1.A passenger train takes 5hrs less for a journey of 396 kmif its speed is increased by 55kmph from its normal speed.The normal speed (to the nearest kmph) is
a)None b)44 c)55 d)66

• let speed=x
  396/x=t ===>396=xt.....1
  396/x+55=t-5===>396=(x+55)(t-5)...2
  equating eq1 and eq2
  we get
  11t-55=x
  heat and trail method (xt=396)
  x=44 then t=9
  xt=396 satisfied
  so
  ANS=: 44
  
• 8 years agoHelpfull: Yes(7) No(0)
• 396/x- 396/x+5 = 5
  since time differecnce is 5 hour
  check by option
  
• 8 years agoHelpfull: Yes(3) No(2)
• distance=(product of speed/diff of speed )* time difference
  so, 396=[x(let)*(x+55)/55]*5
  396*11=x*(x+55)
  now either you will solve it in a proper way or just put the option int it.
  suppose i have chosen option 44 as answer because 44*(44+55) unit digit is 6 so choose option 44 first if you multiply it you will get L.H.S=R.H.S
  so correct answer is : 44
• 8 years agoHelpfull: Yes(2) No(0)
• let normal speed of train is = xkm/hr
  so 396/x=T
  and 396/(x+55)=T-5
  =>396/x-396/(x+55)=5
  =>(396x+396*55-396x)/x(x+55)=5
  =>x^2+55x-4356=0
  =>(x+99)(x-44)=0
  so x=44km/hr
• 8 years agoHelpfull: Yes(2) No(2)
• pls anyone provide the clear solution to this problem asap.
• 8 years agoHelpfull: Yes(0) No(2)
• you can go through option u will get the ans
  
• 8 years agoHelpfull: Yes(0) No(1)
• Ans is 55...
• 8 years agoHelpfull: Yes(0) No(5)
• ans is 36
  None
• 4 years agoHelpfull: Yes(0) No(0)