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Re: Devil's Dozen!!!
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23 Mar 2012, 03:56
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.
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27 Jun 2012, 01:58
shivamayam 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 bunuel  i am unable to find the gap in my logic. Can you pls help? Given: 11m+7b=15  (1) (let's disregard the inequality for the moment)St1: 7m+11b=15 (2) (1)less(2): 4m4b=0 > m=b > substituting m for b in (1), we get 11m+7m=15 > m=5/6. We could substitute 5/6 for m and b in the question to verify whether the equation holds good. So St1 is suff. St2: 10m+8b=15 (3) (1)less(3) gives me m= 5/6. Similar to the above, we could substitute 5/6 for m and b in the question to verify it's validity. So St2 is also suff. So shouldn't the answer be D? You cannot just substitute \(\leq\) sign with \(=\) sign and solve. Those are two very different signs. To see that (2) is not sufficient, consider two cases: A. Muffins and brownies are free  answer YES; B. Muffins are free and each brownie costs $1.8  answer NO. Hope it's clear.
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Re: Devil's Dozen!!!
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20 Mar 2012, 02:08
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? (1) Set A consists of 12 even consecutive integers. (2) n=5.
I didnt solve this one .. but from statement 1 and II since i know the n and K both I can calculate the probability..IMo C Please correct if wrong



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Updated on: 22 Mar 2012, 12:31
Bunuel wrote: The next set of tough and tricky DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!
1. Jules and Jim both invested certain amount of money in bond M for one year, which pays for 12% simple interest annually. If no other investment were made, then Jules initial investment in bond M was how many dollars more than Jim's investment in bond M. (1) In one year Jules earned $24 more than Jim from bond M. (2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M.
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! (2) p is a factor of (n+2)!/n!
3. If x and y are integers, is y an even integer? (1) 4y^2+3x^2=x^4+y^4 (2) y=4x^2
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia? (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. (2) 13 patients of Vertigo Hospital have arachnophobia but not acrophobia.
5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach? (1) Of the astronauts who do NOT listen to Bach 56% are male. (2) Of the astronauts who listen to Bach 70% are female.
6. Is the perimeter of triangle with the sides a, b and c greater than 30? (1) ab=15. (2) The area of the triangle is 50.
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? (1) Set A consists of 12 even consecutive integers. (2) n=5.
8. If p is a positive integer, what is the remainder when p^2 is divided by 12? (1) p is greater than 3. (2) p is a prime.
9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers? (1) The average (arithmetic mean) of the three numbers is 34/3. (2) The largest number of the three distinct numbers is 24.
10. There is at least one viper and at least one cobra in a Pandora's box. How many cobras are there? (1) There are total 99 snakes in Pandora's box. (2) From any two snakes from the a Pandora's box at least one is a viper.
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies? (1) $15 is enough to buy 7 muffins and 11 brownies. (2) $15 is enough to buy 10 muffins and 8 brownies.
12. If x>0 and xy=z, what is the value of yz? (1) \(x^2*y=3\). (2) \(\sqrt{x*y^2}=3\).
13. Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie? (1) Charlie gets to the trailer in 55 minutes. (2) Buster gets to the studio at the same time as Charlie gets to the trailer. 1. D Jules'invested amt  x Jim's invested amt  y Q is xy? a) x+0.12xy0.12y = 24 => xy can be found b) x+0.2xy0.2y = 40 => xy can be found 2. C a) n!(n+2)(n+1)  n! = pK cant say about (n+2)(n+1) being a multiple of p b) it confirms the above missing element as (n+2)(n+1) is a multiple of p 3.C Using both it comes to y^4 = 3x^429x^2 +64 => either x being odd or even 3x^429x^2 is always even hence y^4 is even and hence y is even. 4.A let be n(acrophobia) and c be common to both a) 58 = 45 + x  c + c => x can be determined b) no of patients dont belong to both is not mentioned 5.E No division of how many listen and how many dont is provided 6. A a) ab=15 => c>15 => a+b >15+(c15) => a+b+c >30 b) not relevant 7. C I think the prob will be 1/(K*(K1)*..(K(n1)) so we need both k & n 8. C I know that remainder is always 1 when n^2 is divided by 12 given n is a prime greater than 3. Got from plugging nos 9. B I think we need to find the largest no as xyz=z^2 =>xy=z which is given in second statement. Not sure if I am missing a trap. 10. E As it is not mentioned if there any more varieties of snakesother than cobras and vipers. 11. E Not sure about this. I think each of the statements along with the premise gives that m > b, but by how much can not be concluded and hence cant say. 12. B Again using both also I get yz =9 which is directly implied by statement b. Hence I go with B. Not sure of any trap. 13.E not used to giving explanations hence taking a lot of time
Originally posted by mithun2vrs on 22 Mar 2012, 10:01.
Last edited by mithun2vrs on 22 Mar 2012, 12:31, edited 1 time in total.



