There are 1000 zombies in a research centre, very hungry such that they start eating each other. There are 1 on 4 zombies which can eat two zombies at a time. So, according to the condition that if 1st zombie eat 2nd then 2nd can only eat 1st not any other at same instant, if one of them finished completely then only he can go on next one. If zombies who can eat two at a time takes 1.5 minute to eat both and others who eat one at a time takes 1 minute to eat completely. also zombies who eat 2 at a time have stronger life i.e let 1.5 min. and others have 1minute life , i.e if someone starts eating them. Find the no. of zombies left at last and total time taken till, last zombie left or all finished?
a. 1 left, 562 min.
b. 0 left, 562 min.
c. 1 left, 500 min.
d. 0 left, 500 min.
e. none of these

-by vatan singh
(plz put my name)

• asked in TCS HR round
  firstly read question carefully: see strong zombie has life of 1.5 min. & weak ones has 1 min.
  Now, here are 250 strong zombie( denoted by sz) and 750 weak zombies( denoted by wz) according to question.
  Zombies( denoted by z)
  
  (((concept understanding:
  We need to make the set of z in which they eat each other, if sz 250 start eat 500wz then sz are eliminated by wz, wz left now has life of 0.25 min. Also at same time while they eat each other the left wz also eat each other i.e 750-500=250 left so 125-125 eat each other in 125 min. but 1st set take a time of 250*1.5+((500/2)*0.25) i.e total sz*time take to finish one sz by two wz+ time taken by left wz with life of 0.25 min. eat each other= 437.5min. i.e. not a min. time.
  
  So, obviously it’s a set which is nearer this set: 125 sz eat each other + 125 sz eat 250 wz+ 500 wz eat each other
  In this set the max. time taken by set 3rd ie 500wz take 250 min. to eat each other but in first set there are 2 sz left which eat each other in 0.75 min but can be used to minimize time in such away in next sets
  Can be guessed by thinking about last set, then 2nd last)))
  
  Set is: 114sz eat each other, 136sz eat 272wz or vice versa, 478 wz eat each other.
  1st set- 114 sz , they eat each other such that two at a time i.e if 1 eat 2&3, 2 eat 1&3 , 3 eat 1&2. i.e. a triplet form, time taken to finish each other is 1.5 min. As given that sz eat two in 1.5 min. and there r 38 such triplet; so all 38 triplet finished in 1.5 min.
  2nd set- 136 sz eaten by 272 wz in 136*1.5= 204min., but also sz eat wz, so life of each wz left is 0.25 min. So, when 272wz i.e 136 wz start eating 136 wz they eat in 136*0.25= 34 min. So total time is 204+34= 238 min. Form the triplet in this also like wz-sz-wz.
  3rd set- 478 wz start eating each other i.e 239 start eating 239 other wz so time taken is 239*1= 239 min.
  Max. time taken is by set 3rd so, minimum time taken is 239 min. with 0 z left
  0 left, 239 min.
  
  Ans. None of these
  
• 10 years agoHelpfull: Yes(1) No(1)
• I think the ans is (d). 0 left, 500 min.
• 10 years agoHelpfull: Yes(0) No(2)