Variation | Jamb Mathematics

paragraph

📢 Get ready to ace your exam on Variation in Mathematics! 🌟 Understanding direct, inverse, joint, and partial variations is key to mastering real-world relationships between quantities. Sharpen your problem-solving skills, review key formulas, and practice plenty of questions—because variation is everywhere, and success is in your hands! 🔥📖

paragraph

Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Variation? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Variation together and move one step closer to achieving your exam success! Blissful learning.

paragraph

Calculation Problem involving direct variation

paragraph

Problem 1: Finding the Constant of Variation

y

varies directly as

x

, and

y = 24

when

x = 6

, find the constant of variation.

Solution:

The direct variation equation is $y = kx$ .
Substitute the given values: $24 = k(6)$ .
Solve for $k$ :
$k = \frac{24}{6} = 4$
The constant of variation is 4.

Problem 2: Finding the Missing Value

y

varies directly as

x

, and

y = 15

when

x = 5

, find

y

when

x = 8

Solution:

Use the direct variation formula: $y = kx$ .
Find $k$ :
$15 = k(5) \Rightarrow k = \frac{15}{5} = 3$
Use $k = 3$ to find $y$ when $x = 8$ :
$y = 3(8) = 24$
The missing value is 24.

Problem 3: Determining Direct Variation

Determine if

y = 2x + 3

represents a direct variation.

Solution:

The standard form of direct variation is $y = kx$ .
The given equation has a constant term (+3), which means it is not a direct variation.
The answer is No.

Problem 4: Finding a Value with Fractional Constant

y

varies directly as

x

and

y = 7.5

when

x = 2.5

, find

y

when

x = 10

Solution:

$y = kx$ .
Find $k$ :
$k = \frac{7.5}{2.5} = 3$
Find $y$ when $x = 10$ :
$y = 3(10) = 30$
The answer is 30.

Problem 5: Application in Real-Life Situation

The cost of bananas is directly proportional to their weight. If 4 kg costs $10, how much will 7 kg cost?

Solution:

Let cost be $C$ and weight be $w$ : $C = kw$ .
Find $k$ :
$10 = k(4) \Rightarrow k = \frac{10}{4} = 2.5$
Calculate cost for 7 kg:
$C = 2.5(7) = 17.5$
The cost is $17.50.

Problem 6: Using a Given Ratio

a

is directly proportional to

b

and

a = 18

when

b = 6

, find

a

when

b = 15

Solution:

$a = kb$ .
Find $k$ :
$k = \frac{18}{6} = 3$
Calculate $a$ for $b = 15$ :
$a = 3(15) = 45$
The answer is 45.

Problem 7: Using a Table

The values of

x

and

y

are given:

x = 2, 3, 4

y = 10, 15, 20

Determine if there is a direct variation.

Solution:

Calculate $k = \frac{y}{x}$ for each pair:
$\frac{10}{2} = 5, \quad \frac{15}{3} = 5, \quad \frac{20}{4} = 5$
Since $k$ is constant, it is a direct variation.
The answer is Yes.

Problem 8: Graph Interpretation

A graph passes through

(0,0)

and

(5, 20)

. Does it represent a direct variation?

Solution:

Find $k$ :
$k = \frac{y}{x} = \frac{20}{5} = 4$
The equation $y = 4x$ follows $y = kx$ , so it is a direct variation.
The answer is Yes.

Problem 9: Inverse Calculation

y

is directly proportional to

x

, and

y = 9

when

x = 3

, find

x

when

y = 27

Solution:

$y = kx$ .
Find $k$ :
$k = \frac{9}{3} = 3$
Solve for $x$ when $y = 27$ :
$27 = 3x \Rightarrow x = \frac{27}{3} = 9$
The answer is 9.

Problem 10: Finding a Ratio

x

varies directly as

y

and

x = 16

when

y = 4

, find the ratio

x:y

Solution:

Find $k$ :
$k = \frac{16}{4} = 4$
Since $x = 4y$ , the ratio is 4:1.

Here are 10 more unique calculation problems on direct variation with step-by-step solutions.