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Re: Devil's Dozen!!!
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22 Mar 2012, 11:52
5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach? (1) Of the astronauts who do NOT listen to Bach 56% are male. (2) Of the astronauts who listen to Bach 70% are female.
Answer: A 1)I set up a table of M/F and Listeners/NonListeners. Since 56% are male and there are 35 astronauts, reduce 56/100 where the demoninator is <= 35. The only solution is 14/25. Thus, 25 astronauts do NOT listen to Bach and 10 astronauts listen to Bach. Sufficient.
2)With the same table I already have set up, I can determine that either 10, 20, or 30 listen to Bach; therefore, not sufficient.
C would be a trap answer. Based on (1) our options of 20 and 30 from (2) are eliminated, for a total of 7 female listeners, 3 male listeners, 11 female nonlisteners and 14 male nonlisteners. However, this information is unneccessary since (1) is enough.
Edit: I wanted to add that I came to these conclusions because it's stated "at least one astronaut do NOT listen to Bach." I think that oversight is why many people picked E?



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Re: Devil's Dozen!!!
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24 Mar 2012, 03:22
mithun2vrs wrote: Bunuel wrote: 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?
(1) Set A consists of 12 even consecutive integers; (2) n=5.
We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.
2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).
Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.
Answer: B. I am not sure if I understand correctly. There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k1 integer is 1/(k1) and so on... so selecting n integers would be 1/(k*(k1)*(k2)*...(k(n1)) so we need both k & n. Am I missing something here? We are not interested in the probability of selecting first, second, etc numbers from the set. We are interested in the probability that numbers selected will be in ascending order. Now, any group of n numbers is equally likely to be selected and these n numbers can be selected in n! ways, out of which only 1 will be in ascending order so the probability is 1/n!. Hope it's clear.
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Re: Devil's Dozen!!!
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24 Mar 2012, 03:35
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.
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Re: Devil's Dozen!!!
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24 Mar 2012, 06:43
Bunuel wrote: mithun2vrs wrote: Bunuel wrote: 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?
(1) Set A consists of 12 even consecutive integers; (2) n=5.
We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.
2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).
Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.
Answer: B. I am not sure if I understand correctly. There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k1 integer is 1/(k1) and so on... so selecting n integers would be 1/(k*(k1)*(k2)*...(k(n1)) so we need both k & n. Am I missing something here? We are not interested in the probability of selecting first, second, etc numbers from the set. We are interested in the probability that numbers selected will be in ascending order. Now, any group of n numbers is equally likely to be selected and these n numbers can be selected in n! ways, out of which only 1 will be in ascending order so the probability is 1/n!. Hope it's clear. I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls



