Question
Download Solution PDFA can do 20% of a work in 4 days, and B can do 33\(\frac{1}{3}\)% of the same work 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:
Answer (Detailed Solution Below)
Detailed Solution
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A does total work in 4 × 5 = 20 days
B does total work in 10 × 3 = 30 days
As per question,
| A + B | C | |
| Days of work done | 9 days | 6 days |
| Work done | 9 × 5 = 45 | 60 - 45 =15 |
| Efficiency | 5 | 15/6 = 2.5 |
Efficiency of B + C = 2 + 2.5 = 4.5
75% of work = 60 × (3/4) = 45
Time taken by B and C to complete the 75% work = 45/4.5 = 10 days.
∴ The correct answer is 10 days.
Given:
A can do 20% of a work in 4 days, and B can do 33\(\frac{1}{3}\)% of the same work in 10 days.
They worked together for 9 days and then C completed the remaining work in 6 days.
Concept used:
Efficiency = (Total work / Total time taken)
Efficiency = work done in a single day
Calculation:
A does total work in 4 × 5 = 20 days
B does total work in 10 × 3 = 30 days
Let total work is 60 units ( Lcm of 20 and 30)
The efficiency of A is
⇒ 60 / 20 = 3 unit
The efficiency of B is
⇒ 60 / 30 = 2 unit
They does together in 9 days
9 × (3+2) = 45 units
The remaining 60 - 45 = 15 unit done by C in 6 days
Total work done by C in 6 × 4 = 24 days
Efficiency of C is 60 /24 = 2.5 unit
75% of the total work is 60 × 3/4 = 45 units
Combined efficiency of B and C = 4.5
The number of days required is 45 / 4.5 = 10 days
∴ The correct option is 4
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