Direct, Inverse, Joint and Partial Variation Practice Questions — JAMB UTME

Question 1Question →

A quantity $P$ varies partially as $x$ and partially as the square of $y$ . When $x = 2$ and $y = 3$ , $P = 24$ , and when $x = 5$ and $y = 1$ , $P = 17$ . What is the value of $P$ when $x = 4$ and $y = 3$ ?

30

35

26

21

Show answer & explanation

Answer: 30

Answer

The value of

P

is 30.

The relationship follows the partial variation formula

P = k_1 x + k_2 y^2

. Substituting the given conditions gives

2k_1 + 9k_2 = 24

and

5k_1 + k_2 = 17

. Solving these simultaneous equations yields

k_1 = 3

and

k_2 = 2

. Evaluating

P = 3(4) + 2(3^2)

produces

12 + 18 = 30

.

Step-by-Step Solution

1

Set up the general formula for partial variation.

P = k_1 x + k_2 y^2

, where

k_1

and

k_2

are constants.

Partial variation combines terms linearly with separate variation constants.

2

Substitute the given pairs of values to form simultaneous linear equations.

Equation (1):

2k_1 + 9k_2 = 24

; Equation (2):

5k_1 + k_2 = 17

.

Plugging in

(x=2, y=3, P=24)

and

(x=5, y=1, P=17)

creates a system of equations in terms of

k_1

and

k_2

.

3

Solve the simultaneous linear equations for

k_1

and

k_2

.

From Equation (2),

k_2 = 17 - 5k_1

. Substitute into Equation (1):

2k_1 + 9(17 - 5k_1) = 24 \implies -43k_1 = -129 \implies k_1 = 3

. Then

k_2 = 17 - 5(3) = 2

.

Finding the specific values of the variation constants is required to complete the formula.

4

Calculate

P

for

x = 4

and

y = 3

using the complete formula

P = 3x + 2y^2

.

P = 3(4) + 2(3^2) = 12 + 2(9) = 12 + 18 = 30

.

Evaluating the relationship with the target parameters produces the final answer.

Key Concept

Partial Variation and Simultaneous Linear Equations

Estimated Time:2m 0s

Question 2Question →

The variable $y$ varies inversely as $x$ . If $y = 6$ when $x = 4$ , what is the value of $y$ when $x = 8$ ?

3

12

16

48

Show answer & explanation

Answer: 3

Answer

The value of

y

when

x = 8

is 3.

Since

y

varies inversely as

x

, the formula connecting them is

y = \frac{k}{x}

. Substituting

y = 6

when

x = 4

gives

6 = \frac{k}{4}

, which yields

k = 24

. Using this constant, when

x = 8

,

y = \frac{24}{8} = 3

.

Step-by-Step Solution

1

Write the relationship for inverse variation.

y = \frac{k}{x}

Inverse variation means

y

is inversely proportional to

x

, where

k

is the constant of variation.

2

Substitute given values

y = 6

and

x = 4

to solve for

k

.

6 = \frac{k}{4} \implies k = 6 \times 4 = 24

To complete the equation of variation, the constant

k

must be determined.

3

Substitute

k = 24

and

x = 8

into the variation equation to find

y

.

y = \frac{24}{8} = 3

Evaluating the relationship at

x = 8

gives the requested value.

Key Concept

Inverse Variation

Question 3Question →

A quantity $y$ is partly constant and partly varies directly as $x$ . When $x = 2$ , $y = 10$ , and when $x = 5$ , $y = 19$ . What is the value of $y$ when $x = 8$ ?

28

35

40

25

Show answer & explanation

Answer: 28

Answer

The value of

y

when

x = 8

is 28.

By representing partial variation as

y = c + kx

, substituting the given conditions gives two simultaneous linear equations:

10 = c + 2k

and

19 = c + 5k

. Subtracting the first equation from the second yields

3k = 9

, so

k = 3

. Substituting

k = 3

back into the first equation yields

c = 4

. The general equation is

y = 4 + 3x

. Evaluating at

x = 8

gives

y = 4 + 3(8) = 28

.

