Four person P, Q, R, S were engaged for doing a task, with the condition that P; Q; R; S work respectively,on (Mondays,
Thursdays); (Tuesdays, Fridays); (Wednesdays, Saturdays); (Sundays). The task was began on a Monday,and got completed on
the 15th day, which was also a Monday.If the efficiencies of P, Q, R, S in respect of doing this task were in the proportion 1 : 2 :
3 : 4,then in how many days could R completed the task,working alone without break?
This questions was previously asked in
SSC MTS 12th October 2021 Shift-2
Explanation:
- Day Sequence: The task starts on Monday and finishes on the 15th day, which is a Monday. Thus, two full weeks plus one day are included.
- Workdays: Each person’s working days are scheduled as follows:
- P: Mondays and Thursdays (4 days)
- Q: Tuesdays and Fridays (4 days)
- R: Wednesdays and Saturdays (4 days)
- S: Only on Sundays (2 days)
- Efficiency Proportion: P, Q, R, S work in ratio 1:2:3:4. So, if P completes 1 unit of work per day, R completes 3 units per day.
- Calculating Total Work: Since the task takes 15 days to complete, we calculate the amount of work done by each:
- Total work units done by P in 4 days = 4 units
- Total work units done by Q in 4 days = 8 units
- Total work units done by R in 4 days = 12 units
- Total work units done by S in 2 days = 8 units
- Overall work = 4 + 8 + 12 + 8 = 32 units
- R’s Work: If R completes 3 units per day, to finish 32 units alone, it would take \( \frac{32}{3} \approx 10.67 \) or effectively around 11 days.
- Correct Answer: Option:3, 11 days
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By: santosh