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40 questions at your level
Difficulty
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Level 1
Level 2
Level 3
Level 4
Level 5
Substituting into formulae
15 questions
Lesson
Not started
Mark as done
Work out the value of
3
a
+
2
b
3a + 2b
3
a
+
2
b
when
a
=
4
a = 4
a
=
4
and
b
=
5
b = 5
b
=
5
.
●
●●●●
Level 1
1 mark
Start
→
Mark as done
P
=
2
l
+
2
w
P = 2l + 2w
P
=
2
l
+
2
w
. Find
P
P
P
when
l
=
7
l = 7
l
=
7
and
w
=
3
w = 3
w
=
3
.
●
●●●●
Level 1
1 mark
Start
→
Mark as done
The cost, £
C
C
C
, of hiring a hall for
d
d
d
days is given by
C
=
5
d
+
20
C = 5d + 20
C
=
5
d
+
20
Work out the cost of hiring the hall for
7
7
7
days.
●●
●●●
Level 2
2 marks
Start
→
Mark as done
T
=
2
π
l
g
T = 2\pi\sqrt{\dfrac{l}{g}}
T
=
2
π
g
l
Work out the value of
T
T
T
when
l
=
2.5
l = 2.5
l
=
2.5
and
g
=
9.8
g = 9.8
g
=
9.8
Give your answer correct to 3 significant figures. (2 marks)
●●
●●●
Level 2
2 marks
Start
→
Mark as done
v
=
u
+
a
t
v = u + at
v
=
u
+
a
t
(a)
Work out the value of
v
v
v
when
u
=
−
6
u = -6
u
=
−
6
,
a
=
3
a = 3
a
=
3
and
t
=
4
t = 4
t
=
4
(b)
P
=
3
q
2
P = 3q^2
P
=
3
q
2
Work out the value of
P
P
P
when
q
=
−
5
q = -5
q
=
−
5
●●●
●●
Level 3
3 marks
Start
→
Mark as done
s
=
u
t
+
1
2
a
t
2
s = ut + \dfrac{1}{2}at^2
s
=
u
t
+
2
1
a
t
2
Work out the value of
s
s
s
when
u
=
5
u = 5
u
=
5
,
a
=
−
2
a = -2
a
=
−
2
and
t
=
6
t = 6
t
=
6
●●●
●●
Level 3
2 marks
Start
→
Mark as done
E
=
1
2
m
v
2
E = \dfrac{1}{2}mv^2
E
=
2
1
m
v
2
Work out the value of
E
E
E
when
m
=
8
m = 8
m
=
8
and
v
=
5
v = 5
v
=
5
.
●●●
●●
Level 3
2 marks
Start
→
Mark as done
v
2
=
u
2
+
2
a
s
v^2 = u^2 + 2as
v
2
=
u
2
+
2
a
s
(a)
Work out the value of
v
v
v
when
u
=
3
u = 3
u
=
3
,
a
=
4
a = 4
a
=
4
and
s
=
5
s = 5
s
=
5
, given that
v
>
0
v > 0
v
>
0
. (2 marks)
(b)
Make
s
s
s
the subject of the formula. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
The cost,
£
C
£C
£
C
, for each person going on a coach trip is given by
C
=
12
+
960
n
C = 12 + \frac{960}{n}
C
=
12
+
n
960
where
n
n
n
is the number of people going on the trip.
In March,
40
40
40
people went on the trip.
In April,
64
64
64
people went on the trip.
Each person paid more for the trip in March than in April.
How much more?
(2 marks)
●●●
●●
Level 3
2 marks
Start
→
Mark as done
y
=
5
t
2
−
3
t
y = 5t^2 - 3t
y
=
5
t
2
−
3
t
Work out the value of
y
y
y
when
t
=
−
2
t = -2
t
=
−
2
●●●●
●
Level 4
2 marks
Start
→
Mark as done
v
2
=
u
2
+
2
a
s
v^2 = u^2 + 2as
v
2
=
u
2
+
2
a
s
u
=
3
u = 3
u
=
3
,
a
=
4
a = 4
a
=
4
and
s
=
5
s = 5
s
=
5
.
