A man purchases a shirt and a trouser for ₹ 756. If the cost of trouser is 16%…

Concept: When one quantity is described as p% more than another, take the smaller quantity as the base. If the base is x, the larger quantity is x(1 + p/100). Writing both quantities with the same variable turns a stated total into a single linear equation in x, which is why a percentage comparison plus a total is always enough to fix both values.

Application:

1. Let the cost of the shirt be ₹x.

2. The trouser costs 16% more than the shirt, so the trouser is x(1 + 16/100) = 1.16x.

3. The shirt and the trouser together cost ₹756, so x + 1.16x = 756.

4. Adding the like terms on the left gives 2.16x = 756.

5. Dividing both sides by 2.16 gives x = 756 ÷ 2.16 = 350.

Cross-check: The same result follows from a ratio: shirt : trouser = 100 : 116 = 25 : 29, so the ₹756 bill splits into 25 + 29 = 54 equal parts of 756 ÷ 54 = ₹14 each, and the shirt takes 25 × 14 = ₹350. Substituting back confirms it: a trouser at 1.16 × 350 = ₹406 makes the bill 350 + 406 = ₹756, and the ₹56 difference between the two garments is 56 ÷ 350 = 0.16, that is 16% of the shirt's cost, exactly as stated.

Result: The shirt costs ₹350 (and the trouser ₹406).