Option A: Monthly payment = $4200 for 36 months.
Total: \( 4200 \times 36 = 151200 \).
Answer: \( \$151200 \) (A1 A1 N2)

[2 marks]

Option B: Geometric sequence with first term \( a = 1500 \), common ratio \( r = 1.04 \), \( n = 36 \).
Use sum formula: \( S_n = \frac{a (1 – r^n)}{1 – r} \) (M1)

Substitute: \( S_{36} = \frac{1500 (1 – 1.04^{36})}{1 – 1.04} \).
Calculate: \( 1.04^{36} \approx 4.103906 \),
\( S_{36} = \frac{1500 \times (1 – 4.103906)}{-0.04} = 1500 \times \frac{-3.103906}{-0.04} \approx 1500 \times 77.59765 \approx 116396.475 \).
Correct to 2 decimal places: \( \$116396.48 \) (A1 A1 N2)

[3 marks]

Compound interest formula: \( A = P \left(1 + \frac{r}{k}\right)^{kn} \).
Given: \( P = 160000 \), \( r = 0.05 \), \( k = 12 \), \( n \) = years.
Expression: \( A = 160000 \left(1 + \frac{0.05}{12}\right)^{12n} \) (A1 N1)

[1 mark]

Sorin’s investment after 6 months (\( n = \frac{6}{12} = 0.5 \)):
\( A = 160000 \left(1 + \frac{0.05}{12}\right)^{12 \times 0.5} = 160000 \times 1.0041667^6 \approx 160000 \times 1.025208 \approx 164033.28 \) (A1)

Daniela’s winnings after 6 months (geometric sum, \( a = 1500 \), \( r = 1.04 \), \( n = 6 \)):
\( S_6 = \frac{1500 (1 – 1.04^6)}{1 – 1.04} \), \( 1.04^6 \approx 1.265319 \),
\( S_6 = \frac{1500 \times (1 – 1.265319)}{-0.04} \approx 1500 \times 6.632975 \approx 9949.46 \) (A1)

Total value: \( 164033.28 + 9949.46 = 173982.74 \).
Correct to 2 decimal places: \( \$173982.74 \) (A1 N2)

[3 marks]

Solve: \( 160000 \left(1 + \frac{0.05}{12}\right)^m + \frac{1500 (1 – 1.04^m)}{1 – 1.04} \geq 257000 \) (A1)

Test integer values for \( m \):
For \( m = 28 \): \( 160000 \times 1.0041667^{28} \approx 171135.27 \), \( \frac{1500 \times (1 – 1.04^{28})}{-0.04} \approx 85364.12 \),
Total: \( 171135.27 + 85364.12 \approx 256499.39 < 257000 \).
For \( m = 29 \): \( 160000 \times 1.0041667^{29} \approx 171849.31 \), \( \frac{1500 \times (1 – 1.04^{29})}{-0.04} \approx 89105.11 \),
Total: \( 171849.31 + 89105.11 \approx 260954.42 \geq 257000 \) (A1)

Minimum \( m = 29 \) months (A1 N2)

[3 marks]

Method 1

Compound interest formula: \( A = P \left(1 + \frac{r}{100 \times k}\right)^{k \times n} \).
Given: \( A = 41000 \), \( P = 30000 \), \( k = 4 \), \( n = 6 \).
Substitute: \( 41000 = 30000 \left(1 + \frac{r}{400}\right)^{24} \) (M1)

Solve: \( \left(1 + \frac{r}{400}\right)^{24} = \frac{41000}{30000} \approx 1.3666667 \),
\( 1 + \frac{r}{400} = (1.3666667)^{\frac{1}{24}} \approx 1.0131066 \),
\( \frac{r}{400} \approx 0.0131066 \), \( r \approx 400 \times 0.0131066 \approx 5.24264 \).
Correct to 2 decimal places: \( r = 5.24 \) (A1 A1 N2)

Method 2

Use compound interest with time in quarters: \( A = P \left(1 + \frac{r}{100 \times 4}\right)^{4 \times 6} \).
Substitute: \( 41000 = 30000 \left(1 + \frac{r}{400}\right)^{24} \).
Solve: \( 1 + \frac{r}{400} = \left(\frac{41}{30}\right)^{\frac{1}{24}} \approx 1.0131066 \),
\( r \approx 5.24264 \).
Correct to 2 decimal places: \( r = 5.24 \) (M1 A1 A1 N2)

[3 marks]