Math, asked by sharmashubham8704, 11 months ago

A, b and c were employed to complete a work. Initially all of them worked together, then a left the work and the remaining work is completed by b and c together. 70% of the work is completed by b and c working together and total work is completed in 13 days. Also, a worked with b and c for the initial 3 days and then a left the work. If c is 1/3rd more efficient than b then find the time taken to complete the work when, only a and b work together.

Answers

Answered by sushant2586
1

Answer:

5 days

Step-by-step explanation:

Let A alone take 'a' days to complete the work.

∴ Speed of A to complete the work in on day = 1/a work per day

Similarly, speed of B and C respectively be 1/b and 1/c work per day.

As per given condition,

A, B, C worked togather for 1st 3 days.

Rest of the days, only B and C worked and completed 70% of the work.

∴ in 1st 3 days, 30% of the work completed.

∴ 3 × (1/a + 1/b + 1/c) = 0.3 ....(1)

Total work needed 13 days to complete.

∴ B and C worked for 10 days to complete 70% of the work.

∴ 10 (1/b + 1/c) = 0.7  ....(2)

Now, C is 1/3rd efficient than B,

1/c = 4/3 × 1/b ... (3)

Substituting in eq 2,

10(1/b + 4/3b) = 0.7

∴ 10(3/3b + 4/3b) = 0.7

∴ 10(7 / 3b) = 0.7

∴ 70 / 3b = 0.7

∴ 70 = 2.1b

∴ b = 100/3  

∴ 1/b = 3/100  ...(4)

substituting values from 3 and 4 in eqn 1

1/a + 1/b + 1/c = 0.1

∴ 1/a + 3/100 + 4/3b = 0.1

∴ 1/a + 3/100 + 1/25 = 0.1

∴ 1/a = 10/100 + 3/100 + 4/100

∴ 1/a = 17/100  .... (5)

∴ from eq 4 and 5, when A and B works togather, work completed in one day

Work in one day = 1/a + 1/b

Work in one day = 17/100 + 3/100

Work in one day W = 20/100 = 1/5 work per day

∴ Time to Compler work = 1 ÷ W = 5 days

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