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.

Correct Answer: 116 kg

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

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

Step 1: Formulating the equations from the given clues

1. The total quantity of white rice sold by all three sellers is 230 kg:
$W_A+W_B+W_C=230$ (Equation 1)

2. The quantity of brown rice sold by B is 20% more than the quantity of white rice sold by A:
$B_B=1.20\times W_A$
Which gives:
$W_A=\frac{B_B}{1.20}=\frac{5}{6}{B}_B$ (Equation 2)

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

4. The ratio of brown rice sold by A to B is 2:3:
$\frac{B_A}{B_B}=\frac{2}{3}$
Which gives:
$B_A=\frac{2}{3}{B}_B$ (Equation 4)

5. The average quantity of brown rice sold by all three sellers is 60 kg:
$\frac{B_A+B_B+B_C}{3}=60$
Which means:
$B_A+B_B+B_C=180$ (Equation 5)

6. The total rice (white and brown) sold by C is 75 kg:
$W_C+B_C=75$ (Equation 6)

Step 2: Solving the equations in terms of $B_B$

From Equation 1, we express $W_C$ in terms of $B_B$:
$W_A+W_B+W_C=230$
Substitute Equation 2 and Equation 3:
$\frac{5}{6}{B}_B+(B_B+20)+W_C=230$
$\frac{11}{6}{B}_B+W_C=210$
$W_C=210-\frac{11}{6}{B}_B$ (Equation 7)

From Equation 5, we express $B_C$ in terms of $B_B$:
$B_A+B_B+B_C=180$
Substitute Equation 4:
$\frac{2}{3}{B}_B+B_B+B_C=180$
$\frac{5}{3}{B}_B+B_C=180$
$B_C=180-\frac{5}{3}{B}_B$ (Equation 8)

Now, substitute Equation 7 and Equation 8 into Equation 6 ($W_C+B_C=75$):
$\left(210-\frac{11}{6}{B}_B\right)+\left(180-\frac{5}{3}{B}_B\right)=75$
$390-\left(\frac{11}{6}+\frac{10}{6}\right)B_B=75$
$390-\frac{21}{6}{B}_B=75$
$390-\frac{7}{2}$
$315=\frac{7}{2}$
$B_B=\frac{315\times 2}{7}=45\times 2=90\text{kg}$

Step 3: Finding individual values

Using $B_B=90\text{kg}$:
- Brown rice sold by A ($B_A$):
$B_A=\frac{2}{3}\times 90=60\text{kg}$
- Brown rice sold by C ($B_C$):
$B_C=180-\frac{5}{3}\times 90=180-150=30\text{kg}$

Step 4: Calculations 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:
$\text{Average brown rice (A and C)}=\frac{B_A+B_C}{2}=\frac{60+30}{2}=45\text{kg}$
Therefore:
$W_D=45+5=50\text{kg}$

2. The total quantity of brown rice sold by D ($B_D$) and B is in the ratio of 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. Total quantity of rice (brown and white) sold by D is:
$\text{Total rice sold by D}=W_D+B_D=50+66=116\text{kg}$