Problem 11: Finding the Equation of Direct Variation

y

varies directly as

x

, and

y = 36

when

x = 9

, find the equation relating

x

and

y

Solution:

The equation of direct variation is $y = kx$ .
Find $k$ :
$k = \frac{36}{9} = 4$
The equation is $y = 4x$ .

Problem 12: Word Problem – Car Speed and Distance

A car travels 240 miles in 4 hours at a constant speed. How far will it travel in 7 hours?

Solution:

Let distance be $d$ and time be $t$ : $d = kt$ .
Find $k$ :
$k = \frac{240}{4} = 60$
Calculate distance for 7 hours:
$d = 60(7) = 420$
The car will travel 420 miles.

Problem 13: Checking Direct Variation in a Table

The following table shows values of

x

and

y

. Check if they are directly proportional.

\begin{array}{|c|c|} \hline x & y \\ \hline 2 & 6 \\ 4 & 12 \\ 6 & 20 \\ \hline \end{array}

Solution:

Compute $k = \frac{y}{x}$ $k = \frac{y}{x}$ :
- $\frac{6}{2} = 3$
- $\frac{12}{4} = 3$
- $\frac{20}{6} \neq 3$ (inconsistent)
Since $k$ is not constant, it is not a direct variation.

Problem 14: Finding an Unknown Variable

y

varies directly as

x

, and

y = 27

when

x = 9

, find

x

when

y = 45

Solution:

$y = kx$ .
Find $k$ :
$k = \frac{27}{9} = 3$
Solve for $x$ when $y = 45$ :
$45 = 3x \Rightarrow x = \frac{45}{3} = 15$
The answer is 15.

Problem 15: Physics Application – Hooke’s Law

A spring stretches 12 cm when a force of 6 N is applied. How much will it stretch with a force of 10 N?

Solution:

Let stretch be $s$ and force be $F$ : $s = kF$ .
Find $k$ :
$k = \frac{12}{6} = 2$
Calculate stretch for 10 N:
$s = 2(10) = 20$
The spring will stretch 20 cm.

Problem 16: Ratio of Direct Variation

a

is directly proportional to

b

and

a = 42

when

b = 6

, find the value of

a

when

b = 9

Solution:

$a = kb$ .
Find $k$ :
$k = \frac{42}{6} = 7$
Calculate $a$ when $b = 9$ :
$a = 7(9) = 63$
The answer is 63.

Problem 17: Direct Variation in Finance

If a person earns $450 for working 30 hours, how much will they earn for working 45 hours?

Solution:

Let earnings be $E$ and hours be $h$ : $E = kh$ .
Find $k$ :
$k = \frac{450}{30} = 15$
Calculate earnings for 45 hours:
$E = 15(45) = 675$
The person will earn $675.

Problem 18: Distance and Speed Relationship

If a car moves 90 km in 2 hours at a constant speed, how far will it move in 5 hours?

Solution:

$d = kt$ .
Find $k$ :
$k = \frac{90}{2} = 45$
Calculate distance for 5 hours:
$d = 45(5) = 225$
The car will move 225 km.

Problem 19: Direct Proportion Between Mass and Volume

The mass of a metal cube is 320 g when its volume is 40 cm³. What will be the mass if the volume increases to 70 cm³?

Solution:

$m = kv$ .
Find $k$ :
$k = \frac{320}{40} = 8$
Calculate mass for 70 cm³:
$m = 8(70) = 560$
The mass will be 560 g.

Problem 20: Electrical Resistance and Voltage

The voltage across a resistor varies directly with the current passing through it. If the voltage is 24 V when the current is 3 A, find the voltage when the current is 7 A.

Solution:

$V = kI$ .
Find $k$ :
$k = \frac{24}{3} = 8$
Calculate voltage for 7 A:
$V = 8(7) = 56$
The voltage will be 56 V.

paragraph

calculation problems involving inverse variation

paragraph

Problem 1: Finding the Constant of Inverse Variation

y

varies inversely as

x

, and

y = 12

when

x = 4

, find the constant of variation.

Solution:

The inverse variation equation is $y = \frac{k}{x}$ .
Substitute the given values:
$12 = \frac{k}{4}$
Solve for $k$ :
$k = 12 \times 4 = 48$
The constant of variation is 48.