Step-by-Step Solution

1

Set up the partial variation equation

y = c + kx

, where

c

is the constant part and

k

is the constant of variation.

Partial variation consists of a fixed term plus a variable term.

2

Substitute given values to form simultaneous equations

Equation 1:

10 = c + 2k

Equation 2:

19 = c + 5k

Two pairs of

(x, y)

values are provided to solve for the two unknown constants

c

and

k

.

3

Solve for

k

and

c

Subtract Equation 1 from Equation 2:

9 = 3k \implies k = 3

.
Substitute

k = 3

into Equation 1:

10 = c + 2(3) \implies c = 4

.
Thus,

y = 4 + 3x

.

Eliminating

c

yields the value of

k

, which is then used to find

c

.

4

Calculate

y

for

x = 8

y = 4 + 3(8) = 4 + 24 = 28

.

Substitute the required value of

x

into the established formula.

Key Concept

Partial Variation and Simultaneous Equations

Question 4Question →

Given that $P$ varies directly as the square of $r$ , and $P = 48$ when $r = 4$ , calculate the value of $P$ when $r = 6$ .

Show answer & explanation

Answer: 108

Answer

The value of

P

when

r = 6

is

108

.

Since

P

varies directly as

r^2

, the relationship is expressed as

P = k r^2

. Substituting the given values

P = 48

and

r = 4

yields

48 = 16k

, so

k = 3

. Substituting

k = 3

and

r = 6

into the equation gives

P = 3 \times 6^2 = 3 \times 36 = 108

.

Step-by-Step Solution

1

Set up the variation equation using constant of variation

k

P = k r^2

Direct variation with the square of a variable means

P

is directly proportional to

r^2

.

2

Substitute the initial values

P = 48

and

r = 4

to determine

k

48 = k \times 4^2 \implies 48 = 16k \implies k = 3

Finding the variation constant

k

allows us to establish a specific relationship between

P

and

r

.

3

Calculate

P

for

r = 6

using the specific equation

P = 3 r^2

P = 3 \times 6^2 = 3 \times 36 = 108

Evaluating the formula with the new input

r = 6

yields the required value of

P

.

Key Concept

Direct variation involving square powers

Question 5Question →

Given that $z$ varies directly as $x^2$ and inversely as $\sqrt{y}$ , and $z = 12$ when $x = 2$ and $y = 9$ , what is the value of $z$ when $x = 3$ and $y = 16$ ?

\frac{81}{4}

\frac{27}{2}

\frac{243}{16}

\frac{64}{9}

Show answer & explanation

Answer: $\frac{81}{4}$

Answer

\frac{81}{4}

The joint variation formula is

z = \frac{k x^2}{\sqrt{y}}

. Substituting the given values

x = 2, y = 9, z = 12

gives

12 = \frac{4k}{3}

, which yields

k = 9

. Evaluating

z

for

x = 3

and

y = 16

gives

z = \frac{9 \times 3^2}{\sqrt{16}} = \frac{81}{4}

.

Step-by-Step Solution

1

Set up the joint variation equation

z = \frac{k x^2}{\sqrt{y}}

Direct variation means

x^2

is in the numerator, and inverse variation means

\sqrt{y}

is in the denominator.

2

Substitute the initial values to solve for the constant of variation

k

12 = \frac{k (2)^2}{\sqrt{9}} \implies 12 = \frac{4k}{3} \implies 4k = 36 \implies k = 9

Using

x = 2

,

y = 9

, and

z = 12

allows us to find the constant

k

.

3

Calculate the new value of

z

using

x = 3

and

y = 16

z = \frac{9 (3)^2}{\sqrt{16}} = \frac{9 \times 9}{4} = \frac{81}{4}

Substitute

k = 9

,

x = 3

, and

y = 16

into the variation formula.

Key Concept

Joint Variation involving powers and roots

Estimated Time:1m 30s

Question 6Question →

The electrical resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ . If a wire of length $36\text{ m}$ and diameter $3\text{ mm}$ has a resistance of $16\ \Omega$ , what is the resistance, in ohms, of a wire of the same material with a length of $45\text{ m}$ and a diameter of $5\text{ mm}$ ?

