IB Mathematics: Analysis and Approaches Quiz: Financial Applications
20 questions · exam conditions
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Financial ApplicationsQuestion 1 of 20

Sarah invests $1000 at the start of each year in an account that pays 5% annual interest, compounded annually. Which expression represents the total value of her investment at the end of the 10th year?

1050((1.05)10−1)0.05\frac{1050((1.05)^{10} - 1)}{0.05}
1000((1.05)10−1)0.05\frac{1000((1.05)^{10} - 1)}{0.05}
1000(1.05)101000(1.05)^{10}
10×1000×1.0510 \times 1000 \times 1.05
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Financial Applications

Practice Financial Applications in IB Mathematics: Analysis and Approaches 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 Financial Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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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.

All questions

Question 1

Sarah invests $1000 at the start of each year in an account that pays 5% annual interest, compounded annually. Which expression represents the total value of her investment at the end of the 10th year?

  1. 1050((1.05)10−1)0.05\frac{1050((1.05)^{10} - 1)}{0.05} (correct answer)
  2. 1000((1.05)10−1)0.05\frac{1000((1.05)^{10} - 1)}{0.05}
  3. 1000(1.05)101000(1.05)^{10}
  4. 10×1000×1.0510 \times 1000 \times 1.05
Explanation: When you encounter a problem involving regular payments into an interest-bearing account, you're dealing with an annuity due - a series of equal payments made at the beginning of each period that earn compound interest. The key insight is that each $1000 payment earns interest for a different amount of time. The first payment (start of year 1) earns interest for 10 full years, the second payment earns interest for 9 years, and so on, with the final payment (start of year 10) earning interest for 1 year. This creates the sum: $1000(1.05)10+1000(1.05)9+...+1000(1.05)11000(1.05)^{10} + 1000(1.05)^9 + ... + 1000(1.05)^1 $ This is a geometric series with first term a = 1000(1.05) , common ratio r = 1.05 , and 10 terms. Using the geometric series formula and factoring out 1000, we get: 1000(1.05) \cdot \frac{(1.05)^{10} - 1}{1.05 - 1} = \frac{1050((1.05)^{10} - 1)}{0.05} Answer A is correct because it properly accounts for the annuity due structure with the 1050 factor. Answer B uses 1000 instead of 1050, which would apply if payments were made at the end of each year (ordinary annuity), not the beginning. Answer C calculates only a single lump sum of $1000 compounded for 10 years, ignoring the additional nine payments. Answer D simply multiplies total payments by one year's interest, completely ignoring compound growth over multiple years. Study tip: Remember that annuity due problems involve payments at period beginnings, requiring you to multiply the ordinary annuity formula by $$(1 + r)$$.

Question 2

Maya has $10,000 in a savings account that does not pay any interest. If the average annual rate of inflation is 3%, what is the real value (purchasing power) of her money after 5 years, to the nearest dollar?

  1. $8500
  2. $8587 (correct answer)
  3. $8626
  4. $11593
Explanation: Inflation reduces the purchasing power of money, which is mathematically equivalent to depreciation. We use the formula Vn=V0(1−i)nV_n = V_0(1-i)^n, where V0=10000V_0 = 10000, i=0.03i = 0.03, and n=5n=5. The real value after 5 years is V5=10000(1−0.03)5=10000(0.97)5V_5 = 10000(1 - 0.03)^5 = 10000(0.97)^5. Calculating this gives 10000×0.858734...≈8587.3410000 \times 0.858734... \approx 8587.34. To the nearest dollar, the value is $8587. A: This is the result of using a simple depreciation model: 10000(1−5×0.03)=10000(0.85)=850010000(1 - 5 \times 0.03) = 10000(0.85) = 8500. C: This is the value after 4 years, not 5: 10000(0.97)4≈8852.910000(0.97)^4 \approx 8852.9. Re-calculating: 10000×(0.97)5≈858710000 \times (0.97)^5 \approx 8587. It's possible C arises from another error. Let's check my calculation again. (0.97)5=0.85873...(0.97)^5 = 0.85873.... The calculation is correct. Maybe C is from a different common error. For example 10000/(1.03)5≈862610000 / (1.03)^5 \approx 8626. This is a valid alternative way to think about real value, so it's a strong distractor. D: This is the result of incorrectly treating inflation as growth: 10000(1.03)5≈1159310000(1.03)^5 \approx 11593.

