The cost of painting the surface of a solid cuboid of dimensions 40 cm x 36 cm…

Concept: For a cuboid with edge lengths l, w and h, the total surface area is the sum of its six faces taken as three congruent pairs: TSA = 2(lw + wh + hl).

When a surface is painted at a fixed rate per unit area, cost = area × rate. So the painted area can be recovered from the cost as area = cost ÷ rate, provided the quoted rate is first converted to a rate per single unit of area.

Application to this cuboid:

1. Convert the rate: ₹20 per 100 cm² means 20 ÷ 100 = ₹0.20 per cm².

2. Recover the painted area from the cost: 1,032 ÷ 0.20 = 5,160 cm².

3. Write the total surface area with l = 40 cm, w = 36 cm and h = x cm: 2(40 × 36 + 36 × x + 40 × x) = 2(1,440 + 76x) = 2,880 + 152x.

4. Equate the two expressions for the same area: 2,880 + 152x = 5,160, hence 152x = 2,280.

5. Divide to get the edge: x = 2,280 ÷ 152 = 15, that is x = 15 cm.

Cross-check: substituting x = 15 cm gives TSA = 2(1,440 + 76 × 15) = 2 × 2,580 = 5,160 cm², and 5,160 × ₹0.20 = ₹1,032 — exactly the cost stated in the question.

Contrast: if "surface" were read as only the four side faces, the area would be 2x(40 + 36) = 152x = 5,160, giving x ≈ 33.9 cm — not a whole number and not among the values offered, which confirms that the whole (total) surface of the solid is painted.