Show answer & explanation

Answer: 7.2

Answer

The resistance of the wire is 7.2 ohms.

The equation governing the relation is

R = \frac{kL}{d^2}

. Substituting the initial parameters

R=16\ \Omega

,

L=36\text{ m}

, and

d=3\text{ mm}

gives

16 = \frac{36k}{9} = 4k

, which yields

k = 4

. Using

k = 4

with the new dimensions

L=45\text{ m}

and

d=5\text{ mm}

gives

R = \frac{4 \times 45}{5^2} = \frac{180}{25} = 7.2\ \Omega

.

Step-by-Step Solution

1

Formulate the joint variation equation

R = \frac{kL}{d^2}

Direct variation places length

L

in the numerator and inverse variation of the square of diameter

d

places

d^2

in the denominator.

2

Determine the variation constant

k

k = 4

Substituting

R = 16

,

L = 36

, and

d = 3

gives

16 = \frac{36k}{9} \implies 16 = 4k \implies k = 4

.

3

Calculate the new resistance

R = 7.2\ \Omega

Substituting

k = 4

,

L = 45

, and

d = 5

into

R = \frac{kL}{d^2}

yields

R = \frac{4 \times 45}{25} = 7.2

.

Key Concept

Joint Variation involving direct proportionality and inverse square law

Question 7Question →

The hourly operational cost, $C$ Naira, of an industrial water pump is partly constant and partly varies jointly as the flow rate, $r$ in litres per second, and the square of the pressure head, $h$ in metres. When $r = 10\text{ L/s}$ and $h = 4\text{ m}$ , the operational cost is $\text{N}620$ . When $r = 15\text{ L/s}$ and $h = 2\text{ m}$ , the operational cost is $\text{N}380$ . What is the operational cost in Naira when $r = 20\text{ L/s}$ and $h = 3\text{ m}$ ?

Show answer & explanation

Answer: 668

Answer

The operational cost when

r = 20

and

h = 3

is 668 Naira.

The partial and joint variation relationship is defined by

C = k_1 + k_2 r h^2

. Substituting the two given states gives the simultaneous equations

620 = k_1 + 160k_2

and

380 = k_1 + 60k_2

. Subtracting these equations gives

100k_2 = 240

, so

k_2 = 2.4

. Substituting

k_2 = 2.4

into the second equation yields

k_1 = 236

. Finally, evaluating

C

for

r = 20

and

h = 3

gives

C = 236 + 2.4(20)(3^2) = 236 + 432 = 668

.

Step-by-Step Solution

1

Set up the variation equation

C = k_1 + k_2 r h^2

, where

k_1

is the constant part and

k_2

is the constant of joint variation.

The problem states that

C

is partly constant (

k_1

) and partly varies jointly as

r

and

h^2

(

k_2 r h^2

).

2

Form simultaneous linear equations using the given data points

(1)

620 = k_1 + 160k_2

and (2)

380 = k_1 + 60k_2

Substituting

r = 10, h = 4, C = 620

gives

10 \times 4^2 = 160

. Substituting

r = 15, h = 2, C = 380

gives

15 \times 2^2 = 60

.

3

Solve for the constants

k_1

and

k_2

k_2 = 2.4

and

k_1 = 236

Subtracting equation (2) from (1) eliminates

k_1

, giving

100k_2 = 240 \implies k_2 = 2.4

. Substituting back into equation (2) gives

k_1 = 380 - 60(2.4) = 236

.

4

Calculate the operational cost for the target parameters

C = 236 + 2.4 \times 20 \times 3^2 = 668

Substitute

k_1 = 236

,

k_2 = 2.4

,

r = 20

, and

h = 3

into the variation formula.

Key Concept

Partial and Joint Variation

Question 8Question →

A variable $y$ is partly constant and partly varies directly as $\sqrt{x}$ . Given that $y = 26$ when $x = 16$ , and $y = 38$ when $x = 49$ , what is the value of $y$ when $x = 64$ ?

Show answer & explanation

Answer: 42

Answer

The value of

y

when

x = 64

is

42

.

