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Zorluk: Çok zorWork Rate and Combined Work

Three specialized synthesis columns—Alpha, Beta, and Gamma—are used in a pharmaceutical facility to purify batches of a chemical compound. Operating simultaneously at their respective constant rates, Column Alpha and Column Beta can process a standard batch in 88 hours, while Column Beta and Column Gamma working together can process the exact same batch in 1212 hours.

Column Alpha begins processing a standard batch alone. After 44 hours of operation, Column Alpha experiences a mechanical restriction that reduces its processing rate by 3313%33\frac{1}{3}\%. At that exact moment, Column Gamma is brought online to assist Column Alpha. Working together, Column Alpha (at its reduced rate) and Column Gamma complete the remaining portion of the batch in 8.48.4 hours.

How many hours would it take Column Beta, operating alone at its normal constant rate, to process an entire standard batch of the chemical compound?

  1. A
    15 hours
  2. B
    18 hours
  3. 20 hoursCevap
  4. D
    24 hours
  5. E
    30 hours

Cevap

Column Beta would take 20 hours to complete an entire batch working alone.
The correct response of 20 hours is determined by establishing the rates of the three columns. With a+b=1/8a + b = 1/8 and b+c=1/12b + c = 1/12, subtracting these equations gives c=a−1/24c = a - 1/24. During the first 4 hours, Column Alpha completes 4a4a of the batch. In the second stage, Column Alpha works at 2/3a2/3 a while Column Gamma works at cc, giving a joint rate of 5/3a−1/245/3 a - 1/24. Multiplying this combined rate by 8.48.4 hours and setting it equal to the remaining work 1−4a1 - 4a yields a=3/40a = 3/40. Substituting this back into a+b=1/8a + b = 1/8 yields b=1/20b = 1/20, meaning Column Beta requires 20 hours to complete a batch alone.

Adım Adım Çözüm

1
Define individual work rates in batches per hour for Column Alpha (aa), Column Beta (bb), and Column Gamma (cc).
From the given combined rates: a+b=18a + b = \frac{1}{8} and b+c=112b + c = \frac{1}{12}.
Combined work rates equal the sum of individual work rates.
2
Express Column Gamma's rate (cc) in terms of Column Alpha's rate (aa).
Subtracting the second equation from the first gives (a+b)−(b+c)=18−112  ⟹  a−c=124  ⟹  c=a−124(a + b) - (b + c) = \frac{1}{8} - \frac{1}{12} \implies a - c = \frac{1}{24} \implies c = a - \frac{1}{24}.
Isolating one rate variable simplifies the multi-stage work equation.
3
Calculate the work completed during Stage 1 and express the remaining work.
Work completed in Stage 1 (44 hours at rate aa) is 4a4a. The remaining work to be done is 1−4a1 - 4a.
Total work equals 11 batch.
4
Formulate the combined rate for Stage 2 and set up the equation for the remaining work.
Alpha's reduced rate is (1−13)a=23a\left(1 - \frac{1}{3}\right)a = \frac{2}{3}a. The combined rate with Gamma is 23a+c=23a+(a−124)=53a−124\frac{2}{3}a + c = \frac{2}{3}a + \left(a - \frac{1}{24}\right) = \frac{5}{3}a - \frac{1}{24}. Stage 2 takes 8.4=4258.4 = \frac{42}{5} hours, so 425(53a−124)=1−4a\frac{42}{5}\left(\frac{5}{3}a - \frac{1}{24}\right) = 1 - 4a.
Work completed in Stage 2 equals combined rate multiplied by time spent in Stage 2.
5
Solve the algebraic equation for aa, then find bb.
Expanding the equation: 14a−720=1−4a  ⟹  18a=2720  ⟹  a=34014a - \frac{7}{20} = 1 - 4a \implies 18a = \frac{27}{20} \implies a = \frac{3}{40}. Then b=18−a=18−340=240=120b = \frac{1}{8} - a = \frac{1}{8} - \frac{3}{40} = \frac{2}{40} = \frac{1}{20}.
Since Beta's rate is 120\frac{1}{20} batch/hour, the time taken by Beta alone is 11/20=20\frac{1}{1/20} = 20 hours.

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