The arithmetic mean (average) amount of savings of ten students is Rs 600. Three students have no savings, and each of the remaining seven students has at least Rs 250, including Nihar, who has exactly Rs 1300. Under these conditions, what is the largest possible amount of savings (in Rs) that any one student could have?

Introduction / Context:
This problem is about averages and optimization with constraints. You know the overall average savings, some students have zero savings, others have at least a minimum amount, and one student has a fixed amount. You are asked to find the maximum amount that any one student could have while still satisfying all these conditions.

Given Data / Assumptions:
• There are 10 students in total.• Average savings of all 10 students = Rs 600.• Three students have no savings at all (Rs 0 each).• The remaining 7 students each have at least Rs 250.• Nihar is one of the seven and has exactly Rs 1300.• We want the maximum possible savings of any one student.

Concept / Approach:
The total savings of all students can be found from the average. To maximize the savings of one student, we minimize the savings of the remaining students subject to the given constraints. Since the three students already have zero, we keep them as they are. Among the remaining six students apart from the maximum saver, five should be set at the minimum allowed level (Rs 250) while Nihar is fixed at Rs 1300. The leftover amount then becomes the maximum possible for the remaining student.

Step-by-Step Solution:
Step 1: Total savings of all 10 students = 10 * 600 = Rs 6000.Step 2: Three students have zero savings, so they contribute Rs 0.Step 3: Out of the remaining 7 students, one is Nihar with Rs 1300. Let the student with maximum savings have M rupees.Step 4: The other 5 students each have at least Rs 250. To maximize M, set each of these 5 students at exactly Rs 250.Step 5: Total savings of these 5 students = 5 * 250 = Rs 1250.Step 6: Now total savings = 0 (three students) + 1300 (Nihar) + 1250 (five students) + M (one student) = 6000.Step 7: So 1300 + 1250 + M = 6000.Step 8: Add 1300 and 1250 to get 2550, so 2550 + M = 6000.Step 9: Therefore, M = 6000 - 2550 = Rs 3450.

Verification / Alternative check:
Check that M = 3450 satisfies all constraints. The seven students other than the three with zero have savings: 1300 (Nihar), 3450, and five times 250. Their total is 1300 + 3450 + 1250 = 6000, which matches the overall total. Each of the seven has at least 250, the average is 600, and the three zeros are included. Thus the conditions are satisfied.

Why Other Options Are Wrong:
3250: This is less than the value obtained after minimizing the others, so it is not the maximum possible.3650 and 3850: These are too large; if any student had these amounts while others met the constraints, the total would exceed Rs 6000.

Common Pitfalls:
Students may forget that to maximize one value, the other variable values must be minimized according to the constraints. Another mistake is ignoring the fixed amount for Nihar or wrongly changing the zero savings for the three students. Careful organization of the constraints leads to the correct maximum value.

Final Answer:
The largest possible savings any one student could have is Rs 3450.