CHAPTER 7 TRANSPOSITION OF FORMULAE
EXERCISE 30, Page 60
1. Make d the subject of the formula: a + b = c - d - e
Since a + b = c - d – e then d = c – e – a - b
2. Make x the subject of the formula: y = 7x
Dividing both sides of y = 7x by 7 gives:
x=
y
7
3. Make v the subject of the formula: pv = c
Dividing both sides of pv = c
by p gives:
v=
c
p
4. Make ‘a’ the subject of the formula: v = u + at
Since v = u + at then v – u = at
and dividing both sides by t gives:
v u
v u
= a or a =
t
t
5. Make y the subject of the formula: x + 3y = t
Since x + 3y = t then 3y = t – x
and dividing both sides by 3 gives:
y
tx
1
or y   t  x 
3
3
6. Make r the subject of the formula: c = 2r
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c
c
 r or r =
2
2
Dividing both sides of c = 2r by 2 gives:
7. Make x the subject of the formula: y = mx + c
Since y = mx + c
then y – c = mx
and dividing both sides by m gives:
y c
y c
= x or x =
m
m
8. Make T the subject of the formula: I = PRT
I
I
 T or T =
PR
PR
Dividing both sides of I = PRT by PR gives:
9. Make L the subject of the formula: X L  2 f L
X
XL
 L or L = L
2f
2f
Dividing both sides of X L  2 f L by 2πf gives:
10. Make R the subject of the formula: I =
Multiplying both sides of I =
E
by R gives: I R = E
R
and dividing both sides by I gives:
11. Make x the subject of the formula: y 
Since y 
E
R
R=
E
I
x
3
a
x
x
 3 then y – 3 =
a
a
Multiplying both sides by ‘a’ gives: a(y – 3) = x or x = a(y – 3)
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12. Make C the subject of the formula: F =
Rearranging F =
9
C  32 gives:
5
Multiplying both sides by
i.e.
9
C + 32
5
F – 32 =
9
C
5
5
5 9 
5
gives:
 F  32    
 C 
9
9
 9  5 
5
 F  32   C
9
or
C
5
 F  32 
9
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EXERCISE 31, Page 61
1. Make r the subject of the formula: S =
a
1 r
a
by (1 – r) gives: S(1 – r) = a
1 r
Multiplying both sides of S =
i.e.
S – Sr = a
from which,
S – a = Sr
and dividing both sides by S gives:
Sa
a
Sa
or r = 1 = r i.e. r =
S
S
S
2. Make x the subject of the formula: y =
Multiplying both sides of y =
(x  d)
d
 x  d
by d gives:
d
yd
 xd

Dividing both sides by  gives:
d
and
Alternatively, from the first step,
yd = (x – d)
i.e.
yd = x - d
and
from which,
yd = (x – d)
yd
= x or

x  d
yd

yd + d = x
x=
3. Make f the subject of the formula: A =
yd  d d  y   



i.e. x =
d
y  

3(F  f )
L
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Multiplying both sides of A =
3(F  f )
by L gives:
L
AL = 3(F – f)
AL
 Ff
3
Dividing both sides by 3 gives:
f  F
and
4. Make D the subject of the formula: y =
Multiplying both sides of y =
AL
3
or f =
3F  AL
3
A B2
5CD
A B2
by D gives:
5CD
yD =
A B2
5C
D=
A B2
5Cy
Dividing both sides by y gives:
5. Make t the subject of the formula: R = R0(1 + t)
Removing the bracket in R  R 0 1  t  gives: R  R 0  R 0  t
from which,
R  R0  R0  t
and
R  R0
R  R0
= t or t =
R 0
R 0
6. Make R2 the subject of the formula:
Rearranging
1
1
1


gives:
R R1 R 2
1
1
1
+
=
R1
R2
R
1 1
1


R R1 R 2
i.e.
1
1 1 R1  R
 

R 2 R R1
R R1
Turning both sides upside down gives:
R2 
R R1
R1  R
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7. Make R the subject of the formula: I =
Ee
Rr
Multiplying both sides by (R + r) gives:
I(R + r) = E – e
i.e.
IR+Ir=E–e
and
IR=E–e–Ir
and dividing both sides by I gives:
R=
E e Ir
I
or R =
Ee
r
I
8. Make b the subject of the formula: y = 4ab2c2
y
 b2
4ac2
Dividing both sides by 4a c 2 gives:
or
Taking the square root of both sides gives:
9. Make x the subject of the formula:
Rearranging
b2 
y
4ac 2
y
4ac2
b=
b2
a2
+
=1
y2
x2
a 2 b2

 1 gives:
x 2 y2
a2
b2 y2  b2
1



x2
y2
y2
Turning both sides upside down gives:
x2
y2

a 2 y2  b2
Multiplying both sides by a 2 gives:
 y2 
a 2 y2
x2  a2  2


2
2
2
 y b  y b
Taking the square root of both sides gives:
x=
i.e.
x=
a 2 y2

y2  b2
a 2 y2
y b
2
2

ay
y2  b2
ay
y 2  b2
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L
g
10. Make L the subject of the formula: t = 2
Dividing both sides of t = 2
L
by 2 gives:
g
t
L

