If a boy travels with 5/6th of usual his speed,he is late by 10 mins.What is the usual time taken to travel?

• speed=5/6*x
  time=6/5t becz speed inversely proportional to time
  6t/5-t=10
  t=50min
  
• 10 years agoHelpfull: Yes(38) No(1)
• Let s , t be the usual speed nd time of the boy nd d be the distance
  Usual time taken t = d/s --> (1)
  
  Late by 10mins if speed id 5/6th of s
  10 mins => 600 secs [Convert mins to secs or hours]
  Then d/(s*5/6) = d/s + 600
  6d/5s = t+600 [Since t = d/s... From (1)]
  6t/5 = t+600
  6t = 5t+3000
  t = 3000 secs or 50 mins
  
  Ans : 50 mins
• 10 years agoHelpfull: Yes(37) No(5)
• new speed=5/6 of usual speed
  new time taken=6/5 of usual time
  6/5 of usual time-usual time=10
  => 1/5 of usual time=10
  => usual time= 50 min.
• 8 years agoHelpfull: Yes(5) No(0)
• it was very simple trick
  just difference of 5/6 is 6-5=1
  so now divide this '1' by 10 i.e, 10/1=10
  so now multiply this '10' with numerator is 5 so 5*10=50
  ans is 50.
• 8 years agoHelpfull: Yes(3) No(1)
• numerator/diff of n.r and d.r*time
  5/6-5=5/1*10=50 mins
  
• 9 years agoHelpfull: Yes(2) No(0)
• D=ST
  D=(5/6)S(T+10)
  =>ST=(5/6)S(T+10)
  =>T=10 mins
• 8 years agoHelpfull: Yes(2) No(2)
• =>s*t=5/6s(t+10)
  =>6t=(5t+50)
  =>t=50 minutes
• 8 years agoHelpfull: Yes(1) No(1)
• let usual speed be s
  let usual time be t
  3/4th of the speed=3/4s
  10 minutes more the time =t+10
  s t
  3/4s t+10
  t,s are inversely proportional to each other so take direct product of them
  s*t=3/4s(t+10)
  t=50
  
• 8 years agoHelpfull: Yes(1) No(1)
• 1 hr
  Simple
• 8 years agoHelpfull: Yes(1) No(8)
• fast formula
  t*x/(y-x)
  x/y=5/6
  x=5 y=6
  t=10
  hence 10*5/6-5=50
• 8 years agoHelpfull: Yes(0) No(0)
• First digit be x
  Third digit 2x
  Second digit=(first+third i.e. x+2x=3x)
  Fourth digit=twice if third digie i.i 2*2x=4x
  
  
  Now sum of the n natural number is=n(n+1)/2.
  Therefore, 4(4+1)/2=10
  
  So,x+3x+2x+4x=10
  x=1
  
  Therefore Number will be 1324 or double 2648.
• 8 years agoHelpfull: Yes(0) No(0)
• let usual speed is S and usual time is T.
  1. d=(S/T);
  2. d=((5/6)S/(T+10));
  now equating and solving, we get...
  T=60 min... (usual Time)....
• 8 years agoHelpfull: Yes(0) No(0)