Three neighbors are there. 1st one lends 2nd and 3rd that many no. of tractors that then already each had. After few months, 2nd lends to 1st and 3rd that many tractors then they had. After a few months 3rd lends to 1st and 2nd that many tractors then they had. Now each of them got 24. Find how many they had initially?

• Try solving from the end
  1st 2nd 3rd
  24 24 24 (last)
  12 12 48 (before 3rd lend to 1st and 2nd as many as they had
  6 42 12 (before 2nd's transaction
  39 21 12 (before 1st's transaction)
  
  
• 11 years agoHelpfull: Yes(42) No(10)
• here is a short cut fr this prob
  no. of neighbors =3
  no.of tractors left with each of them=24
  now 24/2^3=24/8=3
  now 13*3 7*3 4*3=39 21 12
  same is da case for any question of dis type.jus divide the number left by 2 power transactions.multiply it with 13,7 and 4 which gives u final result
  
• 10 years agoHelpfull: Yes(28) No(10)
• 1st lends y tractors to 2nd and z tractors to 3rd.
  So now no of Tractors:
  1st: x-(y+z)
  2nd: 2y
  3rd: 2z
  
  2nd lends x-(y+z) tractors to 1st and 2z tractors to 3rd.
  So now no of Tractors:
  1st: 2x-2y-2z
  2nd: 2y-(x-y+z)
  3rd: 4z
  
  3rd lends 2x-2y-2z tractors to 1st and 2y-(x-y+z) tractors to 2nd.
  So now no of Tractors:
  1st: 4x-4y-4z
  2nd: 6y-2x-2z
  3rd: 4z -(x+y-3z)
  4x-4y-4z = 24
  or, x-y-z = 6 ..........(i)
  6y-2x-2z = 24
  or -x+3y-z = 12 ....(ii)
  -x-y+7z = 24 .........(iii)
  
  Solving (i) (ii) and (iii)
  x = 39, y = 21, z = 12
  1ST HAD 39
  2ND HAD 21
  3RD HAD 12
• 10 years agoHelpfull: Yes(23) No(8)
• 1ST HAD 39
  2ND HAD 21
  3RD HAD 12
• 15 years agoHelpfull: Yes(17) No(14)
• 39, 21 and 12 tractors each in order for 1st, 2nd and 3rd neighbours
• 15 years agoHelpfull: Yes(10) No(11)
• 39, 21 and 12 tractors
• 10 years agoHelpfull: Yes(9) No(5)
• from the the last step
  24 24 24 (after 3rd step)
  12 12 48 (after 2nd step)
  6 42 24 (after 1st step)
  39 21 12 (initially)
• 10 years agoHelpfull: Yes(8) No(4)
• we can easily solve this from end
  1st 2nd 3rd
  24 24 24 (finally they got) (3rd has 48,after lends to 1st and 2nd how much they had i.e., 12 and 12 then 48-24=24)
  12 12 48 (before 3rd lends to 2nd and 3rd )(2nd has 42,after lends to 1st and 3rd i.e 6 and 24 then 42-(6+24)=12)
  6 42 24 (before 2nd)(1st has 39,after lends to 2nd and 3rd i.e 21 and 12 then 39-(21+12)= 6)
  39 21 12 (before 1st)
• 10 years agoHelpfull: Yes(7) No(1)
• Please explain,,,, how did u all get this answer!!
• 12 years agoHelpfull: Yes(4) No(1)
• pooja we dont have a full day to give the test .... harjinders ans is perfect
• 10 years agoHelpfull: Yes(1) No(1)
• best solution done by deepika..(y)
• 10 years agoHelpfull: Yes(0) No(1)
• we would get the solution by the equations 4x-4y-4z=24
  -2x+6y-2z=24
  -x-y+7z=24
  we will get x=39,y=21,z=12
• 9 years agoHelpfull: Yes(0) No(1)
• 13*3=39,
  7*3=21,
  4*3=12
• 9 years agoHelpfull: Yes(0) No(1)