The hourly operational cost, C Naira, of an industrial water pump is partly constant and p… — JAMB UTME

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 $h = 3\text{ m}$ ?

Answer: 668 Naira

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

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

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

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

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

Answer

Step-by-Step Solution

Key Concept

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