Solving the Sweet Purchase Problem: Maximizing Sweets for Rs 200

The problem asks us to find the maximum number of sweets Sunil can purchase for exactly Rs. 200, given that there are two types of sweets costing Rs. 7 and Rs. 10 each.

Setting up the Equation

Let's define the variables:

• Let x be the number of sweets costing Rs. 7.
• Let y be the number of sweets costing Rs. 10.

The total cost of purchasing these sweets is given by the equation:

\(7x + 10y = 200\)

We are looking for non-negative integer values for \(x\) and \(y\) (since you can't buy negative or fractional sweets) that satisfy this equation. Additionally, we want to maximize the total number of sweets, which is \(x + y\).

Finding Integer Solutions for \(7x + 10y = 200\)

We need to find integer pairs \((x, y)\) that satisfy the equation. Since \(x\) and \(y\) must be non-negative, we can analyze the equation:

\(10y = 200 - 7x\)

For \(y\) to be a non-negative integer, two conditions must be met:

1. \(200 - 7x\) must be a non-negative number, which means \(200 \ge 7x\). This implies \(x \le \frac{200}{7} \approx 28.57\). So, \(x\) can be any integer from 0 up to 28.
2. \(200 - 7x\) must be divisible by 10. Since 200 is divisible by 10, \(7x\) must also be divisible by 10. Since 7 and 10 have no common factors other than 1, \(x\) must be divisible by 10.

Considering that \(x\) must be an integer between 0 and 28 (inclusive) and must be divisible by 10, the possible values for \(x\) are 0, 10, and 20.

Calculating \(y\) and Total Sweets for Each Possible \(x\) Value

Now, we calculate the corresponding value of \(y\) and the total number of sweets \(x + y\) for each possible value of \(x\):

If we consider \(x = 30\), then \(7x = 210\). \(y = \frac{200 - 210}{10} = \frac{-10}{10} = -1\), which is not a valid number of sweets.

Identifying the Maximum Number of Sweets

Comparing the total number of sweets calculated for each valid possibility:

• When \(x=0\) and \(y=20\), total sweets = 20.
• When \(x=10\) and \(y=13\), total sweets = 23.
• When \(x=20\) and \(y=6\), total sweets = 26.

The maximum number of sweets Sunil can get while spending exactly Rs. 200 is 26.

Conclusion on Maximizing Sweets for Rs 200

By setting up and solving the linear Diophantine equation \(7x + 10y = 200\) for non-negative integers and evaluating \(x+y\) for each solution, we found that the maximum number of sweets is 26.

Revision Table: Sweet Purchase Problem

Additional Information: Linear Diophantine Equations and Optimization

A linear Diophantine equation is an equation of the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are integers, and we are looking for integer solutions for \(x\) and \(y\).

• An equation \(ax + by = c\) has integer solutions if and only if the greatest common divisor (GCD) of \(a\) and \(b\) divides \(c\). In our problem, GCD(7, 10) = 1, and 1 divides 200, so integer solutions exist.
• To find all integer solutions, one usually finds a particular solution using the extended Euclidean algorithm and then uses it to express the general solution. However, for simple cases or when non-negativity constraints limit the number of possibilities, a more direct approach by checking possible values might be easier, as done in this sweet purchase problem.
• When solving problems like this, remember that the context often requires non-negative integer solutions (e.g., number of items, people, etc.).
• Optimization in such problems involves finding the solution among the valid integer solutions that maximizes or minimizes a certain expression (in this case, \(x + y\)).