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What these monkeys are doing boss

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What these monkeys are doing boss

by maihuna » Tue Dec 08, 2009 6:07 am
In an experiment researchers followed the activities of three groups of monkeys in a given 90 days period:

M1: Goes for food A for 3 days in a row and then 2 days for another food B in a row and repeats this pattern.
M2: Goes for 2 day in a row for A and then next two days for B and repeats the pattern.
M3: Goes for food A only when both M1 and M2 goes for it.

How many times M3 eats food A in the given 90 day period.

12
15
23
30
45
Last edited by maihuna on Tue Dec 08, 2009 10:34 am, edited 1 time in total.
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by sunil_snath » Tue Dec 08, 2009 8:48 am
Very good question. I am getting the answer as 27 :(.

Here's my explanation:

every 20th (LCM of (3+2) and (2+2)) day we come to the starting point, and we have 3 common days every 10 days. So extrapolating this info to 90 days would give

9*3 = 27

Maybe I am missing some boundary conditions.
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by Testluv » Tue Dec 08, 2009 9:21 am
I am also getting 27; my reasoning was essentially the same as sunil's. Actually, this question is not well-designed. We know that the third monkey eats food A whenever the other two do. But, we don't know if this is the only time the third monkey eats food A. More technically, we learn that both of the other two monkeys eating food A is a sufficient condition but we have not learned that it is a necessary condition for the third monkey eating food A. To correct for this, the question either should have asked for the minimum number of times the third monkey eats food A OR it should have specified that the third monkey eats food A if and only if both of the other monkeys have. Because this question is not well-designed, I would not trust the source, and I would not worry about the fact that 27 doesn't show up in the answer choices.
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by maihuna » Tue Dec 08, 2009 10:00 am
Testluv wrote:I am also getting 27; my reasoning was essentially the same as sunil's. Actually, this question is not well-designed. We know that the third monkey eats food A whenever the other two do. But, we don't know if this is the only time the third monkey eats food A. More technically, we learn that both of the other two monkeys eating food A is a sufficient condition but we have not learned that it is a necessary condition for the third monkey eating food A. To correct for this, the question either should have asked for the minimum number of times the third monkey eats food A OR it should have specified that the third monkey eats food A if and only if both of the other monkeys have. Because this question is not well-designed, I would not trust the source, and I would not worry about the fact that 27 doesn't show up in the answer choices.
Humm testluv let me put the following: The answer sand data both are correct, and your answer is wrong. Coming back to questions, yes it contains some word issue, like, group vs M3 as monkey and minimum no of days etc, if it helps I will say there are three monkeys, M1 M2 and M3 and M3 eats food A only when both M1 and M2 eats. Now solve it. Do not go for validity of the source, look for reasoning, its quite unseen and very good for a 2 minute Q
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by ChillingHEAT » Tue Dec 08, 2009 11:50 am
Is the answer is 15?

I am getting by taking LCM of
M1 : 3
M2 : 2
as 6.

Since we have total as 90 days and in every 6th day both M1 and M2 will go together therefore M3 will accompany them on every 6th day. Therefore, answer should be 90/6 = 15
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by maihuna » Tue Dec 08, 2009 11:58 am
ChillingHEAT wrote:Is the answer is 15?

I am getting by taking LCM of
M1 : 3
M2 : 2
as 6.

Since we have total as 90 days and in every 6th day both M1 and M2 will go together therefore M3 will accompany them on every 6th day. Therefore, answer should be 90/6 = 15
Humm nope see this pattern:

M1: YYYNNYYYNNYYYNN
M2: YYNNYYNNYYNNYYN
M3: YYNNNY...

Now you can generate the pattern...the match is 2 1 2 sort of... testluv can u recognize flaw in ur logic for...of 27...
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by djkvakin » Tue Dec 08, 2009 2:39 pm
maihuna wrote:
Testluv wrote:I am also getting 27; my reasoning was essentially the same as sunil's. Actually, this question is not well-designed. We know that the third monkey eats food A whenever the other two do. But, we don't know if this is the only time the third monkey eats food A. More technically, we learn that both of the other two monkeys eating food A is a sufficient condition but we have not learned that it is a necessary condition for the third monkey eating food A. To correct for this, the question either should have asked for the minimum number of times the third monkey eats food A OR it should have specified that the third monkey eats food A if and only if both of the other monkeys have. Because this question is not well-designed, I would not trust the source, and I would not worry about the fact that 27 doesn't show up in the answer choices.
Humm testluv let me put the following: The answer sand data both are correct, and your answer is wrong. Coming back to questions, yes it contains some word issue, like, group vs M3 as monkey and minimum no of days etc, if it helps I will say there are three monkeys, M1 M2 and M3 and M3 eats food A only when both M1 and M2 eats. Now solve it. Do not go for validity of the source, look for reasoning, its quite unseen and very good for a 2 minute Q
Well, to put all matters to rest, I created a simple spreadsheet and the answer is undoubtedly 27.
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by Stuart@KaplanGMAT » Tue Dec 08, 2009 2:39 pm
maihuna wrote:Humm nope see this pattern:

M1: YYYNNYYYNNYYYNN
M2: YYNNYYNNYYNNYYN
M3: YYNNNY...

Now you can generate the pattern...the match is 2 1 2 sort of... testluv can u recognize flaw in ur logic for...of 27...
The answer to the question posted (ignoring the grammatical/stylistic issues) is definitely 27. If you were to write out the entire sequence of 90 meals, you'd count 27 matches between groups 1 and 2.

As previously noted, the cycle is 5*4 = 20 meals. For each set of 20, there are 6 points of A intersection, then for the extra 10 there are an additional 3 points of intersection.

6*(4 sets of 20) = 24
24 + 3 = 27
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by Testluv » Tue Dec 08, 2009 4:00 pm
maihuna wrote:
ChillingHEAT wrote:Is the answer is 15?

I am getting by taking LCM of
M1 : 3
M2 : 2
as 6.

Since we have total as 90 days and in every 6th day both M1 and M2 will go together therefore M3 will accompany them on every 6th day. Therefore, answer should be 90/6 = 15
Humm nope see this pattern:

M1: YYYNNYYYNNYYYNN
M2: YYNNYYNNYYNNYYN
M3: YYNNNY...

Now you can generate the pattern...the match is 2 1 2 sort of... testluv can u recognize flaw in ur logic for...of 27...
There was no flaw in my logic (I am Testluv!). As Sunil and Stuart point out, there is a recurring pattern where there are 6 points of intersection for every sequence of 20 meals. 90/20 = 4.5. Therefore, there are 4.5 * 6 or 27 points of intersection. 27 does not appear among the answer choices. Therefore, I would not trust this source.
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