that assumes equal, consecutive payments. The correct approach is to find what equal payment PP in years 1, 2, 4, and 5 will grow to $25,000 by the end of year 5. Each payment compounds differently: the year 1 payment grows for 4 years to $P(1.07)4P(1.07)^4 ,year2growsfor3yearsto, year 2 grows for 3 years to P(1.07)3P(1.07)^3 ,year4growsfor1yearto, year 4 grows for 1 year to P(1.07)1P(1.07)^1 ,andyear5doesn′tgrowatall,remaining, and year 5 doesn't grow at all, remaining PP .Settinguptheequation:. Setting up the equation: P[(1.07)4+(1.07)3+(1.07)1+1]=25,000P[(1.07)^4 + (1.07)^3 + (1.07)^1 + 1] = 25,000 ,so, so P=25000(1.07)4+(1.07)3+(1.07)1+1P = \frac{25000}{(1.07)^4 + (1.07)^3 + (1.07)^1 + 1} $. This matches answer D. Answer A incorrectly tries to modify a standard annuity formula by dividing by 4, but this doesn't account for the different compounding periods. Answer B uses the wrong exponents—it includes (1.07)^5 (which would represent a year 0 payment) and (1.07)^2 (year 3), neither of which applies here. Answer C attempts to adjust the standard annuity formula with a \frac{5}{4} factor, but this oversimplifies the problem and uses the wrong denominator structure. Remember: when payments are missing or irregular in annuity problems, abandon the standard formulas and track each payment's individual growth to the target date.