The relationship is given by the partial variation equation

y = a + b\sqrt{x}

. Substituting the pairs

(16, 26)

and

(49, 38)

yields the linear system

a + 4b = 26

and

a + 7b = 38

. Solving this system gives the constants

a = 10

and

b = 4

. Substituting

x = 64

into

y = 10 + 4\sqrt{64}

results in

y = 10 + 4(8) = 42

.

Step-by-Step Solution

1

Formulate the partial variation equation.

y = a + b\sqrt{x}

, where

a

and

b

are constants of variation.

Partial variation implies

y

is the sum of a constant term

a

and a term directly proportional to

\sqrt{x}

.

2

Set up simultaneous equations using the given pairs of

(x, y)

.

Equation 1:

a + 4b = 26

Equation 2:

a + 7b = 38

Evaluating

\sqrt{16} = 4

and

\sqrt{49} = 7

simplifies the relationship into two linear equations in two unknowns.

3

Solve for constants

a

and

b

.

b = 4

and

a = 10

Subtracting Equation 1 from Equation 2 yields

3b = 12 \implies b = 4

, and substituting

b = 4

back into Equation 1 gives

a = 10

.

4

Calculate

y

when

x = 64

.

y = 10 + 4(8) = 42

Using the specific formula

y = 10 + 4\sqrt{x}

for

x = 64

gives

y = 10 + 32 = 42

.

Key Concept

Partial Variation with Simultaneous Equations

Question 9Question →

The time, $t$ hours, required to complete a road maintenance project varies inversely as the number of workers, $w$ , assigned to the project. If 8 workers can finish the project in 15 hours, calculate the time, in hours, required for 12 workers to finish the same project.

Show answer & explanation

Answer: 10

Answer

10 hours

Because the time

t

varies inversely as the number of workers

w

, the total worker-hours required for the project is constant:

k = 8 \times 15 = 120

worker-hours. Dividing this total work by 12 workers gives

\frac{120}{12} = 10

hours.

Step-by-Step Solution

1

Set up the inverse variation equation

t = \frac{k}{w}

, where

k

is the constant of variation.

Inverse variation implies that as the number of workers increases, the time required decreases proportionally.

2

Determine the value of the constant of variation

k

k = t \times w = 15 \times 8 = 120

.

Substitute the known pair of values (

w = 8, t = 15

) into the equation.

3

Compute the new value of

t

for 12 workers

t = \frac{120}{12} = 10

hours.

Substitute

w = 12

and

k = 120

into

t = \frac{k}{w}

.

Key Concept

Inverse Variation

Question 10Question →

The total energy loss $E$ in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency $f$ in Hz of the alternating current. Given that $E = 120\text{ J}$ when $f = 10\text{ Hz}$ , and $E = 360\text{ J}$ when $f = 20\text{ Hz}$ , what is the value of $E$ in Joules when $f = 15\text{ Hz}$ ?

Show answer & explanation

Answer: 220

Answer

220

The relation describes a partial variation model

E = c + k f^2

. Substituting the two given conditions (

f=10, E=120

and

f=20, E=360

) yields the system of equations

c + 100k = 120

and

c + 400k = 360

. Solving this system gives

k = 0.8

and

c = 40

. Evaluating

E = 40 + 0.8(15)^2

results in

220\text{ J}

.

Step-by-Step Solution

1

Write the general equation for partial variation involving a constant term and a term proportional to

f^2

E = c + k f^2

Partial variation consists of a sum of a constant component and a variable component.

2

Substitute the known conditions into the variation equation to create a system of linear equations

Equation 1:

c + 100k = 120

; Equation 2:

c + 400k = 360

Two pairs of values are provided to determine the two unknown constants

c

and

k

.

3

Solve the system of simultaneous equations for

k

and

c

k = 0.8

and

c = 40

Subtracting Equation 1 from Equation 2 eliminates

c

, allowing direct calculation of

k

, after which

c

is found by substitution.

4

Calculate the required value of

E

when

f = 15

E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220

Applying the discovered constants to the target frequency yields the final energy loss value.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:2m 0s

Question 11Question →

If $y$ varies directly as $x$ , and $y = 20$ when $x = 4$ , what is the value of $y$ when $x = 10$ ?