2
g
2
L
 t 
  
g
 2 
Squaring both sides gives:
or
 t 
L = g

 2 
Multiplying both sides by g gives:
L  t 
 
g  2 
2
2
or L =
g t2
4 2
11. Make u the subject of the formula: v2 = u2 + 2as
Since v2 = u2 + 2as then v2 - 2as = u2
or u2 = v2 - 2as
Taking the square root of each side gives:
12. Make ‘a’ the subject of the formula: N =
Squaring both sides of N =
ax
gives:
y
Multiplying both sides by y gives:
from which,
v
u=
2
 2as 
ax


 y 
N2 
ax
y
N2 y  a  x
or a + x = N 2 y
a = N2y - x
13. The lift force, L, on an aircraft is given by: L =
1 2
 v a c where ρ is the density, v is the
2
velocity, a is the area and c is the lift coefficient. Transpose the equation to make the velocity
the subject.
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Since L =
from which,
1 2
v ac
2
then
2L
 v2
a c
velocity, v =
2L
ac
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EXERCISE 32, Page 62
1. Make ‘a’ the subject of the formula: y =
Multiplying both sides of y =
a 2m  a 2n
x
a 2m  a 2n
by x gives:
x
xy = a 2 m  a 2 n
xy = a 2  m  n 
and factorising gives:
xy
 a2
mn
Dividing both sides by (m – n) gives:
xy
mn
xy
mn
a
Taking the square root of both sides gives:
a2 
or
2. Make R the subject of the formula: M = (R4 - r4)
Dividing both sides of M =   R 4  r 4  by  gives:
M
 R 4  r4

M 4
 r  R4

and rearranging gives:
Taking the fourth root of both sides gives:
3. Make r the subject of the formula: x + y =
Multiplying both sides of x + y =
R=
4
or
R4 
M 4
r

M
 r4

r
3 r
r
by (3 + r) gives: (x + y)(3 + r) = r
3 r
Multiplying the brackets gives:
3x + xr + 3y + yr = r
and rearranging gives:
xr + yr – r = -3x – 3y
Factorising gives:
r(x + y – 1) = -3( x + y)
Dividing both sides by (x + y – 1) gives:
r=
3(x  y)
x  y 1
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Multiplying numerator and denominator by -1 gives:
4. Make L the subject of the formula: m =
Multiplying both sides of m 
r=
3(x  y)
1 x  y
L
L  rCR
L
by (L + rCR) gives:
L  rCR
m(L + rCR) = L
mL + mrCR = L
Removing brackets gives:
and rearranging gives:
mrCR = L - mL
Factorising gives:
mrCR = L( - m)
Dividing both sides by ( - m) gives:
L=
5. Make b the subject of the formula: a2 =
Multiplying both sides by b 2 gives:
b2  c2
b2
a 2 b2  b2  c2
and rearranging gives:
c2  b2  a 2 b2
Factorising gives:
b 2 1  a 2   c2
Dividing both sides by 1  a 2  gives:
b2 
Taking the square root of both sides gives:
b=
Hence,
b=
6. Make r the subject of the formula:
mrCR
m
or
b2  a 2 b2  c2
c2
1 a2
c2
c2

1 a2
1 a2
c
1  a2
x 1 r2
=
y 1 r2
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x 1  r 2   y 1  r 2 
Rearranging by ‘cross-multiplying’ gives:
x  xr 2  y  yr 2
Removing brackets gives:
x  y  yr 2  xr 2 or
and rearranging gives:
yr 2  xr 2  x  y
r2  x  y  x  y
Factorising gives:
xy
xy
Dividing both sides by (x + y) gives:
r2 
Taking the square root of both sides gives:
r=
7. A formula for the focal length, f, of a convex lens is:
1 1
1
= + . Transpose the formula to make
f
u
v
xy
xy
v the subject and evaluate v when f = 5 and u = 6
Rearranging
1 1 1 u f
  
v f u
uf
1 1 1
 
gives:
f u v
Turning each side upside down gives:
When f = 5 and u = 6, then
v=
v=
uf
uf
uf
(6)(5) 30
= 30


