Question
[Maximum mark: 7]
John is building furniture using cylindrical logs of length $1.8 \mathrm{~m}$ and radius $9.2 \mathrm{~cm}$. A wedge is cut from one log and the cross-section of this $\log$ is illustrated in the following diagram.
(a) Find the length of the wedge arc, ABC. [2]
(b) Find the area of the empty sector, OABC. [2]
(c) Find the volume of each log. [3]
Answer/Explanation
(a) Using the length of an arc formula, we get
$
\begin{aligned}
l & =\frac{\theta}{360} 2 \pi r \\
l & =\frac{100}{360} 2 \pi(9.2) \\
& \approx 16.1 \mathrm{~cm}
\end{aligned}
$
(b) Using the area of a sector formula, we get
$
\begin{aligned}
A & =\frac{\theta}{360} \pi r^2 \\
& =\frac{100}{360} \pi\left(9.2^2\right) \\
& \approx 73.9 \mathrm{~cm}^2
\end{aligned}
$
(c) First we find the area of a log’s sector. Using the area of a sector formula, we get
$
\begin{aligned}
A & =\frac{\theta}{360} \pi r^2 \\
& =\frac{360-100}{360} \pi\left(9.2^2\right) \\
& =\frac{260}{360} \pi\left(9.2^2\right) \\
& \approx 192 \mathrm{~cm}^2
\end{aligned}
$
The volume of the log can now be found by
$
\begin{aligned}
& V=A l \\
& V=192(180) \\
& \approx 34600 \mathrm{~cm}^3
\end{aligned}
$
Question
[Maximum mark: 6]
Charles plans to invest in a retirement plan for 30 years. In this plan, he will deposit 1000 British pounds (GBP) at the end of every month and receive a $6.5 \%$ interest rate per annum, compounded monthly.
(a) Find the future value of the investment at the end of the 30 years. Give your answer correct to the nearest pound. [3]
After the 30-year period, Charles will start receiving regular monthly payments of 1500 GBP.
(b) Calculate the number of years it will take Charles’s monthly retirement payments to match the total amount originally invested. [3]
Answer/Explanation
(a) Using the TVM Solver on G.D.C., we have
Hence the value of the investment at the end of the 30 years is $1106178 \mathrm{GBP}$
(b) The amount that Charles invests over 30 years is
$
1000 \times 360=360000 \mathrm{GBP}
$
Hence Charles breaks even with the amount he invested after
$
\frac{360000}{1500 \times 12}=20 \text { years }
$