A cuboid has a total surface area of 94 cm$^2$ and a volume of 60 cm$^3$. Its length, breadth, and height are positive consecutive integers with $l < b < h$. Quantity I: Lateral surface area of the cuboid Quantity II: Total surface area of a cube with side $b$ Which quantity is greater?

Quantity I > Quantity II
Quantity I < Quantity II
Quantity I \ge Quantity II
Quantity I \le Quantity II

Correct answer: Quantity I < Quantity II

Solution

The consecutive integers satisfying the given conditions are $3,4,5$. Then the cuboid’s lateral surface area is $2h(l+b)=2\cdot5\cdot(3+4)=70$, while the cube’s total surface area is $6b^2=6\cdot16=96$. Therefore, Quantity I is less than Quantity II.

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