Question 3

An investment earns interest at a rate of 6% per annum, compounded monthly. Which of the following expressions gives the minimum number of months, mm, required for the investment to at least double in value?

  1. log⁡(2)log⁡(1.06)\frac{\log(2)}{\log(1.06)}
  2. log⁡(2)12log⁡(1.06)\frac{\log(2)}{12\log(1.06)}
  3. 12×log⁡(2)log⁡(1.06)12 \times \frac{\log(2)}{\log(1.06)}
  4. log⁡(2)log⁡(1.005)\frac{\log(2)}{\log(1.005)} (correct answer)
Explanation: When you encounter compound interest problems with different compounding frequencies, the key is identifying the correct periodic interest rate and time units that match each other. For monthly compounding at 6% annual rate, you need the monthly interest rate: 6%12=0.5%=0.005\frac{6\%}{12} = 0.5\% = 0.005. The compound interest formula becomes A=P(1+0.005)mA = P(1 + 0.005)^m where mm is the number of months. To find when an investment doubles, set A=2PA = 2P: 2P=P(1.005)m2P = P(1.005)^m, which simplifies to 2=(1.005)m2 = (1.005)^m. Taking logarithms of both sides: log⁡(2)=mlog⁡(1.005)\log(2) = m \log(1.005), so m=log⁡(2)log⁡(1.005)m = \frac{\log(2)}{\log(1.005)}. Answer D correctly uses the monthly interest rate (1.005) with months as the time unit. Answer A uses log⁡(1.06)\log(1.06), which corresponds to the annual rate of 6%. This would give the number of years to double, not months. Answer B takes the result from A and divides by 12. This attempts to convert years to months but uses the wrong base calculation, compounding the error from A. Answer C multiplies the result from A by 12, which also incorrectly tries to convert between time units while using the wrong interest rate. The critical insight is that your interest rate and time period must match: monthly rate with months, annual rate with years. Always convert to the compounding period when setting up your equation, not afterward through multiplication or division.

Question 4

A machine is valued at $50,000. Its value depreciates by 12% each year (compound depreciation). A different model uses linear depreciation. If both models give the same value for the machine after exactly 2 years, what is the annual linear depreciation amount?

  1. $5,280
  2. $5,640 (correct answer)
  3. $6,000
  4. $11,280
Explanation: First, calculate the value of the machine after 2 years using the compound depreciation model. The value V2V_2 is given by V2=50000(1−0.12)2=50000(0.88)2=50000(0.7744)=38720V_2 = 50000(1 - 0.12)^2 = 50000(0.88)^2 = 50000(0.7744) = 38720. Next, set up the equation for the linear depreciation model. Let dd be the annual depreciation amount. The value after 2 years is VL=50000−2dV_L = 50000 - 2d. Since the values are the same, we set them equal: 38720=50000−2d38720 = 50000 - 2d. Now, solve for dd: 2d=50000−38720=112802d = 50000 - 38720 = 11280. d=112802=5640d = \frac{11280}{2} = 5640. The annual linear depreciation amount is $5,640. A: This is the amount of depreciation in the second year only under the compound model: (50000×0.88)×0.12=5280(50000 \times 0.88) \times 0.12 = 5280. C: This is the depreciation amount if simple depreciation were used on the original value: 50000×0.12=600050000 \times 0.12 = 6000. D: This is the total depreciation over two years, not the annual amount.

Question 5

Liam buys a collectible for $500. For the first 3 years, its value appreciates by 10% annually. For the next 2 years, its value depreciates by 10% annually. What is the value of the collectible after these 5 years?

