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Maths Puzzle
Numerical Ability
Time Distance and Speed
A cyclist left point A for point B and travelled at a constant speed of 25km/h. When he coverred the istance of 8.33km he was overtaken by a car that left point A 12 minute after the cyclist and travelled at constant speed too. When the cyclist travelled another 30 km, he encountered the car returning from B. Assume that the car did not stop at point B. ind the distance between A and B.
Read Solution (Total 1)
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- 62.91 Km.
Time taken for cyclist to cover distance of 8.33 or 25/3 km= (25/3)/25=1/3 Hr
=20 min.
As car left the point A 12 min. after,so distance of 25/3 Km. is covered by car in 8 min or 8/60 Hr.,So the speed of the car= (25/3)/(8/60)=125/2 Km/hr.
Time taken for cyclist to cover another 30 Km= 30/25 or 6/5 hr.
Upto this point total time taken by cyclist= (1/3)+(6/5)=23/15 hr. and
Total distance covered by cyclist= 25*(23/15)=115/3 Km.
If from this point,distance of B is 'x' ,and as cyclist encounters a car returning from B, then car must have covered the distance= (115/3)+2x
Upto this point car has taken time(8 min.less than cyclist)=[(23/15)-(8/60)]=7/5 Hr., So distance covered= (7/5)*(125/2)=875/10 Km.
So (115/3)+2x =875/10 , x=295/12
Distance between A & B= dist. traveled by cyclist upto he encounters car+ x
=(25/3)+ 30 +(295/12)= 755/12=62.91 Km. - 10 years agoHelpfull: Yes(2) No(3)
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