50

8

5

2

Show answer & explanation

Answer: 50

Answer

The value of

y

when

x = 10

is 50.

Direct variation means

y = kx

. Substituting

y = 20

and

x = 4

yields

k = 5

. Substituting

k = 5

and

x = 10

gives

y = 50

.

Step-by-Step Solution

1

Set up the equation for direct variation

y = kx

, where

k

is the constant of variation.

Direct variation implies that

y

is directly proportional to

x

.

2

Substitute the given initial values (

y = 20

,

x = 4

) to find

k

20 = k(4) \implies k = \frac{20}{4} = 5

.

Determining the constant of variation allows us to write the specific relationship equation.

3

Calculate

y

when

x = 10

using the constant

k = 5

y = 5 \times 10 = 50

.

Substituting

x = 10

into

y = 5x

yields the required value.

Key Concept

Direct Variation (

y = kx

)

Estimated Time:45s

Question 12Question →

The power consumption $P$ (in watts) of a variable-speed motor is partly constant and partly varies directly as the square of its operational speed $v$ (in revolutions per second). If $P = 250\text{ W}$ when $v = 10\text{ rev/s}$ and $P = 700\text{ W}$ when $v = 20\text{ rev/s}$ , calculate the value of $P$ (in watts) when $v = 15\text{ rev/s}$ .

Show answer & explanation

Answer: 437.5

Answer

437.5 W

The partial variation equation is

P = k_1 + k_2 v^2

. Setting up simultaneous equations

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

yields

k_1 = 100

and

k_2 = 1.5

. Substituting

v = 15

into

P = 100 + 1.5(15^2)

gives

P = 437.5\text{ W}

.

Step-by-Step Solution

1

Express the partial variation mathematically.

P = k_1 + k_2 v^2

, where

k_1

and

k_2

are constants.

The total power consumption is the sum of a fixed baseline constant

k_1

and a variable component proportional to

v^2

.

2

Form simultaneous linear equations using the provided data points.

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

.

Substituting

v = 10

gives

10^2 = 100

, and substituting

v = 20

gives

20^2 = 400

.

3

Solve for the constants

k_1

and

k_2

.

k_2 = 1.5

and

k_1 = 100

.

Subtracting the two equations eliminates

k_1

, yielding

300 k_2 = 450 \implies k_2 = 1.5

. Substituting

k_2 = 1.5

back into

k_1 + 100 k_2 = 250

gives

k_1 = 100

.

4

Calculate the value of

P

at

v = 15\text{ rev/s}

.

P = 437.5\text{ W}

.

Substitute

v = 15

,

k_1 = 100

, and

k_2 = 1.5

into the governing formula

P = 100 + 1.5(15^2) = 100 + 337.5 = 437.5

.

Key Concept

Partial Variation and Simultaneous Linear Equations

Estimated Time:1m 30s

Question 13Question →

The total cost $C$ per trip of operating a high-speed passenger ferry consists of a fixed administrative overhead cost and an operational cost that varies directly as the cube of its speed $v$ in knots. Given that the total cost per trip is $\$1,400$ when the ferry travels at $10\text{ knots}$ and $\$4,200$ when it travels at $20\text{ knots}$ , what is the total cost per trip when the ferry operates at a speed of $15\text{ knots}$ ?

\\

2,350$

\\

2,800$

\\

3,300$

\\

4,725$

Show answer & explanation

Answer: $\\$ 2,350$

Answer

\\

2,350$

The partial variation formula is

C = k_1 + k_2 v^3

. Substituting the given points yields

k_1 + 1000 k_2 = 1400

and

k_1 + 8000 k_2 = 4200

. Subtracting these equations gives

7000 k_2 = 2800

, so

k_2 = 0.4

and

k_1 = 1000

. Evaluating at

v = 15

gives

C = 1000 + 0.4(3375) = 1000 + 1350 = \\

2,350$.

Step-by-Step Solution

1

Set up the partial variation equation

C = k_1 + k_2 v^3

Partial variation combines a fixed constant

k_1

with a variable term

k_2 v^3

.

