To solve this problem, we'll systematically work through the relationships given between the prices, weights, and quantities of apples and oranges.

Step-by-Step Solution:

1. Define Variables:

   • Let the price of an orange be $P_o$ and the price of an apple be $P_a$.
   • Let the weight of an orange be $W_o$ and the weight of an apple be $W_a$.

2. Set Up Given Ratios:

   • The ratio of the price of an orange to an apple is $5:3$. Therefore: $\frac{P_o}{P_a} = \frac{5}{3} \implies P_o = \frac{5}{3}P_a$
   • The ratio of the weight of an orange to an apple is $4:1$. Thus: $\frac{W_o}{W_a} = \frac{4}{1} \implies W_o = 4W_a$

3. Weight of Packets:

   • The weight of a packet of apples is twice the weight of a packet of oranges. Let the weight of a packet of oranges be $W$. Then, the weight of a packet of apples is $2W$.
   • Each packet of oranges weighs $W$, so it contains $\frac{W}{W_o}$ oranges.
   • Each packet of apples weighs $2W$, so it contains $\frac{2W}{W_a}$ apples.

4. Express the number of fruits in each packet:

   • The number of oranges in a packet: $\text{Number of oranges per packet} = \frac{W}{W_o}$
   • The number of apples in a packet: $\text{Number of apples per packet} = \frac{2W}{W_a}$

5. Calculate the Price of Each Packet:

   • The price of a packet of oranges (denoted as $C_o$) is: $C_o = \left(\frac{W}{W_o}\right) P_o$
   • The price of a packet of apples (denoted as $C_a$) is: $C_a = \left(\frac{2W}{W_a}\right) P_a$

6. Simplify Packet Prices Using Ratios:

   • Substituting $W_o = 4W_a$ and $P_o = \frac{5}{3}P_a$ gives: $C_o = \left(\frac{W}{4W_a}\right) \left(\frac{5}{3}P_a\right) = \frac{5WP_a}{12W_a}$ $C_a = \left(\frac{2W}{W_a}\right) P_a = \frac{2WP_a}{W_a}$

7. Find a Common Weight for Simplification:

   • Assume $W_a = 1$ unit (smallest weight unit), and thus $W_o = 4 \times 1 = 4$.
   • Substitute back: $C_o = \frac{5W}{12}$ $C_a = 2WP_a$

8. Ensure Integer Prices:

   • To ensure integer prices for packets: $\frac{5W}{12} = \text{integer}, \quad 2W = \text{integer}$
   • Let $W = 12k$ for some integer $k$, then: $C_o = \frac{5 \times 12k}{12} = 5k, \quad C_a = 2 \times 12k = 24k$

9. Calculate Total Cost for Jia:

   • Jia buys 4 packets of oranges and 2 packets of apples: $\text{Total cost} = 4C_o + 2C_a = 4(5k) + 2(24k) = 20k + 48k = 68k$
   • To keep prices integer, use $k = 1$, hence: $\text{Total price paid} = 68 \times 1 = 68 \text{ Rs.}$

Thus, Jia could have paid a total of 68 Rs.

Would you like more details or have any questions about the solution steps?

Here are five related questions to consider:

1. How would the total price change if the ratio of weights was different?
2. What if Jia bought a different number of packets?
3. How would the answer change if the price ratios were reversed?
4. Can this method apply to more than two fruits?
5. How do different weights affect pricing in bulk?

Tip: Always double-check ratios and unit conversions to ensure accuracy in your calculations.