(a) Bananas cost $\$1.20$ per kilogram and apples cost $\$2.25$ per bag.
Work out the total cost of 3.5 kg of bananas and 2 bags of apples.
(b) Students receive a 10% discount on their shopping.
Before the discount, the cost of a student’s shopping is $\$16.80$.
Work out the amount of the discount.
(c) The cost of a cabbage increases by 15%.
Calculate the new price if the original price is $\$1.80$.
(d) Some customers have their shopping delivered to their home.
The cost is $\$5$ plus $\$1.50$ for each kilometre travelled from the shop to their home.
(i) Show that the cost for a customer living 10km from the shop is $\$20$.
(ii) On the grid, draw a line to show the cost of having shopping delivered.
(e) A bottle of water costs $\$1.55$.
Suki has $\$20$.
Work out the maximum number of bottles Suki can buy and the change she receives.
(f) A farmer delivers eggs to the shop in trays of 50.
The eggs are then put into boxes of 12.
There are no eggs left in the trays and all of the egg boxes are full.
Work out the smallest possible number of eggs that the farmer delivers.
(g) The shop sells bottles of orange juice in three different sizes.
Work out which bottle is the best value.
Show how you decide.
▶️ Answer/Explanation
(a) Ans: \$8.70
Bananas: $3.5 \times 1.20 = 4.20$
Apples: $2 \times 2.25 = 4.50$
Total: $4.20 + 4.50 = 8.70$
(b) Ans: \$1.68
Discount: $10\% \text{ of } 16.80 = 0.10 \times 16.80 = 1.68$
(c) Ans: \$2.07
Increase: $15\% \text{ of } 1.80 = 0.27$
New price: $1.80 + 0.27 = 2.07$
(d)(i) Ans: \$20
Delivery cost: $5 + (1.50 \times 10) = 5 + 15 = 20$
(d)(ii)
Line from (0,5) to (10,20) showing cost increases \$1.50 per km
(e) Ans: 12 bottles, \$1.40 change
Max bottles: $\left\lfloor \frac{20}{1.55} \right\rfloor = 12$
Change: $20 – (12 \times 1.55) = 1.40$
(f) Ans: 300 eggs
LCM of 50 and 12: $2^2 \times 3 \times 5^2 = 300$
(g) Ans: Bottle A
Price per litre:
A: $\frac{1.30}{0.5} = 2.60$
B: $\frac{3.20}{1.2} \approx 2.67$
C: $\frac{5.25}{2} = 2.625$
Bottle A is cheapest per litre.
(a) In a sport, teams are given points using the formula
number of points = number of wins × 4 + number of draws × 2 + bonus points
One team has 15 wins, 7 draws and 6 bonus points. Calculate the total number of points for this team.
(b) Solve \(\frac{x}{2} = 18\)
(c) Solve \(4x + 12 = 18\)
(d) Expand and simplify \(6(3x – 4) + 5(x – 2)\)
(e) \(T = 5r – 6\)
Make \(r\) the subject of this formula.
(f) Bo has a green bag and a blue bag. Each bag contains some marbles. The green bag has \(x\) marbles. There are 5 times as many marbles in the blue bag than in the green bag.
Bo now adds 6 marbles to each bag. There are now 4 times as many marbles in the blue bag than in the green bag.
Use this information to write down an equation and solve it to find the value of \(x\).
▶️ Answer/Explanation
(a) Ans: 80
Using the points formula:
Points = (15 wins × 4) + (7 draws × 2) + 6 bonus = 60 + 14 + 6 = 80.
(b) Ans: 36
\(\frac{x}{2} = 18\)
Multiply both sides by 2: \(x = 36\).
(c) Ans: 1.5
\(4x + 12 = 18\)
Subtract 12: \(4x = 6\)
Divide by 4: \(x = 1.5\).
(d) Ans: \(23x – 34\)
Expand:
\(6(3x – 4) + 5(x – 2) = 18x – 24 + 5x – 10\)
Combine like terms: \(23x – 34\).
(e) Ans: \(r = \frac{T + 6}{5}\)
\(T = 5r – 6\)
Add 6: \(T + 6 = 5r\)
Divide by 5: \(r = \frac{T + 6}{5}\).
(f) Ans: \(x = 18\)
Initial marbles:
Green bag: \(x\), Blue bag: \(5x\)
After adding 6 to each: \(x + 6\) and \(5x + 6\)
Set up equation: \(5x + 6 = 4(x + 6)\)
Solve: \(5x + 6 = 4x + 24\) → \(x = 18\).