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Re: Devil's Dozen!!!
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24 Mar 2012, 06:51
mithun2vrs wrote: I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order.
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Re: Devil's Dozen!!!
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24 Mar 2012, 14:52
mithun2vrs wrote: Bunuel wrote: mithun2vrs wrote: I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order. Sorry to keep on with the problem, Probability is defined as No of fav outcomes / No of Total possible outcomes With your example, in both cases No of favorable outcomes is 1 and in 1st case No of total possible outcomes is 12c5 and in the second case No of total possible outcomes is 24c5. Pls correct me where I am going wrong. Again: you count the probability of selecting 5 numbers out of 12 or 24, whereas the question asks about the probability of selecting some group of numbers in a certain ORDER. You just interpret the question in a wrong way.
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04 Apr 2012, 21:39
alphabeta1234 wrote: Bunuel wrote: 8. If p is a positive integer, what is the remainder when p^2 is divided by 12?
(1) p is greater than 3. Clearly insufficient: different values of \(p\) will give different values of the remainder. (2) p is a prime. Also insufficient: if \(p=2\) then the remainder is 4 but if \(p=3\) then the remainder is 9.
(1)+(2) You can proceed with number plugging and try several prime numbers greater than 3 to see that the remainder will always be 1 (for example try \(p=5\), \(p=7\), \(p=11\)).
If you want to doublecheck this with algebra you should apply the following property of the prime number: any prime number greater than 3 can be expressed either as \(p=6n+1\) or \(p=6n1\).
If \(p=6n+1\) then \(p^2=36n^2+12n+1\) which gives remainder 1 when divided by 12;
If \(p=6n1\) then \(p^2=36n^212n+1\) which also gives remainder 1 when divided by 12.
Answer: C. Hey Bunuel, I am intrigued by your second method. But if n=6 we get \(p=6n+1=25\), a multiple of 5. Can this be a problem? Moreover, for your method to show that the remaind was equal to 1. Is this what you did ? \(p^2=36n^2+12n+1=12(3n^2+n)+1=12k+1\) ? Thank you! Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3). So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n1\)), where \(n\) is an integer >1. But:Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so viseversa of above property is not correct. For example 25 (for \(n=4\)) yields a remainder of 1 upon division by 6 and it's not a prime number. alphabeta1234 wrote: Moreover, for your method to show that the remaind was equal to 1. Is this what you did ?
\(p^2=36n^2+12n+1=12(3n^2+n)+1=12k+1\) ?
Thank you! Yes, first two terms in \(p^2=36n^2+12n+1\) are divisible by 12 so leave no remainder and 1 divided by 12 yields reminder of 1. Hope it helps.
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Re: Devil's Dozen!!!
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18 Apr 2012, 03:45
some2none wrote: Bunuel wrote: 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?
(1) Set A consists of 12 even consecutive integers; (2) n=5.
We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.
2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).
Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.
Answer: B. Some amount of ambiguity exists, at least to me, in the question. The ambiguity disappears if the question asks what is the probability that SELECTED numbers will be in ascending order? rather than what is the probability that numbers will be selected in ascending order?Let's say k = 7 and n = 3 (selected numbers) The ways in which n (=3) can be selected from k (=7) = 7C3 And probability of selecting n(=3) in ascediting order is 1/7C3. Again, the answer would be the same 1/n!=1/3!=1/6: 1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 2, 6; 1, 2, 7; 1, 3, 4; 1, 3, 5; 1, 3, 6; 1, 3, 7; 1, 4, 5; 1, 4, 6; 1, 4, 7; 1, 5, 6; 1, 5, 7; 1, 6, 7; 2, 3, 4; 2, 3, 5; 2, 3, 6; 2, 3, 7; 2, 4, 5; 2, 4, 6; 2, 4, 7; 2, 5, 6; 2, 5, 7; 2, 6, 7; 3, 4, 5; 3, 4, 6; 3, 4, 7; 3, 5, 6; 3, 5, 7; 3, 6, 7; 4, 5, 6; 4, 5, 7; 4, 6, 7; 5, 6, 7. 35 ways (naturally) to select 3 numbers out of 7 in ascending order. Total # of ways to select 3 numbers out of 7 when order matters is \(P^3_7=210\). Hence the probability is 35/210=1/6. Hope it's clear.
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Re: Devil's Dozen!!!
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26 Jun 2012, 18:34
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 bunuel  i am unable to find the gap in my logic. Can you pls help? Given: 11m+7b=15  (1) (let's disregard the inequality for the moment) St1: 7m+11b=15 (2) (1)less(2): 4m4b=0 > m=b > substituting m for b in (1), we get 11m+7m=15 > m=5/6. We could substitute 5/6 for m and b in the question to verify whether the equation holds good. So St1 is suff. St2: 10m+8b=15 (3) (1)less(3) gives me m= 5/6. Similar to the above, we could substitute 5/6 for m and b in the question to verify it's validity. So St2 is also suff. So shouldn't the answer be D?



