IB Mathematics SL 1.8 Systems of linear equations AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Olivia organizes a concert. The ticket prices for the concert are shown in the following table.
| Ticket Type | Price (in Australian dollars, $) |
|---|---|
| Adult | 15 |
| Child | 10 |
| Student | 12 |
- A total of 600 tickets were sold.
- The total amount of money from ticket sales was \( \$7816 \).
- There were twice as many adult tickets sold as child tickets.
Let the number of adult tickets sold be \( x \), the number of child tickets sold be \( y \), and the number of student tickets sold be \( z \).
(a) Write down three equations that express the information given above. [3]
(b) Find the number of each type of ticket sold. [2]
▶️ Answer/Explanation
Markscheme
(a)
Total tickets:
\( x + y + z = 600 \) A1
Total money:
\( 15 x + 10 y + 12 z = 7816 \) A1
Adult to child ratio:
\( x = 2 y \) A1
[3 marks]
Total tickets:
\( x + y + z = 600 \) A1
Total money:
\( 15 x + 10 y + 12 z = 7816 \) A1
Adult to child ratio:
\( x = 2 y \) A1
[3 marks]
(b)
Substitute \( x = 2 y \) into the first equation:
\[ \begin{aligned} 2 y + y + z &= 600 \\ 3 y + z &= 600 \end{aligned} \]
Substitute \( x = 2 y \) into the second equation:
\[ \begin{aligned} 15 (2 y) + 10 y + 12 z &= 7816 \\ 30 y + 10 y + 12 z &= 7816 \\ 40 y + 12 z &= 7816 \end{aligned} \]
Solve the system: \( z = 600 – 3 y \), then:
\[ \begin{aligned} 40 y + 12 (600 – 3 y) &= 7816 \\ 40 y + 7200 – 36 y &= 7816 \\ 4 y &= 616 \\ y &= 154 \end{aligned} \] M1
Then:
\[ \begin{aligned} x &= 2 \times 154 = 308 \\ z &= 600 – 3 \times 154 = 600 – 462 = 138 \end{aligned} \] A1
Number of tickets: \( x = 308 \), \( y = 154 \), \( z = 138 \). A1
[2 marks]
Substitute \( x = 2 y \) into the first equation:
\[ \begin{aligned} 2 y + y + z &= 600 \\ 3 y + z &= 600 \end{aligned} \]
Substitute \( x = 2 y \) into the second equation:
\[ \begin{aligned} 15 (2 y) + 10 y + 12 z &= 7816 \\ 30 y + 10 y + 12 z &= 7816 \\ 40 y + 12 z &= 7816 \end{aligned} \]
Solve the system: \( z = 600 – 3 y \), then:
\[ \begin{aligned} 40 y + 12 (600 – 3 y) &= 7816 \\ 40 y + 7200 – 36 y &= 7816 \\ 4 y &= 616 \\ y &= 154 \end{aligned} \] M1
Then:
\[ \begin{aligned} x &= 2 \times 154 = 308 \\ z &= 600 – 3 \times 154 = 600 – 462 = 138 \end{aligned} \] A1
Number of tickets: \( x = 308 \), \( y = 154 \), \( z = 138 \). A1
[2 marks]
Total Marks: 5
Question
Sofia uses a set of coordinate axes to draw her design of a window. The base of the window is on the \( x \)-axis, the upper part of the window is in the form of a quadratic curve, and the sides are vertical lines. The curve has end points \( (0, 10) \) and \( (8, 10) \) and its vertex is \( (4, 12) \). Distances are measured in centimetres.
The quadratic curve can be expressed in the form \( y = a x^2 + b x + c \) for \( 0 \leq x \leq 8 \).
(a)(i) Write down the value of \( c \). [1]
(a)(ii) Hence form two equations in terms of \( a \) and \( b \). [2]
(a)(iii) Hence find the equation of the quadratic curve. [2]
(b) Find the area of the shaded region in Sofia’s design. [3]
▶️ Answer/Explanation
Markscheme
(a)(i)
At \( x = 0 \), \( y = c = 10 \). A1
Value of \( c \): 10.
[1 mark]
At \( x = 0 \), \( y = c = 10 \). A1
Value of \( c \): 10.
[1 mark]
(a)(ii)
Using \( c = 10 \):
At \( (8, 10) \):
\[ \begin{aligned} y &= a (8)^2 + b (8) + 10 \\ 10 &= 64 a + 8 b + 10 \end{aligned} \] A1
At \( (4, 12) \):
\[ \begin{aligned} y &= a (4)^2 + b (4) + 10 \\ 12 &= 16 a + 4 b + 10 \end{aligned} \] A1
Equations: \( 64 a + 8 b + 10 = 10 \), \( 16 a + 4 b + 10 = 12 \).