2

Form simultaneous linear equations using given conditions

Equation (1):

k_1 + 1000 k_2 = 1400

; Equation (2):

k_1 + 8000 k_2 = 4200

Substitute

v = 10

,

C = 1400

and

v = 20

,

C = 4200

into the variation model.

3

Solve for the constants

k_1

and

k_2

k_2 = 0.4

and

k_1 = 1000

Subtract Equation (1) from Equation (2):

7000 k_2 = 2800 \implies k_2 = 0.4

. Substitute

k_2 = 0.4

into Equation (1) to get

k_1 = 1400 - 400 = 1000

.

4

Calculate total cost

C

for speed

v = 15\text{ knots}

C = 1000 + 0.4(15)^3 = 1000 + 0.4(3375) = 1000 + 1350 = 2350

Substitute

k_1 = 1000

,

k_2 = 0.4

, and

v = 15

back into the formula.

Key Concept

Partial Variation with non-linear powers solved via simultaneous linear equations

Estimated Time:2m 0s

Question 14Question →

The monthly operating cost $C$ (in naira) of a commercial power generator is partly constant and partly varies directly as the square of its daily operating time $h$ (in hours). When the generator operates for $3$ hours per day, the monthly cost is ₦ $5,500$ . When it operates for $5$ hours per day, the monthly cost is ₦ $13,500$ . What is the daily operating time, in hours, when the monthly operating cost is ₦ $33,000$ ?

8 hours

7 hours

9 hours

10 hours

Show answer & explanation

Answer: 8 hours

Answer

8 hours

The relationship is modeled by partial variation

C = k_1 + k_2 h^2

. Solving the simultaneous equations derived from

h = 3

(

C = 5500

) and

h = 5

(

C = 13500

) yields

k_2 = 500

and

k_1 = 1000

. Substituting

C = 33000

into

33000 = 1000 + 500 h^2

gives

500 h^2 = 32000

, so

h^2 = 64

, which yields

h = 8

hours.

Step-by-Step Solution

1

Set up the general formula for partial variation.

C = k_1 + k_2 h^2

, where

k_1

and

k_2

are constants.

The cost consists of a constant part (

k_1

) and a part that varies directly as the square of daily hours (

k_2 h^2

).

2

Form simultaneous linear equations using the given data points.

Equation (1):

5500 = k_1 + 9 k_2

Equation (2):

13500 = k_1 + 25 k_2

Substitute

h = 3, C = 5500

and

h = 5, C = 13500

into the variation equation.

3

Solve for the variation constants

k_1

and

k_2

.

Subtract Equation (1) from Equation (2):

8000 = 16 k_2 \implies k_2 = 500

.
Substitute

k_2 = 500

into Equation (1):

5500 = k_1 + 9(500) \implies k_1 = 1000

.

Eliminating

k_1

gives

k_2

, which is then used to find the constant part

k_1

.

4

Calculate

h

when

C = 33000

.

33000 = 1000 + 500 h^2 \implies 32000 = 500 h^2 \implies h^2 = 64 \implies h = 8

hours.

Substitute the values of

k_1, k_2,

and target

C

into the relation equation to solve for

h

.

Key Concept

Partial Variation with Simultaneous Equations

Estimated Time:2m 30s

Question 15Question →

The total cost $C$ of printing a school magazine is partly constant and partly varies directly as the number of copies $n$ printed. If it costs $\text{N}70,000$ to print $500$ copies and $\text{N}140,000$ to print $1,200$ copies, what is the cost of printing $2,000$ copies?

\text{N}200,000

\text{N}220,000

\text{N}240,000

\text{N}280,000

Show answer & explanation

Answer: $\text{N}220,000$

Answer

\text{N}220,000

In partial variation, the total cost

C

is represented by

C = k_1 + k_2 n

. Subtracting the two linear equations

70,000 = k_1 + 500 k_2

and

140,000 = k_1 + 1,200 k_2

gives

700 k_2 = 70,000

, so

k_2 = 100

. Substituting

k_2 = 100

yields the fixed constant

k_1 = 20,000

. Substituting

n = 2,000

into

C = 20,000 + 100n

yields

20,000 + 200,000 = \text{N}220,000

.

