A train starts full of passengers. At the first station, it drops one-third of the passengers and takes 280 more. At the second station, it drops one-half of the new total and takes 12 more. On arriving at the third station, it is found to have 248 passengers. Find the number of passengers in the beginning.
(a) 240 (b) 248 (c) 280 (d) 288

• ans:d
  the passenger in the train is =x
  after 1 stop=x-x/3+280
  after 2 stop---- (x-x/3+280)/2+12=248
  solve it and ans =288
• 13 years agoHelpfull: Yes(10) No(0)
• ans:let the passenger be =x
  after 1 stop become =(x-x/3)+280
  after 2 stop become =(x-x/3+280)/2+12=248
  ans =288
• 13 years agoHelpfull: Yes(2) No(0)
• let the initial nimber of passengers be x.
  now accc to the question we have at the 1 station we have x-1/3ofx+280(we have to add 280 here,now at the second station we have( 1/2( x-1/3ofx+280))+12,at the end of 3 station we have...( 1/2( x-1/3ofx+280))+12=248
  calculate the value of x from here...will get 288 as answer
• 13 years agoHelpfull: Yes(0) No(2)
• Let 'x' be the total number of passengers in the train at the beginning.
  At the first station, (x-(x/3))+280 =y (say)
  At the second Station, y/2 + 12
  Hence at the third station y/2 + 12 =248
  upon solving, x=288
  So answer is (d)
• 7 years agoHelpfull: Yes(0) No(0)
• let initial x passenger in the train.
  1st stoppage=(x-x/3)+280
  =2x/3+280
  2nd stoppage=(2x/3+280)/2+12
  =x/3+140+12
  =x/3+152
  at 3rd station x/3+152=288
  x=288
  hence total no of passangers are 288
• 6 years agoHelpfull: Yes(0) No(0)