Problem 2: Finding the Missing Value

y

varies inversely as

x

, and

y = 5

when

x = 8

, find

y

when

x = 10

Solution:

Use the equation $y = \frac{k}{x}$ .
Find $k$ :
$k = 5 \times 8 = 40$
Find $y$ when $x = 10$ :
$y = \frac{40}{10} = 4$
The missing value is 4.

Problem 3: Determining Inverse Variation

Does the equation

xy = 15

represent an inverse variation?

Solution:

Rewrite as $y = \frac{15}{x}$ , which follows the inverse variation form.
The answer is Yes.

Problem 4: Solving for x

y

varies inversely as

x

, and

y = 18

when

x = 2

, find

x

when

y = 6

Solution:

Find $k$ :
$k = 18 \times 2 = 36$
Solve for $x$ when $y = 6$ :
$6x = 36 \Rightarrow x = \frac{36}{6} = 6$
The answer is 6.

Problem 5: Direct vs. Inverse

y

varies inversely as

x

, and

y = 14

when

x = 7

, does

y

increase or decrease as

x

increases?

Solution:

In inverse variation, as $x$ increases, $y$ decreases.
The answer is decreases.

Problem 6: Real-Life Application (Speed and Time)

A car travels 300 km at a speed of 50 km/h. If the speed is increased to 75 km/h, how long will it take?

Solution:

$t$ varies inversely as speed $s$ .
Find $k$ :
$k = 50 \times 6 = 300$
Find $t$ when $s = 75$ :
$t = \frac{300}{75} = 4$
The car will take 4 hours.

Problem 7: Force and Distance

If the force required to move an object varies inversely as the distance, and a force of 12 N moves it 5 m, what force is needed to move it 10 m?

Solution:

$F = \frac{k}{d}$ .
Find $k$ :
$k = 12 \times 5 = 60$
Find $F$ when $d = 10$ :
$F = \frac{60}{10} = 6$
The force is 6 N.

Problem 8: Pressure and Volume (Boyle’s Law)

A gas has a volume of 500 cm³ at a pressure of 100 kPa. What is the volume at 200 kPa?

Solution:

$V = \frac{k}{P}$ .
Find $k$ :
$k = 500 \times 100 = 50000$
Find $V$ when $P = 200$ :
$V = \frac{50000}{200} = 250$
The volume is 250 cm³.

Problem 9: Electrical Resistance and Current

If current

I

varies inversely with resistance

R

, and

I = 10

A when

R = 5 \Omega

, find

I

when

R = 25 \Omega

Solution:

$I = \frac{k}{R}$ .
Find $k$ :

k = 10 \times 5 = 50 $
Find $I$ when $R = 25$ :
$I = \frac{50}{25} = 2$
The current is 2 A.

Problem 10: Work and Workers

If 6 workers complete a job in 12 days, how long will 9 workers take?

Solution:

$d = \frac{k}{w}$ .
Find $k$ :
$k = 6 \times 12 = 72$
Find $d$ when $w = 9$ :
$d = \frac{72}{9} = 8$
The job will take 8 days.

Here are the remaining 10 problems:

Problem 11: Area and Width of a Rectangle

If the area of a rectangle is 48 cm² and its width is 6 cm, find the width when the length is 8 cm.

Solution:

A = l \times w

, and length and width are inversely proportional.

w = \frac{48}{8} = 6

Width is 6 cm.

Problem 12: Gears and Rotations

A gear with 15 teeth makes 20 rotations per minute. How many rotations will a gear with 30 teeth make?

Solution:

k = 15 \times 20 = 300

r = \frac{300}{30} = 10

Rotations per minute = 10.

Problem 13: Pipe Flow Rate

A pipe fills a tank in 10 hours. A larger pipe fills it in 4 hours. How many hours will both take together?

Solution:

\frac{1}{10} + \frac{1}{4} = \frac{2}{20} + \frac{5}{20} = \frac{7}{20}

Total time = $\frac{20}{7}$ hours.

Problem 14: Shadow Length and Object Height

The length of a shadow varies inversely as the height of the object casting it. If a 6-foot pole casts a 10-foot shadow, how long will the shadow be for a 4-foot pole?

