Let's solve the problem step by step:

Part (a): Greatest possible number of biscuits

1. Ram's biscuits are divisible by 5 (no remainder), so $N$ is a multiple of 5.
2. When $N$ is divided by 6, the remainder is 4. This means: $N \equiv 4 \pmod{6}$
3. $N$ must also satisfy $N < 100$.

We need $N$ that satisfies: $N = 5k \quad \text{(multiple of 5)} \quad \text{and} \quad N \equiv 4 \pmod{6}.$

Solve using trial:

1. $N = 5k$ implies $k$ must be an integer.
2. Check multiples of 5 less than 100 that satisfy $N \equiv 4 \pmod{6}$.

Starting with $N = 5k$:

• $k = 1, N = 5$ → $5 \mod 6 = 5$ (not valid).
• $k = 2, N = 10$ → $10 \mod 6 = 4$ (valid).
• $k = 3, N = 15$ → $15 \mod 6 = 3$ (not valid).
• Continue this process until $N < 100$.

The valid numbers are $N = 10, 40, 70$. The greatest possible $N$ is 70.

Part (b): Divide $N = 70$ into 4 bags

To divide the biscuits equally into 4 bags: $70 \div 4 = 17 \quad \text{remainder } 2.$ Each bag will have 17 biscuits, with 2 biscuits left over.

Final Answers:

1. Part (a): The greatest possible number of biscuits is 70.
2. Part (b): There will be 17 biscuits in each bag, with 2 biscuits left over.

Let me know if you need detailed steps! Here are five related questions to expand your understanding:

1. What if Ram divided the biscuits equally into 7 bags? What would be the remainder?
2. How would the solution change if the biscuits were divisible by 4 instead of 5?
3. Can you calculate the total number of valid $N$ values less than 100?
4. Why does the condition $N \equiv 4 \pmod{6}$ matter?
5. How can modular arithmetic be used to solve similar problems efficiently?

Tip: Modular arithmetic is powerful for solving problems involving divisibility and remainders.