  1. $500.00
  2. $520.00
  3. $539.06 (correct answer)
  4. $550.00
Explanation: This is a multi-step calculation. First, calculate the value after 3 years of appreciation. The common ratio is 1+0.10=1.11 + 0.10 = 1.1. Value after 3 years: V3=500(1.1)3=500(1.331)=665.50V_3 = 500(1.1)^3 = 500(1.331) = 665.50. Next, calculate the value after 2 years of depreciation from this new amount. The common ratio is 1−0.10=0.91 - 0.10 = 0.9. Value after 5 years: V5=V3(0.9)2=665.50(0.81)=539.055V_5 = V_3(0.9)^2 = 665.50(0.81) = 539.055. Rounded to two decimal places, the value is $539.06. A: This is a common fallacy where a 10% increase and a 10% decrease are assumed to cancel out. B: This result comes from using simple interest and simple depreciation: 500(1+0.1×3)=650500(1+0.1\times3) = 650, then 650(1−0.1×2)=650(0.8)=520650(1-0.1\times2) = 650(0.8) = 520. D: This result comes from incorrectly netting the interest rates before calculating: 500(1+0.1×3−0.1×2)=500(1.1)=550500(1 + 0.1\times3 - 0.1\times2) = 500(1.1) = 550.

Question 6

An amount PP is invested for one year at a nominal annual rate of 8%. Let A1A_1 be the final amount if compounded annually, A2A_2 if compounded semi-annually, and A4A_4 if compounded quarterly. Which statement is correct?

  1. A1>A2>A4A_1 > A_2 > A_4
  2. A2>A1>A4A_2 > A_1 > A_4
  3. A1=A2=A4A_1 = A_2 = A_4
  4. A1<A2<A4A_1 < A_2 < A_4 (correct answer)
Explanation: When you encounter compound interest problems with different compounding frequencies, remember that more frequent compounding always produces higher returns with the same nominal rate. This tests your understanding of how compounding frequency affects investment growth. Let's compare the three scenarios using the compound interest formula A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt}, where r=0.08r = 0.08, t=1t = 1 year, and nn varies: For annual compounding (n=1n = 1): A1=P(1+0.08)1=1.08PA_1 = P(1 + 0.08)^1 = 1.08P For semi-annual compounding (n=2n = 2): A2=P(1+0.04)2=P(1.04)2=1.0816PA_2 = P(1 + 0.04)^2 = P(1.04)^2 = 1.0816P For quarterly compounding (n=4n = 4): A4=P(1+0.02)4=P(1.02)4=1.08243PA_4 = P(1 + 0.02)^4 = P(1.02)^4 = 1.08243P Since 1.08243>1.0816>1.081.08243 > 1.0816 > 1.08, we have A4>A2>A1A_4 > A_2 > A_1, which means A1<A2<A4A_1 < A_2 < A_4. Option A suggests annual compounding gives the highest return, which contradicts the fundamental principle that more frequent compounding increases returns. Option B incorrectly places semi-annual above both others, missing that quarterly should be highest. Option C assumes all compounding frequencies yield identical results, ignoring that different frequencies compound interest at different intervals throughout the year. The key insight is that quarterly compounding allows interest to earn interest four times per year, while annual compounding only compounds once. Each additional compounding period slightly increases the effective return, even though the nominal rate stays constant at 8%.

Question 7

Maya has $10,000 in a savings account that does not pay any interest. If the average annual rate of inflation is 3%, what is the real value (purchasing power) of her money after 5 years, to the nearest dollar?

  1. $8500
  2. $8587 (correct answer)
  3. $8626
  4. $11593
Explanation: Inflation reduces the purchasing power of money, which is mathematically equivalent to depreciation. We use the formula Vn=V0(1−i)nV_n = V_0(1-i)^n, where V0=10000V_0 = 10000, i=0.03i = 0.03, and n=5n=5. The real value after 5 years is V5=10000(1−0.03)5=10000(0.97)5V_5 = 10000(1 - 0.03)^5 = 10000(0.97)^5. Calculating this gives 10000×0.858734...≈8587.3410000 \times 0.858734... \approx 8587.34. To the nearest dollar, the value is $8587. A: This is the result of using a simple depreciation model: 10000(1−5×0.03)=10000(0.85)=850010000(1 - 5 \times 0.03) = 10000(0.85) = 8500. C: This is the value after 4 years, not 5: 10000(0.97)4≈8852.910000(0.97)^4 \approx 8852.9. Re-calculating: 10000×(0.97)5≈858710000 \times (0.97)^5 \approx 8587. It's possible C arises from another error. Let's check my calculation again. (0.97)5=0.85873...(0.97)^5 = 0.85873.... The calculation is correct. Maybe C is from a different common error. For example 10000/(1.03)5≈862610000 / (1.03)^5 \approx 8626. This is a valid alternative way to think about real value, so it's a strong distractor. D: This is the result of incorrectly treating inflation as growth: 10000(1.03)5≈1159310000(1.03)^5 \approx 11593.