Solution:

Let shadow length be $S$ and height be $H$ , so $S = \frac{k}{H}$ .
Find $k$ :
$k = 6 \times 10 = 60$
Find $S$ when $H = 4$ :
$S = \frac{60}{4} = 15$
The shadow will be 15 feet long.

Problem 15: Music Tempo and Time Taken

The time to play a song varies inversely with the speed of the tempo. If a song lasts 4 minutes at 90 beats per minute (bpm), how long will it last at 120 bpm?

Solution:

Let time be $T$ and tempo be $B$ , so $T = \frac{k}{B}$ .
Find $k$ :
$k = 4 \times 90 = 360$
Find $T$ when $B = 120$ :
$T = \frac{360}{120} = 3$
The song will last 3 minutes.

Problem 16: Number of Students and Food Supply

A food supply that lasts 30 days for 12 students will last how many days for 20 students?

Solution:

$D = \frac{k}{S}$ , where $D$ is days and $S$ is students.
Find $k$ :
$k = 30 \times 12 = 360$
Find $D$ when $S = 20$ :
$D = \frac{360}{20} = 18$
The food will last 18 days.

Problem 17: Water Flow Rate and Filling Time

A tap fills a tank in 5 hours at a flow rate of 12 liters per minute. How long will it take if the flow rate is increased to 20 liters per minute?

Solution:

Let time be $T$ and flow rate be $R$ , so $T = \frac{k}{R}$ .
Find $k$ :
$k = 5 \times 12 = 60$
Find $T$ when $R = 20$ :
$T = \frac{60}{20} = 3$
The tank will fill in 3 hours.

Problem 18: Density and Volume Relationship

The density of a substance varies inversely as its volume. If the density is 8 g/cm³ when the volume is 5 cm³, find the density when the volume is 10 cm³.

Solution:

$D = \frac{k}{V}$ , where $D$ is density and $V$ is volume.
Find $k$ :
$k = 8 \times 5 = 40$
Find $D$ when $V = 10$ :
$D = \frac{40}{10} = 4$
The density is 4 g/cm³.

Problem 19: Speed of a Cyclist and Time Taken

A cyclist travels 120 km in 6 hours. If they increase their speed, completing the journey in 4 hours, what was their new speed?

Solution:

$S = \frac{k}{T}$ , where $S$ is speed and $T$ is time.
Find $k$ :
$k = 120 \times 6 = 720$
Find $S$ when $T = 4$ :
$S = \frac{720}{4} = 180$
The new speed is 180 km/h.

Problem 20: Gear Rotation Problem

A gear with 18 teeth makes 40 rotations per minute. How many rotations per minute will a gear with 24 teeth make?

Solution:

$R = \frac{k}{T}$ , where $R$ is rotations and $T$ is teeth count.
Find $k$ :
$k = 18 \times 40 = 720$
Find $R$ when $T = 24$ :
$R = \frac{720}{24} = 30$
The new gear will make 30 rotations per minute.

paragraph

Calculation problems involving joint variation in jamb mathematics

paragraph

Problem 1: Finding the Constant of Joint Variation

z

varies jointly as

x

and

y

, and

z = 24

when

x = 2

and

y = 3

, find the constant of variation.

Solution:

The joint variation equation is $z = kxy$ .
Substitute the given values:
$24 = k(2)(3)$
Solve for $k$ :
$k = \frac{24}{6} = 4$
The constant of variation is 4.

Problem 2: Finding the Missing Value

z

varies jointly as

x

and

y

, and

z = 36

when

x = 4

and

y = 3

, find

z

when

x = 6

and

y = 5

Solution:

$z = kxy$ .
Find $k$ :
$k = \frac{36}{4 \times 3} = 3$
Find $z$ for $x = 6$ and $y = 5$ :
$z = 3(6)(5) = 90$
The missing value is 90.

Problem 3: Checking Joint Variation

z = 15

when

x = 5

and

y = 2

, check if

z = 30

when

x = 10

and

y = 2

Solution:

Find $k$ from the first case:
$k = \frac{15}{5 \times 2} = 1.5$
Compute $z$ for $x = 10$ and $y = 2$ :
$z = 1.5(10)(2) = 30$
The answer is Yes.

