There is an Arithmetic Progression here,
\(40 + (40+5)+ (40+5+5) .....n \;terms = 385\)
So, sum of terms = 385
first term = 40 and common difference = 5
\(\Rightarrow\;\)\(S = \frac{n}{2}[2a + (n-1)\times d]\)
\(\Rightarrow\;\)\(385 = \frac{n}{2}[80+5n-5)\)
\(\Rightarrow\;\)\(770 = n[75+5n]\)
\(\Rightarrow\;\)\(154 = n(15+n)\)
\(\Rightarrow\;\)\(n = 7\)
So it takes 7 hours to cover a distance of 385 kms.
The correct option is (D): 7 hrs