Direct, Inverse, Joint and Partial Variation Practice Questions — JAMB UTME — Page 2

Question 21Question →

The operating cost $C$ (in Naira per hour) of a speed boat is partly constant and partly varies directly as the square of its speed $v$ (in km/h). At a speed of $20\text{ km/h}$ , the operating cost is $₦18,000$ per hour, and at a speed of $30\text{ km/h}$ , the operating cost is $₦33,000$ per hour. What is the operating cost per hour (in Naira) when the boat travels at a speed of $40\text{ km/h}$ ?

Show answer & explanation

Answer: 54000

Answer

The operating cost per hour when the speed boat travels at

40\text{ km/h}

is

54,000

Naira.

By writing the partial variation relationship as

C = k_1 + k_2 v^2

and substituting the given pairs

(20, 18000)

and

(30, 33000)

, we set up simultaneous equations

k_1 + 400k_2 = 18000

and

k_1 + 900k_2 = 33000

. Solving these yields

k_2 = 30

and

k_1 = 6000

. Evaluating the model at

v = 40\text{ km/h}

gives

C = 6000 + 30(40^2) = 54,000

Naira per hour.

Step-by-Step Solution

1

Set up the general formula for partial variation

C = k_1 + k_2 v^2

The total cost consists of a fixed constant component

k_1

and a variable component

k_2 v^2

that varies directly with speed squared.

2

Form simultaneous linear equations from the given conditions

18,000 = k_1 + 400 k_2

and

33,000 = k_1 + 900 k_2

Substituting

v = 20, C = 18,000

and

v = 30, C = 33,000

creates a solvable system of equations in

k_1

and

k_2

.

3

Solve for the variation constants

k_1

and

k_2

k_2 = 30

and

k_1 = 6,000

Subtracting the first equation from the second eliminates

k_1

, giving

500 k_2 = 15,000 \implies k_2 = 30

. Substituting

k_2 = 30

into the first equation yields

k_1 = 6,000

.

4

Calculate the operating cost at

v = 40\text{ km/h}

C = 54,000

Evaluating

C = 6,000 + 30(40)^2 = 6,000 + 30(1,600) = 6,000 + 48,000 = 54,000

.

Key Concept

Partial Variation with Simultaneous Equations

Question 22Question →

A quantity $Q$ is partly constant and partly varies inversely as $p$ . Given that $Q = 11$ when $p = 2$ , and $Q = 5$ when $p = 5$ , what is the value of $Q$ when $p = 10$ ?

2

3

4

7

Show answer & explanation

Answer: $3$

Answer

The value of

Q

when

p = 10

is

3

.

The relationship for partial inverse variation is

Q = k_1 + \frac{k_2}{p}

. Substituting

(p=2, Q=11)

gives

2k_1 + k_2 = 22

, and substituting

(p=5, Q=5)

gives

5k_1 + k_2 = 25

. Solving these simultaneous equations gives

k_1 = 1

and

k_2 = 20

. Substituting

p = 10

into

Q = 1 + \frac{20}{p}

results in

Q = 1 + 2 = 3

.

Step-by-Step Solution

1

Set up the general formula for partial variation.

Q = k_1 + \frac{k_2}{p}

, where

k_1

and

k_2

are constants.

Partial variation consists of a constant part and a part that varies inversely with

p

.

2

Substitute the given values to form two simultaneous linear equations.

For

p = 2, Q = 11 \implies 11 = k_1 + \frac{k_2}{2} \implies 2k_1 + k_2 = 22

For

p = 5, Q = 5 \implies 5 = k_1 + \frac{k_2}{5} \implies 5k_1 + k_2 = 25

Substituting known data points provides equations to solve for the variation constants.

3

Solve the simultaneous equations for

k_1

and

k_2

.

Subtracting the first equation from the second gives

3k_1 = 3 \implies k_1 = 1

.
Substituting

k_1 = 1

into

2(1) + k_2 = 22

yields

k_2 = 20

.

Determining

k_1

and

k_2

establishes the explicit relationship between

Q

and

p

.

4

Calculate

Q

when

p = 10

.

