There are large number of silver coins weighing 2gm , 5gm , 10gm , 25gm , 50gm each. Consider the following statements: 1. To buy 78gm of coins one must buy at least 7 coins. 2. To weigh 78gm using these coins one can use less than 7 coins. Which of the statements given above is/are correct?

The problem asks us to evaluate two statements related to forming a total weight of 78gm using silver coins of denominations 2gm, 5gm, 10gm, 25gm, and 50gm. Statement 1: "To buy 78gm of coins one must buy at least 7 coins." This statement implies finding the minimum number of coins whose total weight sums up to exactly 78gm. To minimize the number of coins, we should prioritize using coins of larger denominations. Let's try to form 78gm with the fewest coins: Start with the largest coin: 1 x 50gm = 50gm. Remaining weight needed = 78 - 50 = 28gm. (1 coin used) Now we need to make 28gm with the remaining coins (2, 5, 10, 25). Try using 25gm: 1 x 25gm = 25gm. Remaining weight needed = 28 - 25 = 3gm. (2 coins used so far) We cannot make 3gm using only 2gm and 5gm coins. So, this path (50+25) doesn't work directly. Let's try another combination for 28gm: Use 10gm coins for 28gm: 2 x 10gm = 20gm. Remaining weight needed = 28 - 20 = 8gm. (1 + 2 = 3 coins used so far) Now we need to make 8gm using 2gm and 5gm coins. Use 2gm coins: 4 x 2gm = 8gm. (3 + 4 = 7 coins used so far) So, 50gm + 10gm + 10gm + 2gm + 2gm + 2gm + 2gm = 78gm. This combination uses 7 coins. To confirm this is the minimum, we can systematically check if 6 coins are possible. The maximum weight with 6 coins using the available denominations is less than 6x50=300, but we need to check specifically for 78gm. If we use one 50gm coin, we need 28gm from 5 coins. - Using one 25gm coin for 28gm: 25gm + 3gm (from 4 coins). Impossible to make 3gm. - Using two 10gm coins for 28gm: 20gm + 8gm (from 3 coins). Impossible to make 8gm from 3 coins (e.g., 5+2+2=9, 2+2+2=6). If we don't use a 50gm coin, we need 78gm from 6 coins. Max from 25gm coins: 3x25=75. Remaining 3gm from 3 coins. Impossible. A detailed check reveals that 6 coins cannot sum to 78gm. Therefore, the minimum number of coins required is 7. Statement 1 is CORRECT. Statement 2: "To weigh 78gm using these coins one can use less than 7 coins." This statement implies using a balance scale, where coins can be placed on both sides to balance the unknown weight. If the object to be weighed (X) is on one side, and coins (C_left) are on the same side, and other coins (C_right) are on the opposite side, the balance equation is: X + C_left = C_right X = C_right - C_left We need to find a combination of coins (C_right) and another combination (C_left) such that their difference is 78gm, and the total number of coins used (number of coins in C_right + number of coins in C_left) is less than 7. Let's try to find such a combination: We need C_right - C_left = 78. Consider C_right = 50gm + 25gm + 5gm = 80gm. This uses 3 coins. Now, if we set C_left = 2gm. This uses 1 coin. Then C_right - C_left = 80gm - 2gm = 78gm. The total number of coins used is 3 (for C_right) + 1 (for C_left) = 4 coins. Since 4 is less than 7, it is possible to weigh 78gm using less than 7 coins. For example, place the object to be weighed and a 2gm coin on one pan, and 50gm, 25gm, 5gm coins on the other pan. The scale will balance. Statement 2 is CORRECT. Since both statements are correct, option C is the correct answer. The final answer is C