A child buys three different kinds of chocolates costing Rs. 2, Rs. 5 and Rs. 10 each, and spends a total of Rs. 120. He must buy at least one chocolate of each kind. What is the minimum possible total number of chocolates he can buy?

Introduction / Context:
This is a number and optimization problem involving three types of chocolates with different prices. The child spends exactly Rs. 120 and must buy at least one of each kind. We are asked to find the minimum possible number of chocolates, which means we should try to maximize the use of higher priced chocolates while still satisfying all constraints. Problems like this test integer equation handling and logical optimization reasoning.

Given Data / Assumptions:
• Price of chocolate type 1 = Rs. 2 each. • Price of chocolate type 2 = Rs. 5 each. • Price of chocolate type 3 = Rs. 10 each. • Total money spent = Rs. 120. • At least one chocolate of each type must be bought.

Concept / Approach:
Let the child buy x chocolates of Rs. 2, y chocolates of Rs. 5, and z chocolates of Rs. 10. The total cost equation is 2x + 5y + 10z = 120. Also, x ≥ 1, y ≥ 1, z ≥ 1. We want to minimize N = x + y + z, which suggests using as many Rs. 10 chocolates as possible. We then search for integer solutions that satisfy both the cost equation and the minimum count requirement and choose the one with the smallest N.

Step-by-Step Solution:
Step 1: Write the cost equation: 2x + 5y + 10z = 120. Step 2: Divide the whole equation by 1 (not changing it) but focus on maximizing z to reduce the total count. Step 3: Try z = 10 (ten chocolates of Rs. 10 cost Rs. 100). Remaining amount = 120 − 100 = 20. Step 4: Now solve 2x + 5y = 20 with x ≥ 1, y ≥ 1. Step 5: Possible solutions include (x, y) = (5, 2) because 2*5 + 5*2 = 10 + 10 = 20. Step 6: This gives x = 5 chocolates of Rs. 2, y = 2 chocolates of Rs. 5, z = 10 chocolates of Rs. 10. Step 7: Total number of chocolates N = x + y + z = 5 + 2 + 10 = 17. Step 8: Check if fewer chocolates are possible by trying z = 11 or higher; but that would exceed the budget, while smaller z decreases cost per chocolate but increases the number of chocolates needed to reach Rs. 120.

Verification / Alternative check:
Confirm the total amount spent: 5 chocolates at Rs. 2 cost Rs. 10, 2 chocolates at Rs. 5 cost Rs. 10, and 10 chocolates at Rs. 10 cost Rs. 100. Total cost = 10 + 10 + 100 = Rs. 120. Constraints of at least one chocolate of each type are satisfied. Exploring other feasible combinations shows that any solution with fewer than 17 chocolates either fails the cost equation or violates the minimum one-per-type condition.

Why Other Options Are Wrong:
• 15 and 19 correspond to other possible combinations that either do not satisfy the Rs. 120 total or are not the minimum. • 22 clearly uses more chocolates than necessary with the same total cost.

Common Pitfalls:
A common mistake is to ignore the constraint of having at least one of each type, which could give a smaller number of chocolates but an invalid solution. Another error is random guessing of combinations without systematically checking the cost equation and count minimization together.

Final Answer:
The minimum possible total number of chocolates the child can buy is 17.