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A CAR LEFT X FOR Y AND TRAVELLED AT AN AVERAGE SPEED OF 60 KMPH IT REACHED Y 2 HOURS AFTER THE SCHEDULED TIME IF IT HAD LEFT X FOR Y AND TRAVELLED AT AN AVERAGE SPEED OF 80 KMPH IT WOULD HAVE REACHED Y 2 HOURS BEFORE THE SCHEDULED TIME AT WHAT AVERAGE SPEED MUST IT TRAVEL IN ORDER TO REACH Y AT THE SCHEDULED TIME ?(IN KMPH)
A CAR LEFT X FOR Y AND TRAVELLED AT AN AVERAGE SPEED OF 60 KMPH IT REACHED Y 2 HOURS AFTER THE SCHEDULED TIME IF IT HAD LEFT X FOR Y AND TRAVELLED AT AN AVERAGE SPEED OF 80 KMPH IT WOULD HAVE REACHED Y 2 HOURS BEFORE THE SCHEDULED TIME AT WHAT AVERAGE SPEED MUST IT TRAVEL IN ORDER TO REACH Y AT THE SCHEDULED TIME ?(IN KMPH)
A)440/7
B)70
C)500/7
D)480/7
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Hi
let scheduled time be t ,
In both cases it traveled same distance :
so we get 60(t+2) = 80(t-2)
we get 60t+120 = 80t-160
we get 20t = 280
t = 14 hrs
Therefore total distance = 60(t+2) =960
Now 960 has to be covered in 14 hours
so we get average speed = 960/14 = 480/7
