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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 24 Sep 2018, 01:31
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28% (02:16) correct 72% (02:21) wrong based on 155 sessions

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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%
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D
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 02 Oct 2018, 02:12
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Bunuel
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%
Bunuel

Can you please confirm-Even I got the answer as C 20%

=73-61/61=20%
Where, Q multiplies 73 times and P 61 times.
Show more

I too got C at first attempt.

But assuming a year composed by 365 days you get:
P multiplies \(\frac{(365*3)}{18}=60.833=60\) times (P does not complete its 61st cycle)
Q multiplies \(\frac{(365*3)}{15}=73\) times

\(\frac{73}{60}-1=0.2166\) which is closer to 22%.
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 24 Sep 2018, 01:36
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Bunuel
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%
Show more

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
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 01 Oct 2018, 11:44
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Bunuel
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%
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Bunuel

Can you please confirm-Even I got the answer as C 20%

=73-61/61=20%
Where, Q multiplies 73 times and P 61 times.
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 23 Mar 2024, 11:27
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Poorly Quality question . Requires hideous approximations of 365/18 & 365/15 to select b/w 20 & 22%
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri Updated on: 24 Mar 2024, 07:31
Originally posted by 8Harshitsharma on 24 Mar 2024, 05:00.
Last edited by 8Harshitsharma on 24 Mar 2024, 07:31, edited 1 time in total.
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GMATinsight

Bunuel
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%
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
Show more
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.­
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 24 Mar 2024, 06:16
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Given: At a lab, bacteria P multiplies itself in every 18 days, while bacteria Q multiplies itself in every 15 days.
Asked: 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?

3-year = 3*365 = 1095 days

The number of times bacteria P multiplies itself in 3-year period = 1095/18 = 60 times
The number of times bacteria Q multiplies itself in 3-year period = 1095/15 = 73 times

 The percent by which the number of times bacteria Q multiplies itself is more than the number of times bacteria P multiplies itself in a 3-year period = (73/60 - 1)*100% = 1300/60 = 22%

IMO D
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 24 Mar 2024, 06:21
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8Harshitsharma
 
GMATinsight

Bunuel
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%
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
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.­
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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
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 24 Mar 2024, 07:07
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Regor60


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

 
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­Hello friend, 

In the interest of people going through the post replies looking for an explanation of this problem, I want to point out some assumptions I can infer from the statements you used that are unwarranted in a GMAT problem - whether quant or verbal.
Quote:
   the normal understanding that bacterial growth is a continuous process should prevail, with 60 5/6 and 20% being the appropriate answer.
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You are going out of bounds from a GMAT problem's standpoint by bringing in an outside assumption. The question states clearly that bacteria P and bacteria Q respectively multiply "in" every 18 days and 15 days.
Ask the question: What if the bacteria is in the process of chromosome and plasma division for the remaining days in the cycle and the division concludes exclusively on the 18th day and 15th day of the P's and Q's cycle? In that case, can you count those prospective bacteria as separate bacteria when they still share the same body? Obviously not, since they are not physically separated.

Hence, answer choice D is unambiguously correct for this problem.
Quote:
Since the question doesn't clearly stipulate that it is interested in the number of "complete" doublings,
Show more
The question states the respective type of bacteria "multiplies" and does not necessarily double. They may very well have different growth rates for all we know. So the assumption of a direct linear relation between multiplication and days past is unwarranted. This, however, doesn't make a difference in this question, still, it is important to refrain from using explicit assumptions such as this in GMAT questions.

We all should play within the boundary set by the question maker, and the cognizance of this makes such a question hard :)

Hope this helps­
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 03 Apr 2024, 15:15
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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
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 04 Apr 2024, 00:28
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rahul5657
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
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­
The question does not ask to compare the number of bacteria at the end of a 3-year period. Instead, it asks to compare the number of times each colony doubled. ­
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At a lab, bacteria P multiplies itself in every 18 days, while bacteri 04 Apr 2024, 01:43
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Hi rahul5657

You have misinterpreted the question stem.

The question stem is about the percentage difference between the number of times the population of Bacteria P and Q multiply
while you are focussing on the percentage difference in population of bacteria in 360 days


If you just understand that there is a difference between the two languages then you will be one step closer to answer

In my explanation, I have taken the time duration 90 days (LCM of 15 and 18) because the percentage increase remains the same in any fixed duration.

i.e. Number of multiplication of Bacteria Q in 360 days = 360/15 = 24
and Number of multiplication of Bacteria P in 360 days = 360/18 = 20

Percentage of Q greater than that of P = (24-20)*100/20 = 20%

Answer: Option C

I hope this explains your doubts! :)

rahul5657
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
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