Direct and indirect variation problems can be solved by writing the appropriate variation equation based on whether the quantities vary directly, inversely, or jointly. The variation equation introduces a constant of proportionality that can be solved for by substituting known values. Common variation equations include: y = kx for direct variation, y = k/x for inverse variation, and z = kxy or z = kx^2 for joint variation, where k is the constant of proportionality.

1. Direct & Indirect Variation
If two quantities 𝑥 and 𝑦 are directly proportional, we write
𝑦 ∝ 𝑥.
If they are inversely proportional, we write
𝑦 ∝
1
𝑥

2. Joint Variation
In joint variation one quantity varies directly as two
quantities.
If 𝑧 relates directly to both 𝑥 and 𝑦, we write 𝑧 ∝ 𝑥𝑦.
𝑧 = 𝑘𝑥𝑦
E.g 𝐴 = 𝑙𝑤, 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,
𝑦 = 𝑚𝑥 + 𝑐𝑐

4. Solution
𝑥 ∝ 𝑛
𝑥 = 𝑘 𝑛
9 = 𝑘 9
9 = 3𝑘
𝑘 = 3
This implies that:
𝑥 = 3 𝑛
When 𝑛 =
17
9
,
𝑥 = 3
17
9
𝑥 = 17
If 𝑥 varies directly as
𝑛 𝑎𝑛𝑑 𝑥 =
9 𝑤ℎ𝑒𝑛 𝑛 =
9, 𝑓𝑖𝑛𝑑 𝑥 𝑤ℎ𝑒𝑛 𝑛 =
17
9
A. 4
B. 27
C. 17
D. 17
JAMB 2002

5. Solution
According to the variation statement,
𝐿 ∝ 𝑇2
⇒ 𝐿 = 𝑘𝑇2
𝑊ℎ𝑒𝑛 𝑙 = 64, 𝑡 = 4
⇒ 64 = 𝑘(42)
64 = 16𝑘
𝑘 = 4
⇒ 𝐿 = 4𝑇2
When 𝑇 = 9
𝐿 = 4 × 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. 𝑥 varies directly as
square root of 𝑡
and inversely as 𝑠.
When 𝑥 = 4, 𝑡 = 9
and 𝑠 = 18.
i. Express 𝑥 in
terms of 𝑠 and 𝑡
ii. Find 𝑥, when
𝑡 = 81 and
𝑠 = 27.
SSSCE 2002
Solution
According to the variation statement, 𝑥 ∝
𝑡
𝑠
→ 𝑥 =
𝑘 𝑡
𝑠
, where k is the constant of variation.
But 𝑥 = 4, 𝑡 = 9 𝑎𝑛𝑑 𝑠 = 18
→ 4 =
𝑘 9
18
→ 4 =
3𝑘
18
3𝑘 = 4 × 18
𝑘 =
4 × 18
3
𝑘 = 24
Hence,
i. 𝑥 =
24 𝑡
𝑠
ii. When 𝑡 = 81 and 𝑠 = 27
𝑥 =
24 81
27
=
24 × 9
27
= 8

7. 𝑅 varies inversely
as the cube of 𝑠
root of 𝑡 and
inversely as 𝑠. If
R = 9, 𝑤ℎ𝑒𝑛 𝑆 = 3
and
Find 𝑆, when 𝑅 =
243
64
.
WASSCE 2006
Solution
According to the variation statement, 𝑅 ∝
1
𝑠3
→ 𝑅 =
𝑘
𝑠3, where k is the constant of variation.
But R = 9, 𝑎𝑛𝑑 𝑠 = 3
→ 9 =
𝑘
33
→ 9 =
𝑘
27
𝑘 = 27 × 9
𝑘 = 243
Hence,
i. 𝑅 =
243
𝑠3
ii. When 𝑅 =
243
64
,
Then
243
64
=
243
𝑆3
1
64
=
1
𝑠3
𝑠3
= 43
∴ 𝑠 = 4

8. 1. 𝑦 varies directly as 𝑥. Find the constant of variation if y = 21 when
𝑥 = 7.
A. 1
B. 2
C. 3
D. 4
E. 5
The correct answer is 3
Solution
𝑦 ∝ 𝑥
⇒ 𝑦 = 𝑘𝑥
𝑦 = 21 𝑤ℎ𝑒𝑛 𝑥 = 7
⇒ 21 = 7𝑘
𝑘 = 3

9. 2. 𝑦 varies directly as 𝑥. Find the equation of variation if y = 21 when
𝑥 = 7.
A. 𝑦 = 𝑥
B. 𝑦 = 2𝑥
C. 𝑦 = 3𝑥
D. 𝑦 = 4𝑥
E. 𝑦 = 5𝑥
The correct answer is y = 3𝑥
Solution
𝑦 ∝ 𝑥
⇒ 𝑦 = 𝑘𝑥
𝑦 = 21 𝑤ℎ𝑒𝑛 𝑥 = 7
⇒ 21 = 7𝑘
𝑘 = 3
Hence the variation equation is
𝑦 = 3𝑥

