Answer : Option B

Explanation :

Given that distance travelled in 1^st hour = 35 km
and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2^nd hour = 37 km
Hence distance travelled in 3^rd hour = 39 km
...
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term, a=35, number of terms,n = 12 and common difference, d=2.

$MF#%\boxed{\text{The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . . }\\
\text{is called an Arithmetic Progression(AP)}\\
\text{where a is the first term and d is the common difference of the AP}}\\$MF#%

$MF#%\boxed{\text{Sum of the first n terms of an Arithmetic Progression(AP),}\\
S_n = \dfrac{n}{2}[2a + \left(n-1\right)d]\\
\text{where n = number of terms}}$MF#%

$MF#%\begin{align}
&\text{Hence, }[35 + 37 + 39 + \text{... (12 terms)}] \\
&= S_{12} = \dfrac{12}{2}\left[2 \times 35 + \left(12-1\right)2 \right]\\
&= 6 \left[70 + 22\right] = 6 \times 92 = 552
\end{align} $MF#%