Question 8

Investment A starts with $10,000 and earns 5% annual interest compounded annually. Investment B starts with $8,000 and earns 7% annual interest compounded annually. What is the first integer nn such that at the end of year nn, the value of Investment B is greater than the value of Investment A?

  1. 11
  2. 12 (correct answer)
  3. 13
  4. 14
Explanation: Let VA(n)V_A(n) and VB(n)V_B(n) be the values of the investments after nn years. We have VA(n)=10000(1.05)nV_A(n) = 10000(1.05)^n and VB(n)=8000(1.07)nV_B(n) = 8000(1.07)^n. We need to find the smallest integer nn for which VB(n)>VA(n)V_B(n) > V_A(n). 8000(1.07)n>10000(1.05)n8000(1.07)^n > 10000(1.05)^n (1.07)n(1.05)n>100008000\frac{(1.07)^n}{(1.05)^n} > \frac{10000}{8000} (1.071.05)n>1.25\left(\frac{1.07}{1.05}\right)^n > 1.25 Take logarithms of both sides: nln⁡(1.071.05)>ln⁡(1.25)n \ln\left(\frac{1.07}{1.05}\right) > \ln(1.25) n>ln⁡(1.25)ln⁡(1.07/1.05)≈0.22310.01886≈11.83n > \frac{\ln(1.25)}{\ln(1.07/1.05)} \approx \frac{0.2231}{0.01886} \approx 11.83 Since nn must be an integer representing the end of a year, the first integer value for which this inequality holds is n=12n=12. A: This is the result of incorrectly rounding down 11.83. C, D: These would result from calculation errors.

Question 9

An amount PP is invested for one year at a nominal annual rate of 8%. Let A1A_1 be the final amount if compounded annually, A2A_2 if compounded semi-annually, and A4A_4 if compounded quarterly. Which statement is correct?

  1. A1>A2>A4A_1 > A_2 > A_4
  2. A2>A1>A4A_2 > A_1 > A_4
  3. A1=A2=A4A_1 = A_2 = A_4
  4. A1<A2<A4A_1 < A_2 < A_4 (correct answer)
Explanation: When you encounter compound interest problems with different compounding frequencies, remember that more frequent compounding always produces higher returns with the same nominal rate. This tests your understanding of how compounding frequency affects investment growth. Let's compare the three scenarios using the compound interest formula A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt}, where r=0.08r = 0.08, t=1t = 1 year, and nn varies: For annual compounding (n=1n = 1): A1=P(1+0.08)1=1.08PA_1 = P(1 + 0.08)^1 = 1.08P For semi-annual compounding (n=2n = 2): A2=P(1+0.04)2=P(1.04)2=1.0816PA_2 = P(1 + 0.04)^2 = P(1.04)^2 = 1.0816P For quarterly compounding (n=4n = 4): A4=P(1+0.02)4=P(1.02)4=1.08243PA_4 = P(1 + 0.02)^4 = P(1.02)^4 = 1.08243P Since 1.08243>1.0816>1.081.08243 > 1.0816 > 1.08, we have A4>A2>A1A_4 > A_2 > A_1, which means A1<A2<A4A_1 < A_2 < A_4. Option A suggests annual compounding gives the highest return, which contradicts the fundamental principle that more frequent compounding increases returns. Option B incorrectly places semi-annual above both others, missing that quarterly should be highest. Option C assumes all compounding frequencies yield identical results, ignoring that different frequencies compound interest at different intervals throughout the year. The key insight is that quarterly compounding allows interest to earn interest four times per year, while annual compounding only compounds once. Each additional compounding period slightly increases the effective return, even though the nominal rate stays constant at 8%.

Question 10

A machine is valued at $50,000. Its value depreciates by 12% each year (compound depreciation). A different model uses linear depreciation. If both models give the same value for the machine after exactly 2 years, what is the annual linear depreciation amount?

