IBDP Maths MAA SL 1.4 Financial applications HL Paper 1- Exam Style Questions
IB MATHEMATICS AA HL – Practice Questions- Paper1-All Topics
Topic : SL 1.4 Financial applications HL Paper 1
Topic 1- Number and algebra- Weightage : 15 %
All Questions for Topic : SL 1.4 –Financial applications of geometric sequences and series: compound interest, annual depreciation
▶️ Answer/Explanation
(a) Village A, 8% annual increase, current population 1000.
Formula: \( P_n = 1000(1.08)^n \).
(i) After 5 years: \( P_5 = 1000(1.08)^5 \approx 1469 \).
(ii) 5 years ago: \( P_{-5} = 1000(1.08)^{-5} \approx 681 \).
(iii) Solve \( 1000(1.08)^n > 2000 \): \( (1.08)^n > 2 \).
\( n \log 1.08 > \log 2 \implies n > \frac{\log 2}{\log 1.08} \approx 9.006 \).
Full years: \( n = 10 \).
(b) Village B, 8% annual decrease, current population 1000.
Formula: \( P_n = 1000(0.92)^n \).
(i) After 5 years: \( P_5 = 1000(0.92)^5 \approx 659 \).
(ii) 5 years ago: \( P_{-5} = 1000(0.92)^{-5} \approx 1517 \).
(iii) Solve \( 1000(0.92)^n < 500 \): \( (0.92)^n < 0.5 \).
\( n \log 0.92 < \log 0.5 \implies n > \frac{\log 0.5}{\log 0.92} \approx 8.31 \).
Full years: \( n = 9 \).
Markscheme
(a) Formula: \( P_n = 1000(1.08)^n \).
(i) \( 1000(1.08)^5 \approx 1469 \).
(ii) \( 1000(1.08)^{-5} \approx 681 \).
(iii) \( 1000(1.08)^n = 2000 \), \( n \approx 9.006 \), so \( n = 10 \).
(b) Formula: \( P_n = 1000(0.92)^n \).
(i) \( 1000(0.92)^5 \approx 659 \).
(ii) \( 1000(0.92)^{-5} \approx 1517 \).
(iii) \( 1000(0.92)^n = 500 \), \( n \approx 8.31 \), so \( n = 9 \).
▶️ Answer/Explanation
(a)
Formula: \( FV = PV \left(1 + \frac{r}{n}\right)^{nt} \).
\( PV = 25000 \), \( r = 0.036 \), \( n = 12 \), \( t = 5 \).
\( FV = 25000 \left(1 + \frac{0.036}{12}\right)^{12 \cdot 5} = 25000 (1.003)^{60} \approx 29951 \).
(b)
Formula: \( FV = PV \left(1 + \frac{r}{n}\right)^{nt} \).
\( FV = 20000 \), \( r = 0.057 \), \( n = 2 \), \( t = 1.5 \).
\( 20000 = x \left(1 + \frac{0.057}{2}\right)^{2 \cdot 1.5} = x (1.0285)^3 \).
\( x = \frac{20000}{(1.0285)^3} \approx 18383 \).
Markscheme
(a) \( FV = 25000 \left(1 + \frac{3.6}{100 \cdot 12}\right)^{12 \cdot 5} \approx 29951 \).
(b) \( 20000 = x \left(1 + \frac{5.7}{100 \cdot 2}\right)^{2 \cdot 1.5} \), \( x \approx 18383 \).
▶️ Answer/Explanation
(a) Formula: \( FV = PV \left(1 + \frac{r}{k}\right)^{kt} \), \( PV = 10000 \), \( r = 0.12 \), \( t = 5 \).
(i) Yearly (\( k = 1 \)): \( FV = 10000 (1.12)^5 \approx 17623 \).
(ii) Half-yearly (\( k = 2 \)): \( FV = 10000 \left(1 + \frac{0.12}{2}\right)^{2 \cdot 5} \approx 17908 \).
(iii) Quarterly (\( k = 4 \)): \( FV = 10000 \left(1 + \frac{0.12}{4}\right)^{4 \cdot 5} \approx 18061 \).
(iv) Monthly (\( k = 12 \)): \( FV = 10000 \left(1 + \frac{0.12}{12}\right)^{12 \cdot 5} \approx 18167 \).
(b) Solve \( 10000 \left(1 + \frac{r}{k}\right)^{kt} > 20000 \).
(i) Yearly (\( k = 1 \)): \( 10000 (1.12)^n > 20000 \implies (1.12)^n > 2 \).
\( n \log 1.12 > \log 2 \implies n > 6.12 \), so \( n = 7 \).
(ii) Monthly (\( k = 12 \)): \( 10000 (1.01)^{12n} > 20000 \implies (1.01)^{12n} > 2 \).
\( 12n \log 1.01 > \log 2 \implies n > 5.83 \), so \( n = 6 \).
Markscheme
(a)
(i) \( FV = 10000 (1 + \frac{12}{100})^5 \approx 17623 \).
(ii) \( FV = 10000 \left(1 + \frac{12}{100 \cdot 2}\right)^{2 \cdot 5} \approx 17908 \).
(iii) \( FV = 10000 \left(1 + \frac{12}{100 \cdot 4}\right)^{4 \cdot 5} \approx 18061 \).
(iv) \( FV = 10000 \left(1 + \frac{12}{100 \cdot 12}\right)^{12 \cdot 5} \approx 18167 \).
(b)
(i) \( 10000 (1.12)^n = 20000 \), \( n \approx 6.12 \), so \( n = 7 \).
(ii) \( 10000 \left(1 + \frac{12}{100 \cdot 12}\right)^{12n} = 20000 \), \( n \approx 5.83 \), so \( n = 6 \).