A Resource for Free-standing Mathematics Qualifications
Rearrange Formulae
Worksheet
1 The speed of a car after t seconds is given by the formula v  u  at
where u is the starting speed and a is the acceleration.
Rearrange this formula to make the subject (a) u (b) t
2 In an electrical circuit, the formula relating power, P, to the current, I, and resistance, R, is
P = I2R. Rearrange this formula to make the subject (a) R (b) I
v2
2g
where g is the acceleration due to gravity. Rearrange this formula to make the subject v.
3 If a ball is thrown up in the air at a velocity of v, the height it reaches is given by h 
5
F  32
9
where C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit.
Rearrange this formula to make the subject F.
4 A formula for converting temperatures is C 
5 When a car accelerates steadily from a speed u to a speed v in time t, the distance it travels is given
t u  v 
by: s 
. Rearrange this formula to make the subject (a) t (b) v
2
6 The volume of a sphere of radius r is given by the formula V 
4 3
r
3
r
Rearrange this formula to make the subject r
7 When the speed of a car increases from speed u with acceleration a over a distance s, its final speed
v 2  u 2  2as .
is v where:
Rearrange this formula to make the subject (a) s (b) u
8 When the speed of a car increases from speed u with acceleration a, the distance that it travels in
s  ut  12 at 2 .
time t is given by:
Rearrange this formula to make the subject (a) a (b) u
9 The time taken for a pendulum of length l
l
to make one full swing is T  2
g
where g is the acceleration due to gravity.
Rearrange this formula to make the subject l.
string
mass
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l
metres
T seconds
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Rearrange Formulae
10 (a) The volume of a cylinder is given by the formula V  r 2 h
where r is the radius and h is the height.
Rearrange this formula to make the subject r.
r
h
(b) The formula for the total surface area of the cylinder is S  2r r  h
Rearrange this formula to make the subject h.

11 The area of a washer is given by the formula A   R 2  r 2
where r is the inner radius and R is the outer radius.
Rearrange this formula to make the subject r.

R
r
12 (a) The volume of a cone with radius r and height h is given
by V  13 r 2 h . Rearrange this formula to make the subject r.
(b) The formula S  r r 2  h 2 gives the area of the curved surface
of the cone. Rearrange this formula to make the subject h.
h
r
13 When an object of mass m, moves with velocity v at height h its total energy is E  12 mv 2  mgh .
Rearrange this formula to make the subject (a) v (b) m.
14 If a principal £P earns simple interest at a rate of R% per annum, the total amount after
PRT
T years is given by the formula A  P 
.
100
Rearrange this formula to make the subject P.
15 When an amount of money, £P, is left in a building society account that gives r% interest per
n
r 

annum, the amount in the account after n years is given by the formula A  P1 
 .
 100 
Rearrange this formula to make the subject r.
16 If the monthly rate of interest on a loan is m %, the annual percentage rate (APR) is given by:
12


m 
R = 100 1 
  1 .
 100 

Rearrange this formula to make the subject m.

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Rearrange Formulae
Teacher Notes
Unit
Intermediate Level, Using algebra, functions and graphs
Skills used in this activity:
 rearranging formulae
Notes
Ideally this topic should be covered using formulae that are relevant and useful to your students. If
you use the Word versions of these files, you can select the formulae that are most relevant to your
students, delete the rest and add others if you wish.
The Powerpoint presentation gives some examples that can be used to introduce this work. The
worksheets include a wide range of formulae from different contexts that students can use for
practice. Each solution is given on the set of cards on the following pages. These can be laminated
and cut out for use by students who find rearranging formulae difficult. Ask the students to put the
cards in the right order, then write out the solution.
Answers
NB In many cases alternative answers are possible.
(b) t 
1 (a) u  v  at
vu
a
2
5 (a) t 
2s
uv
(b) v 
7 (a) s 
v2  u2
2a
(b) u  v 2  2as
2s
u
t
F
6
r3
(a)
a
2
T 
l 
 g
 2 
11
r R 
13 (a) v 
A

2E  mgh
m
(b)
m
2E
v  2 gh
2
 A 
r  100 n  1
 P

15

P
R
I
9
C  32
5
4
8
(b)
2
3V
4
2s  ut 
t
9
2
P
R
I
v  2 gh
3
(a)
2
V
h
(b) u 
s 1
 at
t 2
(b) h 
S
r
2r
10 (a)
r
12 (a)
3V
r
h
14
P
16
 R

m  100 12
 1  1
 100

2
S
(b) h     r 2
 r 
100 A
100  RT
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v  u  at
v  u  at
P = I 2R
v  at  u
v  u  at
P =R
I2
u  v - at
t v - u
a
R= P
I2
I= P
R
P = I 2R
P
=I
R

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4
P = I2
R
S = 2πr r h
1
V = πr 2h
3
4
V = πr 3
3
S =r+h
2πr
3V = πr 2h
3V = 4 πr 3
S –r=h
2πr
3V
= r2
h
3V
= r3
4π
h= S –r
2πr
3V
=r
h
3 3V = r
4π
3V
h
3V
r=3
4π






r=

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
2
2
A = π R - r 
V = πr 2h
S = r 2  h2
πr
V = r2
h
A = R2 – r2
π




2
S  = r 2  h2
πr 
V =r
h
A + r2 = R2
π




2
S  – r 2 = h2
πr 
V
h
r2 = R2 – A
π
2
S   r 2
πr 
r = R2  A
π











S = πr r 2  h 2
r=
2
S   r 2 = h
πr 
h=





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

T = 2π
l
g
C = 5  F 32
9
s = t u  v 
2
T =
2π
l
g
9C = 5 F 32
2s = t u  v 
2




T
2π
l
=
g
9 C = F  32
5
2s
=t
uv




2
T  g = l
2π 
9 C + 32 = F
5
t=
2

T

l=
 g
2π 
F = 9 C + 32
5









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2s
uv






r

A = P 1

100 
A 
r 
 1

P  100 
n
n
v 2  u2  2as
v 2  u2  2as
v 2  2as  u2
n A  1 r
100
P
v 2  u2 = s
2a
v 2  2as = u
n A 1  r
100
P
2 2
s = v u
2a
u = v 2  2as














A
100 n 1  r
P 

v 2  u2  2as


A
r  100 n 1
P 
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
s = t u  v 
2
s = ut  1 at 2
2
s = ut  1 at 2
2
2s = t u  v 
s – ut = 1 at 2
2
s – 1 at 2 = ut
2
2s
 uv
t
2(s – ut) = at 2
s  1 at = u
t 2
2s
u  v
t
2 s - ut 
=a
t2
u = s  1 at
t 2
2s
v  u
t
2 s - ut 
a=
t2
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9
E = 1 mv 2  mgh
2
2
v
h=
2g
E = 1 mv 2  mgh
2
E – mgh = 1 mv 2
2
2gh = v2
2E = mv 2  2mgh
2 E mgh = mv 2







2 gh = v
2 E  mgh 
= v2
m
v = 2 gh
2 E  mgh 
=v
m
2 E  mgh 
v=
m
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





2
2E = m v  2 gh
2E = m
v 2  2 gh
m=
2E
v 2  2 gh








R = 100 1 
A = P + PRT
100

m 

100 
12
1
100A = 100P + PRT
R  1 m 12 1


100


100
100A = P(100 + RT)
R  1  1  m 12

100 

100
12 R  1  1  m
100
100
100 A = P
100  RT
12 R 1 1  m
100
100
P = 100 A
100  RT

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













R
m  100 12
 1 1

100









R
100 12
1 1  m

100