10. 3. If p varies inversely as 𝑞. Then the equation that describes this
variation is:
A. 𝑝 ∝
1
𝑞
B. 𝑝 ∝ 𝑞
C. 𝑝 =
1
𝑞
D. 𝑝 ∝ −𝑞
E. 𝑝 = −𝑞
Solution
𝑝 ∝
1
𝑞

11. 4. If p varies directly as 𝑞. Then 𝑞 varies----------- as 𝑝.
A. 𝑖𝑛𝑣𝑒𝑟𝑠𝑒𝑙𝑦
B. 𝑖𝑛𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦
C. 𝑗𝑜𝑖𝑛𝑡𝑙𝑦
D. Proportionally
E. 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦
The correct answer is directly
Solution
If p varies directly as 𝑞. Then 𝑞 also
varies directly as 𝑝.

12. 5. The values in the table below represent a/an -------- variation.
A. 𝑖𝑛𝑣𝑒𝑟𝑠𝑒
B. 𝑝𝑎𝑟𝑡𝑖𝑎𝑙
C. 𝑗𝑜𝑖𝑛𝑡
D. Proportional
E. 𝑑𝑖𝑟𝑒𝑐𝑡
The correct answer is direct
Solution
When 𝑥 = −2, 𝑦 = 1
When 𝑥 = −1, 𝑦 = 0.5
As 𝑥 is increasing 𝑦 is is increasing.
The values represent an a direct
variation
𝒙 -2 -3 0 1 2
𝑦 1 0.5 0 -0.5 -1

13. 6. Find the variation equation for the values in the table below.
A. 𝑦 = −2𝑘
B. 𝑦 = 2𝑘
C. 𝑦 =
1
2
D. 𝑦 = −
1
2
𝑘
E. 𝑦 = −𝑘
The correct answer is 𝑦 = −
1
2
𝑘
Solution
When 𝑥 = −2, 𝑦 = 1
When 𝑥 = −1, 𝑦 = 0.5
As 𝑥 is increasing 𝑦 is is increasing.
The values represent an a direct
variation
𝑦𝑥
𝑦 = 𝑘𝑥
1 = −2𝑘
𝑘 = −
1
2
𝒙 -2 -3 0 1 2
𝑦 1 0.5 0 -0.5 -1

14. 7. A variable 𝑛 varies jointly as 𝑥 and 𝑦 and varies inversely as the cube of
𝑧. If 𝑛 = −14 when 𝑥 = 3, y = 7, and z = z = −3, what is 𝑡ℎ𝑒 constant of
variation?
A. 18
B. 19
C. 20
D. 21
E. 22
The correct answer is 18
Solution
𝑛 ∝
𝑥𝑦
𝑧3
𝑛 =
𝑘𝑥𝑦
𝑧3
→ 𝑘 =
𝑛𝑧3
𝑥𝑦
𝑘 =
(−14)(−3)3
3 × 7
= 18

15. 8. A variable 𝑛 varies jointly as 𝑥 and 𝑦 and varies inversely as the cube of
𝑧. If 𝑛 = −14 when 𝑥 = 3, y = 7, and z = z = −3, what is 𝑛when 𝑥 = 4, y =
5 and z = 3?
A.
4
3
B.
40
3
C.
3
40
D. −
40
3
E. −
30
7
The correct answer is
40
3
Solution
𝑛 ∝
𝑥𝑦
𝑧3
𝑛 =
𝑘𝑥𝑦
𝑧3
→ 𝑘 =
𝑛𝑧3
𝑥𝑦
𝑘 =
(−14)(−3)3
3 × 7
= 18
Now when 𝑥 = 4, 𝑦 = 5, 𝑎𝑛𝑑 𝑧 = 3.
𝑛 =
18 × 4 × 5
33
=
40
3

16. 9. If C is partly constant and partly varies jointly as L and B. Then;
A. 𝐶 = 𝐴 + 𝐿𝐵
B. 𝐶 = 𝐴 +
𝐾𝐿
𝐵
C. 𝐶 = 𝐴𝐿𝐵
D. 𝐶 = 𝐴 + 𝐾𝐿𝐵
E. 𝐶 = 𝐴 +
𝐾𝐵
𝐿
The correct answer is 𝐶 = 𝐴 + 𝐾𝐿𝐵
Solution
If C is partly constant and partly varies jointly as L and B.
Then;
𝐶 = 𝐴 + 𝐾𝐿𝐵

17. 10. If C is partly constant and partly varies jointly as L and the square of
B. Then;
A. 𝐶 = 𝐴 + 𝐿𝐵2
B. 𝐶 = 𝐴 +
𝐾𝐿
𝐵2
C. 𝐶 = 𝐴𝐿𝐵2
D. 𝐶 = 𝐴 + 𝐾𝐿𝐵2
E. 𝐶 = 𝐴 +
𝐾𝐵2
𝐿
The correct answer is 𝐶 = 𝐴 + 𝐾𝐿𝐵2
Solution
If C is partly constant and partly varies jointly as L and B.
Then;
𝐶 = 𝐴 + 𝐾𝐿𝐵2

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