A mixture of 30 bags of Type A millet costing Ksh 400 per bag and 50 bags of Type B millet costing Ksh 380 per bag is to be sold at a profit of 20%. Find the selling price of one bag of the mixture. A

Cost of Type A = $30 \text{ bags} \times \text{Ksh } 400/\text{bag} = \text{Ksh } 12000$

Cost of Type B = $50 \text{ bags} \times \text{Ksh } 380/\text{bag} = \text{Ksh } 19000$

Total Cost = $\text{Ksh } 12000 + \text{Ksh } 19000 = \text{Ksh } 31000$

Total Bags = $30 \text{ bags} + 50 \text{ bags} = 80 \text{ bags}$

Cost Price per Bag = $\frac{\text{Total Cost}}{\text{Total Bags}} = \frac{\text{Ksh } 31000}{80} = \text{Ksh } 387.50$

Selling Price = Cost Price $\times (1 + \text{Profit Percentage})$
Selling Price = $\text{Ksh } 387.50 \times (1 + 0.20)$
Selling Price = $\text{Ksh } 387.50 \times 1.20$
Selling Price = $\text{Ksh } 465$

The selling price of one bag of the mixture is $\boxed{\text{Ksh } 465}$.

Given Selling Price (SP) = Ksh 498.55 and Profit = 30%.
Let Cost Price (CP) be the cost per bag.
$SP = CP \times (1 + \text{Profit Percentage})$
$\text{Ksh } 498.55 = CP \times (1 + 0.30)$
$\text{Ksh } 498.55 = CP \times 1.30$
$CP = \frac{\text{Ksh } 498.55}{1.30} = \text{Ksh } 383.50$

Let $x$ be the number of bags of Type A and $y$ be the number of bags of Type B.
Total cost of mixture = $x \times \text{Ksh } 400 + y \times \text{Ksh } 380 = 400x + 380y$
Total number of bags = $x+y$
Cost price per bag = $\frac{400x + 380y}{x+y}$
We equate this to the calculated cost price per bag:

To find the ratio $x:y$:

Multiply numerator and denominator by 100 to remove decimals:

Simplify the fraction by dividing both by 50:

The ratio $x:y$ is $\boxed{7:33}$.

From part (a), the mixture has 30 bags of Type A and 50 bags of Type B, totaling 80 bags.
In one bag of mixture (a):
Amount of Type A = $\frac{30}{80} = \frac{3}{8}$
Amount of Type B = $\frac{50}{80} = \frac{5}{8}$

From part (b), the ratio of Type A to Type B is $7:33$. The total parts are $7+33=40$.
In one bag of mixture (b):
Amount of Type A = $\frac{7}{40}$
Amount of Type B = $\frac{33}{40}$

Total Type A = (Amount of Type A in bag from a) + (Amount of Type A in bag from b)
Total Type A = $\frac{3}{8} + \frac{7}{40}$
To add, find a common denominator (40):
Total Type A = $\frac{3 \times 5}{8 \times 5} + \frac{7}{40} = \frac{15}{40} + \frac{7}{40} = \frac{22}{40}$

Total Type B = (Amount of Type B in bag from a) + (Amount of Type B in bag from b)
Total Type B = $\frac{5}{8} + \frac{33}{40}$
To add, find a common denominator (40):
Total Type B = $\frac{5 \times 5}{8 \times 5} + \frac{33}{40} = \frac{25}{40} + \frac{33}{40} = \frac{58}{40}$

Ratio = Total Type A : Total Type B
Ratio = $\frac{22}{40} : \frac{58}{40}$
This simplifies to $22:58$.
Divide both sides by their greatest common divisor, which is 2:
Ratio = $11:29$

The ratio of Type A millet to Type B millet in the final mixture is $\boxed{11:29}$.