Step-by-Step Solution

1

Set up the partial variation equation

C = k_1 + k_2 n

, where

k_1

is the fixed cost and

k_2

is the rate per copy

Partial variation consists of a constant term and a term that varies directly with the independent variable.

2

Substitute given values to form simultaneous equations

Equation 1:

70,000 = k_1 + 500 k_2

Equation 2:

140,000 = k_1 + 1,200 k_2

Using the two given data points

(n=500, C=70,000)

and

(n=1,200, C=140,000)

creates a system of linear equations.

3

Solve for the variation constants

k_1

and

k_2

Subtracting Equation 1 from Equation 2 gives

70,000 = 700 k_2 \implies k_2 = 100

.
Substituting

k_2 = 100

into Equation 1 gives

70,000 = k_1 + 500(100) \implies k_1 = 20,000

.

Determines the specific values for the fixed overhead and rate per copy.

4

Calculate the total cost for

2,000

copies

C = 20,000 + 100(2,000) = 20,000 + 200,000 = \text{N}220,000

Evaluates the completed formula

C = 20,000 + 100n

at

n = 2,000

.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:1m 30s

Question 16Question →

The electrical resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ . A wire of length $50\text{ m}$ and diameter $2\text{ mm}$ has a resistance of $5\text{ ohms}$ . Calculate the resistance (in ohms) of a wire made of the same material with a length of $80\text{ m}$ and a diameter of $4\text{ mm}$ .

Show answer & explanation

Answer: 2

Answer

The resistance of the wire is

2\text{ ohms}

.

The variation model is

R = \frac{k L}{d^2}

. Substituting

R = 5\text{ ohms}

,

L = 50\text{ m}

, and

d = 2\text{ mm}

gives

5 = \frac{50k}{4}

, so

k = 0.4

. Substituting

k = 0.4

,

L = 80\text{ m}

, and

d = 4\text{ mm}

yields

R = \frac{0.4 \times 80}{4^2} = \frac{32}{16} = 2\text{ ohms}

.

Step-by-Step Solution

1

Set up the variation formula

R = \frac{k L}{d^2}

Resistance varies directly as length

L

and inversely as the square of diameter

d

.

2

Calculate the constant of variation

k

k = 0.4

Substitute the initial values

R = 5

,

L = 50

, and

d = 2

into the variation equation.

3

Compute the new resistance

R

R = 2\text{ ohms}

Substitute

k = 0.4

,

L = 80

, and

d = 4

into the formula.

Key Concept

Direct and Inverse Joint Variation

Estimated Time:1m 30s

Question 17Question →

The rate of heat transfer $Q$ across a building wall varies directly as the surface area $A$ of the wall and the temperature difference $\Delta T$ between the interior and exterior, and inversely as the wall thickness $d$ . When the surface area is $4\text{ m}^2$ , the temperature difference is $15^\circ\text{C}$ , and the thickness is $0.05\text{ m}$ , the heat transfer rate is $1200\text{ W}$ . What is the heat transfer rate in watts when the surface area is $6\text{ m}^2$ , the temperature difference is $20^\circ\text{C}$ , and the thickness is $0.08\text{ m}$ ?

Show answer & explanation

Answer: 1500

Answer

The heat transfer rate is

1500\text{ W}

.

Establishing the variation constant

k = 1

using the initial given values and substituting the new parameters yields

Q = \frac{1 \times 6 \times 20}{0.08} = 1500\text{ W}

.

Step-by-Step Solution

1

Formulate the variation equation

Q = \frac{k A \Delta T}{d}

Direct variation means multiplying by

A

and

\Delta T

, while inverse variation means dividing by

d

.

2

Calculate the constant of variation

k

k = 1

Substituting

Q=1200

,

A=4

,

\Delta T=15

, and

d=0.05

gives

1200 = \frac{60k}{0.05} = 1200k

, so

k = 1

.

3

Compute the target heat transfer rate

Q

1500\text{ W}

Substituting

k=1

,

A=6

,

\Delta T=20

, and

d=0.08

gives

Q = \frac{1 \times 6 \times 20}{0.08} = \frac{120}{0.08} = 1500\text{ W}

.

