Question : A can do 20% of a task in 4 days and B can do $33 \frac{1}{3}$% of the same task in 10 days. They worked together for 9 days and then, C completed the remaining work in 6 days. B and C together will complete 75% of the same work in:
Option 1: 12 days
Option 2: 9 days
Option 3: 11 days
Option 4: 10 days
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Correct Answer: 10 days
Solution :
Time A needs to do 100% of the task alone = 4 × 5 = 20 days
Time B needs to do 100% of the task alone = 10 × 3 = 30 days
Let the total work be LCM of (20, 30) = 60 units
Total work = Time × Efficiency
A's efficiency = $\frac{60}{20}$ = 3 units per day
B's efficiency = $\frac{60}{30}$ = 2 units per day
Work completed by A and B together in 9 days is (3 + 2) × 9 = 45 units
Work still to be done is (60 – 45) = 15 units
The remaining work can be finished by C in 6 days.
C's efficiency = $\frac{15}{6}$ = 2.5 units
B and C's combined efficiency = (2 + 2.5) = 4.5 units
Now, 75% of the total work is 75% of 60 = 45 units
Required time for B and C working together = $\frac{45}{4.5}$ = 10 days
Hence, the correct answer is 10 days.
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