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Re: Devil's Dozen!!!
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 23 Mar 2012, 03:53
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23 Mar 2012, 03:56
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23 Mar 2012, 03:57
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. No info about Buster. Not sufficient. (2) Buster gets to the studio at the same time as Charlie gets to the trailer --> Charlie needed 20 minutes less than Buster to cover the same distance, which means that the rate of Charlie is higher than that of Buster. Since after they pass each other they need the same time to get to their respective destinations (they get at the same time to their respective destinations) then Buster had less distance to cover ahead (at lower rate) than he had already covered (which would be covered by Charlie at higher rate). Sufficient. Answer: B.
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Re: Devil's Dozen!!!
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23 Mar 2012, 10:18
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. I didnot understand why 1 is not sufficient, can you elaborate on that ? what i did is - I took n! common * coprime number So if we have n! everytime the factors for n! and n!* coprime should always be same.. What am i missing? Will Vote for A Given n = +ve Int P = Prime (A) p is a factor of (n+2)!-n! Let n = 4 = 6! - 4! = 4![(6*5) - 1] = 4! (29) therefore n! = 4! will have sufficient factors as 4! (29) data sufficient Let n = 17 = (19)! - 17! = 17! [(19 * 18) - 1] data sufficient Let n = 11 = (13!) - 11! = 11! [(13*12) - 1] data sufficient (B) p is a factor of (n+2)!/n! Let n = 2 = (4!) / 2! = 12 = 2*2*3 = data sufficient Let n = 4 = (6!) / 4! = 6 * 5 = 2*3*5 = data not sufficient
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Re: Devil's Dozen!!!
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23 Mar 2012, 18:39
Bunuel wrote: 7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, 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 k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) so we need both k & n. Am I missing something here?
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Re: Devil's Dozen!!!
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23 Mar 2012, 18:44
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 the a 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.
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Re: Devil's Dozen!!!
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24 Mar 2012, 03:16
kuttingchai 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. I didnot understand why 1 is not sufficient, can you elaborate on that ? There are two examples given which give two different answers to the question whether p is a factor of n!. (1) p is a factor of (n+2)!-n! If \(n=2\) and \(p=2\) (notice that in this case \((n+2)!-n!=22\) and \(p=2\) is a factor of 22), then since \(p=2\) is a factor of \(n!=2!\) the answer to the question is YES; If \(n=2\) and \(p=11\) (notice that in this case \((n+2)!-n!=22\) and \(p=11\) is a factor of 22), then since \(p=11\) is NOT a factor of \(n!=2!\) the answer to the question is NO. Two different answers, hence not sufficient.
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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 one-by-one, 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 k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) 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 one-by-one, 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 k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) 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
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Re: Devil's Dozen!!!
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24 Mar 2012, 08:41
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.
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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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Re: Devil's Dozen!!!
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04 Apr 2012, 17:28
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 double-check 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 \(p=6n-1\).
If \(p=6n+1\) then \(p^2=36n^2+12n+1\) which gives remainder 1 when divided by 12;
If \(p=6n-1\) then \(p^2=36n^2-12n+1\) 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!
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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 double-check 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 \(p=6n-1\).
If \(p=6n+1\) then \(p^2=36n^2+12n+1\) which gives remainder 1 when divided by 12;
If \(p=6n-1\) then \(p^2=36n^2-12n+1\) 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\(p=6n+5\) (\(p=6n-1\)), 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 vise-versa 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, 02:52
Bunuel wrote: 7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, 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 (=n) will be in ascending order? rather than what is the probability that numbers will be selected in ascending order?The ambiguity is caused by the fact that question clearly says that n out of k ARE SELECTED one by one. So, any further reference of selection, unless clearly stated, naturally considers n out k possibilities.
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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 one-by-one, 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): 4m-4b=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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Re: Devil's Dozen!!!
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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): 4m-4b=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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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
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Re: Devil's Dozen!!! &nbs
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27 Jun 2012, 01:58
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