Given that
v
v
v
is positive, work out the value of
v
v
v
.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
P
=
2
a
−
b
a
+
b
P = \dfrac{2a - b}{a + b}
P
=
a
+
b
2
a
−
b
Work out the value of
P
P
P
when
a
=
4
a = 4
a
=
4
and
b
=
−
2
b = -2
b
=
−
2
●●●●
●
Level 4
3 marks
Start
→
Mark as done
a
=
3
b
2
−
2
c
3
a = 3b^2 - 2c^3
a
=
3
b
2
−
2
c
3
Work out the value of
a
a
a
when
b
=
−
4
b = -4
b
=
−
4
and
c
=
2
c = 2
c
=
2
.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
k
=
2
(
l
−
m
)
2
k = 2(l - m)^2
k
=
2
(
l
−
m
)
2
Work out the value of
k
k
k
when
l
=
3
l = 3
l
=
3
and
m
=
7
m = 7
m
=
7
.
●●●●
●
Level 4
3 marks
Start
→
Mark as done
The formula for converting a temperature
C
C
C
(in °C) to
F
F
F
(in °F) is
F
=
9
C
5
+
32
F = \dfrac{9C}{5} + 32
F
=
5
9
C
+
32
(a)
Work out the value of
F
F
F
when
C
=
−
40
C = -40
C
=
−
40
.
(b)
Work out the value of
C
C
C
when
F
=
212
F = 212
F
=
212
.
●●●●●
Level 5
5 marks
Start
→
Rearranging formulae
16 questions
Lesson
Not started
Mark as done
Make
x
x
x
the subject of
y
=
x
+
5
y = x + 5
y
=
x
+
5
●
●●●●
Level 1
1 mark
Start
→
Mark as done
Make
a
a
a
the subject of
v
=
3
a
v = 3a
v
=
3
a
●
●●●●
Level 1
1 mark
Start
→
Mark as done
Make
t
t
t
the subject of
s
=
4
t
−
1
s = 4t - 1
s
=
4
t
−
1
●●
●●●
Level 2
2 marks
Start
→
Mark as done
Make
x
x
x
the subject of
y
=
4
x
−
7
\;y = 4x - 7
y
=
4
x
−
7
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Make
t
t
t
the subject of
v
=
u
+
a
t
\;v = u + at
v
=
u
+
a
t
●●●
●●
Level 3
2 marks
Start
→
Mark as done
Make
m
m
m
the subject of
E
=
m
g
h
\;E = mgh
E
=
m
g
h
●●●
●●
Level 3
2 marks
Start
→
Mark as done
v
2
=
u
2
+
2
a
s
v^2 = u^2 + 2as
v
2
=
u
2
+
2
a
s
(a)
Work out the value of
v
v
v
when
u
=
3
u = 3
u
=
3
,
a
=
4
a = 4
a
=
4
and
s
=
5
s = 5
s
=
5
, given that
v
>
0
v > 0
v
>
0
. (2 marks)
(b)
Make
s
s
s
the subject of the formula. (2 marks)
●●●
●●
Level 3
4 marks
Start
→
Mark as done
Make
x
x
x
the subject of
5
x
−
3
=
y
(
x
+
4
)
5x - 3 = y(x + 4)
5
x
−
3
=
y
(
x
+
4
)
(3 marks)
●●●
●●
Level 3
3 marks
Start
→
Mark as done
Leah is trying to make
w
w
w
the subject of
y
=
w
4
−
6
y = \dfrac{w}{4} - 6
y
=
4
w
−
6
Here is her working.
y
+
6
=
w
4
y + 6 = \frac{w}{4}
y
+
6
=
4
w
4
×
y
+
6
=
w
4 \times y + 6 = w
4
×
y
+
6
=
w
w
=
4
y
+
6
w = 4y + 6
w
=
4
y
+
6
Leah's answer is wrong.
(a)
What mistake has Leah made?
