A & B traveling from X to Y. A starts at 12 pm at a speed of 63m/hr. B at 1:30 pm at a speed of 84m/hr.At what time will B be 34m ahead of A?

• Speed f A=63m/hr=6321/20m/min.
  speed f B=84m/hr=7/5m/min.
  bcoz B start at 1.30 & A start at 12.00
  so dstnc traveled by A in extra 90 min.=(21/20)*90=189/2
  spose in x min. B wl b 34 m ahead of A
  so, (21/20)*x+189/2 +34=(28/20)*x
  so,x=2570/7=367.14min.
  so time wil b 7.37.14 pm
• 12 years agoHelpfull: Yes(15) No(4)
• at 1.30 p.m. A will be at (63+(63/2)) = 94.5 m
  B will be at 0
  now if we consider the distance travelled according to the question..
  94.5+ x(63)+34 = 0+ x(84)....where x be the no hrs they travel to meet the condition....
  now x=6.1 hr...so the time will be 6 hr 6 minute after 1.30 pm which is "7.36 pm"
• 12 years agoHelpfull: Yes(10) No(1)
• more precise ans =7:36:36
  equation =84t-(63t+94.5)=34
  after solving t=6.11 hr
  t=6 hr 6minutes 36 seocond
  
• 12 years agoHelpfull: Yes(4) No(0)
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• 12 years agoHelpfull: Yes(2) No(11)