A stadium was to be built in 1500 days. The contractor employed 200 men, 300 women and 750 robotic machines. After 600 days, 75% of the work was still to be done. Fearing delay, the contractor removed all women and 500 robotic machines. Also, he employed some more men having the same efficiency as earlier employed men. This led to a speedup in work and the stadium got built 50 days in advance. Find the additional number of men employed if in one day, six men, ten women and fifteen robotic machines have same work output.

**(A)** 1100

**(B)** 1340

**(C)** 1300

**(D)** 1140

**Answer:** **(D)** **Explanation:** Let the total work be 4 units.

=> Work done in first 600 days = 25% of 4 = 1 unit

=> Work done in next 850 days = 75% of 4 = 3 unit

Also, we are given that the daily work output of 6 men, 10 women and 15 robotic machines are same.

=> 6 Em = 10 Ew = 15 Er

=> Em : Ew : Er = 5 : 3 : 2, where ‘Em’ is the efficiency of 1 man, ‘Ew’ is the efficiency of 1 woman and ‘Er’ is the efficiency of 1 robotic machine.

Therefore, ratio of efficiency of man, woman and robotic machine = 5:3:2.

If ‘k’ is the constant of proportionality, Em = 5k, Ew = 3k and Er = 2k.

Here, we need to apply the formula

**∑(M _{i} E_{i}) D_{1} H_{1} / W_{1} = ∑**, where

(M_{j} E_{j}) D_{2} H_{2} / W_{2}

∑(M

_{i}E

_{i}) = (200 x 5k) + (300 x 3k) + (750 x 2k)

∑(M

_{j}E

_{j}) = (200 x 5k) + (m x 5k) + (250 x 2k), where ‘m’ is the additional men employed

D

_{1}= 600 days

D

_{2}= 850 days

H

_{1}= H

_{2}= Daily working hours

W

_{1}= 1 unit

W

_{2}= 3 units

So, we have

3400k x 600 / 1 = (1500 + 5m)k x 850 / 3

=> 3400k x 1800 = (1500 + 5m)k x 850

=> 1500 + 5m = 7200

=> 5m = 5700

=> m = 1140

Therefore, additional men employed = 1140

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