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There are 10 bacteria in a flask. Every hour 3 bacteria die and the remaining ones are each divided into 2 after 1 day, how many bacteria will live there?
Assume that the flask is large enough to contain any number of bacteria?

  1.   226
  2.   226 + 6
  3.   224 + 6
  4.   224
asked in Combinatory by Veteran (10.2k points)   | 75 views

226 + 6 ?

yes,your answer is correct,,how did you get that??

1 Answer

+1 vote
Best answer
hours bacteria alive difference
1 (10-3)*2 = 14  
2 (14-3)*2 = 22 22-14 = 8
3 (22-3)*2 = 38 38-22 = 16
4 (38-3)*2 = 70 70-38 = 32

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from above table we found that

after 2 hours bacteria = 14 + 2= 22

after 3 hours bacteria = 14 + 23 + 24 = 38

after 4 hours bacteria = 14 + 23 + 24 + 25= 70

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so after k hours bacteria = 14 + 23 + 24 + 2+ ........ + 2k+1

put value of k as 24

baacteria = 14 + 23 + 24 + 2+ ........ + 225

                   = 7 + 1 + 2 + 22 + 23 + 24 + 2+ ........ + 225

              = 7 + 226 - 1

              = 226 + 6

answered by Boss (7.4k points)  
selected by

thanks for the above solution..

can we solve this using recursion?

like i was solving like this but could 'nt proced further-

an+1 = 2(an - 3)

so,an =( an+1/2) + 3

an+1 = (an+2/2) + 3 putting this in an,we get a= (an+2 / 22) + 3/2 + 3

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solving o..we get

an = a n + k /2 + 3( 1+ 1/2 + 1/22 + ....... + 1/ 2k-1)

now,i dun know how to solve further

i was taking base condition as a10 = 14..but i did nt lead me anywhere..can u pls help

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