A man spends Rs. 810 on buying trousers at Rs. 70 each and shirts at Rs. 30 each. If he purchases the maximum possible number of trousers within this budget and spends the remainder on shirts, what is the ratio of the number of trousers to shirts bought?

Introduction / Context:
This question tests integer equation solving and optimization within a budget. A fixed total amount is spent on two types of items with known prices, and we must maximize the quantity of one item while ensuring that the remaining money is exactly enough to buy an integer number of the other item. Then we compute the ratio of items purchased.

Given Data / Assumptions:
• Total money spent = Rs. 810.
• Price of one trouser = Rs. 70.
• Price of one shirt = Rs. 30.
• The man wants to buy as many trousers as possible and then spend the remainder on shirts.
• All quantities of trousers and shirts must be whole numbers.

Concept / Approach:
Let the number of trousers be t and shirts be s. Then the budget equation is 70t + 30s = 810. We must choose t as large as possible such that the remaining amount 810 − 70t is non-negative and divisible by 30, ensuring that s is an integer. We systematically test values of t from the maximum possible downward until the remainder condition is satisfied. Then we compute the resulting ratio t : s and simplify it.

Step-by-Step Solution:
Step 1: Write the budget equation: 70t + 30s = 810.Step 2: The maximum possible t must satisfy 70t ≤ 810, so t ≤ 810 / 70 = 11.57..., hence t can be at most 11.Step 3: Try t = 11: 70 * 11 = 770, remainder for shirts = 810 − 770 = 40, which is not divisible by 30, so discard.Step 4: Try t = 10: 70 * 10 = 700, remainder = 110, not divisible by 30, discard.Step 5: Try t = 9: 70 * 9 = 630, remainder = 810 − 630 = 180, which is divisible by 30, giving s = 180 / 30 = 6.Step 6: So the feasible maximum trousers is t = 9 and shirts s = 6.Step 7: Ratio of trousers to shirts = 9 : 6, which simplifies to 3 : 2.

Verification / Alternative check:
Check the total cost with t = 9 and s = 6: cost of trousers = 9 * 70 = 630; cost of shirts = 6 * 30 = 180; total = 630 + 180 = 810, so the budget is exactly used. Attempting a higher t like 10 or 11 fails to produce integer s as seen earlier, so 9 is indeed the maximum possible number of trousers. This confirms that the ratio 3 : 2 is correct.

Why Other Options Are Wrong:
Ratio 1 : 3 would mean far fewer trousers than shirts and would not correspond to the maximum-t scenario under this budget equation.
Ratios 2 : 1 and 2 : 3 correspond to other combinations of t and s that do not both satisfy the cost equation and the requirement of maximizing trousers, so they are not valid answers for this question.

Common Pitfalls:
A frequent mistake is to forget the integer constraint and choose a value of t for which s becomes fractional. Another error is to simply divide the total amount by one of the prices and ignore that money must be split between two items. Systematically checking feasible integer values near the upper limit of t is the safest method for such problems.

Final Answer:
The ratio of the number of trousers to shirts purchased is 3 : 2.