1. 1
   A courier company charges a call-out fee plus a rate per parcel. One customer's bill in dollars satisfies $5x+8=12$, where $x$ is a scaling factor. A second customer's bill satisfies $7(x+3)=5x-8$. A third satisfies $\frac{4x+5}{2}=3$.

   (a)

   Solve $5x+8=12$.

   [1 mark]

   • A$x=0.4$
   • B$x=0.8$
   • C$x=4$
   • D$x=1$

   (b)

   Solve $7(x+3)=5x-8$ for the second customer.

   [1 mark]

   • A$x=-14.5$
   • B$x=14.5$
   • C$x=-2.5$
   • D$x=2.5$

   (c)

   Solve $\frac{4x+5}{2}=3$ for the third customer.

   [1 mark]

   • A$x=2.75$
   • B$x=-0.25$
   • C$x=0.25$
   • D$x=-2.75$

   Total for question 1: 3 marks

2. 2
   A taxi company charges a fixed fee plus a rate per kilometre. A second journey with a different taxi firm has a fare, in dollars, given by $3(2k-1)=27$, where $k$ is the number of kilometres travelled.

   (a)

   A taxi journey costs a fixed 4 dollars plus 2 dollars per kilometre. If the total fare is 18 dollars, form and solve an equation to find the number of kilometres travelled, $k$.

   [2 marks]

   (b)

   A second taxi firm's fare satisfies $3(2k-1)=27$. Solve this equation to find $k$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A mobile phone company sells two data plans. The first plan's monthly cost in dollars, $C$, satisfies $\frac{3C-6}{4}=9$. The second plan's cost, $D$, satisfies $9(D-2)=4(D+8)$.

   (a)

   A phone plan costs a monthly fee plus a charge per gigabyte of data. The total monthly cost in dollars, $C$, satisfies $\frac{3C-6}{4}=9$. Solve this equation to find $C$, showing all steps.

   [3 marks]

   (b)

   The second plan's cost satisfies $9(D-2)=4(D+8)$. Solve this equation to find $D$, showing all steps.

   [3 marks]

   Total for question 3: 6 marks

4. 4
   A leisure centre is building a rectangular swimming pool whose length is 3 metres more than its width, $w$, with a perimeter of 46 metres. Around the pool, a path of equal width $p$ metres is built such that the total surrounding perimeter (pool plus path) satisfies $2((w+3+2p)+(w+2p))=70$. A separate poolside equation involves fractional coefficients: $\frac{2}{3}(x-6)=\frac{1}{4}(x+8)$.

   (a)

   A rectangular swimming pool has a length that is 3 metres more than its width, $w$. The perimeter of the pool is 46 metres. Form an equation in terms of $w$ and solve it to find the width of the pool.

   [4 marks]

   (b)

   Using the width $w$ found in part (a), form an equation from $2((w+3+2p)+(w+2p))=70$ and solve it to find the path width, $p$.

   [4 marks]

   (c)

   A separate poolside sign uses the equation $\frac{2}{3}(x-6)=\frac{1}{4}(x+8)$. Solve this equation fully to find $x$, showing all working with fractions cleared.

   [5 marks]

   Total for question 4: 13 marks

End of questions