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A private company entrusts work to 20 men working 12 hours a day. This group can complete the work in 24 Days. The company now wants to entrust twice the work to 60 men working 4 hours a day. Assume that 2 men of the first group do as much work in one hour as 3 men of the second group in 1 and 1/2 hours. How much number of days will the second group of men take to complete this work?

A private company entrusts work to 20 men working 12 hours a day. This group can complete the work in 24 Days. The company now wants to entrust twice the work to 60 men working 4 hours a day. Assume that 2 men of the first group do as much work in one hour as 3 men of the second group in 1 and 1/2 hours. How much number of days will the second group of men take to complete this work? Correct Answer 108

Let efficiency of men in group 1 be E1 and that of second group be E2.

Ratio of efficiency of men in 1 group to that of 2nd group can be found by using the formula,

E1/E2 = Time taken by men in 2nd group to do certain amount of work/Time taken by men in 1 group to do the same amount of work as that of men in 2nd group

⇒ (3 × 1.5) : (2 × 1)

⇒ 4.5 : 2

Now, M1 × D1 × T1 × E1 × W2 = M2 × D2 × T2 × E2 × W1

Where,M1 and M2 = number of men in group 1st and 2nd respectively

D1 and D2 = number of days required by the Group 1 and group 2 respectively to complete work

T1 and T2 = working hours per day by group 1 and group 2.

W1 and W2 = amount of work by group 1 and group2.

Since we have to calculate the time taken by group 2 to complete twice the amount of work as that of group 1

W2 = 2 × W1

We had earlier calculated E1/E2 = 4.5/2

Also from the question we can infer that

M1 = 20, M2 = 60

T1 = 12, T2 = 4

D1 = 24 and D2 is what we need to find

Substituting all the values in eq 1, we can find D2 as follows,

∴ D2 = (20 × 24 × 12 × 4.5 × 2) / (60 × 4 × 2 × 1) = 108 days

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