Question

A cuboid has edges of x cm, 1 cm and 2 cm. The total surface area of the cuboid has a numerical value which is some integral multiple of the numerical value of its volume. What is the value of x for minimum possible positive integral multiple ? 

(a) 5 cm 

(b) 2 cm 

(c) 3 cm 

(d) 4 cm

Answer 1

(b) 2 cm

Total surface of the cuboid 

= 2 (lb + bh + lh) 

= 2 (x + 2 + 2x) = 6x + 4 

Volume of the cuboid = l × b × h = x × 1 × 2 = 2x

Given, 6x + 4 = n × 2x, where n is an integer 

⇒ (2n – 6)x = 4 

Since x is a length of the cuboid; x must be positive. 

For positive volume of x, (2n – 6) > 0 or 2n > 6 or n > 3 

∴ n is a positive integer, minimum value of n = 4 

∴ 6x + 4 = 2 × 4 × x ⇒ x = 2 cm.