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In India we have an exam called NEST. I gave it today, and this was a question I encountered:

Lactobacillus sp. and Streptococcus sp. are two bacterial species responsible for curdling milk. One quantum of each of these species was introduced to a very large container of milk. One quantum of either species can curdle 10 ml of milk in 26 minutes, which is also the doubling time for Streptococcus sp.. The doubling time for Lactobacillus sp., however is 78 minutes. What will be the ratio of the total milk curdled by Streptococcus sp. to Lactobacillus sp. at the end of 156 minutes.

Now I am a biology student. This came in the general section of the paper which means it can be solved by using basic mathematics. Here's what I did:

Let $$N_S=number\;of\;quanta\;Streptococcus\;sp.\;in\;the\;milk\\N_L=number\;of\;quanta\;Lactobacillus\;sp.\;in\;the\;milk\\Then,\;N_S=2^{t/26}\;and\;N_L=2^{t/78} ;where\;t\;is\;the\;time\;in\;minutes $$

The rate at which milk is being curdled is directly proportional to the number of bacteria in the container. Thus, the rate at which milk is being curdled will vary as well. Initally, I imagined a scenario in which, say Streptococcus sp. curdles 10 ml of milk in 26 minutes, 20 ml of milk in the next 26 minutes, 40 ml of milk inthe the next 26 minutes and so on. However, this approach seemed slightly wrong to me (gut feeling, nothing else). So in the end, I did not attempt the question(there is negative marking).

How would one solve the question.Thoughts please. Thanks.

P.S:I was not quite sure what tags to add to this question. So add any as you feel appropriate.

Martin Sleziak
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TKRao
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3 Answers3

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Your gut feeling was incorrect (and so your initial idea was right). In the Streptococcus sp. curdles $10ml$, then $20ml$, $40ml$, etc. for 6 doubling times, giving a total of $630ml$. The Lactobacillus sp. has only $2$ doubling times, giving a total amount of curdled milk as $3*30ml=90ml$. Therefore the ratio is $7$.

preferred_anon
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    The options in the paper were: (a)3, (b)3.5, (c)5.25, (d)7. I would like to point out that although there is a likelihood that the options are wrong, its minuscule. – TKRao May 31 '14 at 11:03
  • Sorry, this was me being stupid. As the other answerers have pointed out, the lacto bacteria curdle milk 3 times per doubling time, which gives a ratio of $7$. – preferred_anon Jun 10 '15 at 12:05
  • What happens if we integrate the $\mathrm{2^{t/26}}$ and $\mathrm{2^{t/78}}$ with respect to time from 0 to 156 minutes? That method gives different answer.Is it wrong? – S R Maiti Jun 01 '18 at 10:05
  • @ShoubhikRajMaiti It depends on the interpretation of the term "doubling time". The discrete model that I and the asker used presumes that at the end of each doubling time, the number of bacteria instantly doubles in value. Using integration, the amount of bacteria increases continuously throughout the time, still doubling in the same time period. I think the simpler discrete model is intended in the question, while the continuous model is probably more realistic. – preferred_anon Jun 01 '18 at 11:55
  • Yes, that explains it.. – S R Maiti Jun 01 '18 at 13:24
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as was said by daniel littlewood, 630 ml milk is curdles by Streptomyces sp. at the end of 156 minutes,as with the Lactobacillus sp. it doubles every 78 minutes,but still it is able to curdle 10 ml each upto 78 minutes becuase there is 1 quanta, after that since no. of quanta doubles therafter after every 26 minutes 20 ml is curdled, so total milk curdled by it is 10+10+10 IN first 78 minutes plus 20+20+20 ml in next 76 minutes, so total 90 ml curdles by Lactobacillus sp. in 156 minutes , now take out the ration of total milk curdled, it turns out to be 630/90= 7, one of the given options, and it turns out to be right! glad you asked this intriguing question, there you go!

Asif
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I am using acronym L and S for both the bacterias, and x be 1 quanta of S and y be 1 quanta of L.

S curdles 10 ml milk using x and doubles up in 26 minutes and becomes 2x. In 156 minutes you have 6 blocks of 26 minutes. So, S goes from

$x->2x->4x->8x->16x->32x$

2x will curdle 20ml,4x -40ml,8x -80ml,16x -160 ml,32x -320 ml

Total milk curdled 630 ml by S in 156 minutes

L curdles 10 ml milk in 26 minutes, but doubles up at the 78th minute. So, L goes from y -> 2y at 78 minute.

Total milk curdled by y=10+10+10+20+20+20=90 ml

Ratio: $\frac{630}{90}$=7

MonK
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