Two cities x and y are 400km apart. Q leaves x 8 hours after P. Both P and Q arrive simultaneously.
Find the time the slower person spent on the trip if the speed of one of them was 15kmph higher than that of
other.

• 19.14 hrs.
  
  Let speed of P = p kmph, then speed of Q=(p+15) kmph.
  Let slower person P takes 't' hrs. then Q takes (t-8) hrs. to reach y
  We have, pt = (p+15)(t-8)
  => 8p - 15t + 120 = 0
  As p=400/t, substituting in above equation, 8x(400/t) -15t + 120 = 0
  => 3t² - 24t - 640 = 0
  On solving t=19.14
• 6 years agoHelpfull: Yes(3) No(6)