Please explain this question how Delta T = Delta T(Urea) + Delta T(Glucose)

Solution -
As we know,
urea = 5g, molecular weight = 60g
glucose= 10g , molecular weight= 180g
So, weight of solvent= 100- 15 = 85g
m₁ and m₂ are for urea and glucose respectively 
∆ T_f = K_f(m₁ +m₂) 
= 1.86( $\frac{5\times 1000}{60\times 85}$ + $\frac{10\times 1000}{180\times 85}$)
= 1.86(0.98+ 0.653)
= 3.03
so, freezing point = 0 - 3.03 = -3.03°C