Q = 1 + \frac{20}{10} = 1 + 2 = 3

.

Substitute

p = 10

into the established variation equation

Q = 1 + \frac{20}{p}

.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:1m 30s

Question 23Question →

The pressure $P$ of a given mass of gas varies directly as its absolute temperature $T$ and inversely as its volume $V$ . Given that $P = 50\text{ kPa}$ when $T = 300\text{ K}$ and $V = 10\text{ m}^3$ , what is the value of $P$ when $T = 360\text{ K}$ and $V = 8\text{ m}^3$ ?

48\text{ kPa}

60\text{ kPa}

75\text{ kPa}

90\text{ kPa}

Show answer & explanation

Answer: $75\text{ kPa}$

Answer

75\text{ kPa}

The relationship is governed by

P = \frac{kT}{V}

. Substituting

P = 50

,

T = 300

, and

V = 10

yields

k = \frac{5}{3}

. Using

k = \frac{5}{3}

with

T = 360

and

V = 8

gives

P = \frac{(5/3) \times 360}{8} = 75\text{ kPa}

.

Step-by-Step Solution

1

Set up the general formula for joint and inverse variation.

P = \frac{kT}{V}

, where

k

is the constant of variation.

Pressure varies directly as temperature

T

and inversely as volume

V

.

2

Substitute initial conditions to determine

k

.

50 = \frac{k \times 300}{10} \implies 50 = 30k \implies k = \frac{5}{3}

.

The initial values

P = 50\text{ kPa}

,

T = 300\text{ K}

, and

V = 10\text{ m}^3

allow solving for

k

.

3

Calculate the new pressure with updated temperature and volume values.

P = \frac{\frac{5}{3} \times 360}{8} = \frac{600}{8} = 75\text{ kPa}

.

Substitute

k = \frac{5}{3}

,

T = 360\text{ K}

, and

V = 8\text{ m}^3

into the variation formula.

Key Concept

Joint and inverse variation in algebraic relationships

Estimated Time:1m 30s

Question 24Question →

If $P$ varies directly as the square root of $q$ and inversely as the square of $r$ , what is the percentage change in $P$ when $q$ is increased by $44\%$ and $r$ is decreased by $20\%$ ?

50\%

increase

125\%

increase

87.5\%

increase

23.2\%

decrease

Show answer & explanation

Answer: $87.5\%$ increase

Answer

An increase of

87.5\%

The variation relation is

P = k \frac{\sqrt{q}}{r^2}

. Increasing

q

by

44\%

scales

\sqrt{q}

by

\sqrt{1.44} = 1.2

. Decreasing

r

by

20\%

scales

r^2

by

(0.8)^2 = 0.64

. Dividing

1.2

by

0.64

gives

1.875

, meaning the new value

P'

is

187.5\%

of the original

P

, which corresponds to an increase of

87.5\%

.

Step-by-Step Solution

1

Set up the joint variation equation

P = k \frac{\sqrt{q}}{r^2}

, where

k

is a constant of variation.

Direct variation puts

\sqrt{q}

in the numerator, and inverse variation puts

r^2

in the denominator.

2

Express new values

q'

and

r'

in terms of original variables

q' = 1.44q

and

r' = 0.80r

.

An increase of

44\%

gives

1 + 0.44 = 1.44

, and a decrease of

20\%

gives

1 - 0.20 = 0.80

.

3

Substitute new variables into the variation formula to find the new value

P'

P' = k \frac{\sqrt{1.44q}}{(0.80r)^2} = k \frac{1.2\sqrt{q}}{0.64r^2} = \frac{1.2}{0.64} \left(k \frac{\sqrt{q}}{r^2}\right) = 1.875P

.

Evaluating

\sqrt{1.44} = 1.2

and

(0.80)^2 = 0.64

gives the multiplier for

P

.

4

Calculate the percentage change in

P

\text{Percentage Change} = (1.875 - 1) \times 100\% = 87.5\%

increase.

Subtracting

1

converts the multiplier to a relative increase.

Key Concept

Joint Variation with Percentage Changes

Estimated Time:1m 30s

Direct, Inverse, Joint and Partial Variation

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes