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A, B, and C can do a piece of work in 24 days, 30 days, and 40 days respectively. They start to work together and after four days, 50% of the completed work is destroyed by the air thereafter, they work for 6 days and after that, 16.66% of the total completed work is destroyed by rain. Thereafter, B left the work and A, C worked with twice their efficiency, after how many days remaining work was completed by A and C.

A, B, and C can do a piece of work in 24 days, 30 days, and 40 days respectively. They start to work together and after four days, 50% of the completed work is destroyed by the air thereafter, they work for 6 days and after that, 16.66% of the total completed work is destroyed by rain. Thereafter, B left the work and A, C worked with twice their efficiency, after how many days remaining work was completed by A and C. Correct Answer 5 / 2 days

Given:

A can do a piece of work = 24 days

B can do a piece of work = 30 days

C can do a piece of work = 40 days

Destroy work by air = 50%

Destroy work by rain = 16.33%

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

∴ Efficiency of A = total work / total days = x / 24

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

∴ Efficiency of C = total work / total days = x / 40

∵ They starts the work together

∴ The work done them in one day = (x / 24) + (x / 30) + (x / 40) = 12x / 120 = x / 10

∴ A, B, and C work for four days together so work completed by them = 4 × (x / 10) = 2x / 5

∵ After four days, work destroy by air = 50%

∴ The destroy work = (2x / 5) × (50 / 100) = x / 5

∴ Undestroyed work after air = (2x / 5) – (x / 5) = x / 5

∴ A, B, and C work for six days together so work completed by them = 6 × (x / 10) = 3x / 5

∴ Total completed work = (x / 5) + (3x / 5) = 4x / 5

∵ After six days, work destroy by air = 16.33% of completed work

∴ The destroy work = (4x / 5) × (16.66 / 100) = (4x / 5) × 1 / 6 = 2x / 15

∴ Undestroyed work after rain = (4x / 5) – (2x / 15) = 10x / 15 = 2x / 3

∴ Remaining work = x – (2x / 3) = x / 3

∵ For remaining work B left and A and C work together with twice their efficiency

∴ Work done by A and C in one days = (2x / 24) + (2x / 40) = 2x / 15

Hence required number of days for completing the work = Remaining work / total efficiency

⇒ (x / 3) ÷ (2x / 15) = 5 / 2 days

Alternate method:

Total work = LCM = 120

∴ Efficiency of A = total work / total days = 120 / 24 = 5

∴ Efficiency of B = total work / total days = 120 / 30 = 4

∴ Efficiency of C = total work / total days = 120 / 40 = 3

∵ They starts the work together

∴ The work done them in one day = 5 + 4 + 3 =12

∴ A, B, and C work for four days together so work completed by them = 4 × 12 = 48

∵ After four days, work destroy by air = 50%

∴ The destroy work = 48 × (50 / 100) = 24

∴ Undestroyed work after air = 48 – 24 = 24

∴ A, B, and C work for six days together so work completed by them = 6 × 12 = 72

∴ Total completed work = 72 + 24 =96

∵ After six days, work destroy by air = 16.33% of completed work

∴ The destroy work = 96 × (16.66 / 100) = 16

∴ Undestroyed work after rain = 96 – 16 = 80

∴ Remaining work = 120 – 80 = 40

∵ For remaining work B left and A and C work together with twice their efficiency

∴ Work done by A and C in one days = 10 + 6 = 16

Hence required number of days for completing the work = Remaining work / total efficiency

⇒ 40 / 16 = 5 / 2 days

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