Step-by-step Solution:

Quick Solution

The goal is to pay three separate amounts using the smallest possible total number of coins. To achieve this, we should use a "greedy" approach for each amount: use as many of the largest value coins as possible for each payment.

Available coin denominations: 50 paise, 25 paise, 10 paise, 5 paise, and 2 paise.

1. Paying 78 paise

To make 78 paise with the minimum number of coins:

• 1 × 50p (Leaves 28p)
• 2 × 10p (Leaves 8p)
• 4 × 2p (Leaves 0p)

Total coins for 78p = 1 + 2 + 4 = 7 coins.

2. Paying 69 paise

To make 69 paise with the minimum number of coins:

• 1 × 50p (Leaves 19p)
• 1 × 10p (Leaves 9p)
• 1 × 5p (Leaves 4p)
• 2 × 2p (Leaves 0p)

Total coins for 69p = 1 + 1 + 1 + 2 = 5 coins.

3. Paying Rs. 1.01 (or 101 paise)

To make 101 paise with the minimum number of coins:

• 1 × 50p (Leaves 51p)
• 1 × 25p (Leaves 26p)
• 2 × 10p (Leaves 6p)
• 3 × 2p (Leaves 0p)

Total coins for 101p = 1 + 1 + 2 + 3 = 7 coins.

4. Total Number of Coins Required

The smallest total number of coins required is the sum of the coins for each separate payment.

7 (for 78p) + 5 (for 69p) + 7 (for 101p) = 19 coins

Final Answer: The smallest number of coins required is 19.