Time and Work Question 4 - CAT 2025
Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is:
Step-by-Step Explanation
- Set up relative efficiency variables:
Let Chandan's daily work output be x units/day.
Bipin is twice as efficient as Chandan = 2x units/day.
Ankita is twice as efficient as Bipin = 4x units/day.
- Track the two distinct phases of the job:
Phase 1 (First 20 days): All three work together.
Combined Daily Efficiency = 4x + 2x + x = 7x units/day
Work completed in Phase 1 = 20 days Ă— 7x units/day = 140x units
Phase 2 (Remaining 40 days): Bipin leaves; Ankita and Chandan continue.
Remaining Daily Efficiency = 4x + x = 5x units/day
Work completed in Phase 2 = 40 days Ă— 5x units/day = 200x units
- Calculate the total work and Chandan's standalone timeframe:
Total Work = 140x + 200x = 340x units
Chandan's standalone time = Total Work/Chandan's Efficiency
= 340x/x
= 340 days
Correct Answer: 340
đź’ˇ FundaMakers Tip:
Convert fractional efficiency profiles into integer work allocations early on. Setting the baseline value of the slowest worker to 1 unit per day lets you treat the total workload as a simple sum of daily units, avoiding variable equations entirely.