Teams A, B, and C consist of five, eight, and ten members, respectively, such that every member within a team is equally productive. Working separately, teams A, B, and C can complete a certain job in 40 hours, 50 hours, and 4 hours, respectively. Two members from team A, three members from team B, and one member from team C together start the job, and the member from team C leaves after 23 hours. The number of additional member(s) from team B, that would be required to replace the member from team C, to finish the job in the next one hour, is
Show Hint
Compute total work done in each phase.
Subtract from 1 (or total work) to find the remaining part.
Use the required time and remaining work to find how many extra workers are needed.
- \(1\)
- \(2\)
- \(3\)
- \(4\)
The Correct Option is B
Approach Solution - 1
Step 1 (per-member rates): Team A (\(5\) members) finishes in \(40\) h, so one A-member \(=\dfrac1{40\times5}=\dfrac1{200}\) job/h. Team B (\(8\) members, \(50\) h): \(\dfrac1{50\times8}=\dfrac1{400}.\) Team C (\(10\) members, \(4\) h): \(\dfrac1{4\times10}=\dfrac1{40}.\)
Step 2 (starting crew): \(2\) A \(+\ 3\) B \(+\ 1\) C give rate \(\dfrac{2}{200}+\dfrac{3}{400}+\dfrac1{40}=\dfrac{4}{400}+\dfrac{3}{400}+\dfrac{10}{400}=\dfrac{17}{400}\) job/h.
Step 3 (after 23 hours): Work done \(=23\times\dfrac{17}{400}=\dfrac{391}{400}.\) Remaining \(=1-\dfrac{391}{400}=\dfrac{9}{400}.\)
Step 4 (C leaves): The C-member exits. Remaining crew \(2\) A \(+\ 3\) B has rate \(\dfrac{4}{400}+\dfrac{3}{400}=\dfrac{7}{400}\) job/h. Add \(n\) more B-members at \(\dfrac1{400}\) each to finish \(\dfrac{9}{400}\) in one hour:
\[\frac{7}{400}+\frac{n}{400}=\frac{9}{400}\Rightarrow 7+n=9\Rightarrow n=2.\]
Final answer: \(2\) extra members from team B.
Approach Solution -2
Step 1: Find individual efficiencies of team members. Let the total job be \(1\) unit. \[ \text{Team A: } 5 \text{ members finish in } 40 \text{ hours} \Rightarrow \text{rate of team A} = \frac{1}{40}. \] So, each member of team A has rate \[ a = \frac{1}{40 \times 5} = \frac{1}{200}. \] \[ \text{Team B: } 8 \text{ members finish in } 50 \text{ hours} \Rightarrow \text{rate of team B} = \frac{1}{50}. \] So, each member of team B has rate \[ b = \frac{1}{50 \times 8} = \frac{1}{400}. \] \[ \text{Team C: } 10 \text{ members finish in } 4 \text{ hours} \Rightarrow \text{rate of team C} = \frac{1}{4}. \] So, each member of team C has rate \[ c = \frac{1}{4 \times 10} = \frac{1}{40}. \]
Step 2: Work done in the first 23 hours. Workers initially: 2 from A, 3 from B, 1 from C. Their combined rate: \[ 2a + 3b + c = 2 \cdot \frac{1}{200} + 3 \cdot \frac{1}{400} + \frac{1}{40} = \frac{1}{100} + \frac{3}{400} + \frac{1}{40}. \] Convert to denominator 400: \[ \frac{1}{100} = \frac{4}{400},\quad \frac{1}{40} = \frac{10}{400}. \] So: \[ 2a + 3b + c = \frac{4}{400} + \frac{3}{400} + \frac{10}{400} = \frac{17}{400}. \] Work done in 23 hours: \[ \text{Work}_1 = 23 \times \frac{17}{400} = \frac{391}{400}. \] Remaining work: \[ \text{Work}_{\text{rem}} = 1 - \frac{391}{400} = \frac{9}{400}. \]
Step 3: Replace team C member with extra members from team B. After 23 hours, the member from C leaves. Remaining workers: 2 from A, 3 from B, plus \(x\) additional members from B. New combined rate: \[ 2a + (3 + x)b = 2 \cdot \frac{1}{200} + (3 + x)\cdot \frac{1}{400} = \frac{1}{100} + \frac{3 + x}{400} = \frac{4}{400} + \frac{3 + x}{400} = \frac{7 + x}{400}. \] We want the remaining work \(\dfrac{9}{400}\) to be finished in the next 1 hour: \[ \frac{7 + x}{400} \times 1 = \frac{9}{400} \quad \Rightarrow \quad 7 + x = 9 \quad \Rightarrow \quad x = 2. \] So, the number of additional members from team B required is: \[ \boxed{2}. \]
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