RRB Group D Model Paper 7 with solutions

(4) Let the length and breadth of the parallelepiped be 5a cm and 3a cm respectively, then, Total surface area of parallelepiped
= 2 (lb + bh + lh)
558 = 2(15 ${\mathit{a}}^2$ + 18a + 30a)
⇒ 15${\mathit{a}}^2$ + 48a = $\frac{558}{2}$ = 279
So
15 ${\mathit{a}}^2$ + 48a - 279 = 0
⇒ 5${\mathit{a}}^2$ + 16a - 93 = 0
⇒ (5a - 31)(a - 3) = 0
⇒ a = $-\frac{31}{5}$ , 3
So, the length of the rectangle is 5 × 3 = 15 cm