18 sharpeners are bought for a total of Rs. 100. A white sharpener costs Rs. 1…

Concept: When a word problem gives one linear equation relating two unknown quantities (counts) and one linear equation relating cost, and the unit prices are themselves related but not both given directly, express everything in terms of a single unknown. Such problems treat both the counts AND the rupee prices as whole numbers, so if the system is still underdetermined, use that whole-number constraint to pin down the unique integer solution.

Application:

1. Let the brown sharpener's price be a rupees, so the white sharpener's price is (a+1) rupees. Let m be the number of brown sharpeners and n be the number of white sharpeners bought.

2. From the total count: m + n = 18.

3. From the total cost: a·m + (a+1)·n = 100, which expands to a·(m+n) + n = 100.

4. Substituting m + n = 18 gives 18a + n = 100, so n = 100 − 18a.

5. Since n must be a whole number with 0 ≤ n ≤ 18 (so that m = 18 − n is also a non-negative whole number) and a itself must be a whole number of rupees, test integer values of a: a = 4 gives n = 28 (invalid, exceeds 18); a = 5 gives n = 10 (valid, with m = 8); a = 6 gives n = −8 (invalid).

6. So a = 5 is the only value that works, and the white sharpener's price is a + 1 = 6 rupees.

Cross-check: 8 brown sharpeners at Rs. 5 cost Rs. 40, and 10 white sharpeners at Rs. 6 cost Rs. 60 — totaling Rs. 100, with 8 + 10 = 18 sharpeners in all, confirming the answer.