In a box, there are a certain number of coins which amount to ₹ 25. There are 2-rupee coins, 50-paise coins and 25-paise coins. If there are at least one 2-rupee coin, two 50-paise coins and two 25-paise coins, then what could be the minimum number of coins in the box?

### Concept & Logic To minimize the total number of coins required to hit a specific monetary target, you must maximize the use of the highest denomination coin available, while carefully fulfilling any mandatory minimum usage constraints for the other coins. ### Step-by-Step Solution * **Given:** Total value = ₹ 25. Denominations = ₹ 2, 50p (₹ 0.50), 25p (₹ 0.25). * Constraints: At least one ₹ 2 coin ($x \ge 1$), two 50p coins ($y \ge 2$), and two 25p coins ($z \ge 2$). * Goal: Minimize total coins $(x + y + z)$. * **Calculation:** Let the number of ₹ 2, 50p, and 25p coins be $x, y,$ and $z$. * Equation: $2x + 0.50y + 0.25z = 25$ * To minimize the total count, we must make $x$ as large as possible. * **Attempt 1:** Let's try the absolute maximum possible $x$. Could $x=12$? This gives $2 \times 12 = 24$. Remaining value = ₹ 1. However, we must satisfy $y \ge 2$ (which is ₹ 1) and $z \ge 2$ (which is ₹ 0.50). This requires a minimum of ₹ 1.50. ₹ 24 + ₹ 1.50 = ₹ 25.50 (Exceeds ₹ 25). So, $x$ cannot be 12. * **Attempt 2:** Let's try $x=11$. This gives $2 \times 11 = 22$. Remaining value = ₹ 3. Now we need to make ₹ 3 using the fewest possible coins of 50p and 25p, maintaining $y \ge 2, z \ge 2$. To minimize coins here, use as many 50p coins as possible. $0.50y + 0.25z = 3$ If we maximize $y \to y=5$ (value ₹ 2.50). The remaining value is ₹ 0.50, which requires exactly $z=2$ coins of 25p. This perfectly fits all constraints! * We have: $x=11, y=5, z=2$. * Total coins = $11 + 5 + 2 = 18$. ### Exam Strategy & Shortcut Work strictly from the highest denomination down. Total target is ₹ 25. 1. Put aside the mandatory minimums first: one ₹ 2, two 50p, two 25p = ₹ 3.50. (Total coins used = 5). 2. Remaining value to fulfill = $25 - 3.50 = 21.50$. 3. Pack this remainder with the highest coin (₹ 2). You can fit ten more ₹ 2 coins = ₹ 20. (Total coins used = $5 + 10 = 15$). 4. Remaining value = ₹ 1.50. 5. Pack this with the next highest (50p). You can fit three 50p coins = ₹ 1.50. (Total coins used = $15 + 3 = 18$). This greedy algorithm works perfectly. ### Common Pitfall A common trap is attempting to maximize $x=12$, finding that it leaves only ₹ 1, and incorrectly assuming you can just use two 50p coins ($y=2$) and zero 25p coins ($z=0$), which violates the strict constraint that there must be at least two 25p coins. ### Final Answer **Therefore, the correct answer is 18.**