Problem 4: Physics Application

The force

F

varies jointly as mass

m

and acceleration

a

. If

F = 60

when

m = 5

kg and

a = 4

m/s², find

F

when

m = 8

kg and

a = 6

m/s².

Solution:

$F = kma$ .
Find $k$ :
$k = \frac{60}{5 \times 4} = 3$
Find $F$ for $m = 8$ , 4 a = 6 $:$ F = 3(8)(6) = 144 $
The force is 144 N.

Problem 5: Electrical Power Calculation in Variation

The electrical power

P varies jointly as current

and voltage

. If

P = 120

when

I = 4

A and

V = 30

V, find

when

I = 6

A and

V = 40 $ V.

Solution:

$P = kIV$ .
Find $k$ :
$k = \frac{120}{4 \times 30} = 1$
Compute $P$ :
$P = 1(6)(40) = 240$
The power is 240 W.

Problem 6: Area of a Triangle

The area

A

of a triangle varies jointly as its base

b

and height

h

. If

A = 30

when

b = 5

and

h = 12

, find

A

when

b = 8

and

h = 15

Solution:

$A = kbh$ .
Find $k$ :
$k = \frac{30}{5 \times 12} = 0.5$
Compute $A$ :
$A = 0.5(8)(15) = 60$
The area is 60 cm².

Problem 7: Work Done by Multiple People

If work

W

varies jointly as the number of worker

N

and time

T

, and

W = 240

when

N = 4

and

T = 6

, find

W

when

N = 5

and

T = 8

Solution:

$W = kNT$ .
Find $k$ :
$k = \frac{240}{4 \times 6} = 10$
Compute $W$ :
$W = 10(5)(8) = 400$
The work is 400 units.

Problem 8: Hooke’s Law (Spring)

The force required to stretch a spring varies jointly as the displacement

x

and the spring constant

k

. If

F = 80

when

x = 4

and

k = 10

, find

F

when

x = 6

and

k = 15

Solution:

$F = kxk_s$ .
Find $k$ :
$k = \frac{80}{4 \times 10} = 2$
Compute $F$ :
$F = 2(6)(15) = 180$
The force is 180 N.

Problem 9: Sound Intensity

The loudness of a sound

L

varies jointly as the intensity

I

and the distance

d

. If

L = 75

when

I = 5

and

d = 3

, find

L

when

I = 8

and

d = 6

Solution:

$L = kId$ .
Find $k$ :
$k = \frac{75}{5 \times 3} = 5$
Compute $L$ :
$L = 5(8)(6) = 240$
The loudness is 240 dB.

Problem 10: Gravitational Force

The gravitational force

F

between two masses varies jointly as their product and inversely as the square of the distance. If

F = 100

when

m_1 = 10

m_2 = 20

, and

d = 5

, find

F

when

m_1 = 15

m_2 = 30

, and

d = 10

Solution:

$F = \frac{k(m_1 m_2)}{d^2}$ .
Find $k$ :
$k = \frac{100 \times 5^2}{10 \times 20} = 5$
Compute $F$ :
$F = \frac{5(15)(30)}{10^2} = 22.5$
The force is 22.5 N.

Problem 11: Velocity, Time, and Distance

The distance

d

traveled by a vehicle varies jointly as its velocity

v

and time

t

. If

d = 240

when

v = 60

km/h and

t = 4

hours, find

d

when

v = 80

km/h and

t = 6

hours.

Solution:

$d = kvt$ .
Find $k$ :
$k = \frac{240}{60 \times 4} = 1$
Compute $d$ :
$d = 1(80)(6) = 480$
The distance is 480 km.

Problem 12: Volume of a Cylinder

The volume

V

of a cylinder varies jointly as the square of the radius

r

and height

h

. If

V = 314

cm³ when

r = 5

cm and

h = 4

cm, find

V

when

r = 7

cm and

h = 6

cm.

Solution:

$V = kr^2h$ .
Find $k$ :
$k = \frac{314}{5^2 \times 4} = \frac{314}{100} = 3.14$
Compute $V$ :
$V = 3.14(7^2)(6) = 3.14(49)(6) = 923.88$
The volume is 923.88 cm³.

