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Difficulty: HardAlgebraic Word Problems and Modeling

An agricultural facility uses two automated irrigation systems, System P and System Q, to water a field. System P pumps water at a constant rate of 120120 gallons per hour after requiring an initial setup overhead of 3030 minutes (0.50.5 hours). System Q pumps water at a constant rate of 180180 gallons per hour after requiring an initial setup overhead of 4545 minutes (0.750.75 hours). Both systems begin their setup process at the exact same time and operate continuously until a combined total of 1,6051,605 gallons of water has been pumped. Which of the following statements must be true? Select all such statements.

  1. The total elapsed time from when setup began until the target volume was reached is 66 hours.Answer
  2. System Q pumped 285285 more gallons of water than System P.Answer
  3. C
    System P was actively pumping water for exactly 55 hours and 1515 minutes.
  4. D
    System Q accounted for more than 60%60\% of the total water pumped.
  5. The ratio of the volume of water pumped by System P to that pumped by System Q is 44:6344:63.Answer

Answer

The correct statements are: the total elapsed time from when setup began is 6 hours, System Q pumped 285 more gallons than System P, and the ratio of water pumped by System P to System Q is 44:63.
Solving the combined linear rate model yields a total elapsed time of 6 hours. With this duration, System P operates for 5.5 hours producing 660 gallons, and System Q operates for 5.25 hours producing 945 gallons. This confirms that the total time is 6 hours, System Q produces 285 more gallons than System P (945 - 660 = 285), and the ratio of System P's volume to System Q's volume is 660:945, which simplifies to 44:63.

Step-by-Step Solution

1
Define the variable for total time and establish time expressions for active pumping.
Let tt be the total elapsed time in hours since both systems began setup (t≥0.75t \geq 0.75). System P actively pumps for (t−0.5)(t - 0.5) hours, and System Q actively pumps for (t−0.75)(t - 0.75) hours.
Setup overhead delays the start of pumping, so active pumping duration equals total elapsed time minus setup time.
2
Set up and solve the linear rate equation for combined total volume.
120(t−0.5)+180(t−0.75)=1605  ⟹  120t−60+180t−135=1605  ⟹  300t−195=1605  ⟹  300t=1800  ⟹  t=6120(t - 0.5) + 180(t - 0.75) = 1605 \implies 120t - 60 + 180t - 135 = 1605 \implies 300t - 195 = 1605 \implies 300t = 1800 \implies t = 6 hours.
The sum of the volumes produced by both systems must equal the target total of 1,605 gallons.
3
Calculate individual pumping times and volumes produced by each system.
System P: active time =5.5= 5.5 hours, volume =120×5.5=660= 120 \times 5.5 = 660 gallons. System Q: active time =5.25= 5.25 hours, volume =180×5.25=945= 180 \times 5.25 = 945 gallons.
Individual volumes are required to evaluate statements regarding volume differences, percentages, and ratios.
4
Evaluate each given statement against the calculated values.
Elapsed time is 66 hours (True). Volume difference is 945−660=285945 - 660 = 285 gallons (True). Active pumping time for P is 5.55.5 hours, not 5.255.25 hours (False). Percentage for Q is 945/1605≈58.88%<60%945 / 1605 \approx 58.88\% < 60\% (False). Ratio P to Q is 660:945=44:63660 : 945 = 44 : 63 (True).
Determines which of the statements must be selected.

Key Concept

Linear Modeling with Combined Work Rates and Staggered Start Times
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Topics

Algebraic Word Problems and Modeling

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