IBDP Maths SL 1.4 Financial applications AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) (i) Falicia’s investment: $9,000 at 7% per annum compounded annually.
\( 9000 \times (1 + \frac{7}{100})^5 \) M1
\( \approx 12622.965 \) A1
To the nearest hundred dollars: $12,600 A1
[3 marks]
(a) (ii) Solve for years \( x \) to reach $20,000:
\( 9000 \times (1 + \frac{7}{100})^x = 20000 \) M1
\( (1.07)^x = \frac{20000}{9000} \approx 2.22222 \) A1
\( x \ln(1.07) = \ln(2.22222) \), so \( x \approx \frac{\ln(2.22222)}{\ln(1.07)} \approx 11.879 \) A1
Since \( x \) is years, round up to 12 years A2
[5 marks]
(b) Dominic’s investment: $9,000 at \( r\% \) per annum compounded monthly, target $20,000 after 10 years.
\( 9000 \times (1 + \frac{r}{100 \times 12})^{12 \times 10} = 20000 \) M1
\( (1 + \frac{r}{1200})^{120} = \frac{20000}{9000} \approx 2.22222 \) A1
\( 1 + \frac{r}{1200} = (2.22222)^{1/120} \approx 1.006676 \), so \( \frac{r}{1200} \approx 0.006676 \), \( r \approx 8.0117 \) A1
To two decimal places: \( r = 8.02\% \) A1
[3 marks]
(c) (i) Aayush’s deposits form a geometric series, initial deposit $9,000, first additional deposit $4,500, each subsequent deposit half the previous.
Sum of additional deposits: \( 4500 \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k = \frac{4500}{1 – \frac{1}{2}} = 9000 \) M1
Total with initial deposit: \( 9000 + 9000 = 18000 \) A1
Since \( 18000 < 20000 \), Aayush cannot reach the target A1 AG
[3 marks]
(c) (ii) Initial deposit \( u_1 \), additional deposits \( \frac{u_1}{2}, \frac{u_1}{4}, \frac{u_1}{8}, \frac{u_1}{16} \) over 4 years (since 5 years includes initial deposit).
Total sum: \( u_1 + \frac{u_1}{2} + \frac{u_1}{4} + \frac{u_1}{8} + \frac{u_1}{16} = u_1 \cdot \frac{1 – (0.5)^5}{1 – 0.5} = u_1 \cdot \frac{0.96875}{0.5} = u_1 \cdot 1.9375 \) M1 A1
Set equal to $20,000: \( u_1 \cdot 1.9375 = 20000 \), so \( u_1 \approx \frac{20000}{1.9375} \approx 10322.58 \) A1
Nearest dollar: $10,323 A1
[8 marks]
▶️ Answer/Explanation
(a) Loan amount = $150,000, annual interest rate = 3.5%. Amount owed after 20 years: \( 150,000 \times 1.035^{20} \) M1 A1.
Calculate: \( 1.035^{20} \approx 1.98979 \), so \( 150,000 \times 1.98979 \approx 298,468.50 \). To the nearest dollar: $298,469 A1.
[3 marks]
(b) Phil’s savings: deposit $ \( P \) at the end of each year for 20 years, interest rate 2%. First deposit earns interest for 19 years, second for 18 years, etc. Total value: \( P (1.02^{19} + 1.02^{18} + \dots + 1.02^0) \) M1.
This is a geometric series with first term \( P \), common ratio 1.02, \( n = 20 \): \( S_{20} = P \sum_{k=0}^{19} 1.02^k = P \cdot \frac{1.02^{20} – 1}{1.02 – 1} = \frac{(1.02^{20} – 1)P}{0.02} \) M1 A1 AG.
[3 marks]
(c) To own the house, savings must equal the loan amount after 20 years: \( \frac{(1.02^{20} – 1)P}{0.02} = 298,469 \) M1. Compute: \( 1.02^{20} \approx 1.485947 \), so \( 1.02^{20} – 1 \approx 0.485947 \), and \( \frac{0.485947}{0.02} \approx 24.29735 \). Thus, \( P = \frac{298,469}{24.29735} \approx 12,283.63 \) A1. Nearest dollar: $12,284 A1.
[3 marks]
(d) (i) David’s deposit \( Q \) earns 2.8% annually. Withdraw $5,000 at the end of each year for \( n \) years. Present value of withdrawals: \( Q = \frac{5000}{1.028} + \frac{5000}{1.028^2} + \dots + \frac{5000}{1.028^n} \). This is the sum of present values of each withdrawal M1 A1. Alternatively, \( Q \cdot 1.028^n = 5000 (1 + 1.028 + \dots + 1.028^{n-1}) \), so \( Q = \frac{5000 (1 – 1.028^n) / (1 – 1.028)}{1.028^n} = \sum_{k=1}^n \frac{5000}{1.028^k} \) A1 AG.
[3 marks]
(d) (ii) For indefinite withdrawals, sum to infinity: \( Q = \frac{5000 / 1.028}{1 – 1/1.028} = \frac{5000 / 1.028}{0.028 / 1.028} = \frac{5000}{0.028} \approx 178,571.4286 \) M1 A1. Nearest dollar: $178,571 A1.
[3 marks]