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Zorluk: ZorAlgebraic Word Problems and Modeling

A water purification facility uses a primary filtration system and a secondary filtration system to process untreated water. The primary system operates at a constant rate that is 25%25\% faster than the secondary system. Working together at their normal constant rates, both systems can process a full reservoir of 36,00036,000 gallons in 88 hours.

On a day when the primary system operates at only 80%80\% of its normal rate due to maintenance while the secondary system operates at its normal rate, both systems work together for 66 hours. At that point, the primary system is shut down completely. How many additional hours will it take the secondary system, working alone at its normal rate, to process the remainder of the reservoir?

  1. A
    4.54.5 hours
  2. B
    5.05.0 hours
  3. 6.06.0 hoursCevap
  4. D
    7.27.2 hours
  5. E
    8.08.0 hours

Cevap

6.06.0 hours
The correct answer of 6.06.0 hours is derived by establishing that the combined normal processing rate is 4,5004,500 gal/hr. Since the primary system is 25%25\% faster than the secondary system (P=1.25SP = 1.25S), the secondary rate is 2,0002,000 gal/hr and the primary rate is 2,5002,500 gal/hr. At 80%80\% efficiency, the primary system operates at 2,0002,000 gal/hr, making the joint rate 4,0004,000 gal/hr. In 66 hours, 24,00024,000 gallons are processed, leaving 12,00012,000 gallons. The secondary system working alone at 2,0002,000 gal/hr processes the remaining volume in exactly 6.06.0 hours.

Adım Adım Çözüm

1
Determine the combined normal operating rate and set up individual rates.
Combined rate = 4,5004,500 gal/hr; Secondary rate = 2,0002,000 gal/hr; Primary rate = 2,5002,500 gal/hr.
The combined rate is 36,000 gallons8 hours=4,500 gal/hr\frac{36,000\text{ gallons}}{8\text{ hours}} = 4,500\text{ gal/hr}. Let SS be the secondary rate. The primary rate is 1.25S1.25S. Thus, S+1.25S=2.25S=4,500S + 1.25S = 2.25S = 4,500, yielding S=2,000 gal/hrS = 2,000\text{ gal/hr} and P=2,500 gal/hrP = 2,500\text{ gal/hr}.
2
Calculate the reduced primary rate and the total water processed in the first 6 hours.
24,00024,000 gallons processed in the first 6 hours.
During maintenance, the primary system operates at 80%80\% of 2,500 gal/hr2,500\text{ gal/hr}, which is 0.80×2,500=2,000 gal/hr0.80 \times 2,500 = 2,000\text{ gal/hr}. The combined rate during this period is 2,000+2,000=4,000 gal/hr2,000 + 2,000 = 4,000\text{ gal/hr}. Over 6 hours, the volume processed is 4,000×6=24,000 gallons4,000 \times 6 = 24,000\text{ gallons}.
3
Find the remaining volume of water to process.
12,00012,000 gallons remaining.
Subtract the volume processed from the total reservoir capacity: 36,000−24,000=12,000 gallons36,000 - 24,000 = 12,000\text{ gallons}.
4
Calculate the time required for the secondary system alone to process the remainder.
6.06.0 hours.
Divide the remaining volume by the normal secondary system rate S=2,000 gal/hrS = 2,000\text{ gal/hr}: 12,000 gallons2,000 gal/hr=6.0 hours\frac{12,000\text{ gallons}}{2,000\text{ gal/hr}} = 6.0\text{ hours}.

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Linear rate modeling and percentage rate adjustment in combined work problems
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Algebraic Word Problems and Modeling

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