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Changing the subject Quiz

Grade 5 • 5F • 10 Questions (33 Marks)

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Changing the subject — Notes & Worked Examples Show / Hide

Explanation:

To change the subject of a formula, use inverse operations step by step to isolate the target letter on one side of the equals sign (multiply out fractions first, expand brackets, and undo powers with roots or vice versa).

If the letter you want appears twice in the formula, expand any brackets, collect all terms containing the target letter on one side and all other terms on the opposite side, then factorise out the target letter before dividing.

Example 1: Make v the subject of E = 12mv2.

Solution:

Multiply both sides by 2: 2E = mv2
Divide both sides by m: 2Em = v2
Square root both sides to isolate v:
Answer: v = ±√(2Em)

Example 2: Make x the subject of y = 3x + 2x − 4.

Solution:

Multiply by (x − 4): y(x − 4) = 3x + 2 → xy − 4y = 3x + 2
Collect x terms on left: xy − 3x = 4y + 2
Factorise out x: x(y − 3) = 4y + 2
Answer: x = 4y + 2y − 3
Question 1
[1 mark]
Make x the subject of the formula:
y = 5x + c
x =
Worked Solution & Answer
Use inverse operations step by step to get the target letter on its own.
x = y - c5
Answer: x = x = y - c5
Question 2
[1 mark]
Make p the subject of the formula:
t = 5p - r
p =
Worked Solution & Answer
Use inverse operations step by step to get the target letter on its own.
p = t + r5
Answer: p = p = t + r5
Question 3
[3 marks]
(a) Ben makes m subject of p = m/3 + 5: p – 5 = m/3; 3 × p – 5 = m; m = 3p – 5. What mistake has Ben made? (1)
(b) Factorise fully 2x³y + 4xy² (2)
(a) What mistake has Ben made? [1 mark]
(b) Factorise fully 2x3 y + 4xy2 [2 marks]
Worked Solution & Answer
(a)
When multiplying both sides by 3, the entire left hand side must be multiplied: 3(p – 5) = 3p – 15, not 3p – 5 B1
Answer (a): He didn't multiply the –5 by 3 (should be 3(p – 5) or 3p – 15)
(b)
Common factor is 2xy: 2xy(x² + 2y) M1 A1
Answer (b): 2xy(x2 + 2y)
Question Paper (PDF) • Mark Scheme (PDF)
Question 4
[3 marks]
(a) Make q the subject of p = 6q + 7 (2) (b) Simplify (m⁻²)⁻³ (1)
(a) Make q the subject of p = 6q + 7 [2 marks]
q =
(b) Simplify (m-2)-3 [1 mark]
Worked Solution & Answer
(a)
p – 7 = 6q M1
⇒ q = (p – 7)/6 A1
Answer (a): q = q = p - 76
(b)
m {}-2 × -3) = m⁶ B1
Answer (b): m6
Question Paper (PDF) • Mark Scheme (PDF)
Question 5 Eduqas Jun 2024 1F Q15
[3 marks]
(a) Factorise 3a + 7ab.
(b) Make w the subject of the formula y = 5w - 4.
(a) Factorise 3a + 7ab. [1 mark]
(b) Make w the subject of the formula y = 5w - 4. [2 marks]
w =
Worked Solution & Answer
(a)
Common factor is a: a(3 + 7b). B1
Answer (a): a(3 + 7b)
(b)
y + 4 = 5w ⇒ w = y + 45. M1 A1
Answer (b): w = w = (y + 4) / 5
Question Paper (PDF) • Mark Scheme (PDF)
Question 6 IGCSE Jan 2023 1F Q13
[4 marks]
p = t - ac
t = 18   a = -3   c = 5
(a) Work out the value of p [2 marks]
p =
(b) Make x the subject of the formula d = 3x + 10 [2 marks]
x =
Worked Solution & Answer
(a)
p = 18 − (−3)(5) = 18 − (−15) = 33 M1 A1
Answer (a): p = 33
(b)
d − 10 = 3x M1
⇒ x = (d − 10)/3 A1
Answer (b): x = x = d - 103
Question Paper (PDF) • Mark Scheme (PDF)
Question 7
[4 marks]
T = 4m2 - 11
(a) Work out the value of T when m = -3 [2 marks]
T =
(b) Make p the subject of the formula d = 3p + 4 [2 marks]
p =
Worked Solution & Answer
(a)
4(–3)² – 11 = 36 – 11 = 25 M1 A1
Answer (a): T = 25
(b)
d – 4 = 3p M1
⇒ p = (d – 4)/3 A1
Answer (b): p = p = d - 43
Question Paper (PDF) • Mark Scheme (PDF)
Question 8
[4 marks]
v2 = u2 + 2as
u = 12   a = -3   s = 18
(a) Work out a value of v. [2 marks]
(b) Make s the subject of v2 = u2 + 2as [2 marks]
s =
Worked Solution & Answer
(a)
v² = 144 + 2(–3)(18) = 144 – 108 = 36 M1
⇒ v = ±6 A1
Answer (a): 6 (or –6)
(b)
v² – u² = 2as M1
⇒ s = (v² – u²)/(2a) A1
Answer (b): s = s = v2 - u22a
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 IGCSE Jun 2025 2F Q12
[5 marks]
(a) Make y the subject of w = 3y – d. (2) (b) Diego buys b boxes (12 crayons each) and p packets (3 crayons each) for a total of T crayons. Write down a formula for T in terms of b and p. (3)
(a) Make y the subject of w = 3y - d [2 marks]
y =
(b) Write down a formula for T in terms of b and p [3 marks]
T =
Worked Solution & Answer
(a)
w + d = 3y M1
⇒ y = (w + d)/3 A1
Answer (a): y = y = w + d3
(b)
Use + for more/older, - for less/fewer/younger, and multiply/divide for rates. M1
e.g. 4 less than m is m - 4 M1
3 times m is 3m
Answer: T = 12b + 3p A1
Answer (b): T = T = 12b + 3p
Question Paper (PDF) • Mark Scheme (PDF)
Question 10 IGCSE Nov 2020 2F Q14
[5 marks]
(a) Factorise 25f − 10 (1) (b) Make y subject of c = 5y − h (2) (c) Solve 4x + 7 > 2 (2)
(a) Factorise 25f - 10 [1 mark]
(b) Make y the subject of the formula c = 5y - h [2 marks]
y =
(c) Solve the inequality 4x + 7 > 2 [2 marks]
Worked Solution & Answer
(a)
5(5f − 2)
Answer (a): 5(5f - 2)
(b)
c + h = 5y M1
⇒ y = (c + h) / 5 A1
Answer (b): y = y = c + h5
(c)
4x > −5 M1
⇒ x > −1.25 A1
Answer (c): x > -1.25
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 33 marks)
Changing the subject Worksheets Next: Quadratic expressions →

Grade 5 Quizzes

25 Topics
5A - Combining ratios5B - Direct and inverse proportion
5C - Reverse percentages5D - Standard form
5E - Speed5F - Changing the subject
5G - Quadratic expressions5H - Solving quadratic equations
5J - Drawing quadratic graphs5K - Cubic/reciprocal graphs
5L - Simultaneous equations5M - Sector areas and arc lengths
5N - Sim. equ. graphically5P - Midpoint of a line segment
5Q - Gradient of a line5R - Equation of a line
5S - Spheres and cones5T - Similar shapes (lengths)
5U - Trigonometry (SOHCAHTOA)5V - Exact trig. values
5W - Probability trees5X - Vectors
5Y - Venn diagrams5Z - Density
5AA - Pressure