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GCSE · Algebra

5.6 Changing the Subject of a Formula

EDEXCEL 1MA1 1MA1 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Make x the subject of each formula.

(a)y = x + 9(1)

(b)y = x - 6(1)

(c)p = t + x(1)

(Total for Question 1 is 3 marks)

Make x the subject of each formula.

(a)y = 4x(1)

(b)y = x/7(1)

(c)w = kx(1)

(Total for Question 2 is 3 marks)

Make x the subject of each formula.

(a)y = 3x + 5(2)

(b)y = 6x - 11(2)

(Total for Question 3 is 4 marks)

Make t the subject of each formula.

(a)p = 8 - 3t(2)

(b)q = 5 - t/2(2)

(Total for Question 4 is 4 marks)

Given that y = 2x + 3, which of the following correctly makes x the subject of the formula?

A) x = (y - 3)/2
B) x = (y + 3)/2
C) x = 2y - 3
D) x = (y - 2)/3

(Total for Question 5 is 1 mark)

Make the given letter the subject of each formula.

(a)P = 2(l + w). Make w the subject.(2)

(b)A = 3(2x - 5). Make x the subject.(2)

(Total for Question 6 is 4 marks)

The formula connecting x and y is y = (2x - 8)/5. Show that x = (5y + 8)/2

(Total for Question 7 is 2 marks)

To make x the subject of y = 4x - 3, Ben writes:
y + 3 = 4x
x = 4y + 3
Explain the error in Ben's method and write down the correct expression for x in terms of y.

(Total for Question 8 is 2 marks)

An old recipe book gives a formula for converting a temperature C degrees Celsius into F degrees Fahrenheit: F = 9C/5 + 32

(a)Make C the subject of the formula.(3)

(b)Use your formula to work out the value of C when F = 68. Give your answer in degrees Celsius.(1)

(Total for Question 9 is 4 marks)

In these formulae, u and v are speeds, a is acceleration, t is time and s is distance.

(a)v = u + at. Make a the subject.(2)

(b)v² = u² + 2as. Make s the subject.(2)

(Total for Question 10 is 4 marks)

The area of a trapezium with parallel sides a and b and height h is given by A = (1/2)(a + b)h

(a)Make h the subject of the formula.(2)

(b)Make a the subject of the formula.(2)

(Total for Question 11 is 4 marks)

The circumference of a circle of radius r is given by the formula C = 2 * π * r

Diagram NOT accurately drawn

(a)Rearrange the formula to make r the subject.(1)

(b)A circular pond has a circumference of 18.84 metres. Work out the radius of the pond. Give your answer correct to 1 decimal place.(2)

(Total for Question 12 is 3 marks)

Make x the subject of each formula.

(a)y = (x + 2)/3(2)

(b)y = 5/(x - 1)(3)

(Total for Question 13 is 5 marks)

A taxi firm uses the formula C = 3 + 2m to work out the cost, C pounds, of a journey of m miles.

(a)Find the cost of a journey of 12 miles.(1)

(b)Rearrange the formula to make m the subject.(2)

(c)Preeti pays 17 pounds sterling for her taxi journey. Work out how many miles she travelled.(2)

(Total for Question 14 is 5 marks)

These formulae involve a square or a square root.

(a)The area of a circle of radius r is A = π * r^2. Make r the subject of the formula.(3)

(b)v² = u² + 2as. Make u the subject of the formula.(3)

(Total for Question 15 is 6 marks)

Make p the subject of the formula: y(p - 3) = 2(5 - p)

(Total for Question 16 is 4 marks)

Make x the subject of the formula: y = (3x + 2)/(x - 4)

(Total for Question 17 is 4 marks)

The time period T of a pendulum of length l is given by T = 2 * π * √l/g, where g is the acceleration due to gravity. Make l the subject of the formula.

(Total for Question 18 is 4 marks)

The volume of a cone is V = (1/3) * π * r² * h. Make r the subject of the formula.

(Total for Question 19 is 4 marks)

The volume of a sphere of radius r is V = (4/3) * π * r³

Diagram NOT accurately drawn

(a)Make r the subject of the formula.(4)

(b)A sphere has volume 500 cm^3. Work out its radius. Give your answer correct to 3 significant figures.(2)

(Total for Question 20 is 6 marks)

Mark scheme · 5.6 Changing the Subject of a Formula

Question 1

(a) B1 x = y - 9 oe
(a) Answer: x = y - 9
(b) B1 x = y + 6 oe
(b) Answer: x = y + 6
(c) B1 x = p - t oe
(c) Answer: x = p - t

Question 2

(a) B1 x = y/4 oe
(a) Answer: x = y/4
(b) B1 x = 7y oe
(b) Answer: x = 7y
(c) B1 x = w/k oe
(c) Answer: x = w/k

Question 3

(a) M1 subtracts 5 from both sides, e.g. 3x = y - 5
(a) A1 x = (y - 5)/3 oe, cao
(a) Answer: x = (y - 5)/3
(b) M1 adds 11 to both sides, e.g. 6x = y + 11
(b) A1 x = (y + 11)/6 oe, cao
(b) Answer: x = (y + 11)/6

Question 4

(a) M1 rearranges to isolate the term in t, e.g. 3t = 8 - p
(a) A1 t = (8 - p)/3 oe, cao
(a) Answer: t = (8 - p)/3
(b) M1 rearranges to isolate t/2, e.g. t/2 = 5 - q
(b) A1 t = 10 - 2q oe, cao
(b) Answer: t = 10 - 2q

Question 5

B1 A
Answer: A

Question 6

(a) M1 divides both sides by 2 or expands the bracket, e.g. P/2 = l + w
(a) A1 w = P/2 - l oe, e.g. w = (P - 2l)/2, cao
(a) Answer: w = P/2 - l
(b) M1 expands and rearranges, e.g. A = 6x - 15 then 6x = A + 15
(b) A1 x = (A + 15)/6 oe, cao
(b) Answer: x = (A + 15)/6

