Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
23 May 2014, 05:14
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 4m4b = 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.
_________________



Intern
Joined: 06 May 2014
Posts: 6
Concentration: Marketing, Technology

Re: Devil's Dozen!!!
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
03 Jul 2014, 05:57
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.
_________________



Manager
Joined: 13 Aug 2012
Posts: 90

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Aug 2014, 08:18
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 doubleset matrix: As you can see # of patients who has acrophobia is 5845=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?
Attachments
sets.png [ 3.34 KiB  Viewed 1780 times ]



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Aug 2014, 10:08
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 doubleset matrix: As you can see # of patients who has acrophobia is 5845=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 = 58and 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.
_________________



Manager
Joined: 13 Aug 2012
Posts: 90

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Aug 2014, 10:22
Bunuel wrote: 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 doubleset matrix: As you can see # of patients who has acrophobia is 5845=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 = 58and 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



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Aug 2014, 10:29
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 doubleset matrix: As you can see # of patients who has acrophobia is 5845=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 = 58and 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.
_________________



Intern
Joined: 04 Oct 2014
Posts: 5

Re: Devil's Dozen!!!
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
22 Oct 2014, 09:49
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.
_________________



Senior Manager
Joined: 27 Aug 2014
Posts: 356
Location: Netherlands
Concentration: Finance, Strategy
GPA: 3.9
WE: Analyst (Energy and Utilities)

Re: Devil's Dozen!!!
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
07 Nov 2014, 04:28
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.
_________________



Manager
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 188
Location: India
MISSION : 800
WE: Design (Manufacturing)

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Mar 2015, 00:58
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. Hi plz explain Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper. Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!! Now what if dr are 10 cobras and 10 viper Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!!
_________________
Thank you
+KUDOS
> I CAN, I WILL <



Math Expert
Joined: 02 Aug 2009
Posts: 8150

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Mar 2015, 01:42
dpo28 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. Hi plz explain Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper. Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!! Now what if dr are 10 cobras and 10 viper Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!! hi, the reason there is only one cobra byu statement 2.... if you pick 2 snakes atleast one is viper... say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper.. so this statement will not be true... but we know this statement is true, so we can have only one cobra.. and when we pick up two snakes it could be two vipers or one viper and one cobra... hope it helped
_________________



Manager
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 188
Location: India
MISSION : 800
WE: Design (Manufacturing)

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Mar 2015, 02:39
chetan2u wrote: dpo28 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. Hi plz explain Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper. Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!! Now what if dr are 10 cobras and 10 viper Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!! hi, the reason there is only one cobra byu statement 2.... if you pick 2 snakes atleast one is viper... say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper.. so this statement will not be true... but we know this statement is true, so we can have only one cobra.. and when we pick up two snakes it could be two vipers or one viper and one cobra... hope it helped Hi chetan2u Thanks for the reply but am not able to get it The questions asks the no of cobras right..!! and for each viper dr is one cobra. Thn dr must be an equal no of both of them. But we cannot sat what's the number. This is where am confused...!! ? I know i sound silly but sometimes your are stuck at a point in some questions..!
_________________
Thank you
+KUDOS
> I CAN, I WILL <



Math Expert
Joined: 02 Aug 2009
Posts: 8150

Re: Devil's Dozen!!!
[#permalink]
Show Tags
18 Mar 2015, 03:41
dpo28 wrote: [quote="chetan2u"
hi, the reason there is only one cobra byu statement 2.... if you pick 2 snakes atleast one is viper... say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper.. so this statement will not be true... but we know this statement is true, so we can have only one cobra.. and when we pick up two snakes it could be two vipers or one viper and one cobra... hope it helped Hi chetan2u Thanks for the reply but am not able to get it The questions asks the no of cobras right..!! and for each viper dr is one cobra. Thn dr must be an equal no of both of them. But we cannot sat what's the number. This is where am confused...!! ? I know i sound silly but sometimes your are stuck at a point in some questions..![/quote] hi, the question never tells us that there is one cobra for each viper.... it says there are atleast one cobra and one viper... so it could be 1 cobra and 10 viper or 1 viper and 10 cobras or 10 viper and 100 cobras.... statement B the statement ll tells us atleast one is viper.. so for more than one cobra, atleast one viper in two snakes will not stand. it is possible to have both cobra if the cobras are more than 2 ... and this statement will not be true... but we know this statement is true, so we can have only one cobra..and when we pick up two snakes it could be two vipers or one viper and one cobra
_________________



Intern
Joined: 06 Jul 2015
Posts: 7

Re: Devil's Dozen!!!
[#permalink]
Show Tags
28 Sep 2015, 11:00
Bunuel wrote: mithun2vrs wrote: Bunuel wrote: 10. There is at least one viper and at least one cobra in a 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. How do we know that there are no other varieties of snakes other than cobra and viper. This is precisely why I chose E. Answer to the question is B, not E. If there is some other snake, then group of {other, cobra} will be possible and the statement (2) will be violated (basically the possibility of other variety of snake is ruled out by the same logic as the possibility of second cobra). Hope it's clear. Dear, How do we know that there are only 2 snakes in the box? In the statement 2, it says " From any two snakes from Pandora's box at least one is a viper.". Doesn't it mean that when picking any 2 snakes out of more numbers of the snakes? Does it imply that there are only two snakes in the box? Thanks! Regards Andy



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
28 Sep 2015, 22:55
andy2whang wrote: Bunuel wrote: Bunuel wrote: 10. There is at least one viper and at least one cobra in a 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. Answer to the question is B, not E. If there is some other snake, then group of {other, cobra} will be possible and the statement (2) will be violated (basically the possibility of other variety of snake is ruled out by the same logic as the possibility of second cobra). Hope it's clear. Dear, How do we know that there are only 2 snakes in the box? In the statement 2, it says " From any two snakes from Pandora's box at least one is a viper.". Doesn't it mean that when picking any 2 snakes out of more numbers of the snakes? Does it imply that there are only two snakes in the box? Thanks! Regards Andy We don't know whether there is only 2 snakes in the box. All we know from the second statement is that there must be one cobra and more than or equal to 1 vipers.
_________________



Manager
Joined: 21 Feb 2017
Posts: 70

Re: Devil's Dozen!!!
[#permalink]
Show Tags
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) ab=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.



Math Expert
Joined: 02 Sep 2009
Posts: 58989

Re: Devil's Dozen!!!
[#permalink]
Show Tags
04 Jun 2017, 06:16
goalMBA1990 wrote: Bunuel wrote: 6. Is the perimeter of triangle with the sides a, b and c greater than 30?
700+ question.
(1) ab=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.
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 8150

Re: Devil's Dozen!!!
[#permalink]
Show Tags
05 Jun 2017, 03:33
7. Set A consists of k distinct numbers. If n numbers are selected from the set onebyone, where n<=k, what is the probability that numbers will be selected in ascending order?[/b] (1) Set A consists of 12 even consecutive integers; (2) n=5. Hi, The distinct number selected can have ONLY one way in which they are in ascending order or for that matter, one way in descending order. 1)Since n <=k and we are selecting n integers, selection can be n! Ways..2) out of this, only one will be in ascending order. So PROBABILITY is 1/n! So we are looking for value of n.. Statement II gives us value of n, hence sufficient Ans B
_________________




Re: Devil's Dozen!!!
[#permalink]
05 Jun 2017, 03:33



Go to page
Previous
1 2 3 4 5
Next
[ 87 posts ]



