Two coffees and one tea cost $\$9$. One coffee and two teas cost $\$7.50$. Find the price of one coffee and the price of one tea.
Question HKC78-AISL-SYS-01
[Maximum mark: 5]
Two coffees and one tea cost $\$9$. One coffee and two teas cost $\$7.50$. Find the price of one coffee and the price of one tea.
Mark Scheme
Let $c$ be the coffee price and $t$ the tea price, in dollars. Then $2c + t = 9$ and $c + 2t = 7.5$. [1]
Double the second equation: $2c + 4t = 15$. Subtract the first: $3t = 6$, so $t = 2$. [2]
$c + 4 = 7.5$, so $c = 3.5$. [1]
Check: $2(3.5) + 2 = 9$ and $3.5 + 4 = 7.5$. A coffee costs $$3.50$ and a tea costs $$2$. [1]
Total: 5 marks
$\$2000$ is invested at $3.5\%$ per year, compounded annually. Find the value of the investment after $4$ years, to the nearest dollar.
Question HKC78-AISL-FIN-01
[Maximum mark: 4]
$\$2000$ is invested at $3.5\%$ per year, compounded annually. Find the value of the investment after $4$ years, to the nearest dollar.
Mark Scheme
Value $= 2000(1.035)^{4}$. [1]
$(1.035)^{4} = 1.147523ldots$. [1]
$2000 times 1.147523ldots = 2295.05ldots$. [1]
To the nearest dollar, the value is $$2295$. [1]
Total: 4 marks
Let $z = 5 – 12i$. Find $|z|$ and the real part of $z^{2}$.
Question HKC78-AISL-CPX-01
[Maximum mark: 5]
Let $z = 5 – 12i$. Find $|z|$ and the real part of $z^{2}$.
Mark Scheme
$|z| = sqrt{5^{2} + (-12)^{2}} = sqrt{25 + 144} = sqrt{169} = 13$. [2]
$z^{2} = (5 – 12i)^{2} = 25 – 120i + 144i^{2} = 25 – 120i – 144 = -119 – 120i$. [2]
The real part is $-119$. [1]
Total: 5 marks
Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}$. Find the matrix $AB$.
Question HKC78-AISL-MAT-01
[Maximum mark: 4]
Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}$. Find the matrix $AB$.
Mark Scheme
Row-by-column products: top-left $2cdot 1 + 1cdot 2 = 4$, top-right $2cdot 4 + 1cdot(-1) = 7$. [2]
Bottom-left $0cdot 1 + 3cdot 2 = 6$, bottom-right $0cdot 4 + 3cdot(-1) = -3$. [1]
$AB = begin{pmatrix} 4 & 7 \ 6 & -3 end{pmatrix}$. [1]
Total: 4 marks
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Question Q1
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 10$.
Step 2: Use formula $u_n = u_1 + (n-1)d$.
$u_{10} = 5 + (10-1)3$
Step 3: Calculate.
$u_{10} = 5 + (9)3 = 5 + 27 = 32$.
Answer: 32
A salary starts at $\$28000$ in year $1$ and increases by $\$1200$ at the start of each later year. Find the salary in year $6$, and the total earned over the first $6$ years.
Question HKC78-AISL-SEQ-01
[Maximum mark: 5]
A salary starts at $\$28000$ in year $1$ and increases by $\$1200$ at the start of each later year. Find the salary in year $6$, and the total earned over the first $6$ years.
Mark Scheme
This is arithmetic with $u_{1} = 28000$ and $d = 1200$.
Year $6$: $u_{6} = 28000 + 5 cdot 1200 = 34000$. [2]
Total: $S_{6} = dfrac{6}{2}(28000 + 34000) = 3 cdot 62000 = 186000$. [3]
Total: 5 marks
A loan of $\$8000$ is repaid by $5$ equal payments, one at the end of each year. Interest is $6\%$ per year, compounded annually. Find the annual payment, to the nearest dollar.
Question HKC78-AISL-FIN-02
[Maximum mark: 5]
A loan of $\$8000$ is repaid by $5$ equal payments, one at the end of each year. Interest is $6\%$ per year, compounded annually. Find the annual payment, to the nearest dollar.
Mark Scheme
This is an ordinary annuity. With $r = 0.06$ and $n = 5$, the payment is $A = 8000 cdot dfrac{0.06(1.06)^{5}}{(1.06)^{5} – 1}$. [2]
$(1.06)^{5} = 1.3382256ldots$, so $A = 8000 cdot dfrac{0.0802935ldots}{0.3382256ldots} = 1899.17ldots$. [2]
To the nearest dollar, the payment is $$1899$. [1]
Total: 5 marks
The complex number $z$ satisfies $z + \overline{z} = 6$ and $z\overline{z} = 13$. Find $z$ in the form $a + bi$ with $b > 0$.
Question HKC78-AISL-CPX-02
[Maximum mark: 5]
The complex number $z$ satisfies $z + \overline{z} = 6$ and $z\overline{z} = 13$. Find $z$ in the form $a + bi$ with $b > 0$.
Mark Scheme
Write $z = a + bi$. Then $z + overline{z} = 2a = 6$, so $a = 3$. [2]
$zoverline{z} = a^{2} + b^{2} = 13$, so $9 + b^{2} = 13$ and $b^{2} = 4$. [2]
$b > 0$, so $b = 2$ and $z = 3 + 2i$. [1]
Total: 5 marks
Let $A = \begin{pmatrix} 4 & 1 \\ 2 & 1 \end{pmatrix}$.
(a) Find $A^{-1}$.
(b) Hence solve $4x + y = 9$ and $2x + y = 5$.
Question HKC78-AISL-MAT-02
[Maximum mark: 6]
Let $A = \begin{pmatrix} 4 & 1 \\ 2 & 1 \end{pmatrix}$.
(a) Find $A^{-1}$.
(b) Hence solve $4x + y = 9$ and $2x + y = 5$.
Mark Scheme
(a) $det A = 4 – 2 = 2$. $A^{-1} = dfrac{1}{2}begin{pmatrix} 1 & -1 \ -2 & 4 end{pmatrix}$. [3]
(b) $begin{pmatrix} x \ y end{pmatrix} = dfrac{1}{2}begin{pmatrix} 1 & -1 \ -2 & 4 end{pmatrix}begin{pmatrix} 9 \ 5 end{pmatrix} = dfrac{1}{2}begin{pmatrix} 4 \ 2 end{pmatrix} = begin{pmatrix} 2 \ 1 end{pmatrix}$. [2]
So $x = 2$ and $y = 1$. Check: $8 + 1 = 9$ and $4 + 1 = 5$. [1]
Total: 6 marks
