The formula v = u + at connects speeds u and v, acceleration a and time t. Make t the subject of the formula.
(Total for Question 2 is 2 marks)
3
Make x the subject of the formula: y = (x - 5)/3
(Total for Question 3 is 2 marks)
4
The circumference of a circle of radius r is given by C = 2 * π * r. Which of the following correctly makes r the subject of the formula?
A) r = C - 2 * π
B) r = C/(2 * π)
C) r = 2 * π * C
D) r = C/2 - π
(Total for Question 4 is 1 mark)
5
The volume of a cone is given by V = (1/3) * π * r2 * h. Make h the subject of the formula.
(Total for Question 5 is 2 marks)
6
An annulus is the flat ring-shaped region between two concentric circles of radius R (outer) and r (inner). Its area is given by A = π * (R2 - r2). Make R the subject of the formula.
(Total for Question 6 is 3 marks)
7
The distance travelled, s, by an object with initial speed u, acceleration a, after time t is given by s = ut + (1/2)at2. Make a the subject of the formula.
(Total for Question 7 is 3 marks)
8
The time period, T, of a simple pendulum of length l is given by T = 2 * π * √l/g, where g is the gravitational field strength.
(a)Make g the subject of the formula.(4)
(b)A pendulum has length 1.5 m and time period 2.46 seconds. Calculate the value of g given by the formula in part (a). Give your answer correct to 3 significant figures.(2)
(Total for Question 8 is 6 marks)
9
Make x the subject of the formula: y = x2/4 + 3
(Total for Question 9 is 3 marks)
10
Make x the subject of the formula: 3(x + 2y) = 5(x - y)
(Total for Question 10 is 3 marks)
11
The kinetic energy of a moving object is given by E = (1/2)mv2, where E is the kinetic energy in joules, m is the mass in kilograms and v is the speed in metres per second. Make v the subject of the formula.
(Total for Question 11 is 3 marks)
12
Make x the subject of the formula: y = 5 - 2x2
(Total for Question 12 is 3 marks)
13
The surface area of a sphere of radius r is given by S = 4 * π * r2
Diagram NOT accurately drawn
(a)Make r the subject of the formula.(3)
(b)A sphere has surface area 350 cm2. Calculate the radius of the sphere. Give your answer correct to 3 significant figures.(2)
(Total for Question 13 is 5 marks)
14
Two resistors, with resistances a ohms and b ohms, are connected in parallel. Their combined resistance, x ohms, satisfies the formula 1/x = 1/a + 1/b. Make x the subject of the formula.
(Total for Question 14 is 3 marks)
15
Make x the subject of the formula: y = (x + 3)/2 - (x - 1)/5
(Total for Question 15 is 4 marks)
16
The total surface area of a closed cylinder with base radius r and height h is given by A = 2 * π * r * h + 2 * π * r2. Make h the subject of the formula.
(Total for Question 16 is 3 marks)
17
Make x the subject of the formula: y = (4x - 1)/(x + 3)
(Total for Question 17 is 4 marks)
18
Make x the subject of the formula: p = √(x + 3)/(x - 2)
(Total for Question 18 is 5 marks)
19
The formula connecting w and x is w = (5 - 2x)/(x + 4). Show that x = (5 - 4w)/(w + 2)
(Total for Question 19 is 4 marks)
20
Make x the subject of the formula: y = (p - 3x)/(2x + q), where p and q are constants.