  1. $5,280
  2. $5,640 (correct answer)
  3. $6,000
  4. $11,280
Explanation: First, calculate the value of the machine after 2 years using the compound depreciation model. The value V2V_2 is given by V2=50000(1−0.12)2=50000(0.88)2=50000(0.7744)=38720V_2 = 50000(1 - 0.12)^2 = 50000(0.88)^2 = 50000(0.7744) = 38720. Next, set up the equation for the linear depreciation model. Let dd be the annual depreciation amount. The value after 2 years is VL=50000−2dV_L = 50000 - 2d. Since the values are the same, we set them equal: 38720=50000−2d38720 = 50000 - 2d. Now, solve for dd: 2d=50000−38720=112802d = 50000 - 38720 = 11280. d=112802=5640d = \frac{11280}{2} = 5640. The annual linear depreciation amount is $5,640. A: This is the amount of depreciation in the second year only under the compound model: (50000×0.88)×0.12=5280(50000 \times 0.88) \times 0.12 = 5280. C: This is the depreciation amount if simple depreciation were used on the original value: 50000×0.12=600050000 \times 0.12 = 6000. D: This is the total depreciation over two years, not the annual amount.

Question 11

Investment A starts with $10,000 and earns 5% annual interest compounded annually. Investment B starts with $8,000 and earns 7% annual interest compounded annually. What is the first integer nn such that at the end of year nn, the value of Investment B is greater than the value of Investment A?

  1. 11
  2. 12 (correct answer)
  3. 13
  4. 14
Explanation: Let VA(n)V_A(n) and VB(n)V_B(n) be the values of the investments after nn years. We have VA(n)=10000(1.05)nV_A(n) = 10000(1.05)^n and VB(n)=8000(1.07)nV_B(n) = 8000(1.07)^n. We need to find the smallest integer nn for which VB(n)>VA(n)V_B(n) > V_A(n). 8000(1.07)n>10000(1.05)n8000(1.07)^n > 10000(1.05)^n (1.07)n(1.05)n>100008000\frac{(1.07)^n}{(1.05)^n} > \frac{10000}{8000} (1.071.05)n>1.25\left(\frac{1.07}{1.05}\right)^n > 1.25 Take logarithms of both sides: nln⁡(1.071.05)>ln⁡(1.25)n \ln\left(\frac{1.07}{1.05}\right) > \ln(1.25) n>ln⁡(1.25)ln⁡(1.07/1.05)≈0.22310.01886≈11.83n > \frac{\ln(1.25)}{\ln(1.07/1.05)} \approx \frac{0.2231}{0.01886} \approx 11.83 Since nn must be an integer representing the end of a year, the first integer value for which this inequality holds is n=12n=12. A: This is the result of incorrectly rounding down 11.83. C, D: These would result from calculation errors.

Question 12

Sarah invests $1000 at the start of each year in an account that pays 5% annual interest, compounded annually. Which expression represents the total value of her investment at the end of the 10th year?

  1. 1050((1.05)10−1)0.05\frac{1050((1.05)^{10} - 1)}{0.05} (correct answer)
  2. 1000((1.05)10−1)0.05\frac{1000((1.05)^{10} - 1)}{0.05}
  3. 1000(1.05)101000(1.05)^{10}
  4. 10×1000×1.0510 \times 1000 \times 1.05
Explanation: When you encounter a problem involving regular payments into an interest-bearing account, you're dealing with an annuity due - a series of equal payments made at the beginning of each period that earn compound interest. The key insight is that each $1000 payment earns interest for a different amount of time. The first payment (start of year 1) earns interest for 10 full years, the second payment earns interest for 9 years, and so on, with the final payment (start of year 10) earning interest for 1 year. This creates the sum: $1000(1.05)10+1000(1.05)9+...+1000(1.05)11000(1.05)^{10} + 1000(1.05)^9 + ... + 1000(1.05)^1 $ This is a geometric series with first term a = 1000(1.05) , common ratio r = 1.05 , and 10 terms. Using the geometric series formula and factoring out 1000, we get: 1000(1.05) \cdot \frac{(1.05)^{10} - 1}{1.05 - 1} = \frac{1050((1.05)^{10} - 1)}{0.05} Answer A is correct because it properly accounts for the annuity due structure with the 1050 factor. Answer B uses 1000 instead of 1050, which would apply if payments were made at the end of each year (ordinary annuity), not the beginning. Answer C calculates only a single lump sum of $1000 compounded for 10 years, ignoring the additional nine payments. Answer D simply multiplies total payments by one year's interest, completely ignoring compound growth over multiple years. Study tip: Remember that annuity due problems involve payments at period beginnings, requiring you to multiply the ordinary annuity formula by $$(1 + r)$$.

