A boy walking at a speed of 10, km/hr reaches | vedclass

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$(A)$ Let the distance be $d\, km$ and the scheduled time be $t\, hr$.Case $1$: Speed $= 10\, km/hr$, Time taken $= t + \frac{15}{60} = t + \frac{1}{4

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$10$

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$(A)$ Let the distance be $d\, km$ and the scheduled time be $t\, hr$.Case $1$: Speed $= 10\, km/hr$, Time taken $= t + \frac{15}{60} = t + \frac{1}{4}\, hr$.Using $d = 	ext{speed} 	imes 	ext{time}$, we get $d = 10(t + \frac{1}{4}) \Rightarrow d = 10t + 2.5$ ... $(I)$.Case $2$: Speed $= 10 + 2 = 12\, km/hr$, Time taken $= t + \frac{5}{60} = t + \frac{1}{12}\, hr$.Using $d = 	ext{speed} 	imes 	ext{time}$, we get $d = 12(t + \frac{1}{12}) \Rightarrow d = 12t + 1$ ... $(II)$.Equating $(I)$ and $(II)$:$10t + 2.5 = 12t + 1$$2.5 - 1 = 12t - 10t$$1.5 = 2t \Rightarrow t = 0.75\, hr$.Substituting $t$ in $(I)$:$d = 10(0.75) + 2.5 = 7.5 + 2.5 = 10\, km$.

$A$ boy walking at a speed of $10\, km/hr$ reaches his school $15\, minutes$ late. Next time he increases his speed by $2\, km/hr$, but still he is late by $5\, minutes$. Find the distance of his school from his house (in $km$).

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