(1 mark)
(b)
Factorise fully
6
a
2
b
−
15
a
b
2
6a^2b - 15ab^2
6
a
2
b
−
15
a
b
2
(2 marks)
●●●
●●
Level 3
3 marks
Start
→
Mark as done
Make
r
r
r
the subject of
A
=
π
r
2
\;A = \pi r^2
A
=
π
r
2
, where
r
r
r
is positive.
●●●●
●
Level 4
2 marks
Start
→
Mark as done
Make
x
x
x
the subject of
y
=
3
x
+
2
5
\;y = \dfrac{3x + 2}{5}
y
=
5
3
x
+
2
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
y
=
5
−
2
x
3
\;y = 5 - \dfrac{2x}{3}
y
=
5
−
3
2
x
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
r
r
r
the subject of
V
=
4
3
π
r
3
\;V = \dfrac{4}{3}\pi r^3
V
=
3
4
π
r
3
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
T
=
2
x
−
5
\;T = 2\sqrt{x} - 5
T
=
2
x
−
5
●●●●●
Level 5
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
y
=
12
x
+
3
\;y = \dfrac{12}{x + 3}
y
=
x
+
3
12
●●●●●
Level 5
3 marks
Start
→
Mark as done
Make
L
L
L
the subject of
T
=
2
π
L
g
\;T = 2\pi\sqrt{\dfrac{L}{g}}
T
=
2
π
g
L
●●●●●
Level 5
3 marks
Start
→
Changing the subject when it appears twice
11 questions
Lesson
Not started
Mark as done
Make
x
x
x
the subject of
5
x
−
3
=
y
(
x
+
4
)
5x - 3 = y(x + 4)
5
x
−
3
=
y
(
x
+
4
)
(3 marks)
●●●
●●
Level 3
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
a
x
+
7
=
b
x
+
c
\;ax + 7 = bx + c
a
x
+
7
=
b
x
+
c
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
n
n
n
the subject of
m
=
3
(
n
−
5
)
n
\;m = \dfrac{3(n - 5)}{n}
m
=
n
3
(
n
−
5
)
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
a
(
x
+
2
)
=
b
(
x
−
5
)
\;a(x + 2) = b(x - 5)
a
(
x
+
2
)
=
b
(
x
−
5
)
●●●●
●
Level 4
4 marks
Start
→
Mark as done
Make
t
t
t
the subject of
s
=
2
t
+
3
t
\;s = \dfrac{2t + 3}{t}
s
=
t
2
t
+
3
●●●●
●
Level 4
3 marks
Start
→
Mark as done
Make
x
x
x
the subject of
y
=
3
x
+
2
x
−
4
\;y = \dfrac{3x + 2}{x - 4}
y
=
x
−
4
3
x
+
2
●●●●●
Level 5
3 marks
Start
→
Mark as done
Make
p
p
p
the subject of
p
−
q
=
p
q
\;p - q = pq
p
−
q
=
pq
●●●●●
Level 5
3 marks
Start
→
Mark as done
y
=
5
x
−
1
x
+
3
y = \dfrac{5x - 1}{x + 3}
y
=
x
+
3
5
x
−
1
(a)
Make
x
x
x
the subject.
(b)
Hence find the value of
x
x
x
when
y
=
4
y = 4
y
=
4
●●●●●
Level 5
4 marks
Start
→
Mark as done
Make
x
x
x
the subject of
T
=
a
+
x
x
\;T = \sqrt{\dfrac{a + x}{x}}
T
=
x
a
+
x
where
x
x
x
and
T
T
T
are positive.
●●●●●
Level 5
4 marks
Start
→
Mark as done
Make
x
x
x
the subject of
v
=
u
−
2
x
x
+
5
\;v = \dfrac{u - 2x}{x + 5}
v
=
x
+
5
u
−
2
x
●●●●●
Level 5
4 marks
Start
→
Mark as done
y
=
2
x
+
7
x
−
1
y = \dfrac{2x + 7}{x - 1}
y
=
x
−
1
2
x
+
7
(a)
Make
x
x
x
the subject.
(b)
Hence find the value of
x
x
x
when
y
=
5
y = 5
y
=
5
.
●●●●●
Level 5
5 marks
Start
→