Question 13

Liam buys a collectible for $500. For the first 3 years, its value appreciates by 10% annually. For the next 2 years, its value depreciates by 10% annually. What is the value of the collectible after these 5 years?

  1. $500.00
  2. $520.00
  3. $539.06 (correct answer)
  4. $550.00
Explanation: This is a multi-step calculation. First, calculate the value after 3 years of appreciation. The common ratio is 1+0.10=1.11 + 0.10 = 1.1. Value after 3 years: V3=500(1.1)3=500(1.331)=665.50V_3 = 500(1.1)^3 = 500(1.331) = 665.50. Next, calculate the value after 2 years of depreciation from this new amount. The common ratio is 1−0.10=0.91 - 0.10 = 0.9. Value after 5 years: V5=V3(0.9)2=665.50(0.81)=539.055V_5 = V_3(0.9)^2 = 665.50(0.81) = 539.055. Rounded to two decimal places, the value is $539.06. A: This is a common fallacy where a 10% increase and a 10% decrease are assumed to cancel out. B: This result comes from using simple interest and simple depreciation: 500(1+0.1×3)=650500(1+0.1\times3) = 650, then 650(1−0.1×2)=650(0.8)=520650(1-0.1\times2) = 650(0.8) = 520. D: This result comes from incorrectly netting the interest rates before calculating: 500(1+0.1×3−0.1×2)=500(1.1)=550500(1 + 0.1\times3 - 0.1\times2) = 500(1.1) = 550.

Question 14

An investment earns interest at a rate of 6% per annum, compounded monthly. Which of the following expressions gives the minimum number of months, mm, required for the investment to at least double in value?

  1. log⁡(2)log⁡(1.06)\frac{\log(2)}{\log(1.06)}
  2. log⁡(2)12log⁡(1.06)\frac{\log(2)}{12\log(1.06)}
  3. 12×log⁡(2)log⁡(1.06)12 \times \frac{\log(2)}{\log(1.06)}
  4. log⁡(2)log⁡(1.005)\frac{\log(2)}{\log(1.005)} (correct answer)
Explanation: When you encounter compound interest problems with different compounding frequencies, the key is identifying the correct periodic interest rate and time units that match each other. For monthly compounding at 6% annual rate, you need the monthly interest rate: 6%12=0.5%=0.005\frac{6\%}{12} = 0.5\% = 0.005. The compound interest formula becomes A=P(1+0.005)mA = P(1 + 0.005)^m where mm is the number of months. To find when an investment doubles, set A=2PA = 2P: 2P=P(1.005)m2P = P(1.005)^m, which simplifies to 2=(1.005)m2 = (1.005)^m. Taking logarithms of both sides: log⁡(2)=mlog⁡(1.005)\log(2) = m \log(1.005), so m=log⁡(2)log⁡(1.005)m = \frac{\log(2)}{\log(1.005)}. Answer D correctly uses the monthly interest rate (1.005) with months as the time unit. Answer A uses log⁡(1.06)\log(1.06), which corresponds to the annual rate of 6%. This would give the number of years to double, not months. Answer B takes the result from A and divides by 12. This attempts to convert years to months but uses the wrong base calculation, compounding the error from A. Answer C multiplies the result from A by 12, which also incorrectly tries to convert between time units while using the wrong interest rate. The critical insight is that your interest rate and time period must match: monthly rate with months, annual rate with years. Always convert to the compounding period when setting up your equation, not afterward through multiplication or division.

Question 15

A new car is purchased for $30,000. Its value depreciates by 15% each year. Find the minimum number of full years after which the car's value will be less than half of its original price.