Question 7

M1 multiplies both sides by 5, e.g. 5y = 2x - 8
A1 correct rearrangement with all steps shown leading to x = (5y + 8)/2, cso
Answer: x = (5y + 8)/2 (printed answer, shown)

Question 8

C1 identifies the error, e.g. Ben divided only the y term by 4 instead of dividing the whole expression (y + 3) by 4
B1 x = (y + 3)/4 oe, cao
Answer: x = (y + 3)/4

Question 9

(a) M1 subtracts 32 from both sides, e.g. F - 32 = 9C/5
(a) M1 multiplies both sides by 5 (or by 5/9), e.g. 5(F - 32) = 9C
(a) A1 C = 5(F - 32)/9 oe, e.g. C = (5F - 160)/9, cao
(a) Answer: C = 5(F - 32)/9
(b) B1 C = 20 (deg C), ft from part a
(b) Answer: C = 20 deg C

Question 10

(a) M1 subtracts u from both sides, e.g. v - u = at
(a) A1 a = (v - u)/t oe, cao
(a) Answer: a = (v - u)/t
(b) M1 subtracts u² from both sides, e.g. v² - u² = 2as
(b) A1 s = (v² - u²)/(2a) oe, cao
(b) Answer: s = (v² - u²)/(2a)

Question 11

(a) M1 multiplies both sides by 2, e.g. 2A = (a + b)h
(a) A1 h = 2A/(a + b) oe, cao
(a) Answer: h = 2A/(a + b)
(b) M1 multiplies by 2 and divides by h, e.g. 2A/h = a + b
(b) A1 a = 2A/h - b oe, cao
(b) Answer: a = 2A/h - b

Question 12

(a) B1 r = C/(2 * π) oe
(a) Answer: r = C/(2 * π)
(b) M1 substitutes C = 18.84 into r = C/(2 * π), ft from part a
(b) A1 awrt 3.0 (m)
(b) Answer: r = 3.0 m (1 dp)

Question 13

(a) M1 multiplies both sides by 3, e.g. 3y = x + 2
(a) A1 x = 3y - 2 oe, cao
(a) Answer: x = 3y - 2
(b) M1 multiplies both sides by (x - 1), e.g. y(x - 1) = 5
(b) M1 expands and isolates the term in x, e.g. yx = 5 + y
(b) A1 x = (5 + y)/y oe, e.g. x = 5/y + 1, cao
(b) Answer: x = (5 + y)/y

Question 14

(a) B1 27 (pounds)
(a) Answer: 27 pounds sterling
(b) M1 subtracts 3 from both sides, e.g. C - 3 = 2m
(b) A1 m = (C - 3)/2 oe, cao
(b) Answer: m = (C - 3)/2
(c) P1 substitutes C = 17 into the rearranged formula from part b, ft
(c) A1 7 (miles), cao/ft
(c) Answer: 7 miles

Question 15

(a) M1 divides both sides by π, e.g. r² = A/π
(a) dM1 square roots both sides, dependent on the previous method mark
(a) A1 r = √A/&π; oe, cao (positive root only, since r is a length)
(a) Answer: r = √A/&π;
(b) M1 subtracts 2as from both sides, e.g. u² = v² - 2as
(b) dM1 square roots both sides, dependent on the previous method mark
(b) A1 u = √v² - 2as oe, cao
(b) Answer: u = √v² - 2as

Question 16

M1 expands both brackets, e.g. yp - 3y = 10 - 2p
M1 collects all terms in p on one side, e.g. yp + 2p = 10 + 3y
dM1 factorises out p, e.g. p(y + 2) = 10 + 3y, dependent on the previous method mark
A1 p = (10 + 3y)/(y + 2) oe, cao
Answer: p = (10 + 3y)/(y + 2)

Question 17

M1 multiplies both sides by (x - 4), e.g. y(x - 4) = 3x + 2
M1 expands and collects all terms in x on one side, e.g. yx - 3x = 2 + 4y
dM1 factorises out x, e.g. x(y - 3) = 2 + 4y, dependent on the previous method mark
A1 x = (2 + 4y)/(y - 3) oe, cao
Answer: x = (2 + 4y)/(y - 3)

Question 18

M1 divides both sides by 2 * π, e.g. T/(2*π) = √l/g
M1 squares both sides, e.g. T²/(4*π²) = l/g
dM1 multiplies both sides by g, dependent on the previous method mark
A1 l = g * T²/(4 * π²) oe, cao
Answer: l = g * T²/(4 * π²)

Question 19

M1 multiplies both sides by 3, e.g. 3V = π * r² * h
M1 divides both sides by π * h, e.g. r² = 3V/(π * h)
dM1 square roots both sides, dependent on the previous method mark
A1 r = √3V/(&π; * h) oe, cao (positive root only, since r is a length)
Answer: r = √3V/(&π; * h)

Question 20

(a) M1 multiplies both sides by 3, e.g. 3V = 4 * π * r³
(a) M1 divides both sides by 4 * π, e.g. r³ = 3V/(4 * π)
(a) dM1 cube roots both sides, dependent on the previous method mark
(a) A1 r = (3V/(4 * π))^1/3 oe, cao
(a) Answer: r = (3V/(4 * π))^1/3
(b) M1 substitutes V = 500 into the formula from part a, ft, e.g. r = (1500/(4*π))^1/3
(b) A1 awrt 4.92 (cm)
(b) Answer: r = 4.92 cm (3 sf)

Changing the Subject of a Formula - Worksheets, Questions and Revision

5.6 Changing the Subject of a Formula

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20