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2. 1.The variable m varies directly as the variable n. When n = 14,
m = 42. What is the value of n when m = 16?
Solution
Given:
n = 14, m= 42
variable m varies directly as the
variable n
Asked: What is the value of n when
m = 16?
m = kn
42 = k(14)
42
14
= k ; 3 = k
m = kn
16 = (3)n
5. 33 = n
The value of n is 5.33 when m = 16

3. 2.The variables m varies inversely as the variable n when
n = 21, m = 8.What is the value of n when m = 28?
Solution
Given:
n = 21, m = 8
m varies inversely as the variable n
Asked: What is the value of n when m =
28?
m =
𝑘
𝑛
8 =
𝑘
21
; 168 = k
What is the value of n when m =
28?
m =
𝑘
𝑛
; 28 =
168
𝑛
28n = 168
n = 6
The value of n = 6 when m = 28.

4. Let’s Learn
The electric current (I), in amperes varies directly as the
voltage(V), in volts in a circuit. When 20 volts are applied in a
circuit, the current is 5 amperes. What is the current when the
24 volts are applied?

5. In solving problems involving variations, follow these steps:
1. Identify the given.
2. Determine what is asked for.
3. Determine the type of variation.
4. Identify the formula to be used.
5. Substitute the given in the formula and solve for the constant
of variation.
6. Find the equation of variation.
7. Use the equation of variation to solve the problem.
8. State the answer.

6. Using the outlined procedure, we have the solution to the problem.
Steps Solution
1. Identify the given. The voltage is 20 volts when the
current is 5 amperes.
The electric current varies directly
as the voltage.
2. Determine what is asked for. Find the current when the voltage
is 24 volts .
3. Determine the type of
variation.
Direct variation
4. Identify the formula to be used. I = kV (1)

7. Steps Solution
5. Substitute the given in the
formula and solve for the constant
of variation.
Solve for k when I = 5 andV = 20
I = kV
5 = k(20)
5
20
=
1
4
= 0. 25 = k
6. Find the equation of variation. I = kV ; I =
1
4
V (2)
7. Use the equation of variation to
solve the problem.
SubstituteV = 24 in (2)
I =
1
4
V
I =
1
4
(24)
= 6 amperes
8. State the answer. The current is 6 amperes when the
voltage is 24 volts.

8. Let’sTryThese
• Example 1. The number of dolls(N) a machine can produce
varies directly as the amount of time (t) the machine is operating.
The machine produces 500 dolls in 8 hours. How long will it take
to produce 1,250 dolls?

9. Steps Solution
1. Identify the given. The machine produces 500 dolls I 8
hours.The number of dolls (N)
varies directly with time, t.
2. Determine what is asked for. Find the time will it take to
produce 1,250 dolls.
3. Determine the type of variation. Direct variation
4. Identify the formula to be used. N = kt (1)

10. Steps Solution
5. Substitute the given in the
formula and solve for the constant
of variation.
Solve for k when N = 500 and t = 8
N = kt
500 = k(8)
500
8
=
125
2
= 62. 5 = k
6. Find the equation of variation. N = kt
N =
125
2
t (2)
7. Use the equation of variation to
solve the problem.
N =
125
2
t
1,250 =
125
2
t
t = 20 hours
8. State the answer. It will take 20 hours to make 1,250
dolls.

11. Example 2. The weight of water in a human body varies directly as
the total weight. A person weighing 80 kg contains 50 kg of water.
How many kg of water are in 70 kg person?
W = kT
50 = k(80kg)
0.625 = k
W = kT
W = 0. 625(70kg)
W = 43. 75kg
The person weighing 70kilograms contains 43.75 kg of water.

12. Example 3. The time (t) needed to do a job varies inversely with
the number of workers (W) assigned on the job. It takes 6 weeks
for 14 workers to build a house. How many weeks would it take
for 6 workers to build the house?
Solution:
Identify the given:
It takes 6 weeks for 14 workers to
build a house.
The time (t) needed to do a job
varies inversely with the number
of workers (W)
Asked:
The number of weeks it would
take for 6 workers to build a
house.

13. Solution
Find first the constant of
variation using the formula of
inverse variation
t =
𝑘
𝑊
6 =
𝑘
14
84 = k (constant of variation)
Find the value of t whenW = 6.
t =
84
6
t = 14 (weeks)
It will take 14 weeks for 6
workers to build the house.

14. Example 4. The luminous intensity (I) of a light bulb varies
inversely as the square of the distance (d) from the light bulb.The
luminous intensity is 100 candelas when the distance is 5 meters.
Find the luminous intensity at 10 meters.
Solution
Identify the given:
The luminous intensity is 100
candelas when the distance is 5
meters.
The luminous intensity (I) of a light
bulb varies inversely with the
square of the distance (d).
Asked:
Find the luminous intensity at 10
meters.

15. Solution
Solve first the constant of
variation using the formula of
inverse variation
I =
𝑘
𝑑2
100 =
𝑘
52
100 =
𝑘
25
2,500 = k (constant of variation)
I =
𝑘
𝑑2
I =
2,500
102
I =
2,500
100
I = 25 candelas
The luminous intensity at 10
meters is 25 candelas.