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
∑(Mi Ei) D1 H1 / W1 = ∑
(Mj Ej) D2 H2 / W2, where
∑(Mi Ei) = (200 x 5k) + (300 x 3k) + (750 x 2k)
∑(Mj Ej) = (200 x 5k) + (m x 5k) + (250 x 2k), where ‘m’ is the additional men employed
D1 = 600 days
D2 = 850 days
H1 = H2 = Daily working hours
W1 = 1 unit
W2 = 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