IB MATHEMATICS AA HL – Practice Questions- Paper1-All Topics
Topic : SL 1.4 Financial applications of geometric sequences and series
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
Question
(a) The population of village A increases by 8% every year. If the population today is 1000 people find
(i) the population after 5 years;
(ii) the population 5 years ago;
(iii) the number of full years after the population will exceed 2000.
▶️Answer/Explanation
Ans: \(FV=1000(1+\frac {8}{100})^n=1000(1.08)^n\)
(i) FV = 1000(1.08)5 = 1469
(ii) FV = 1000(1.08)-5 = 681
(iii) Solve for n : 1000( .1 08)n = 2000 n , n=9.0064 so n=10(b) \(FV = 1000(1-\frac {8}{100})^n=1000(0.92)^n
(i) FV = 1000(0.92)5 = 659
(ii) FV = 1000(0.92)-5 = 1517
(iii) Solve for n: 1000(0.92)n = 500, n = 8.31 s0 n = 9METHOD B: By using GDC-financial mode
(b) The population of village B decreases by 8% every year. If the population today is 1000 people find
(i) the population after 5 years;
(ii) the population 5 years ago;
(iii) the number of full years after the population will fall under 500.
▶️Answer/Explanation
Ans: I% = 8, PV= –1000, [PMT=0, P/Y=1, C/Y=1]
(i) Set n=5 then FV=1469
(ii) Set n= –5 then FV=681
(iii) Set FV = 2000 then n = 8.31 and so n = 9
Question:
Give your answers in this question correct to the nearest whole number.
Shahid invested 25 000 Singapore dollars (SGD) in a fixed deposit account with a nominal
annual interest rate of 3.6 %, compounded monthly.
a. Calculate the value of Shahid’s investment after 5 years. [3]
▶️Answer/Explanation
Ans: \(FV=25000\times (1+ \frac{3.5}{100\times 12})^{12\times 5}\)
OR
N= 5
I%=3.6
PV= \(\mp 25000\)
\(P/Y\) = 1
C/Y = 12
OR
N= 60
I%=3.6
PV = \(\mp 25000\)
P/V = 12
C/Y = 12
FV=29922(SGD)
At the end of the 5 years, Shahid withdrew x SGD from the fixed deposit account and reinvested this into a super-savings account with a nominal annual interest rate of 5.7 %, compounded half-yearly. The value of the super-savings account increased to 20 000 SGD after 18 months.
b. Find the value of x . [3]
▶️Answer/Explanation
Ans: 2000= PV\times \((1+\frac{5.7}{100\times 2})^{2\times 1.5}\)
OR
N= 1.5
I%=5.7
FV=\(\pm 20000\)
\(P/Y\) = 1
C/Y= 2
OR
N=3
I%= 5.7
FV = \(\pm 20000\)
P/V=2
C/Y = 2
X= 18383 (SGD)
Question:
An amount of £ 10 000 is invested at an annual interest rate of 12%.
(a) Find the value of the investment after 5 years
(i) if the interest rate is compounded yearly;
(ii) if the interest rate is compounded half-yearly;
(iii) if the interest rate is compounded quarterly;
(iv) if the interest rate is compounded monthly.
▶️Answer/Explanation
Ans: METHOD A: By using the formula
(i) FV = \(10000(1+\frac {12}{100})^5=17623.42\)
(ii) FV = \(10000(1+\frac {12}{100\chi 2})^{5\chi 2}=17908.48\)
(iii) FV = \(10000(1+\frac {12}{100\chi 4})^{5\chi 4}=18061.11\)
(iv) FV = \(10000(1+\frac {12}{100\chi 12})^{5\chi 12}=18166.97\)
METHOD B: By using GDC-financial mode
(i) n=5, I% = 12, PV = –10000, P/Y=1, C/Y=1. Then press FV = 17623.42
(ii) C/Y=2, then FV = 17908.48
(iii) C/Y=4, then FV = 18061.11
(iv) C/Y=12, then FV = 18166.97
(b) The value of the investment will exceed $ 20 000 after n full years. Calculate the minimum value of n
(i) if the interest rate is compounded yearly;
(ii) if the interest rate is compounded monthly.