(Total for Question 20 is 4 marks)
Mark scheme · 7.7 Rearranging Harder Formulae
Question 1
M1 adds 9 to both sides, e.g. y + 9 = 4x
A1 x = (y + 9)/4 oe, cao
Answer: x = (y + 9)/4
Question 2
M1 subtracts u from both sides, e.g. v - u = at
A1 t = (v - u)/a oe, cao
Answer: t = (v - u)/a
Question 3
M1 multiplies both sides by 3, e.g. 3y = x - 5
A1 x = 3y + 5 oe, cao
Answer: x = 3y + 5
Question 4
B1 B cao
Answer: B
Question 5
M1 multiplies both sides by 3 and divides by π * r2, e.g. 3V = π * r2 * h
A1 h = 3V/(π * r2) oe, cao
Answer: h = 3V/(π * r2)
Question 6
M1 divides both sides by π, e.g. A/π = R2 - r2
M1 adds r2 to both sides, e.g. R2 = A/π + r2
A1 R = √A/&π; + r2 oe, cao (positive root only, since R is a length)
Answer: R = √A/&π; + r2
Question 7
M1 subtracts ut from both sides, e.g. s - ut = (1/2)at2
M1 multiplies both sides by 2, e.g. 2(s - ut) = at2
A1 a = 2(s - ut)/t2 oe, cao
Answer: a = 2(s - ut)/t2
Question 8
(a) M1 divides both sides by 2 * π, e.g. T/(2*π) = √l/g
(a) M1 squares both sides, e.g. T2/(4*π2) = l/g
(a) dM1 rearranges to make g the subject, e.g. g * T2 = 4 * π2 * l, dependent on the previous method mark
(a) A1 g = 4 * π2 * l/T2 oe, cao
(a) Answer: g = 4 * π2 * l/T2
(b) M1 substitutes l = 1.5 and T = 2.46 into the formula from part (a), ft from part (a)
(b) A1 awrt 9.79 (m/s2)
(b) Answer: g = 9.79 m/s2 (3 sf)
Question 9
M1 subtracts 3 from both sides, e.g. y - 3 = x2/4
M1 multiplies both sides by 4 and square roots, e.g. x2 = 4(y - 3)
A1 x = 2 * √y - 3 oe, e.g. x = √4y - 12, cao (allow ±)
Answer: x = 2 * √y - 3
Question 10
M1 expands both brackets, e.g. 3x + 6y = 5x - 5y
M1 collects terms in x on one side, e.g. 11y = 2x
A1 x = 11y/2 oe, cao
Answer: x = 11y/2
Question 11
M1 multiplies both sides by 2 and divides by m, e.g. v2 = 2E/m
dM1 square roots both sides, dependent on the previous method mark
A1 v = √2E/m oe, cao (positive root only, since v is a speed)
Answer: v = √2E/m
Question 12
M1 rearranges to isolate the term in x2, e.g. 2x2 = 5 - y
M1 divides by 2 and square roots, e.g. x2 = (5 - y)/2
A1 x = √(5 - y)/2 oe, cao (allow ±)
Answer: x = √(5 - y)/2
Question 13
(a) M1 divides both sides by 4 * π, e.g. r2 = S/(4 * π)
(a) dM1 square roots both sides, dependent on the previous method mark
(a) A1 r = √S/(4 * &π;) oe, cao (positive root only, since r is a length)
(a) Answer: r = √S/(4 * &π;)
(b) M1 substitutes S = 350 into the formula from part (a), ft, e.g. r = √350/(4*&π;)
(b) A1 awrt 5.28 (cm)
(b) Answer: r = 5.28 cm (3 sf)
Question 14
M1 combines the right-hand side over a common denominator, e.g. 1/x = (b + a)/(ab)
dM1 takes the reciprocal of both sides, dependent on the previous method mark
A1 x = ab/(a + b) oe, cao
Answer: x = ab/(a + b)
Question 15
M1 multiplies every term by 10 (the LCM of 2 and 5), e.g. 10y = 5(x + 3) - 2(x - 1)
M1 expands both brackets correctly, e.g. 10y = 5x + 15 - 2x + 2
M1 collects like terms and isolates x, e.g. 3x = 10y - 17
A1 x = (10y - 17)/3 oe, cao
Answer: x = (10y - 17)/3
Question 16
M1 subtracts 2 * π * r2 from both sides, e.g. A - 2 * π * r2 = 2 * π * r * h
M1 divides both sides by 2 * π * r
A1 h = (A - 2 * π * r2)/(2 * π * r) oe, cao
Answer: h = (A - 2 * π * r2)/(2 * π * r)
Question 17
M1 multiplies both sides by (x + 3), e.g. y(x + 3) = 4x - 1
M1 expands and collects all terms in x on one side, e.g. yx - 4x = -1 - 3y
dM1 factorises out x, e.g. x(y - 4) = -1 - 3y, dependent on the previous method mark
A1 x = (3y + 1)/(4 - y) oe, cao
Answer: x = (3y + 1)/(4 - y)
Question 18
M1 squares both sides, e.g. p2 = (x + 3)/(x - 2)
M1 multiplies both sides by (x - 2), e.g. p2(x - 2) = x + 3
M1 expands and collects all terms in x on one side, e.g. p2 x - x = 3 + 2p2
dM1 factorises out x, e.g. x(p2 - 1) = 3 + 2p2, dependent on the previous method mark
A1 x = (3 + 2p2)/(p2 - 1) oe, cao
Answer: x = (3 + 2p2)/(p2 - 1)
Question 19
M1 multiplies both sides by (x + 4), e.g. w(x + 4) = 5 - 2x
M1 expands and collects all terms in x on one side, e.g. wx + 2x = 5 - 4w
dM1 factorises out x, e.g. x(w + 2) = 5 - 4w, dependent on the previous method mark
A1 correct rearrangement with all steps shown leading to x = (5 - 4w)/(w + 2), cso