1. Direct & Indirect Variation
If two quantities π₯ and π¦ are directly proportional, we write
π¦ β π₯.
If they are inversely proportional, we write
π¦ β
1
π₯
2. Joint Variation
In joint variation one quantity varies directly as two
quantities.
If π§ relates directly to both π₯ and π¦, we write π§ β π₯π¦.
π§ = ππ₯π¦
E.g π΄ = ππ€, The area of a rectangle is varies directly as its
length and width
3. Partial Variation
In partial variation there is a fixed constant in addition to
the constant of proportionality.
The equation is of the form,
π¦ = ππ₯ + ππ
4. Solution
π₯ β π
π₯ = π π
9 = π 9
9 = 3π
π = 3
This implies that:
π₯ = 3 π
When π =
17
9
,
π₯ = 3
17
9
π₯ = 17
If π₯ varies directly as
π πππ π₯ =
9 π€βππ π =
9, ππππ π₯ π€βππ π =
17
9
A. 4
B. 27
C. 17
D. 17
JAMB 2002
5. Solution
According to the variation statement,
πΏ β π2
β πΏ = ππ2
πβππ π = 64, π‘ = 4
β 64 = π(42)
64 = 16π
π = 4
β πΏ = 4π2
When π = 9
πΏ = 4 Γ 92
= 324
The length L of a simple pendulum
varies
directly as the square of its period T . If
a
pendulum with period 4 sec. is 64cm
long,
find the length of a pendulum whose
period is 9sec.
A. 96
B. 324
C. 36
D. 144
JAMB 2004
6. π₯ varies directly as
square root of π‘
and inversely as π .
When π₯ = 4, π‘ = 9
and π = 18.
i. Express π₯ in
terms of π and π‘
ii. Find π₯, when
π‘ = 81 and
π = 27.
SSSCE 2002
Solution
According to the variation statement, π₯ β
π‘
π
β π₯ =
π π‘
π
, where k is the constant of variation.
But π₯ = 4, π‘ = 9 πππ π = 18
β 4 =
π 9
18
β 4 =
3π
18
3π = 4 Γ 18
π =
4 Γ 18
3
π = 24
Hence,
i. π₯ =
24 π‘
π
ii. When π‘ = 81 and π = 27
π₯ =
24 81
27
=
24 Γ 9
27
= 8
7. π varies inversely
as the cube of π
root of π‘ and
inversely as π . If
R = 9, π€βππ π = 3
and
Find π, when π =
243
64
.
WASSCE 2006
Solution
According to the variation statement, π β
1
π 3
β π =
π
π 3, where k is the constant of variation.
But R = 9, πππ π = 3
β 9 =
π
33
β 9 =
π
27
π = 27 Γ 9
π = 243
Hence,
i. π =
243
π 3
ii. When π =
243
64
,
Then
243
64
=
243
π3
1
64
=
1
π 3
π 3
= 43
β΄ π = 4
8. 1. π¦ varies directly as π₯. Find the constant of variation if y = 21 when
π₯ = 7.
A. 1
B. 2
C. 3
D. 4
E. 5
The correct answer is 3
Solution
π¦ β π₯
β π¦ = ππ₯
π¦ = 21 π€βππ π₯ = 7
β 21 = 7π
π = 3
9. 2. π¦ varies directly as π₯. Find the equation of variation if y = 21 when
π₯ = 7.
A. π¦ = π₯
B. π¦ = 2π₯
C. π¦ = 3π₯
D. π¦ = 4π₯
E. π¦ = 5π₯
The correct answer is y = 3π₯
Solution
π¦ β π₯
β π¦ = ππ₯
π¦ = 21 π€βππ π₯ = 7
β 21 = 7π
π = 3
Hence the variation equation is
π¦ = 3π₯
10. 3. If p varies inversely as π. Then the equation that describes this
variation is:
A. π β
1
π
B. π β π
C. π =
1
π
D. π β βπ
E. π = βπ
Solution
π β
1
π
11. 4. If p varies directly as π. Then π varies----------- as π.
