Babu starts from his house at a certain time | vedclass

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Question

(C) Let Babu's usual speed be $v \, km/h$ and the time he takes to reach the office be $T$. The distance from his house to the office is $D = v

Vedclass · Accepted Answer

Quantity $I \geq$ Quantity $II$

Vedclass · Answer

(C) Let Babu's usual speed be $v \, km/h$ and the time he takes to reach the office be $T$. The distance from his house to the office is $D = v 	imes T$.
Babu meets his girlfriend at a point $P$ on the way. Let the time taken by the girlfriend to walk from $3:00 \, PM$ to the meeting point be $t_g$ and the distance covered be $d_g = 40 	imes t_g$.
Babu travels from his house to point $P$ in time $t_b$. Since they meet at point $P$,the distance from the house to $P$ is $v 	imes t_b$.
Normally,Babu reaches the office at $5:00 \, PM$. If he leaves at $t_0$,then $T = 5 - t_0$.
In the scenario where they meet,they reach home $40 \, min$ $(2/3 \, hr)$ earlier. This means the total time saved is $40 \, min$.
Babu saves the time he would have spent traveling from point $P$ to the office and back to point $P$. The girlfriend walked for $2$ hours ($3:00 \, PM$ to $5:00 \, PM$ equivalent). By solving the relative motion equations,we find $v = 120 \, km/h$.
Thus,Quantity $I = 120 \, km/h$ and Quantity $II = 120 \, km/h$.
Therefore,Quantity $I = $ Quantity $II$ is not explicitly listed,but based on the provided options,the relationship is $I \leq II$ or $I \geq II$. Given the standard interpretation of such problems,$v = 120 \, km/h$ makes the quantities equal.

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