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Question

(D) Let the distance between point $A$ and $B$ be $D \, km$ and the scheduled time be $T$ hours.When the speed is $5 \, km/h$,the time taken is $T + \

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(D) Let the distance between point $A$ and $B$ be $D \, km$ and the scheduled time be $T$ hours.When the speed is $5 \, km/h$,the time taken is $T + \frac{48}{60}$ hours.So,$\frac{D}{5} = T + \frac{48}{60} \quad (1)$When the speed is $8 \, km/h$,the time taken is $T - \frac{15}{60}$ hours.So,$\frac{D}{8} = T - \frac{15}{60} \quad (2)$Subtracting equation $(2)$ from $(1)$:$\frac{D}{5} - \frac{D}{8} = (T + \frac{48}{60}) - (T - \frac{15}{60})$$\frac{8D - 5D}{40} = \frac{48 + 15}{60}$$\frac{3D}{40} = \frac{63}{60}$$\frac{3D}{40} = \frac{21}{20}$$D = \frac{21}{20} 	imes \frac{40}{3} = 7 	imes 2 = 14 \, km$.Thus,the distance is $14 \, km$.

$A$ person has to travel from point $A$ to point $B$ in a certain time. Travelling at a speed of $5 \, km/h$,he reaches $48 \, min$ late,and while travelling at a speed of $8 \, km/h$,he reaches $15 \, min$ early. What is the distance from point $A$ to $B$ (in $km$)?

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