Ravi and Ajay start simultaneously from a pla | vedclass

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(C) Let the speed of Ravi be $x \, km/h$. Then,the speed of Ajay is $(x+4) \, km/h$.Since they start simultaneously,the time taken by both to meet is

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let the speed of Ravi be $x \, km/h$. Then,the speed of Ajay is $(x+4) \, km/h$.Since they start simultaneously,the time taken by both to meet is the same.Ravi travels a distance of $(60 - 12) = 48 \, km$ to reach the meeting point.Ajay travels from $A$ to $B$ $(60 \, km)$ and then back $12 \, km$ to the meeting point,covering a total distance of $(60 + 12) = 72 \, km$.Since time = distance / speed,we have:$\frac{48}{x} = \frac{72}{x+4}$$48(x + 4) = 72x$$48x + 192 = 72x$$72x - 48x = 192$$24x = 192$$x = \frac{192}{24} = 8 \, km/h$.Thus,Ravi's speed is $8 \, km/h$.

Ravi and Ajay start simultaneously from a place $A$ towards $B$,$60 \, km$ apart. Ravi's speed is $4 \, km/h$ less than that of Ajay. Ajay,after reaching $B$,turns back and meets Ravi at a place $12 \, km$ away from $B$. Ravi's speed is (in $km/h$):

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