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Devil's Dozen!!!

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Re: Devil's Dozen!!! [#permalink] New post   23 May 2014, 05:14
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PathFinder007 wrote:
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?

700+ question.

Given: \(11m+7b\leq{15}\), where \(m\) and \(b\) are prices of one muffin and one brownie respectively.
Question: is \(27m+27b\leq{45}\)? --> \(9m+9b\leq{15}\). Question basically asks whether we can substitute 2 muffins with 2 brownies.

Now if \(m>b\) we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m<b\) we won't know this for sure.

But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m>b\) or \(m<b\)) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.

(1) $15 is enough to buy 7 muffins and 11 brownies --> \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient.
(1) $15 is enough to buy 10 muffins and 8 brownies --> \(10m+8b\leq{15}\): we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.

Answer: A.


Hi Bunnel,

I just want to know why statement 2 is not sufficient. following is my logic

from question 11m+7b =15

st1- 7m+11b = 15

subtract st2 from 1

4m-4b = 0
m=b

so in question we can say 18b = 15
b=15/18=5/6. using this we can get the price for 27m and 27b

now same i can get from st2 then why this is not sufficient?

please clarify

Thanks.


There is a huge difference between \(11m+7b\leq{15}\) and \(11m+7={15}\). "Enough" should be translated as \(\leq\) only.
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Re: Devil's Dozen!!! [#permalink] New post   02 Jul 2014, 22:20
Bunuel wrote:
priyavenugopal wrote:
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.

Given,
From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..)
From (1) stmt - 7m + 11b = 15
From (2) stmt - 10m + 8b = 15
Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap)
Solving qn stem n stmt 2 eqns, got m and b as 5/6

Having m and b values, we can find whthr 45$ is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D.


Not even to analyze the rest of it, I must say that there is a huge difference between \(11m+7b\leq{15}\) and \(11m+7={15}\). You can not just write = sing instead of <= just because it's more convenient and "enough" should be translated as <= only.


Bunuel
the statement says 15$ is enough, so it implies <= 15$ , consider max situation that it takes 15$ to buy 10m and 8b, which implies 45$ is sufficient . Can we not imply that for any amount less than 15$ , the solution will hold?.

Please help
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Re: Devil's Dozen!!! [#permalink] New post   03 Jul 2014, 05:57
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ayushee01 wrote:
Bunuel wrote:
priyavenugopal wrote:
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.

Given,
From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..)
From (1) stmt - 7m + 11b = 15
From (2) stmt - 10m + 8b = 15
Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap)
Solving qn stem n stmt 2 eqns, got m and b as 5/6

Having m and b values, we can find whthr 45$ is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D.


Not even to analyze the rest of it, I must say that there is a huge difference between \(11m+7b\leq{15}\) and \(11m+7={15}\). You can not just write = sing instead of <= just because it's more convenient and "enough" should be translated as <= only.


Bunuel
the statement says 15$ is enough, so it implies <= 15$ , consider max situation that it takes 15$ to buy 10m and 8b, which implies 45$ is sufficient . Can we not imply that for any amount less than 15$ , the solution will hold?.

Please help


$15 is enough to buy 10 muffins and 8 brownies.
15*3=$45 is enough to buy 10*3=30 muffins and 8*3=24 brownies.

But this is not sufficient to say whether $45 enough to buy 27 muffins and 27 brownies.
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Re: Devil's Dozen!!! [#permalink] New post   18 Aug 2014, 10:08
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mahendru1992 wrote:
Bunuel wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:

As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Answer: A.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?


There might be a group of patients who has neither arachnophobia nor acrophobia. So, it should be a + b + c + neither = 58.
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Re: Devil's Dozen!!! [#permalink] New post   18 Aug 2014, 10:29
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mahendru1992 wrote:
Bunuel wrote:
mahendru1992 wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:

As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Answer: A.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?


There might be a group of patients who has neither arachnophobia nor acrophobia. So, it should be a + b + c + neither = 58.

Okay but b+c=13, D can still be the answer? The question is asking us for b+c


The question asks for the number of patients who has acrophobia. Yellow box in matrix in my solution.

IF the number of patients with neither arachnophobia nor acrophobia is 0, then there will be 26 patients with acrophobia but IF the number of patients with neither arachnophobia nor acrophobia is 13, then there will be 13 patients with acrophobia.
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Re: Devil's Dozen!!! [#permalink] New post   22 Oct 2014, 09:43
hey bunuel, are we assuming that y=1 in question 12? If so, why? I saw that A was not sufficient but chose C because I thought we could simply plug in different values for X (36, 81) so that y could be a number of different fractions
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Re: Devil's Dozen!!! [#permalink] New post   22 Oct 2014, 09:49
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alexanthony813 wrote:
12. If x>0 and xy=z, what is the value of yz?

(1) \(x^2*y=3\). If \(x=1\) then \(y=z=3\) and \(yz=9\) but if \(x=3\) then \(y=\frac{1}{3}\), \(z=1\) and \(yz=\frac{1}{3}\). Not sufficient.

(2) \(\sqrt{x*y^2}=3\) --> \(x*y^2=9\) --> \((xy)*y=9\) --> since \(xy=z\) then: \(z*y=9\). Sufficient.

Answer: B.

hey bunuel, are we assuming that y=1 in question 12? If so, why? I saw that A was not sufficient but chose C because I thought we could simply plug in different values for X (36, 81) so that y could be a number of different fractions


We are not assuming that y is 1.