Problem 13: Financial Investment Returns

The return on an investment

R

varies jointly as the principal

P

and the time

t

. If

R = 900

when

P = 3000

and

t = 3

years, find

R

when

P = 5000

and

t = 4

years.

Solution:

$R = kPt$ .
Find $k$ :
$k = \frac{900}{3000 \times 3} = 0.1$
Compute $R$ :
$R = 0.1(5000)(4) = 2000$
The return is $2000$ .

Problem 14: Gas Law (Boyle’s Law)

The pressure

P

of a gas varies jointly as the temperature

T

and inversely as the volume

V

. If

P = 200

kPa when

T = 300

K and

V = 10

L, find

P

when

T = 400

K and

V = 8

Solution:

$P = \frac{kT}{V}$ .
Find $k$ :
$k = \frac{P \times V}{T} = \frac{200 \times 10}{300} = 6.67$
Compute $P$ :
$P = \frac{6.67(400)}{8} = 333.6$
The pressure is 333.6 kPa.

Problem 15: Population Growth

The population

P

of a colony varies jointly as its initial size

P_0

and the time

t

. If

P = 800

when

P_0 = 200

and

t = 4

years, find

P

when

P_0 = 300

and

t = 5

years.

Solution:

$P = k P_0 t$ .
Find $k$ :
$k = \frac{800}{200 \times 4} = 1$
Compute $P$ :
$P = 1(300)(5) = 1500$
The population is 1500.

Problem 16: Energy Consumption

The energy consumption

E

varies jointly as the power

P

and the time

t

. If

E = 1200

kWh when

P = 100

kW and

t = 12

hours, find

E

when

P = 150

kW and

t = 10

hours.

Solution:

$E = kPt$ .
Find $k$ :
$k = \frac{1200}{100 \times 12} = 1$
Compute $E$ :
$E = 1(150)(10) = 1500$
The energy consumption is 1500 kWh.

Problem 17: Heat Transfer involving variation

The heat energy

Q

transferred varies jointly as the mass

m

and the temperature change

\Delta T

. If

Q = 5000

J when

m = 2

kg and

\Delta T = 50

°C, find

Q

when

m = 4

kg and

\Delta T = 75

°C.

Solution:

$Q = km \Delta T$ .
Find $k$ :
$k = \frac{5000}{2 \times 50} = 50$
Compute $Q$ :
$Q = 50(4)(75) = 15000$
The heat energy is 15000 J.

Problem 18: Flow Rate of Liquid

The volume

V

of liquid flowing through a pipe varies jointly as the speed

s

and the cross-sectional area

A

. If

V = 100

L/min when

s = 5

m/s and

A = 2

m², find

V

when

s = 8

m/s and

A = 3

m².

Solution:

$V = k s A$ .
Find $k$ :
$k = \frac{100}{5 \times 2} = 10$
Compute $V$ :
$V = 10(8)(3) = 240$
The volume is 240 L/min.

Problem 19: Spring Force (Hooke's Law) involving joint variation

The force

F

in a spring varies jointly as the displacement

x

and the spring constant

k

. If

F = 50

N when

x = 5

cm and

k = 2

, find

F

when

x = 8

cm and

k = 3

Solution:

$F = kxk_s$ .
Find $k$ :
$k = \frac{50}{5 \times 2} = 5$
Compute $F$ :
$F = 5(8)(3) = 120$
The force is 120 N.

Problem 20: Momentum Calculation

Momentum

p

varies jointly as mass

m

and velocity

v

. If

p = 600

kg·m/s when

m = 20

kg and

v = 30

m/s, find

p

when

m = 25

kg and

v = 40

m/s.

Solution:

$p = kmv$ .
Find $k$ :
$k = \frac{600}{20 \times 30} = 1$
Compute $p$ :
$p = 1(25)(40) = 1000$
The momentum is 1000 kg·m/s.

paragraph

Calculation problem involving partial Variation

paragraph

Understanding Partial Variation

Partial variation occurs when a quantity is related to another in two ways: partly directly and partly constant. The general form of a partial variation equation is:

y = kx + c

where:

$kx$ represents the direct variation component,
$c$ represents the constant part.