  1. 5 years (correct answer)
  2. 4 years
  3. 6 years
  4. 7 years
Explanation: This is an exponential decay problem where you need to find when a continuously decreasing value drops below a specific threshold. When you see "depreciates by X% each year," you're dealing with exponential decay with the formula: Final Value = Initial Value × (1−rate)time(1 - \text{rate})^{\text{time}}. Here, the car starts at $30,000 and loses 15% annually, so it retains 85% of its value each year. You need to find when the value drops below $15,000 (half of $30,000). Set up the inequality: $30000×(0.85)t<1500030000 \times (0.85)^t < 15000 $. Dividing both sides by 30,000 gives: (0.85)^t < 0.5 . Taking the natural logarithm of both sides: t \times \ln(0.85) < \ln(0.5) . Since \ln(0.85) is negative, dividing flips the inequality: t > \frac{\ln(0.5)}{\ln(0.85)} \approx 4.27 years. Since you need the minimum number of full years, you round up to 5 years. Let's verify by checking each option: After 4 years (B), the value is 30000 \times (0.85)^4 \approx $15,700 , which is still above $15,000. After 5 years (A), the value is $30000 \times (0.85)^5 \approx \13,345$$, which is below $15,000. Options C and D (6 and 7 years) would also work mathematically but aren't the minimum. Study tip: In "minimum time" problems with exponential decay, always check whether the question asks for when the threshold is first crossed versus the minimum number of complete time periods. The phrase "full years" is your cue to round up to the next integer.

Question 16

An investment of $2000 grows to $2205 in two years, with interest compounded annually. What would be the value of the investment after a third year?

  1. $2307.50
  2. $2310.00
  3. $2315.25 (correct answer)
  4. $2320.00
Explanation: Let the annual interest rate be ii and the common ratio be r=1+ir = 1+i. The value after nn years is Vn=V0rnV_n = V_0 r^n. We have V0=2000V_0 = 2000 and V2=2205V_2 = 2205. So, 2205=2000⋅r22205 = 2000 \cdot r^2. This gives r2=22052000=441400r^2 = \frac{2205}{2000} = \frac{441}{400}. Taking the square root, r=441400=2120=1.05r = \sqrt{\frac{441}{400}} = \frac{21}{20} = 1.05. The annual interest rate is 5%. The value after the third year, V3V_3, is V2×r=2205×1.05V_2 \times r = 2205 \times 1.05. 2205×1.05=2205×(1+0.05)=2205+110.25=2315.252205 \times 1.05 = 2205 \times (1 + 0.05) = 2205 + 110.25 = 2315.25. A: This uses simple interest logic. The total interest over 2 years is $205, so the average annual interest is $102.50. Adding this to $2205 gives $2307.50. B: This assumes the interest amount in year 3 is the same as the interest amount in year 2. Interest in year 2 was V2−V1=2205−2000(1.05)=2205−2100=105V_2 - V_1 = 2205 - 2000(1.05) = 2205 - 2100 = 105. Adding this gives 2205+105=23102205+105=2310. D: This is a miscalculation.

Question 17

An investment earns 5% annual interest, compounded annually. After 4 years, the total interest earned is $1077.53. What was the initial principal, to the nearest dollar?

  1. $4500
  2. $5000 (correct answer)
  3. $5388
  4. $9977
Explanation: Let the initial principal be PP. The value after 4 years is A=P(1.05)4A = P(1.05)^4. The interest earned is I=A−P=P(1.05)4−P=P((1.05)4−1)I = A - P = P(1.05)^4 - P = P((1.05)^4 - 1). We are given I=1077.53I = 1077.53. So, 1077.53=P((1.05)4−1)1077.53 = P((1.05)^4 - 1). We can solve for PP: P=1077.53(1.05)4−1P = \frac{1077.53}{(1.05)^4 - 1}. Calculating the denominator: (1.05)4−1≈1.21550625−1=0.21550625(1.05)^4 - 1 \approx 1.21550625 - 1 = 0.21550625. Then P=1077.530.21550625≈4999.98P = \frac{1077.53}{0.21550625} \approx 4999.98. To the nearest dollar, the principal was $5000. A: This would be an underestimate, perhaps from a calculation error. C: This result comes from incorrectly using the simple interest formula: I=P⋅r⋅t⇒1077.53=P⋅0.05⋅4=0.2P⇒P=5387.65I = P \cdot r \cdot t \Rightarrow 1077.53 = P \cdot 0.05 \cdot 4 = 0.2P \Rightarrow P = 5387.65. D: This is the principal required to earn $1077.53 in interest in just one year: P=1077.53/1.05−1077.53P = 1077.53/1.05 - 1077.53? No. Let's try 1077.53=P(1.05)41077.53 = P(1.05)^4, confusing interest with final amount. That would give P≈8864P \approx 8864. Let's try final amount - interest = P. P(1.05)4−1077.53=PP(1.05)^4 - 1077.53 = P. P((1.05)4−1)=1077.53P((1.05)^4-1)=1077.53. This is the correct equation. It seems D is from some other error.

