a student reached his school late by 20mins by travelling at a speed of 9kmph.if he had travelled at a speed of 12kmph,he would have reached his school 20mins early.what is the distance between house and school?

• distance = distance either student reaches school before time or after time
  9(t+20/60) = 12(t-20/60)
  i.e
  9(t+1/3) = 12(t-1/3)
  9(3t+1) = 12(3t-1)
  t=7/3
  put t value in any of distance equation
  put t= 7/3 in 9(t+1/3)
  we get 24 km
• 11 years agoHelpfull: Yes(58) No(4)
• let the distance is x,
  so,(x/9-x/12)=(20+20)/60 [as 20 min earlier and 20min late]
  solving,x=24
• 11 years agoHelpfull: Yes(17) No(3)
• ans=24km.
  sol:
  9=D/(t+20/60): 20 mins converted into hours by dividing 60!
  &
  12=D/(t-20/60)
  now
  by solving this we will get
  t=7/3 hours
  now
  D=9*{(7/3)+(20/60)}
  D= 24 km
• 11 years agoHelpfull: Yes(10) No(0)
• i dont think it wud be ever asked in tcs.....
  
• 11 years agoHelpfull: Yes(7) No(8)
• d/9-d/12=40/60
  solving ,3d=2*9*12
  so, d=24
• 11 years agoHelpfull: Yes(6) No(0)
• Let, the actual time be x;
  student reach 20min late,so time=(x+(20/60))
  so,distance=9*(x+(20/60))...(1)
  again,at speed rate of 12km/hr we have,
  student reach 20min early,so time=(x-(20/60))
  so,distane=12*(x-(20/60))...(2)
  from eqn 1& 2,
  9*(x+1/3)=12*(x-1/3)
  27x+9=36x-12
  9x=21
  x=7/3 hrs...(3),actual time to reach school
  thus,distance between house and school is-
  d=9*(7/3+20/60)=9*(7/3+1/3)=9*(8/3)=72/3="24 km" answer
  
  so,distance=9*(x-20/60)...(1)
  
• 11 years agoHelpfull: Yes(4) No(3)