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Question

Difficulty: HardSystems of Linear Equations

Two water pumps, Pump A and Pump B, working simultaneously at their respective constant rates, can fill an empty storage tank in 44 hours. If Pump A operates alone at its constant rate for 22 hours and then Pump B operates alone at its constant rate for 77 hours, the tank is also filled completely. How many hours would it take Pump A, working alone at its constant rate, to fill the entire storage tank?

  1. A
    103\frac{10}{3} hours
  2. 203\frac{20}{3} hoursAnswer
  3. C
    1010 hours
  4. D
    1212 hours
  5. E
    1515 hours

Answer

203\frac{20}{3} hours (or 6236\frac{2}{3} hours)
The correct answer is derived by setting up two linear equations representing the total work accomplished: 4rA+4rB=14r_A + 4r_B = 1 and 2rA+7rB=12r_A + 7r_B = 1. Solving this system gives Pump A's rate rA=320r_A = \frac{3}{20} tanks per hour. The time required for Pump A working alone is the reciprocal of its rate, which equals 203\frac{20}{3} hours.

Step-by-Step Solution

1
Define variables and formulate the system of linear equations
Let rAr_A be the rate of Pump A (tanks/hour) and rBr_B be the rate of Pump B (tanks/hour).
Combined work equation: 4(rA+rB)=1  ⟹  4rA+4rB=14(r_A + r_B) = 1 \implies 4r_A + 4r_B = 1
Sequential work equation: 2rA+7rB=12r_A + 7r_B = 1
Work done equals rate multiplied by time, and completing one full tank corresponds to total work =1= 1.
2
Express rBr_B in terms of rAr_A using the first equation
rA+rB=14  ⟹  rB=14−rAr_A + r_B = \frac{1}{4} \implies r_B = \frac{1}{4} - r_A
Simplifying the combined rate equation allows substitution into the second linear equation.
3
Substitute rBr_B into the second equation and solve for rAr_A
2rA+7(14−rA)=1  ⟹  2rA+74−7rA=1  ⟹  −5rA=1−74=−34  ⟹  rA=3202r_A + 7\left(\frac{1}{4} - r_A\right) = 1 \implies 2r_A + \frac{7}{4} - 7r_A = 1 \implies -5r_A = 1 - \frac{7}{4} = -\frac{3}{4} \implies r_A = \frac{3}{20}
Eliminating rBr_B isolates rAr_A as a single-variable linear equation.
4
Calculate the time required for Pump A to fill the tank alone
TimeA=1rA=1320=203 hours\text{Time}_A = \frac{1}{r_A} = \frac{1}{\frac{3}{20}} = \frac{20}{3}\text{ hours}
The total time to complete 1 unit of work is the reciprocal of the unit work rate.

Key Concept

Formulating and solving systems of two linear equations in two variables derived from work-rate relationships.
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Topics

Systems of Linear Equations

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