Answer : Option B

Explanation :

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Solution 1 (Recommended)
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$MF#%\boxed{\text{Speed and time are inversely proportional (when distance is constant) }\\
\Rightarrow \text{Speed ∝ }\dfrac{1}{\text{Time}}\text{ (when distance is constant)}}$MF#%

$MF#%\begin{align}
&\text{Here distance is constant and hence speed and time are inversely proportional}\\
&\text{Speed ∝ }\dfrac{1}{\text{Time}}\\\\
&\Rightarrow \dfrac{\text{Speed1}}{\text{Speed2}} = \dfrac{\text{Time2}}{\text{Time1}}\\\\
&\Rightarrow \dfrac{7}{\text{8}} = \dfrac{4}{\text{Time1}}\\\\
&\Rightarrow \text{Time1 = }\dfrac{4 \times 8}{7}\text{ hr}\\\\
&\Rightarrow \text{Speed of the first train = }\dfrac{\text{Distance}}{\text{Time1}}=\dfrac{400}{\left(\dfrac{4 \times 8}{7}\right)} \\& = \dfrac{100 \times 7}{8} = 12.5\times 7 = 87.5\text{ km/hr}\\\\
\end{align} $MF#%

$MF#%\begin{align}
&\Rightarrow \text{Speed of the 2nd train = }\dfrac{\text{Distance}}{\text{Time}} = \dfrac{400}{4} = 100\text{ km/hr}\\
&\Rightarrow 8x = 100\\
&\Rightarrow x = \dfrac{100}{8} = 12.5\\
&\Rightarrow \text{Speed of the first train = }7x = 7 \times 12.5 = 87.5 \text{ km/hr}\\
\end{align} $MF#%