Question 18

The value of an asset, VnV_n, in dollars at the end of year nn is modelled by the sequence Vn=25000×(0.88)nV_n = 25000 \times (0.88)^n, for n≥0n \ge 0. Which of the following statements is correct?

  1. The asset appreciates by 12% per year.
  2. The value of the asset depreciates by $3000 each year.
  3. The annual rate of depreciation is 12%. (correct answer)
  4. The asset's value is 88% of its original value after 2 years.
Explanation: The model is of the form Vn=V0(r)nV_n = V_0(r)^n, where V0V_0 is the initial value and rr is the common ratio. Here, V0=25000V_0 = 25000 and r=0.88r = 0.88. Since r<1r < 1, the asset is depreciating. The common ratio for depreciation is r=1−ir = 1 - i, where ii is the annual depreciation rate. So, 0.88=1−i0.88 = 1 - i, which gives i=1−0.88=0.12i = 1 - 0.88 = 0.12. The annual rate of depreciation is 12%. A: This is incorrect; the value is decreasing, so it depreciates. B: This describes linear (simple) depreciation. The depreciation in the first year is 25000×0.12=300025000 \times 0.12 = 3000, but in the second year it is (25000×0.88)×0.12=2640(25000 \times 0.88) \times 0.12 = 2640. The amount is not constant. D: The value after 2 years is 25000×(0.88)2=25000×0.774425000 \times (0.88)^2 = 25000 \times 0.7744, which is 77.44% of its original value, not 88%.

Question 19

A car depreciates by 20% in its first year. For each subsequent year, it depreciates by 10%. If the car was purchased for $40,000, what is its value at the end of the third year?

  1. $25,600
  2. $25,920 (correct answer)
  3. $28,800
  4. $32,000
Explanation: This is a multi-stage depreciation problem. Value after the first year (20% depreciation): V1=40000(1−0.20)=40000(0.8)=32000V_1 = 40000(1 - 0.20) = 40000(0.8) = 32000. For the next two years, the depreciation is 10% per year. We calculate the value after year 2 and year 3 based on the new value. Value after the second year: V2=V1(1−0.10)=32000(0.9)=28800V_2 = V_1(1 - 0.10) = 32000(0.9) = 28800. Value after the third year: V3=V2(1−0.10)=28800(0.9)=25920V_3 = V_2(1 - 0.10) = 28800(0.9) = 25920. The value at the end of the third year is $25,920. A: This results from incorrectly applying 20% depreciation for all three years: 40000(0.8)3=2560040000(0.8)^3 = 25600. C: This is the value after two years, not three. D: This is the value after one year, not three.

Question 20

An item depreciates by rr% per year. After 10 years, its value is 25% of its original value. Find the value of rr, to the nearest whole number.

  1. 13 (correct answer)
  2. 14
  3. 15
  4. 16
Explanation: Let the original value be V0V_0. The value after 10 years is V10=0.25V0V_{10} = 0.25 V_0. The annual rate of depreciation is i=r/100i = r/100. The formula for depreciation is V10=V0(1−i)10V_{10} = V_0(1-i)^{10}. Substituting the given information: 0.25V0=V0(1−i)100.25 V_0 = V_0(1-i)^{10}. Divide by V0V_0: 0.25=(1−i)100.25 = (1-i)^{10}. To solve for ii, we take the 10th root of both sides: 1−i=(0.25)1/101-i = (0.25)^{1/10}. i=1−(0.25)1/10i = 1 - (0.25)^{1/10}. Using a calculator, (0.25)0.1≈0.87055(0.25)^{0.1} \approx 0.87055. So, i=1−0.87055=0.12945i = 1 - 0.87055 = 0.12945. The percentage rate rr is i×100≈12.945i \times 100 \approx 12.945%. To the nearest whole number, r=13r=13. B, C, D: These would result from calculation errors. For example, a common error is to use simple depreciation: 1−10i=0.25⇒10i=0.75⇒i=0.0751 - 10i = 0.25 \Rightarrow 10i = 0.75 \Rightarrow i = 0.075, or 7.5%, which is not an option.