Question difficulty level:
Basic
Moderate
Challenging
Difficult
Genius :)
Basic
Moderate
Challenging
Difficult
Genius :)
Paper 2 – Calculator
[Maximum mark: 6]
In this question, all answers must be given to two decimal places.
Anushka decides to invest \$ 3500 in an account for eight years and makes no further deposits or withdrawals. The annual interest rate paid by the bank is 1.87\, \%, compounded monthly.
(a) Find how much money Anushka will have in the account after eight years.
[3]
(b) Write down the interest earned by Anushka.
[1]
Rayaan decides to invest \$ P into an account also for eight year and makes no further deposits or withdrawals. The annual interest rate paid by the bank is 2.05\, \%, compounded quarterly.
At the end of the eight years, Rayaan’s account and Anushka’s account will have the same amount of money.
(c) Find the value of P.
[2]
(a)
Using the compound interest formula (you can also use the TVM Solver inside the Finance app), we get:
FV = 3500\left ( 1 + \frac{1.87}{100(12)} \right )^{(12)(8)} \approx
\boldsymbol{{\color{Purple} \approx \$4064.32}}
(b)
4064.32 - 3500 =
\boldsymbol{{\color{Purple} = \$ 564.32}}
(c)
Using the TVM Solver inside the Finance app (you can also use the compound interest formula), we get:
N = 8 \times 4 = 32
I\, \% = 2.05
FV = 4064.32
P/Y = 4
C/Y = 4
Solving for PV we get that:
\boldsymbol{{\color{Purple} P \approx 3451.00}}
.
Paper 2 – Calculator
[Maximum mark: 9]
There are 700 raspberry bushes on a farm at the end of 2023. The number of raspberry bushes decreases by 7\, \% each year, unless new bushes are planted. The owner of the farm plants 80 new bushes each year.
(a) Find the approximate number of raspberry bushes on the farm at the end of 2030.
[4]
(b) Find in what year the number of raspberry bushes on the farm first exceeds 1100.
[5]
(a)
We can calculate a 7\, \% decrease by multiplying the initial number of raspberry bushes by 0.93.
So for the number of raspberry bushes at the end of 2024 we get:
700 \times 0.93 + 80
For the number of raspberry bushes at the end of 2025 we get:
(700 \times 0.93 + 80) \times 0.93 + 80 = 700 \times 0.93^{2} + 80 \times 0.93 + 80 = 700 \times 0.93^{2} + 80(0.93 + 1)
For the number of raspberry bushes at the end of 2026 we get:
(700 \times 0.93^{2} + 80(0.93 + 1)) \times 0.93 + 80 = 700 \times 0.93^{3} + 80(0.93^{2} + 0.93 + 1)
Following this pattern, for the number of raspberry bushes at the end of 2030 we get:
700 \times 0.93^{7} + 80(0.93^{6} + 0.93^{5} + 0.93^{4} + 0.93^{3} + 0.93^{2} + 0.93 + 1) \approx
\boldsymbol{{\color{Purple} \approx 876 \: \textbf{bushes}}}
(b)
Let’s write an expression for the number of raspberry bushes n years after 2023:
700 \times 0.93^{n} + 80(0.93^{n-1} + 0.93^{n-2} + ... + 0.93 + 1)
The expression in the brackets is the sum of the first n terms of a geometric sequence with u_{1} = 1 and r = 0.93, so for the number of raspberry bushes n years after 2023 we get:
700 \times 0.93^{n} + 80\left ( \frac{1(1 - 0.93^{n})}{1 - 0.93} \right )
We are looking for the smallest value of n for which the number of raspberry bushes exceeds 1100, so simplifying and equating the expression to 1100 we get:
700 \times 0.93^{n} + \frac{8000(1 - 0.93^{n})}{7} = 1100
Solving for n either by graphing or using the Solver function on the calculator, we get that n \approx 32.2, therefore the first year in which the number of raspberry bushes exceeds 1100 is 2023 + 33, so:
\boldsymbol{{\color{Purple} 2056}}
.
Paper 2 – Calculator
[Maximum mark: 16]
Ella invests $6000 into an account that pays an 8.5\: \% interest rate per annum, compounded semi-annually.
(a) Find the value of Ella’s investment after 7 years. Give your answer to the nearest dollar.
[3]
After n full years the value of Ella’s investment is $13000.
(b) Find the smallest possible value of n.
[2]
Claudia also invests $6000 into an account that pays an r\: \% interest rate per annum, compounded quarterly. After 11 years the value of Claudia’s investment is $14000.
(c) Find the smallest possible value of r.
[3]
Instead of investing her money into an account, Alma puts her savings into a box and then each year she adds x\: \% of what she added to the box the previous year. Alma’s goal is to have a total of $17500 in the box. The first year she puts $4500 into the box.
(d) Let x = 70. Determine whether Alma will achieve her goal.
[5]
(e) Find the smallest possible value of x so that Alma reaches her goal in 20 years.
[3]
(a)
Using the compound interest formula (you can also use the Finance app on your calculator):
6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times 7}\approx
\boldsymbol{{\color{Purple} \approx 10745}}
(b)
Using the compound interest formula (you can also use the Finance app on your calculator):
13000 = 6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times n}
Using the calculator to solve for n:
n = 9.288, therefore
\boldsymbol{{\color{Purple} n = 10}}
(c)
Using the compound interest formula (you can also use the Finance app on your calculator):
14000 = 6000\left ( 1 + \frac{r}{100 \times 4} \right )^{4 \times 11}
Using the calculator to solve for r we get that:
\boldsymbol{{\color{Purple} r \approx 7.78}}
(d)
Adding 70\: \% of what was added the previous year means that the amounts added each year form a geometric sequence where r = 0.7.
Using the sum of an infinite geometric sequence formula:
S_{\infty} = \frac{4500}{1 - 0.7} = 15000
\boldsymbol{{\color{Purple} \textbf{Since}\: 15000 < 17500, \textbf{Alma will not reach her goal.}}}
(e)
Using the sum of n terms of a geometric sequence formula:
17500 = \frac{4500(1 - r^{20})}{1 - r}
Using the calculator to solve for r:
r \approx 0.743543, so:
\boldsymbol{{\color{Purple} x \approx 74.4}}
.