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(C) Let the distance between $P$ and $Q$ be $d$.Speed of the first man $(v_1)$ $= \frac{d}{4} \, 	ext{km/h}$.Speed of the second man $(v_2)$ $= \frac

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(C) Let the distance between $P$ and $Q$ be $d$.Speed of the first man $(v_1)$ $= \frac{d}{4} \, 	ext{km/h}$.Speed of the second man $(v_2)$ $= \frac{d}{4} \, 	ext{km/h}$.The second man starts $2 \, 	ext{hours}$ earlier. Let the first man meet the second man after $x \, 	ext{hours}$ from the time he starts from $P$.In $x \, 	ext{hours}$,the first man covers a distance of $v_1 	imes x = \frac{d}{4}x$.The second man has been traveling for $(x + 2) \, 	ext{hours}$ when they meet. The distance covered by the second man is $v_2 	imes (x + 2) = \frac{d}{4}(x + 2)$.The sum of the distances covered by both men must equal the total distance $d$:$\frac{d}{4}x + \frac{d}{4}(x + 2) = d$Dividing both sides by $d$ (assuming $d 
eq 0$):$\frac{x}{4} + \frac{x + 2}{4} = 1$$x + x + 2 = 4$$2x = 2$$x = 1 \, 	ext{hour}$.

$A$ man leaves a point $P$ and reaches the point $Q$ in $4 \, hours$. Another man leaves the point $Q$,$2 \, hours$ earlier and reaches the point $P$ in $4 \, hours$. Find the time in which the first man meets the second man (in $hours$).

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