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A, B, and C can do a piece of work in 20 days, 30 days, and 15 days respectively. They start the work together but after 3 days they leave the work and the work is stopped for the next 15 days. After that D, E and F joined the work and did the work alternately and D started the remaining work. If D’s efficiency is twice A, E’s efficiency is half of B and F is equally efficient as C. So find in how many days D, E, and F complete the remaining work.

A, B, and C can do a piece of work in 20 days, 30 days, and 15 days respectively. They start the work together but after 3 days they leave the work and the work is stopped for the next 15 days. After that D, E and F joined the work and did the work alternately and D started the remaining work. If D’s efficiency is twice A, E’s efficiency is half of B and F is equally efficient as C. So find in how many days D, E, and F complete the remaining work. Correct Answer 9 days

Given:

A can do the work = 20 days

B can do the work = 30 days

C can do the work = 15 days

Efficiency of D = Twice the efficiency of A

Efficiency of E = Half the efficiency of B

Efficiency of F = Equal the efficiency of C

Concept used:

Total work = LCM

Work done in unit value of time is known as efficiency

Formula used:

Efficiency = total work / total days

Calculations:

Let the total work be x

So, Efficiency of A = total work / total days = x / 20

Efficiency of B = total work / total days = x / 30

Efficiency of C = total work / total days = x / 15

∵ They starts the work together

∴ The work done them in one day = (x / 20) + (x / 30) + (x / 15) = 3x / 20

∴ A, B, and C work for three days together so work completed by them = 3 × (3x / 20) = 9x / 20

∴ The remaining work = x – (9x / 20) = 11x / 20

∵ A, B, and C left the work and D, E and F joined at the place of them and work alternatively

∴ Work done by D in a day = 2 × work done by A in a day

⇒ 2 × (x / 20) = x / 10

Next day work done by E = (1 / 2) × work done by B in a day

⇒ (1 / 2) × (x / 30) = x / 60

Next day work done by F = Work done by C in a day

⇒ x / 15

Then the work done by D, E and F alternatively = (x / 10) +(x / 60) + (x / 15) = 11x / 60

Hence they repeat the process for two more time

∴ The work done by D, E and F = (11x / 60) × 3 = 11x / 20

Hence the remaining work is completed by D, E and F = 3 + 3 + 3 = 9 days

Alternate method

A can do the work = 20 days

B can do the work = 30 days

C can do the work = 15 days

Total work = LCM = 60

∴ Efficiency of A = total work / total days = 60 / 20 = 3

∴ Efficiency of B = total work / total days = 60 / 30 = 2

∴ Efficiency of C = total work / total days = 60 / 15 = 4

∵ They starts the work together

∴ The work done them in one day = 3 + 2 + 4 = 9

∴ A, B, and C work for three days so work completed by them = 3 × 9 = 27

∴ The remaining work = 60 – 27 = 33

∵ A, B, and C left the work and D, E and F joined at the place of them and work alternatively

∴ Work done by D in a day = 2 × work done by A in a day

⇒ 2 × 3 = 6

Next day work done by E = (1 / 2) × work done by B in a day

⇒ (1 / 2) × 2 = 1

Next day work done by F = Work done by C in a day

⇒ 4

Then the work done by D, E and F alternatively = 6 + 1 + 4 = 11

Hence they repeat the process for two more time

∴ The work done by D, E and F = 11 × 3 = 33

Hence the remaining work is completed by D, E and F = 3 + 3 + 3 = 9 days

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