Problem 1: Finding the Equation of Partial Variation

A quantity

y

is partly constant and partly varies directly as

x

. When

x = 2

y = 8

, and when

x = 5

y = 14

. Find the equation relating

y

and

x

Solution:

Assume $y = kx + c$
Use the given values:
- $8 = 2k + c$ → (Equation 1)
- $14 = 5k + c$ → (Equation 2)
Subtract Equation 1 from Equation 2: $(14 - 8) = (5k - 2k)$ $6 = 3k \Rightarrow k = 2$
Substitute $k = 2$ into Equation 1: $8 = 2(2) + c$ $c = 4$
The equation is $y = 2x + 4$ .

Problem 2: Finding y for a Given x

Using the equation

y = 2x + 4

, find

y

when

x = 10

Solution:

Substitute $x = 10$ into the equation: $y = 2(10) + 4$ $y = 20 + 4 = 24$
The answer is $y = 24$ .

Problem 3: Checking if a Relationship is Partial Variation

Is the equation

y = 3x + 5

an example of partial variation?

Solution:

The equation is in the form $y = kx + c$
Since it has both a variable term ( $3x$ ) and a constant ( $5$ ), it represents partial variation.
The answer is Yes.

Problem 4: Finding the Constant and Variable Terms

A taxi service charges a fixed fare of

5 plus

2 per kilometer traveled. Write the equation and find the fare for 8 km.

Solution:

Let $y$ be the total fare and $x$ be the distance traveled.
Equation: $y = 2x + 5$ .
For $x = 8$ : $y = 2(8) + 5 = 16 + 5 = 21$
The fare is $21.

Problem 5: Determining x for a Given y

Using

y = 2x + 4

, find

x

when

y = 30

Solution:

Solve for $x$ : $30 = 2x + 4$ $30 - 4 = 2x$ $x = \frac{26}{2} = 13$
The answer is $x = 13$ .

Problem 6: Real-Life Application (Internet Subscription)

An internet provider charges a fixed monthly fee of

10 plus

0.05 per MB of data used. Write the equation and find the total charge for using 200 MB.

Solution:

Equation: $y = 0.05x + 10$ .
For $x = 200$ : $y = 0.05(200) + 10$ $y = 10 + 10 = 20$
The charge is $20.

Problem 7: Finding the Rate of Variation

y = 4x + 6

, what is the rate of variation?

Solution:

The coefficient of $x$ (4) represents the rate of variation.
The rate is 4.

Problem 8: Identifying Constant

For

y = 3x + 7

, identify the constant part.

Solution:

The constant term is $7$ .
The answer is 7.

Problem 9: Solving for x

y = 5x + 9

and

y = 29

, find

x

Solution:

Solve for $x$ : $29 = 5x + 9$ $29 - 9 = 5x$ $x = \frac{20}{5} = 4$
The answer is 4.

Problem 10: Graph Interpretation

If a line has equation

y = 2x + 3

, does it represent partial variation?

Solution:

The equation is in the form $y = kx + c$ , indicating partial variation.
The answer is Yes.

Problem 11: Partial Variation in Salary

A worker’s total earnings include a base salary of

500 plus

15 per hour worked. Write an equation and find earnings for 40 hours.

Solution: Equation:

y = 15x + 500

,
For

x = 40

y = 15(40) + 500 = 1100

.
Earnings = $1100.

Problem 12: Partial Variation in Fuel Cost

A car rental charges

50 plus

0.20 per mile. Find the cost for 150 miles.

Solution: Equation:

y = 0.2x + 50

,
For

x = 150

y = 0.2(150) + 50 = 80

.
Cost = $80.

paragraph

Thank you for taking the time to read my blog post! Your interest and engagement mean so much to me, and I hope the content provided valuable insights and sparked your curiosity. Your journey as a student is inspiring, and it’s my goal to contribute to your growth and success.

paragraph

If you found the post helpful, feel free to share it with others who might benefit. I’d also love to hear your thoughts, feedback, or questions—your input makes this space even better. Keep striving, learning, and achieving! 😊📚✨

paragraph

I recommend you check my Post on the following:

paragraph

- Jamb Mathematics- Lesson notes on Inequalities for utme Success

paragraph

This is all we can take on "Jamb Mathematics - Lesson Notes on Variation for UTME Candidate"

paragraph