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Zorluk: Çok zorFractions and Rational Numbers

An industrial chemical reservoir is emptied by three pumps, P1P_1, P2P_2, and P3P_3, operating independently at different constant rates. Working alone, P1P_1 can empty the full reservoir in 66 hours, P2P_2 can empty it in 88 hours, and P3P_3 can empty it in 1212 hours. Initially, the reservoir is completely full. First, P1P_1 and P2P_2 work together for 22 hours. Then P1P_1 is turned off, and P3P_3 is turned on to work alongside P2P_2. Additionally, while P2P_2 and P3P_3 are working together, a defect causes liquid to leak out of the bottom of the reservoir at a constant rate equal to 16\frac{1}{6} of the combined emptying rate of P2P_2 and P3P_3. How many additional hours will it take to completely empty the remaining liquid from the reservoir?

  1. 127\frac{12}{7} hoursCevap
  2. B
    22 hours
  3. C
    109\frac{10}{9} hours
  4. D
    125\frac{12}{5} hours
  5. E
    14449\frac{144}{49} hours

Cevap

127\frac{12}{7} hours
The option stating '127\frac{12}{7} hours' is correct because during the first 2 hours, P1P_1 and P2P_2 empty 712\frac{7}{12} of the reservoir, leaving 512\frac{5}{12} remaining. Then, P2P_2 and P3P_3 have a combined rate of 524\frac{5}{24}, and the defect adds an extra 5144\frac{5}{144} per hour, giving a total emptying rate of 35144\frac{35}{144} reservoirs per hour. Dividing 512\frac{5}{12} by 35144\frac{35}{144} yields 127\frac{12}{7} hours.

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1
Calculate the combined emptying rate of P1P_1 and P2P_2 and determine the fraction emptied in the first 22 hours.
Rate of P1=16P_1 = \frac{1}{6} reservoir/hr, Rate of P2=18P_2 = \frac{1}{8} reservoir/hr. Combined rate = \frac{1}{6} + \frac{1}{8} = \frac{7}{24} reservoir/hr. In 22 hours, amount emptied = 2×724=7122 \times \frac{7}{24} = \frac{7}{12} of the reservoir.
Determines how much work was completed during the initial stage.
2
Find the remaining fraction of liquid in the reservoir.
Remaining fraction = 1712=5121 - \frac{7}{12} = \frac{5}{12} of the reservoir.
Establishes the remaining volume that must be emptied in the second stage.
3
Calculate the combined rate of P2P_2, P3P_3, and the defect rate.
Combined rate of P2P_2 and P3=18+112=524P_3 = \frac{1}{8} + \frac{1}{12} = \frac{5}{24} reservoir/hr. Defect rate = \frac{1}{6} \times \frac{5}{24} = \frac{5}{144} reservoir/hr. Total emptying rate = \frac{5}{24} + \frac{5}{144} = \frac{30 + 5}{144} = \frac{35}{144} reservoir/hr.
Combines all simultaneous emptying processes during the second stage.
4
Divide the remaining fraction by the total emptying rate to find the additional time needed.
Time = \frac{5/12}{35/144} = \frac{5}{12} \times \frac{144}{35} = \frac{5}{35} \times \frac{144}{12} = \frac{1}{7} \times 12 = \frac{12}{7} hours.
Applies the rate formula Time=WorkRate\text{Time} = \frac{\text{Work}}{\text{Rate}}.

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Compound Fraction Work Rates and Rational Operations
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