uf
65
1
8. The quantity of heat, Q, is given by the formula Q = mc(t2 - t1). Make t2 the subject of the formula
and evaluate t2 when m = 10, t1 = 15, c = 4 and Q = 1600
Removing the brackets in Q = m c  t 2  t1  gives:
Q = mct 2  mct1
Q + mct1  mct 2
and rearranging gives:
mct 2  Q  mvt1
or
Dividing both sides by mc gives:
t2 
Q  mvt1
mc
or
t2 
Q
t
mc 1
or
t 2  t1 
Q
mc
When m = 10, t1 = 15, c = 4 and Q = 1600,
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t 2 = t1 
Q
1600
1600
 15 
 15 
 15  40 = 55
mc
(10)(4)
40
9. The velocity, v, of water in a pipe appears in the formula h =
0.03Lv 2
. Express v as the subject
2dg
of the formula and evaluate v when h = 0.712, L = 150, d = 0.30 and g = 9.81
Multiplying both sides of h =
0.03L v 2
by 2dg gives: 2dgh = 0.03L v 2
2dg
2dgh
 v2
0.03L
Dividing both sides by 0.03L gives:
Taking the square root of each side gives:
or
v2 
2dgh
0.03L
2dgh
0.03L
v=
When h = 0.712, L = 150, d = 0.30 and g = 9.81,
v=
2dgh
2(0.30)(9.81)(0.712)

 0.931296 = 0.965
0.03L
0.03(150)
10. The sag S at the centre of a wire is given by the formula: S =
 3d(l  d) 


8


Make l the subject of the formula and evaluate l when d = 1.75 and S = 0.80
Squaring both sides of S =
3d (l  d)
gives:
8
S2 
3d  l  d 
8
Multiplying both sides by 8 gives:
8S2  3d  l  d 
Removing the bracket gives:
8S2  3dl  3d 2
8S2  3d 2  3dl
Rearranging gives:
3dl  8S2  3d 2
or
Dividing both sides by 3d gives:
l=
8S2  3d 2 8S2 3d 2


3d
3d 3d
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© John Bird Published by Taylor and Francis
i.e.
l=
When d = 1.75 and S = 0.80,
l=
8S 2
d
3d
8S2
8(0.80) 2
d 
 1.75  0.975  1.75 = 2.725
3d
3(1.75)
11. An approximate relationship between the number of teeth, T, on a milling cutter, the diameter
of cutter, D, and the depth of cut, d, is given by: T 
12.5 D
D  4d
Determine the value of D when T = 10 and d = 4 mm.
Multiplying both sides of T 
12.5 D
by D + 4d gives:
D  4d
Removing brackets gives:
T(D + 4d) = 12.5D
TD + 4dT = 12.5D
Rearranging gives:
4dT = 12.5D – TD
or
12.5D – TD = 4dT
Factorising gives:
D(12.5 – T) = 4dT
Dividing both sides by (12.5 – T) gives:
When T = 10 and d = 4 mm, then D =
D=
4dT
12.5  T
4dT
4(4)(10) 160


= 64 mm
12.5  T 12.5  10 2.5
12. A simply supported beam of length L has a centrally applied load F and a uniformly distributed
load of w per metre length of beam. The reaction at the beam support is given by:
R=
1
 F  wL 
2
Rearrange the equation to make w the subject. Hence determine the value of w when L = 4 m,
F = 8 kN and R = 10 kN
Since
R=
1
 F  wL  then
2
2R = F + wL
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© John Bird Published by Taylor and Francis
and
2R – F = wL
from which,
w=
2R  F
L
When L = 4 m, F = 8 kN and R = 10 kN,
w=
2(10)  8 12
 = 3 kN/m
4
4
13. The rate of heat conduction through a slab of material, Q, is given by the formula
Q
kA(t1  t 2 )
where t1 and t 2 are the temperatures of each side of the material, A is the area
d
of the slab, d is the thickness of the slab, and k is the thermal conductivity of the material.
Rearrange the formula to obtain an expression for t 2
Since
Q
kA(t1  t 2 )
then
d
Qd = kA  t1  t 2 
Qd
= t1  t 2
kA
i.e.
from which,
t 2 = t1 -
Qd
kA
 r
14. The slip, s, of a vehicle is given by: s = 1 
  100% where r is the tyre radius, ω is the
v 

angular velocity and v the velocity. Transpose to make r the subject of the formula.
Since
 r
s = 1 
  100%
v 

and
from which,
then
s
r
=1100
v
r
s
=1v
100
r=
v
s 
1

  100 
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15. The critical load, F newtons, of a steel column may be determined from the formula
L
F
 n where L is the length, EI is the flexural rigidity, and n is a positive integer.
EI
Transpose for F and hence determine the value of F when n = 1, E = 0.25 1012 N / m 2 ,
I = 6.92 106 m4 and L = 1.12 m
Since
F
 n
EI
L
F
n

EI
L
then
F n


EI  L 
and
2
 n 
F = EI 

 L 
i.e.
2
When n = 1, E = 0.25 1012 N / m2 , I = 6.92 106 m4 and L = 1.12 m,
2
2
n
6
12
6  1  
load, F = EI 
   0.25 10  6.92 10  
 = 13.61 10 N = 13.61 MN
 L 
 1.12 
16. The flow of slurry along a pipe on a coal processing plant is given by: V 
pr 4
8
Transpose the equation for r
Since V 
and
from which,
pr 4
8
then
8V  pr 4
8V
 r4
p
r=
4
 8  V 


 p 
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