At a lab, bacteria P multiplies itself in every 18 days, while bacteri

Let, Total days = LCM (15 and 18) = 90

P multiplies 90/18 = 5 times in 90 days
Q multiplies 90/15 = 6 times in 90 days

i.e. in any given time (i.e. 3 years as well) the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself = (6-5)*100/5 = 20%

Answer: Option C

At a lab, bacteria P multiplies itself in every 18 days, while bacteria Q multiplies itself in every 15 days. Approximately by what percent is the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period?

(A) 12%
(B) 16%
(C) 20%
(D) 22%
(E) 33%

At a lab, bacteria P multiplies itself in every 18 days, while bacteria Q multiplies itself in every 15 days. Approximately by what percent is the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period?

(A) 12%
(B) 16%
(C) 20%
(D) 22%
(E) 33%

Bacteria P multiplies itself in every 18 days =>
at 0 days = x number of bacteria P
after 18 days = 2x
after 36 days = 2* (2x) = 4x
after 56 days = 2*(4x) = 2^(56/18) times of x = 2^3 times of x
...
after 3 yrs = 2^(365*3/18) times of x= 2^(60.83) times of x ......(bacteria P)

Similarly, bacteria Q multiplies itself in every 15 days =>
at 0 days = y number of bacteria Q
after 3 yrs = 2^(365*3/15) times of y = 2^73 times of y .......(bacteria Q)

Here the number of times the bacteria increases is exponential not linear. Which will give different answer.

Can some pls explain?


Bunuel GMATinsight

 

The assumption above in red is absolutely not warranted, the cyclicity GMATinsight claimed will only be true when the number of days (d) falls within the range \( 90*k - 15 < d < 90*k + 15 \), where k is a positive integer (i.e. k= 1, 2, 3, ....). Outside this range, the distribution of the no. of cycles of P to that of Q will not be in the ratio 5:6 because either Q would not have completed its final cycle or would have completed its cycle earlier than P (I know this sounds overcomplicated, but I lack the comm to make it any simpler than this. When you read the explanation with the example below you'll get what I am saying) 

Imagine we are on the day 105, P would have completed 105/18 = 5 cycles, BUT Q would have completed 105/15 = 7 cycles (Note that d = 105 is of the form 90*1 + 15 which falls outside the range that we found above, and hence, the cycles are uneven i.e. not in the ration 5:6)

Now coming to our problem, since 3 years have 365*3 = 1095 days which is 90*12 + 15, it falls juusst outside of the range we found above hence the cycles will not be a normal 5:6 ratio cycle (if you do the calculation by hand you'll see P undergoes 60 complete cycles (1095/18) whereas Q undergoes 73 complete cycles (1095/15) AND NOT 60*(6/5) i.e. 72, and hence, the answer turns out to be \(\frac{73-60}{60}*100\)­ => 1300/60 ≅ 21.67% closer to 22%. CHOICE D is correct.

Nice test of LCM and uneven cycles in the question.

I hope this helps someone.­

Q doubles 73 times in 3 years.

P doubles 60 5/6 times over the same period.

Truncating 60 5/6 to 60 is the difference between 20% and 22%.

Since the question doesn't clearly stipulate that it is interested in the number of "complete" doublings, the normal understanding that bacterial growth is a continuous process should prevail, with 60 5/6 and 20% being the appropriate answer.

Posted from my mobile device

 

   the normal understanding that bacterial growth is a continuous process should prevail, with 60 5/6 and 20% being the appropriate answer.

Since the question doesn't clearly stipulate that it is interested in the number of "complete" doublings,

Bacteria P multiplies itself in every 18 days =>
at 0 days = x number of bacteria P
after 18 days = 2x
after 36 days = 2* (2x) = 4x
after 56 days = 2*(4x) = 2^(56/18) times of x = 2^3 times of x
...
after 3 yrs = 2^(365*3/18) times of x= 2^(60.83) times of x ......(bacteria P)

Similarly, bacteria Q multiplies itself in every 15 days =>
at 0 days = y number of bacteria Q
after 3 yrs = 2^(365*3/15) times of y = 2^73 times of y .......(bacteria Q)

Here the number of times the bacteria increases is exponential not linear. Which will give different answer.

Can some pls explain?


Bunuel GMATinsight