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Difficulty: HardDirect, Inverse, Joint and Partial Variation

The monthly operating cost CC (in naira) of a commercial power generator is partly constant and partly varies directly as the square of its daily operating time hh (in hours). When the generator operates for 33 hours per day, the monthly cost is ₦5,5005,500. When it operates for 55 hours per day, the monthly cost is ₦13,50013,500. What is the daily operating time, in hours, when the monthly operating cost is ₦33,00033,000?

  1. 8 hoursAnswer
  2. B
    7 hours
  3. C
    9 hours
  4. D
    10 hours

Answer

8 hours
The relationship is modeled by partial variation C=k1+k2h2C = k_1 + k_2 h^2. Solving the simultaneous equations derived from h=3h = 3 (C=5500C = 5500) and h=5h = 5 (C=13500C = 13500) yields k2=500k_2 = 500 and k1=1000k_1 = 1000. Substituting C=33000C = 33000 into 33000=1000+500h233000 = 1000 + 500 h^2 gives 500h2=32000500 h^2 = 32000, so h2=64h^2 = 64, which yields h=8h = 8 hours.

Step-by-Step Solution

1
Set up the general formula for partial variation.
C=k1+k2h2C = k_1 + k_2 h^2, where k1k_1 and k2k_2 are constants.
The cost consists of a constant part (k1k_1) and a part that varies directly as the square of daily hours (k2h2k_2 h^2).
2
Form simultaneous linear equations using the given data points.
Equation (1): 5500=k1+9k25500 = k_1 + 9 k_2
Equation (2): 13500=k1+25k213500 = k_1 + 25 k_2
Substitute h=3,C=5500h = 3, C = 5500 and h=5,C=13500h = 5, C = 13500 into the variation equation.
3
Solve for the variation constants k1k_1 and k2k_2.
Subtract Equation (1) from Equation (2): 8000=16k2  ⟹  k2=5008000 = 16 k_2 \implies k_2 = 500.
Substitute k2=500k_2 = 500 into Equation (1): 5500=k1+9(500)  ⟹  k1=10005500 = k_1 + 9(500) \implies k_1 = 1000.
Eliminating k1k_1 gives k2k_2, which is then used to find the constant part k1k_1.
4
Calculate hh when C=33000C = 33000.
33000=1000+500h2  ⟹  32000=500h2  ⟹  h2=64  ⟹  h=833000 = 1000 + 500 h^2 \implies 32000 = 500 h^2 \implies h^2 = 64 \implies h = 8 hours.
Substitute the values of k1,k2,k_1, k_2, and target CC into the relation equation to solve for hh.

Key Concept

Partial Variation with Simultaneous Equations
Estimated Time:2m 30s
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Topics

Direct, Inverse, Joint and Partial Variation

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