Page 8 - 1202 Question Bank Mathematics Form 5

1.3 Combined Variation Therefore, p = 9! r
s
1. Joint variation is a relation between three or more When p = 18 and s = 2
variables. The relation can be
(a) two direct variation 18 = 9! r
Example: 2
p varies directly as s and t. ! r = 18 × 2
p fi s and p fi t 9
that is, ! r = 4

2
p fi st (! r ) = 4 2
p = ©PAN ASIA PUBLICATIONS
p = kst r = 16
where k is a constant.
(b) two inverse variation METHOD 2
Example:
y varies inversely as the cube of x and the square p = 9! r
of z. s
1 1 ps
y fi and y fi k =
x 3 x 2 ! r
that is,
p s p s
Hence, 1 1 = 2 2
1
y fi ! r ! r
x x 1 2
3 2
k 6(3) 18(2)
y = =
3 2
x x ! 4 ! r
2
where k is a constant. 36 × ! 4
(c) One direct variation and one inverse variation ! r = 18
2
Example:
! r = 4
2
u varies directly as m and inversely as the square r = 4 = 16
2
root of n. 2
1
u fi m and u fi 3. Distance, s travelled by a bike is varies
! n
that is, directly as the square of the speed, v, and
varies inversely as acceleration, a. Given that
m
u fi s = 120 m, v = 6 m s , and a = 0.5 m s . Calculate the
–1
–2
! n –1
value of a when s = 360 m and v = 9 m s .
km
u = Solution:
! n 2
s fi v
where k is a constant. a 2
s = kv
2. Solving problem in joint variation a
Example: 120 = k(6) 2
Given that p varies directly as ! r and inversely as s. 0.5
p = 6 when r = 4 and s = 3. Find the value of r when k = 120 × 0.5
p = 18 and s = 2. 5 36
=
3
METHOD 1 5
( ) (9) 2
! r 360 = 3
p fi 36
s
135
k! r a = 360
s
= 0.375 m s −2
k! 4
6 =
3
6 × 3
k = = 9
! 4
2

C01 1202BS Maths F5.indd 2 26/01/2022 3:43 PM