Forming and solving equations Quiz (Grade 4)

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Topic:

Score: 0 / 10

Forming and solving equations — Notes & Worked Examples Show / Hide

Explanation:

To form and solve an equation from a word problem or geometry diagram, write an algebraic expression for each part using the given letter, then use a known mathematical fact to set up an equation with an = sign.

Common facts used in exams include: perimeter = sum of all outer sides, opposite sides of a rectangle are equal, angles on a straight line = 180°, angles in a triangle = 180°, and angles in a quadrilateral = 360°. Always read the end of the question to check if you need to substitute x back in!

Example 1: Find the size of the largest angle in the triangle below.

(2x + 15) + (3x − 5) + (x + 20) = 180 → 6x + 30 = 180 → x = 25

Largest angle = 3(25) − 5 = 70°

Example 2: A rectangle has sides (2x + 5) cm and (x + 1) cm, and perimeter 42 cm. Find its length.

2(2x + 5) + 2(x + 1) = 42 → 6x + 12 = 42 → x = 5

Length = 2(5) + 5 = 15 cm

One Page Guide (PDF) Forming and solving equations Worksheets

Question 1

[2 marks]

The perimeter of the rectangle is 56
Find the value of x

x =

Worked Solution & Answer

Use the given geometric or word rule (e.g. angles in a triangle add to 180°, or perimeter is the sum of all sides) to write an equation, then solve for b.
x = 4

Answer: x = x = 4

Question 2

[2 marks]

Form an equation and find x

x =

Worked Solution & Answer

Use the given geometric or word rule (e.g. angles in a triangle add to 180°, or perimeter is the sum of all sides) to write an equation, then solve for b.
x = 12

Answer: x = x = 12

Question 3

[3 marks]

The diagram shows triangle AOB.

Angle AOB is **not** an obtuse angle.
Find the greatest value of x.
You must show all your working.

Worked Solution & Answer

Not obtuse ⇒ Angle AOB ≤ 90° M2
2x + 3x + 10 ≤ 90 ⇒ 5x ≤ 80 ⇒ x ≤ 16, so greatest value of x is 16. A1

Answer: 16

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[3 marks]

Angles around a point are 90°, 2x° and 3x°. Find the value of x.

Find the value of x.

x =

Worked Solution & Answer

2x + 3x + 90 = 360 M1
⇒ 5x = 270 M1
⇒ x = 54 A1

Answer: x = 54

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[4 marks]

Here is a quadrilateral ABCD.

All the measurements are in centimetres.
The perimeter of ABCD is 52 centimetres.
Work out the length of DC.

centimetres

Worked Solution & Answer

(x + 1) + (x – 1) + 2x + (2x – 5) = 6x – 5 = 52 M2
⇒ 6x = 57 M1
⇒ 2x = 19 cm A1

Answer: 19 centimetres

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 AQA Nov 2022 2F Q27 / 2H Q12

[4 marks]

The length of this rectangle is 6 times the width.

Two of these rectangles are joined, with no overlap, to make this L-shape.
The perimeter of the L-shape is 98.8 cm
Work out the value of the perimeter of **one** of the rectangles.

Answer cm

Worked Solution & Answer

Perimeter of L-shape = 6x + x + (6x – x) + 6x + x + 6x + x = 26x. M2
26x = 98.8 ⇒ x = 3.8. M1
Perimeter of one rectangle = 14x = 14 × 3.8 = 53.2 cm. A1

Answer: Answer 53.2 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 AQA Nov 2023 3F Q15

[4 marks]

Here are two number machines.
The outputs are the same.

Work out the value of a.

a =

Worked Solution & Answer

Output of machine 1: 30 ÷ 6 + 18 = 5 + 18 = 23 M1
Output of machine 2: 30 × 3 - a = 90 - a M1
Since the outputs are equal: M1
90 - a = 23 ⇒ a = 90 - 23 = 67. A1

Answer: a = 67

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[4 marks]

ABCD is a kite.

AB = (4x - 2) cm
Jasper says that x could be 0.5

(a) Explain why Jasper cannot be correct. [1 mark]

(b) Find the value of x. [3 marks]

x =

Worked Solution & Answer

(a)

4(0.5) – 2 = 0 C1

Answer (a): If x = 0.5, AB = 4(0.5) – 2 = 0 cm, and a side length cannot be 0

(b)

Perimeter = 2AB + 2(3AB) = 8AB = 64 ⇒ AB = 8 M2
4x – 2 = 8 ⇒ 4x = 10 ⇒ x = 2.5 A1

Answer (b): x = 2.5

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 AQA Nov 2024 2F Q26

[5 marks]

Here is a rectangle.
All measurements are in centimetres.

AB : BC = 1 : 3
Work out the value of the area.

Answer cm²

Worked Solution & Answer

(The area of a rectangle is length × width)
Opposite sides are equal: 4x + 1 = 2x + 17 ⇒ 2x = 16 ⇒ x = 8 M2
Width AB = 4(8) + 1 = 33 cm M1
Length BC = 3 × 33 = 99 cm M1
Area = 33 × 99 = 3267 cm². A1

Answer: Answer 3267 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 10

[5 marks]

The diagram shows a pentagon.
The pentagon has one line of symmetry.

AE = 4x
AB = 2x + 1
BC = x + 2
All these measurements are given in centimetres.
The perimeter of the pentagon is 18 cm.

(a) Show that 10x + 6 = 18 [3 marks]

(b) Find the value of x. [2 marks]

x =

Worked Solution & Answer

(a)

4x + (2x + 1) + (x + 2) + (x + 2) + (2x + 1) = 10x + 6 = 18 M2 C1

Answer (a): By symmetry, ED = 2x + 1 and CD = x + 2; Perimeter = 4x + 2(2x + 1) + 2(x + 2) = 4x + 4x + 2 + 2x + 4 = 10x + 6 = 18

(b)

10x = 12 M1
⇒ x = 1.2 A1

Answer (b): x = 1.2

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 36 marks)

Forming and solving equations Worksheets Next: nth term of a linear sequence →

Grade 4 Quizzes

24 Topics

4A - Compound interest	4B - Indices (index laws)
4C - Prime factors HCF and LCM	4D - Real life graphs
4E - Inequalities on a number line	4F - Solving inequalities
4G - Forming and solving equations	4H - nth term of a linear sequence
4J - Expanding and factorising	4K - Pythagoras theorem
4L - Angles in parallel lines	4M - Angles in polygons
4N - Surface area	4P - Volume of a cuboid
4Q - Volume of a prism	4R - Bearings
4S - Cylinders	4T - Plans and elevations
4U - Scatter graphs	4V - Probability
4W - Averages from tables	4X - Reverse mean
4Y - Area of a trapezium	4Z - Loci and construction