A. πππ£πππ πππ¦
B. ππππππππ‘ππ¦
C. πππππ‘ππ¦
D. Proportionally
E. ππππππ‘ππ¦
The correct answer is directly
Solution
If p varies directly as π. Then π also
varies directly as π.
12. 5. The values in the table below represent a/an -------- variation.
A. πππ£πππ π
B. ππππ‘πππ
C. πππππ‘
D. Proportional
E. ππππππ‘
The correct answer is direct
Solution
When π₯ = β2, π¦ = 1
When π₯ = β1, π¦ = 0.5
As π₯ is increasing π¦ is is increasing.
The values represent an a direct
variation
π -2 -3 0 1 2
π¦ 1 0.5 0 -0.5 -1
13. 6. Find the variation equation for the values in the table below.
A. π¦ = β2π
B. π¦ = 2π
C. π¦ =
1
2
D. π¦ = β
1
2
π
E. π¦ = βπ
The correct answer is π¦ = β
1
2
π
Solution
When π₯ = β2, π¦ = 1
When π₯ = β1, π¦ = 0.5
As π₯ is increasing π¦ is is increasing.
The values represent an a direct
variation
π¦π₯
π¦ = ππ₯
1 = β2π
π = β
1
2
π -2 -3 0 1 2
π¦ 1 0.5 0 -0.5 -1
14. 7. A variable π varies jointly as π₯ and π¦ and varies inversely as the cube of
π§. If π = β14 when π₯ = 3, y = 7, and z = z = β3, what is π‘βπ constant of
variation?
A. 18
B. 19
C. 20
D. 21
E. 22
The correct answer is 18
Solution
π β
π₯π¦
π§3
π =
ππ₯π¦
π§3
β π =
ππ§3
π₯π¦
π =
(β14)(β3)3
3 Γ 7
= 18
15. 8. A variable π varies jointly as π₯ and π¦ and varies inversely as the cube of
π§. If π = β14 when π₯ = 3, y = 7, and z = z = β3, what is πwhen π₯ = 4, y =
5 and z = 3?
A.
4
3
B.
40
3
C.
3
40
D. β
40
3
E. β
30
7
The correct answer is
40
3
Solution
π β
π₯π¦
π§3
π =
ππ₯π¦
π§3
β π =
ππ§3
π₯π¦
π =
(β14)(β3)3
3 Γ 7
= 18
Now when π₯ = 4, π¦ = 5, πππ π§ = 3.
π =
18 Γ 4 Γ 5
33
=
40
3
16. 9. If C is partly constant and partly varies jointly as L and B. Then;
A. πΆ = π΄ + πΏπ΅
B. πΆ = π΄ +
πΎπΏ
π΅
C. πΆ = π΄πΏπ΅
D. πΆ = π΄ + πΎπΏπ΅
E. πΆ = π΄ +
πΎπ΅
πΏ
The correct answer is πΆ = π΄ + πΎπΏπ΅
Solution
If C is partly constant and partly varies jointly as L and B.
Then;
πΆ = π΄ + πΎπΏπ΅
17. 10. If C is partly constant and partly varies jointly as L and the square of
B. Then;
A. πΆ = π΄ + πΏπ΅2
B. πΆ = π΄ +
πΎπΏ
π΅2
C. πΆ = π΄πΏπ΅2
D. πΆ = π΄ + πΎπΏπ΅2
E. πΆ = π΄ +
πΎπ΅2
πΏ
The correct answer is πΆ = π΄ + πΎπΏπ΅2
Solution
If C is partly constant and partly varies jointly as L and B.
Then;
πΆ = π΄ + πΎπΏπ΅2
18. SUMMARY
To solve variation problems,
1. Write the variation equation
2. Introduce an equal sign with the constant of proportionality
3. Substitute all given values and solve