A person travels from Chennai to Pondicherry in cycle at 7.5 Kmph. Another person travels the same distance in train at a speed of 30 Kmph and reached 30 mins earlier. Find the distance.
A)5 Km
B)10 Km
C)15km
D)20km

• t=d/s
  (d/7.5)-(d/30)=(30/60)
  (20d-5d)/150=1/2
  d=5.
  Ans is 5.
• 6 years agoHelpfull: Yes(24) No(0)
• Let, time taken by 🚲 =t
  //ly, time taken by train 🚉 =t-30
  We know that.....speed=distance /time
  Speed of bicycle 🚲, 7.5=d/t....(1)
  Speed of 🚉, 30=d/(t-30/60)....(2)
  Sol 1&2,we get t=0.666
  By sub and value in equal(1)
  We d=4.999~5km
• 6 years agoHelpfull: Yes(7) No(1)
• let the distance be x kms.
  then A/Q,
  x/(7.5) - x/(30) =30/60 => x= 5kms. (Option A)
• 6 years agoHelpfull: Yes(1) No(0)
• d/7.5-d/30=0.5h
  22.5*d/(30*7.5)=0.5
  d=5 km
• 6 years agoHelpfull: Yes(1) No(0)
• b) 10 km
  speed= distance/time
  equate two equations:
  let time be t hour
  1)7.5=d/t
  2)30=d/t-1/2
  by equating 1 & 2
  t=2/3
  therefore
  d=10
• 5 years agoHelpfull: Yes(1) No(2)
• d1 = d2
  s1 * t1 = s2*t2
  7.5*(x + 1/2) = 30*x ....30mins --> 1/2hr
  x = 1/6
  
  d1 = d2 = 30x = 30*1/6 = 5km
• 4 years agoHelpfull: Yes(1) No(0)
• x/7.5-x/30=30/60
  x=5
• 5 years agoHelpfull: Yes(0) No(0)
• let time taken by cycle = t
  30(t-30/60)= 7.5t
  4(t-1/2)= t
  3t =2
  t= 2/3hour
  distance = 7.5*2/3
  distance = 5km
• 5 years agoHelpfull: Yes(0) No(0)
• Let, time taken by 🚲 =t
  //ly, time taken by train 🚉 =t-30
  We know that.....speed=distance /time
  Speed of bicycle 🚲, 7.5=d/t....(1)
  Speed of 🚉, 30=d/(t-30/60)....(2)
  Sol 1&2,we get t=0.666
  By sub and value in equal(1)
  We d=4.999~5km
• 4 years agoHelpfull: Yes(0) No(0)
• 7.5x=30(x-30)
  7.5x=30x-900
  -22.5x=-900
  x=40mis
  d=30*(10/60)
  d=5kmh
• 4 years agoHelpfull: Yes(0) No(0)
• 7.5x=30(x-30)
  7.5x=30x-900
  -22.5x=-900
  x=40mis
  d=30*(10/60)
  d=5kmh
• 4 years agoHelpfull: Yes(0) No(0)
• Since the distance differs the answer would be 10
• 4 years agoHelpfull: Yes(0) No(0)