[2 marks]
Using \( c = 10 \):
At \( (8, 10) \):
\[ \begin{aligned} y &= a (8)^2 + b (8) + 10 \\ 10 &= 64 a + 8 b + 10 \end{aligned} \] A1
At \( (4, 12) \):
\[ \begin{aligned} y &= a (4)^2 + b (4) + 10 \\ 12 &= 16 a + 4 b + 10 \end{aligned} \] A1
Equations: \( 64 a + 8 b + 10 = 10 \), \( 16 a + 4 b + 10 = 12 \).
[2 marks]
(a)(iii)
Simplify equations:
\[ \begin{aligned} 64 a + 8 b &= 0 \\ 16 a + 4 b &= 2 \end{aligned} \]
Divide second equation by 4: \( 4 a + b = \frac{1}{2} \). M1
Solve: \( b = -4 a \), substitute into \( 4 a + b = \frac{1}{2} \):
\[ \begin{aligned} 4 a – 4 a &= \frac{1}{2} \\ a &= -\frac{1}{8}, \quad b = 1 \end{aligned} \] A1
Equation: \( y = -\frac{1}{8} x^2 + x + 10 \). A1
[2 marks]
Simplify equations:
\[ \begin{aligned} 64 a + 8 b &= 0 \\ 16 a + 4 b &= 2 \end{aligned} \]
Divide second equation by 4: \( 4 a + b = \frac{1}{2} \). M1
Solve: \( b = -4 a \), substitute into \( 4 a + b = \frac{1}{2} \):
\[ \begin{aligned} 4 a – 4 a &= \frac{1}{2} \\ a &= -\frac{1}{8}, \quad b = 1 \end{aligned} \] A1
Equation: \( y = -\frac{1}{8} x^2 + x + 10 \). A1
[2 marks]
(b)
Area is the integral of the curve from \( x = 0 \) to \( x = 8 \):
\[ \begin{aligned} \text{Area} &= \int_{0}^{8} \left( -\frac{1}{8} x^2 + x + 10 \right) dx \end{aligned} \] M1
Antiderivative:
\[ \begin{aligned} \int \left( -\frac{1}{8} x^2 + x + 10 \right) dx &= -\frac{1}{8} \cdot \frac{x^3}{3} + \frac{x^2}{2} + 10 x \\ &= -\frac{1}{24} x^3 + \frac{1}{2} x^2 + 10 x \end{aligned} \] A1
Evaluate:
\[ \begin{aligned} \left[ -\frac{1}{24} x^3 + \frac{1}{2} x^2 + 10 x \right]_0^8 &= \left( -\frac{1}{24} (8^3) + \frac{1}{2} (8^2) + 10 (8) \right) – 0 \\ &= \left( -\frac{512}{24} + \frac{64}{2} + 80 \right) \\ &= \left( -\frac{64}{3} + 32 + 80 \right) \\ &= -\frac{64}{3} + \frac{96}{3} + \frac{240}{3} = \frac{272}{3} \approx 90.67 \text{ cm}^2 \end{aligned} \] A1
Area: \( \frac{272}{3} \approx 90.67 \text{ cm}^2 \).
[3 marks]
Area is the integral of the curve from \( x = 0 \) to \( x = 8 \):
\[ \begin{aligned} \text{Area} &= \int_{0}^{8} \left( -\frac{1}{8} x^2 + x + 10 \right) dx \end{aligned} \] M1
Antiderivative:
\[ \begin{aligned} \int \left( -\frac{1}{8} x^2 + x + 10 \right) dx &= -\frac{1}{8} \cdot \frac{x^3}{3} + \frac{x^2}{2} + 10 x \\ &= -\frac{1}{24} x^3 + \frac{1}{2} x^2 + 10 x \end{aligned} \] A1
Evaluate:
\[ \begin{aligned} \left[ -\frac{1}{24} x^3 + \frac{1}{2} x^2 + 10 x \right]_0^8 &= \left( -\frac{1}{24} (8^3) + \frac{1}{2} (8^2) + 10 (8) \right) – 0 \\ &= \left( -\frac{512}{24} + \frac{64}{2} + 80 \right) \\ &= \left( -\frac{64}{3} + 32 + 80 \right) \\ &= -\frac{64}{3} + \frac{96}{3} + \frac{240}{3} = \frac{272}{3} \approx 90.67 \text{ cm}^2 \end{aligned} \] A1
Area: \( \frac{272}{3} \approx 90.67 \text{ cm}^2 \).
[3 marks]
Total Marks: 8