▶️Answer/Explanation
Ans: METHOD A: By using the formula
(i) 20000 = \(10000(1+\frac {12}{100\chi 12})^n\) by GDC the solution is n = 6.11, so n = 7
(ii) 20000 = \(10000(1+\frac {12}{100\chi 12})^12n\) by GDC the solution is n = 5.81, so n = 6
METHOD B: By using GDC-financial mode
(i) I% = 12, PV= –10000, P/Y=1, C/Y=1, FV=20000. Then press n=6.11 so n = 7
(ii) I% = 12, PV= –10000, P/Y=1, C/Y=12, FV=20000. Then press n=5.81 so n = 6
Question
Give your answers in this question correct to the nearest whole number.
Shahid invested 25 000 Singapore dollars (SGD) in a fixed deposit account with a nominal
annual interest rate of 3.6 %, compounded monthly.
Calculate the value of Shahid’s investment after 5 years. [3]
At the end of the 5 years, Shahid withdrew x SGD from the fixed deposit account and
reinvested this into a super-savings account with a nominal annual interest rate of
5.7 %, compounded half-yearly.
The value of the super-savings account increased to 20 000 SGD after 18 months.
Find the value of x . [3]
▶️Answer/Explanation
Ans:
(a)
\(FV=25000\times (1+ \frac{3.5}{100\times 12})^{12\times 5}\)
OR
N= 5
I%=3.6
PV= \(\mp 25000\)
\(P/Y\) = 1
C/Y = 12
OR
N= 60
I%=3.6
PV = \(\mp 25000\)
P/V = 12
C/Y = 12
FV=29922(SGD)
(b)
2000= PV\times \((1+\frac{5.7}{100\times 2})^{2\times 1.5}\)
OR
N= 1.5
I%=5.7
FV=\(\pm 20000\)
\(P/Y\) = 1
C/Y= 2
OR
N=3
I%= 5.7
FV = \(\pm 20000\)
P/V=2
C/Y = 2
X= 18383 (SGD)
Question
An amount of £ 10 000 is invested at an annual interest rate of 12%.
(a) Find the value of the investment after 5 years
(i) if the interest rate is compounded yearly;
(ii) if the interest rate is compounded half-yearly;
(iii) if the interest rate is compounded quarterly;
(iv) if the interest rate is compounded monthly.
(b) The value of the investment will exceed $ 20 000 after n full years. Calculate the minimum value of n
(i) if the interest rate is compounded yearly;
(ii) if the interest rate is compounded monthly.
▶️Answer/Explanation
Ans
METHOD A: By using the formula
(a) (i) FV = \(10000(1+\frac {12}{100})^5=17623.42\)
(ii) FV = \(10000(1+\frac {12}{100\chi 2})^{5\chi 2}=17908.48\)
(iii) FV = \(10000(1+\frac {12}{100\chi 4})^{5\chi 4}=18061.11\)
(iv) FV = \(10000(1+\frac {12}{100\chi 12})^{5\chi 12}=18166.97\)
(b)(i) 20000 = \(10000(1+\frac {12}{100\chi 12})^n\) by GDC the solution is n = 6.11, so n = 7
(ii) 20000 = \(10000(1+\frac {12}{100\chi 12})^12n\)
by GDC the solution is n = 5.81, so n = 6
METHOD B: By using GDC-financial mode
(a) (i) n=5, I% = 12, PV = –10000, P/Y=1, C/Y=1. Then press FV = 17623.42
(ii) C/Y=2, then FV = 17908.48
(iii) C/Y=4, then FV = 18061.11
(iv) C/Y=12, then FV = 18166.97
(b) (i) I% = 12, PV= –10000, P/Y=1, C/Y=1, FV=20000. Then press n=6.11 so n = 7 (ii) I% = 12, PV= –10000, P/Y=1, C/Y=12, FV=20000. Then press n=5.81 so n = 6