Key Concept

Joint and Inverse Variation

Question 18Question →

The maximum safe load $L$ supported by a horizontal wooden beam varies directly as its width $w$ and the square of its depth $d$ , and inversely as its length $l$ . A beam of width $6\text{ cm}$ , depth $10\text{ cm}$ , and length $4\text{ m}$ can support a maximum safe load of $900\text{ kg}$ . What is the maximum safe load that can be supported by a beam of the same material having a width of $4\text{ cm}$ , depth $12\text{ cm}$ , and length $6\text{ m}$ ?

576\text{ kg}

480\text{ kg}

1,296\text{ kg}

864\text{ kg}

Show answer & explanation

Answer: $576\text{ kg}$

Answer

576\text{ kg}

The relationship is modeled by

L = \frac{k w d^2}{l}

. Using the initial parameters (

w=6

,

d=10

,

l=4

,

L=900

), we find

k = 6

. Substituting

w=4

,

d=12

, and

l=6

into the equation gives

L = \frac{6 \times 4 \times 144}{6} = 576\text{ kg}

.

Step-by-Step Solution

1

Formulate the variation equation

L = \frac{k \cdot w \cdot d^2}{l}

Direct variation means multiplying factors in the numerator, while inverse variation places the variable in the denominator.

2

Calculate the constant of variation

k

using initial conditions

900 = \frac{k \cdot 6 \cdot 10^2}{4} \implies 900 = \frac{600 k}{4} = 150 k \implies k = 6

Substitute

L = 900

,

w = 6

,

d = 10

, and

l = 4

to solve for

k

.

3

Calculate the new load

L

for the new dimensions

L = \frac{6 \cdot 4 \cdot 12^2}{6} = \frac{6 \cdot 4 \cdot 144}{6} = 576\text{ kg}

Substitute

k = 6

,

w = 4

,

d = 12

, and

l = 6

into the variation formula.

Key Concept

Joint and Inverse Variation

Estimated Time:1m 30s

Question 19Question →

If $y$ varies inversely as $x$ , and $y = 8$ when $x = 3$ , what is the value of $y$ when $x = 6$ ?

4

16

24

2

Show answer & explanation

Answer: 4

Answer

4

For inverse variation,

y = \frac{k}{x}

. Given

y = 8

when

x = 3

, the constant of variation

k = 8 \times 3 = 24

. Substituting

x = 6

gives

y = \frac{24}{6} = 4

.

Step-by-Step Solution

1

Set up the equation for inverse variation

y = \frac{k}{x}

Inverse variation means

y

is inversely proportional to

x

with constant

k

.

2

Solve for the variation constant

k

using

y = 8

and

x = 3

k = 8 \times 3 = 24

Multiply both sides of the equation by

x

.

3

Calculate

y

when

x = 6

y = \frac{24}{6} = 4

Substitute the constant

k = 24

and

x = 6

into the inverse variation equation.

Key Concept

Inverse Variation

Question 20Question →

The electric power $P$ dissipated in a resistor varies directly as the square of the current $I$ flowing through it. If a current of $3\text{ A}$ produces a power of $45\text{ W}$ , what is the power dissipated, in watts, when the current is $5\text{ A}$ ?

Show answer & explanation

Answer: 125

Answer

The power dissipated when the current is 5 A is 125 W.

Because electric power varies directly as the square of the current, the formula is

P = k I^2

. Substituting the initial conditions gives

45 = k(3^2) = 9k

, so

k = 5

. Evaluating at

I = 5\text{ A}

gives

P = 5(5^2) = 125\text{ W}

.

Step-by-Step Solution

1

Set up the variation equation

P = k I^2

Power varies directly as the square of current.

2

Find the constant of variation

k

k = 5

Substitute

P = 45

and

I = 3

into the variation equation:

45 = k(3^2) \implies 45 = 9k \implies k = 5

.

3

Calculate the required power for

I = 5\text{ A}

P = 125\text{ W}

Substitute

k = 5

and

I = 5

into

P = k I^2

:

P = 5(5^2) = 125

.

Key Concept

Direct variation involving a squared quantity

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