From (2) we have that \((xy)*y=9\) and from the stem we know that \(xy=z\). Now, simply substitute xy with z in \((xy)*y=9\) to get \(z*y=9\).

Hope it's clear.
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Re: Devil's Dozen!!! [#permalink] New post   06 Nov 2014, 14:15
Dear Bunuel,

I have a question to your solution on the following problem:

2. If n is a positive integer and p is a prime number, is p a factor of n!?

you specify for statement 2 the following sentence:

(2) p is a factor of (n+2)!/n! --> \frac{(n+2)!}{n!}=(n+1)(n+2) --> if n=2 then (n+1)(n+2)=12 and for p=2 then answer will be YES but for p=3 the answer will be NO. Not sufficient.

now consider (n+2)!/n!, this can be reduced to (n+1)(n+2), for n is odd, the expression is even, further for n is even, the expression is even again.
the statement specifies that p is a factor of an even number and is prime, thus p should be even and the only even prime number is 2. How did you consider p= 3? in your solution.

Thanks in advance,
Sant
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Re: Devil's Dozen!!! [#permalink] New post   07 Nov 2014, 04:28
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santorasantu wrote:
Dear Bunuel,

I have a question to your solution on the following problem:

2. If n is a positive integer and p is a prime number, is p a factor of n!?

you specify for statement 2 the following sentence:

(2) p is a factor of (n+2)!/n! --> \frac{(n+2)!}{n!}=(n+1)(n+2) --> if n=2 then (n+1)(n+2)=12 and for p=2 then answer will be YES but for p=3 the answer will be NO. Not sufficient.

now consider (n+2)!/n!, this can be reduced to (n+1)(n+2), for n is odd, the expression is even, further for n is even, the expression is even again.
the statement specifies that p is a factor of an even number and is prime, thus p should be even and the only even prime number is 2. How did you consider p= 3? in your solution.

Thanks in advance,
Sant


The red part is not correct. An even number can have an odd factor. For example, 12 is even and it has an odd factor 3.
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Re: Devil's Dozen!!! [#permalink] New post   04 Jun 2017, 05:35
Bunuel wrote:
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ question.

(1) a-b=15. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D.


hi Bunuel,
I didn't get your point mentioned in "For a given perimeter equilateral triangle has the largest area." because in statement B nothing is mentioned about type of triangle. Also please correct me if I am wrong for the statement "can't we do like this if we are assuming triangle as equilateral: (50= sqrt of 3 * a^2 ) / 4 and find what is a? Then a+a+a will give definite answer. So B is also sufficient to answer this question."

Thanks.
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Re: Devil's Dozen!!! [#permalink] New post   04 Jun 2017, 06:16
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goalMBA1990 wrote:
Bunuel wrote:
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ question.

(1) a-b=15. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D.


hi Bunuel,
I didn't get your point mentioned in "For a given perimeter equilateral triangle has the largest area." because in statement B nothing is mentioned about type of triangle. Also please correct me if I am wrong for the statement "can't we do like this if we are assuming triangle as equilateral: (50= sqrt of 3 * a^2 ) / 4 and find what is a? Then a+a+a will give definite answer. So B is also sufficient to answer this question."

Thanks.


From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30. So, if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.
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Re: Devil's Dozen!!! [#permalink] New post   27 Oct 2018, 14:08
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Hi Bunuel! I am a bit confused for the case n=1.

If n=1 then statement one equates to 5 and statement 2 equates to 6. Hence the possible values of P are 5,2,3. none of which is a factor of 1?

If each statement is true, then P should then be a prime factor of 5 and either 2 or 3, but that's impossible?

Will be grateful if you could clarify. Thank you.

Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.

(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)-1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.
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Re: Devil's Dozen!!! [#permalink] New post   12 Oct 2019, 16:03
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


What if there are two pairs of VC VC, then you would have two cobras? Question is asking for absolute number of cobras, but you could have infinite pair of vipre cobras right? thus E.. Maybe I am missing something.. thank you
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Re: Devil's Dozen!!! [#permalink] New post   13 Oct 2019, 00:48
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DanielMx wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


What if there are two pairs of VC VC, then you would have two cobras? Question is asking for absolute number of cobras, but you could have infinite pair of vipre cobras right? thus E.. Maybe I am missing something.. thank you


We cannot have more than 1 cobra. If there are 2 cobras and 2 vipers then the second statement will NOT hold. (2) says from ANY two snakes from Pandora's box at least one is a viper. If there are 2 cobras and 2 vipers then we could have two snakes from which BOTH are cobras.
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Re: Devil's Dozen!!! [#permalink] New post   27 Jun 2020, 09:36
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.

(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)-1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.




Hi Bunuel,
As per 1, the following would be true
n+2!-n!= P*(some number)

Simplifying this we get-

n!(n2+3n+1)=p*(some number). The expression n2+3n+1 is an integer, so isn't p a factor of n!

Thanks,
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Re: Devil's Dozen!!! [#permalink] New post   27 Jun 2020, 11:43
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srikarkali wrote:
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.

(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)-1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.




Hi Bunuel,
As per 1, the following would be true
n+2!-n!= P*(some number)

Simplifying this we get-

n!(n2+3n+1)=p*(some number). The expression n2+3n+1 is an integer, so isn't p a factor of n!

Thanks,


Why can't p be a factor so n^2 + 3n + 1 instead ? There are two cases given in the solution, in one p is a factor of n! and in another p is not a factor of n! (it's a factor of n^2 + 3n + 1).
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Re: Devil's Dozen!!! [#permalink] New post   07 Jun 2023, 10:12
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