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14 Jul 2012, 21:30
Bunuel wrote: 3. If x and y are integers, is y an even integer?
(1) 4y^2+3x^2=x^4+y^4 > rearrange: \(3x^2x^4=y^44y^2\) > \(x^2(3x^2)=y^2(y^24)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3x^2=oddodd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^24y)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^24y)=odd*(oddeven)=odd*odd=odd\). Sufficient.
(2) y=4x^2 > if \(x=odd\) then \(y=evenodd=odd\) but if \(x=even\) then \(y=eveneven=even\). Not sufficient.
Answer: A. Regarding statement (1), I understood how we got \(x^2(3x^2)=y^2(y^24)\). Later, we are examining \(y^2(y^24y)\)  question: how did we get this extra 'y' here? Am getting a lot of DS questions wrong, would appreciate if you could explain. Thanks.



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14 Feb 2013, 02:59
Sachin9 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 coprime, 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. Bunuel, I did not understand the below: 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: \(n!\) Please clarify. . We got that prime number p is NOT a factor of \((n+1)(n+2)1\) but it IS a factor of \(n!*((n+1)(n+2)1)\), thus it must be a factor of n!. For example, if we are told that 3 IS a factor of xy (where x and y are positive integers) and is NOT a factor of y, then it wold mean that 3 IS a factor of x. Hope it's clear.
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09 Apr 2013, 09:46
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: Attachment: Vertigo.png 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. Sorry to disturb you Bunuel, but why haven't you considered in your matrix the case where people can have acrofobia but not arachnofobia? If you add a new incognite there (let's say "y"), first statement is insufficient and you would need the second one to solve the problem. Where is my argument failing? Thank you in advance



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Re: Devil's Dozen!!!
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24 May 2013, 02:53
cumulonimbus 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 did it as below, please confirm if this method can be used in all the cases. is 45 =27m+27b i.e. 5 =3m+3b 1. 15>=11m+7b (given in question stem) 15>=7m+11b add both equations: 30>=18m+18b 5>=6m+6bsufficient2. 15>=10m+8b => 8/3*(5=3m+3b) => 13.33=8m+8b here we don't know for sure whether 1.7 will be sufficient to buy 2 muffins hence insufficientYes, it's a valid approach, though the red parts are not correct. The question asks whether \(27m+27b\leq{45}\) (\(3m+3b\leq{5}\)) not whether \(27m+27b={45}\) (should be \(\leq\) instead of =). Next, when reducing \(18m+18b\leq{30}\) you get \(3m+3b\leq{5}\) or \(6m+6b\leq{10}\) not \(6m+6b\leq{5}\). Hope it helps.
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Re: Devil's Dozen!!!
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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.
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Re: Devil's Dozen!!!
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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
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Re: Devil's Dozen!!!
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28 Oct 2018, 01:31
gmat800live wrote: 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 coprime, 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. When we consider two statements together then n cannot be 1 because p cannot be a prime which is simultaneous a factor of 5 and 6.
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Re: Devil's Dozen!!!
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