Budget & Quadratic Equations Worksheet
Expressions · Forming Equations · Solving Quadratics — 5 Variations
Instructions: Show all your working clearly. For parts that say "show", you must derive the given result step by step to earn full marks (M1, A1). Round answers to 2 decimal places where necessary.
Key Concept — Budget Problems
- If an item costs $x, the number of items you can buy with a budget of $B is Bx.
- When two items have different prices, the difference in quantity purchased equals the difference between their individual quantities.
VARIATION A
Pastries — Donuts & Cupcakes
[8 marks]Question 1
- (a) The price of a donut is $x. Write down, in terms of x, an expression for the number of donuts the students can buy if they decide to buy donuts only. [1]
- (b) The price of a cupcake is $(x + 1). If they spend $36 on donuts only, they can buy 25 more donuts than when they spend $36 on cupcakes only. Write down an equation and show that it simplifies to 25x2 + 25x − 36 = 0. [3]
- (c) Solve 25x2 + 25x − 36 = 0. [2]
- (d) If the students decide to buy equal number of donuts and cupcakes with $36, find the maximum number of each type of pastry they can get. [2]
VARIATION B
School Supplies — Pencils & Pens
[8 marks]Question 2
- (a) The price of a pencil is $x. Write down, in terms of x, an expression for the number of pencils the students can buy if they decide to buy pencils only. [1]
- (b) The price of a pen is $(x + 2). If they spend $36 on pencils only, they can buy 15 more pencils than when they spend $36 on pens only. Write down an equation and show that it simplifies to 15x2 + 30x − 72 = 0. [3]
- (c) Solve 15x2 + 30x − 72 = 0. [2]
- (d) If the students decide to buy equal number of pencils and pens with $36, find the maximum number of each type of stationery they can get. [2]
VARIATION C
Sports Day — Water Bottles & Energy Bars
[8 marks]Question 3
- (a) The price of a water bottle is $x. Write down, in terms of x, an expression for the number of water bottles the committee can buy if they decide to buy water bottles only. [1]
- (b) The price of an energy bar is $(x + 1). If they spend $50 on water bottles only, they can buy 25 more water bottles than when they spend $50 on energy bars only. Write down an equation and show that it simplifies to x2 + x − 2 = 0. [3]
- (c) Solve x2 + x − 2 = 0. [2]
- (d) If the committee decides to buy equal number of water bottles and energy bars with $50, find the maximum number of each item they can get. [2]
VARIATION D
Ticket Sales — Movie Tickets & Popcorn
[8 marks]Question 4
- (a) The price of one movie ticket is $x. Write down, in terms of x, an expression for the number of movie tickets the club can buy if they decide to buy tickets only. [1]
- (b) The price of a popcorn box is $(x + 3). If they spend $72 on movie tickets only, they can buy 4 more movie tickets than when they spend $72 on popcorn boxes only. Write down an equation and show that it simplifies to x2 + 3x − 54 = 0. [3]
- (c) Solve x2 + 3x − 54 = 0. [2]
- (d) If the club decides to buy equal number of movie tickets and popcorn boxes with $72, find the maximum number of each item they can get. [2]
VARIATION E
Fruit Stand — Apples & Oranges
[8 marks]Question 5
- (a) The price of an apple is $x. Write down, in terms of x, an expression for the number of apples the family can buy if they decide to buy apples only. [1]
- (b) The price of an orange is $(x + 2). If they spend $48 on apples only, they can buy 4 more apples than when they spend $48 on oranges only. Write down an equation and show that it simplifies to x2 + 2x − 24 = 0. [3]
- (c) Solve x2 + 2x − 24 = 0. [2]
- (d) If the family decides to buy equal number of apples and oranges with $48, find the maximum number of each type of fruit they can get. [2]
Answer Key
Variation A — Donuts & Cupcakes
(a) 36x [1]
(b) 36x − 36x+1 = 25 → 36(x+1) − 36x = 25x(x+1) → 36x+36−36x = 25x2+25x → 25x2 + 25x − 36 = 0 [M1, A1, A1]
(c) Factorise: (5x+9)(5x−4) = 0 → x = −1.8 or x = 0.8 [A1, A1]
(d) Price of each = $x + $(x+1) = $(2x+1). Using x=0.8: 2(0.8)+1 = $2.60. Number = ⌊36/2.60⌋ = 13 each (costs $33.80) [A1, A1]
Variation B — Pencils & Pens
(a) 36x [1]
(b) 36x − 36x+2 = 15 → 36(x+2)−36x = 15x(x+2) → 72 = 15x2+30x → 15x2 + 30x − 72 = 0 [M1, A1, A1]
(c) Divide by 3: 5x2+10x−24=0. Factorise: (5x−6)(x+4)=0 → x=1.2 or x=−4 [A1, A1]
(d) Price of each = $(2x+2). Using x=1.2: 2(1.2)+2 = $4.40. Number = ⌊36/4.40⌋ = 8 each (costs $35.20) [A1, A1]
Variation C — Water Bottles & Energy Bars
(a) 50x [1]
(b) 50x − 50x+1 = 25 → 50(x+1)−50x = 25x(x+1) → 50 = 25x2+25x → x2 + x − 2 = 0 [M1, A1, A1]
(c) Factorise: (x+2)(x−1) = 0 → x = −2 or x = 1 [A1, A1]
(d) Price of each = $(2x+1). Using x=1: 2(1)+1 = $3. Number = ⌊50/3⌋ = 16 each (costs $48) [A1, A1]
Variation D — Movie Tickets & Popcorn
(a) 72x [1]
(b) 72x − 72x+3 = 4 → 72(x+3)−72x = 4x(x+3) → 216 = 4x2+12x → x2 + 3x − 54 = 0 [M1, A1, A1]
(c) Factorise: (x+9)(x−6) = 0 → x = −9 or x = 6 [A1, A1]
(d) Price of each = $(2x+3). Using x=6: 2(6)+3 = $15. Number = ⌊72/15⌋ = 4 each (costs $60) [A1, A1]
Variation E — Apples & Oranges
(a) 48x [1]
(b) 48x − 48x+2 = 4 → 48(x+2)−48x = 4x(x+2) → 96 = 4x2+8x → x2 + 2x − 24 = 0 [M1, A1, A1]
(c) Factorise: (x+6)(x−4) = 0 → x = −6 or x = 4 [A1, A1]
(d) Price of each = $(2x+2). Using x=4: 2(4)+2 = $10. Number = ⌊48/10⌋ = 4 each (costs $40) [A1, A1]