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.

The total quantity of white rice sold by D is 5 kg more than the average quantity of brown rice sold by A and C. If the total quantity of brown rice sold by D and B is in the ratio of 11:15 respectively, then find the total quantity of rice (brown and white) sold by D.

Let us break down the given information step-by-step to find the quantities of rice sold by sellers A, B, C, and D.

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

We are given the following relations:

1. The quantity of brown rice sold by B is 20% more than the quantity of white rice sold by A:
$B_B=1.2\times W_A$

2. The ratio of brown rice sold by A to B is 2:3:
$\frac{B_A}{B_B}=\frac{2}{3}\Rightarrow B_A=\frac{2}{3}{B}_B$

Substituting $B_B=1.2\times W_A$ into the ratio:
$B_A=\frac{2}{3}\left(1.2W_A\right)=0.8W_A$

3. The quantity of white rice sold by B is 20 kg more than that of brown rice sold by B:
$W_B=B_B+20=1.2W_A+20$

4. The average quantity of brown rice sold by all three sellers is 60 kg:
$\frac{B_A+B_B+B_C}{3}=60\Rightarrow B_A+B_B+B_C=180$

5. The total rice (white and brown) sold by C is 75 kg:
$W_C+B_C=75\Rightarrow W_C=75-B_C$

6. The total quantity of white rice sold by all three sellers is 230 kg:
$W_A+W_B+W_C=230$

Let us substitute $W_B$ and $W_C$ into this equation:
$W_A+(1.2W_A+20)+(75-B_C)=230$
$2.2W_A+95-B_C=230\Rightarrow 2.2W_A-B_C=135$

From the brown rice equation:
$B_A+B_B+B_C=180$
Substitute $B_A=0.8W_A$ and $B_B=1.2W_A$:
$0.8W_A+1.2W_A+B_C=180\Rightarrow 2W_A+B_C=180\Rightarrow B_C=180-2W_A$

Now, substitute $B_C$ into the equation $2.2W_A-B_C=135$:
$2.2W_A-(180-2W_A)=135$
$4.2W_A-180=135$
$4.2W_A=315\Rightarrow W_A=\frac{315}{4.2}=75\text{kg}$

Now we can compute the quantities for A, B, and C:

- For A:
$W_A=75\text{kg}$
$B_A=0.8\times 75=60\text{kg}$

- For B:
$B_B=1.2\times 75=90\text{kg}$
$W_B=90+20=110\text{kg}$

- For C:
$B_C=180-2\left(75\right)=30\text{kg}$
$W_C=75-30=45\text{kg}$

Now we analyze the requirements for seller D:

1. The total quantity of white rice sold by D ($W_D$) is 5 kg more than the average quantity of brown rice sold by A and C:
Average of brown rice by A and C = $\frac{B_A+B_C}{2}=\frac{60+30}{2}=45\text{kg}$
Therefore, $W_D=45+5=50\text{kg}$

2. The ratio of brown rice sold by D and B is 11:15 respectively:
$\frac{B_D}{B_B}=\frac{11}{15}$
Since $B_B=90\text{kg}$:
$B_D=\frac{11}{15}\times 90=11\times 6=66\text{kg}$

3. Find the total quantity of rice (brown and white) sold by D:
Total rice sold by D = $W_D+B_D=50+66=116\text{kg}$

Thus, the correct answer is 116 kg.