Anisha bought 24 pencils for Rs 128. She paid 1 rupee more for each white…

When two related unit prices differ by a fixed amount and the combined count and total cost are known, express both prices in terms of one variable, form one equation for the total count and one for the total cost, then solve them together (by substitution) for the integer values that satisfy both constraints.

1. Let the price of a black pencil be Rs x, so the price of a white pencil is Rs (x + 1).

2. Let w be the number of white pencils and b the number of black pencils, so b + w = 24 (total pencils).

3. The total cost is bx + w(x + 1) = 128, which expands to x(b + w) + w = 128.

4. Substituting b + w = 24 gives 24x + w = 128, so w = 128 − 24x.

5. Since w must be a whole number between 0 and 24, testing x = 5 gives w = 128 − 120 = 8, and b = 24 − 8 = 16.

6. The white-pencil price is x + 1 = 5 + 1 = Rs 6.

7. Check: 16 × 5 + 8 × 6 = 80 + 48 = 128, which matches the total cost given in the question.

Testing the neighbouring integer values confirms this is the only valid solution: x = 4 gives w = 128 − 96 = 32, which exceeds the total of 24 pencils, and x = 6 gives w = 128 − 144 = −16, a negative count — both impossible. So x = 5, giving a white-pencil price of Rs 6 and 8 white pencils, is the unique solution.