Read the following information carefully and answer the questions given below. The information about three sellers who sold two types of rice, i.e., brown and white.

The quantity of white rice sold by all three seller is 230 kg. The quantity of brown rice sold by B is 20% more than the quantity of white rice sold by A. The quantity of white rice sold by B is 20 kg more than that of brown rice. The ratio of brown rice sold by A to B is 2:3. The average quantity of brown rice sold by all three sellers is 60 kg, and the total rice (white and brown) sold by C is 75 kg.

Find the ratio of the total quantity of white rice sold by B to the total quantity of brown rice sold by A and B together.

The correct answer is 11:15.

Let us break down the given information step-by-step to find the quantities of rice sold by each seller.

Let the quantities of white rice sold by sellers A, B, and C be $W_A$, $W_B$, and $W_C$ respectively.
Similarly, let the quantities of brown rice sold by sellers A, B, and C be $B_A$, $B_B$, and $B_C$ respectively.

Step 1: Expressing the brown rice quantities
The quantity of brown rice sold by B is 20% more than the quantity of white rice sold by A:
$B_B=1.2\times W_A$    — (Equation 1)

The ratio of brown rice sold by A to B is 2:3:
$\frac{B_A}{B_B}=\frac{2}{3}$
Substituting Equation 1 into this ratio:
$B_A=\frac{2}{3}\times 1.2\times W_A=0.8\times W_A$    — (Equation 2)

Step 2: Expressing the white rice sold by B
The quantity of white rice sold by B is 20 kg more than that of brown rice sold by B:
$W_B=B_B+20$
Using Equation 1:
$W_B=1.2\times W_A+20$    — (Equation 3)

Step 3: Using the total quantities of white and brown rice
The total quantity of white rice sold by all three sellers is 230 kg:
$W_A+W_B+W_C=230$
Substituting Equation 3 into this total:
$W_A+(1.2\times W_A+20)+W_C=230$
$2.2\times W_A+W_C=210$
$W_C=210-2.2\times W_A$    — (Equation 4)

The total rice (white and brown) sold by C is 75 kg:
$W_C+B_C=75$
Substituting Equation 4 into this relation:
$(210-2.2\times W_A)+B_C=75$
$B_C=2.2\times W_A-135$    — (Equation 5)

Step 4: Solving for white rice sold by A ($W_A$)
The average quantity of brown rice sold by all three sellers is 60 kg, which means their sum is:
$B_A+B_B+B_C=3\times 60=180$
Substituting Equations 1, 2, and 5 into the sum:
$(0.8\times W_A)+(1.2\times W_A)+(2.2\times W_A-135)=180$
$4.2\times W_A-135=180$
$4.2\times W_A=315$
$W_A=\frac{315}{4.2}=75\text{kg}$

Step 5: Calculating the required quantities
Now we determine the specific values needed for the final ratio:
1. White rice sold by B ($W_B$):
$W_B=1.2\times 75+20=90+20=110\text{kg}$

2. Brown rice sold by A and B together ($B_A+B_B$):
$B_A=0.8\times 75=60\text{kg}$
$B_B=1.2\times 75=90\text{kg}$
$B_A+B_B=60+90=150\text{kg}$

Step 6: Finding the required ratio
The ratio of the total quantity of white rice sold by B to the total quantity of brown rice sold by A and B together is:
$\text{Ratio}=\frac{W_B}{B_A+B_B}=\frac{110}{150}=\frac{11}{15}$

Thus, the required ratio is 11:15.