| Author | 4 Posts |
sam_step1
Forum Junior
Topics: 23
Posts: 38
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Hi can any one answer following nbme qs for me-
Disfectant kills 90% of bacterium in an inoculumin 10 min. Whats shortest time in min. required to reduce no. of bacterium form 10 power 9 to less than 10?
10/ 50/ 100/ 150/ 140.
Thanks
~Sam.
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| kiterunner
Forum Junior
Topics: 14
Posts: 65
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To me it seems like after every 10 min, (n-1)th power of 10 bacteria are left, so to reduce the # of bacteria from '10 power 9' to '10 power 1' or less, would be (9-0) or (9-1) times 10 min.
I`d go for 90 since we want the # of bacteria to be less than 10. So I guess the answer ud be C)100
What`s the answer ????
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| drgho
Forum Junior
Topics: 1
Posts: 115
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i think 100 mins too.
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| romano
Forum Newbie
Topics: 3
Posts: 18
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Every 10 minutes the number decrease with 1 power. So after 90 minutes this gives 10 bacteria left, but it should be less than 10 so you'll have to go with 100 minutes.
Hope I doesn't get similar questions on the exam when the options are like, 90, 100, 110... That makes it a bit harder.
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