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3 men, 2 women and 4 children can complete a work in 18 days. A women and 2 children can complete it in 40 days and 3 men and 5 children can complete the work in 36 days, then find in how much time will 10W + 20C to complete the work half work.

3 men, 2 women and 4 children can complete a work in 18 days. A women and 2 children can complete it in 40 days and 3 men and 5 children can complete the work in 36 days, then find in how much time will 10W + 20C to complete the work half work. Correct Answer 2 days

Given:

3M + 2W + 4C = 18 days

W + 2C = 40 days

3M + 5C = 36 days

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 3M + 2W + 4C = x / 18      ...1)

Efficiency of W + 2C = x / 40      ...2)

Efficiency of 3M + 5C = x / 36      ...3)

Multiple equation (2) with 2

W + 2C = x / 40

⇒ 2W + 4C = 2x / 40

⇒ 2W + 4C = x / 20      ...4)

Subtract equation (4) from equation (1)

⇒ 3M + 2W + 4C – (2W + 4C) = (x / 18) – (x / 20)

⇒ 3M = (10x – 9x) / 180

⇒ M = x / 540

Efficiency of one man = x / 540

Put the value of efficiency of one man in equation (3)

3M + 5C = x / 36

⇒ 3(x / 540) + 5C = x / 36

⇒ x / 180 + 5C = x / 36

⇒ 5C = (x / 36) – (x / 180)

⇒ 5C = (5x – x) / 180

⇒ 5C = 4x / 180

⇒ C = x / 225

Efficiency of one child = x / 225

Put the value of efficiency of one child in equation (2)

W + 2C = x / 40

⇒ W + 2(x / 225) = x / 40

⇒ W = (x / 40) – (2x / 225)

⇒ W = (45x – 16x) / 1800

⇒ W = 29x / 1800

Efficiency of one woman = 29x / 1800

Efficiency of 10W + 20C = 10(29x / 1800) + 20(x / 225)

⇒ (29x / 180) + (4x / 45)

⇒ 45x / 180

⇒ x / 4

Half work = x / 2

Time Taken by 10W + 20C = Half work / Efficiency

⇒ (x / 2) / (x / 4)

⇒ 4 / 2

⇒ 2 days

∴ Time taken by 10W + 20C is 2 days

Alternate solution:

Total work = LCM

Total work = 360

Efficiency of 3M + 2W + 4C = 360 / 18 = 20      ...1)

Efficiency of W + 2C = 360 / 40 = 9      ...2)

Efficiency of 3M + 5C = 360 / 36 = 10      ...3)

Multiple equation (2) with 2

W + 2C = 9

⇒ 2W + 4C = 18      ...4)

Subtract equation (4) from equation (1)

⇒ 3M + 2W + 4C – (2W + 4C) = 20 – 18

⇒ 3M = 2

⇒ M = 2 / 3

Efficiency of one man = 2 / 3

Put the value of efficiency of one man in equation (3)

3M + 5C = 10

⇒ 3(2 / 3) + 5C = 10

⇒ 5C = 8

⇒ C = 8 / 5

Efficiency of one child = 8 / 5

Put the value of efficiency of one child in equation (2)

W + 2C = 9

⇒ W + 2(8 / 5) = 9

⇒ W = (45 – 16)5

⇒ W = 29 / 5

Efficiency of one woman = 29 / 5

Efficiency of 10W + 20C = 10(29 / 5) + 20(8 / 5)

⇒ 58 + 32 = 90

Half work = 360 / 2 = 180

Time Taken by 10W + 20C = Half work / Efficiency

⇒ 180 / 90

⇒ 2 days

∴ Time taken by 10W + 20C is 2 days

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