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Devil's Dozen!!! [#permalink]
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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. Solution: devilsdozen129312.html#p10638462. 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! Solution: devilsdozen129312.html#p10638473. If x and y are integers, is y an even integer?(1) 4y^2+3x^2=x^4+y^4 (2) y=4x^2 Solution: devilsdozen129312.html#p10638484. 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) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Solution: devilsdozen129312.html#p10638635. 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. Solution: devilsdozen129312.html#p10638676. 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. Solution: devilsdozen129312.html#p10638717. 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. Solution: devilsdozen12931220.html#p10638748. 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. Solution: devilsdozen12931220.html#p10638849. 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. Solution: devilsdozen12931220.html#p106388610. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?(1) There are total 99 snakes in Pandora's box. (2) From any two snakes from Pandora's box at least one is a viper. Solution: devilsdozen12931220.html#p106388811. 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. Solution: devilsdozen12931220.html#p106389212. If x>0 and xy=z, what is the value of yz?(1) \(x^2*y=3\). (2) \(\sqrt{x*y^2}=3\). Solution: devilsdozen12931220.html#p106389413. 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. Solution: devilsdozen12931220.html#p1063897
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Re: Devil's Dozen!!! [#permalink]
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12 May 2013, 02:58
up4gmat 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. Hi Bunuel, I have a question.From 1),cant we have : as p is a factor of (n+2)!n! so, n![(n+2)(n+1)1]. .so p is a factor of n! as it is some value * n! ?? Please clarify. p is a factor of \((n+2)!n!=n!((n+1)(n+2)1)\) does not necessarily means that p is a factor of n! it could be a factor of another multiple: ((n+1)(n+2)1). Check examples in (1) to verify. Hope it helps.
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Re: Devil's Dozen!!! [#permalink]
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21 May 2013, 05:46
Bunuel wrote: 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. Responding to a pm: This is not a weighted average question since there is no 'average' speed to be considered. This is more apt for relative speed concepts though I would think in terms of ratio of speeds (who is faster and who is slower) since we don't need to give values  only whether he is closer to trailer or studio. So we only need to figure out their speeds relative to each other (more/less).
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Re: Devil's Dozen!!! [#permalink]
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24 May 2013, 02:38
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+6b sufficient2. 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 insufficient



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Re: Devil's Dozen!!! [#permalink]
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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!!! [#permalink]
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20 Jun 2013, 08:20
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!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. Solution: devilsdozen12931220.html#p1063892 After getting it wrong initially I applied the following technique: 11m +7b <=15 question asks if 27m+27b<=45 stmnt 1: 7m+11b<=15 add this and question stem. we get 18m+18b<=30 i.e (m+b)<=30/18 27(m+b)<=30/18 *27 i.e. 27m+27b<=45 Stmnt 2: after any addition or subtraction does not yeild conclusive results.



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Re: Devil's Dozen!!! [#permalink]
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23 Jun 2013, 21:16
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. One doubt dear Lets assume k=1 2 3 4 5 6 7 8 9 10 Now if first select the number 4 then the total number i.e. k is definitely going to affect the total probability Please clear.



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Re: Devil's Dozen!!! [#permalink]
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23 Jun 2013, 21:32
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. You wrote in your explanation that " n numbers can be selected in n!ways" but I think it should be kCn Please explain Thanks



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Re: Devil's Dozen!!! [#permalink]
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23 Jun 2013, 22:04
Vinayprajapati 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. One doubt dear Lets assume k=1 2 3 4 5 6 7 8 9 10 Now if first select the number 4 then the total number i.e. k is definitely going to affect the total probability Please clear. Great Question Bunuel! As for your doubt Vinay, let me try to explain it. The question does not care about the number of ways of selecting n numbers from k numbers. It is only asking for the probability of selecting numbers in increasing order. Probability of selecting numbers in increasing order + Probability of selecting numbers in some other order = 1 Say out of the 10 numbers, you select 3 numbers. 1, 5, and 7 You can select them in increasing order = 1, 5, 7  total 1 way You can select them in some other order = 1, 7, 5/ 5, 1, 7/5, 7, 1/7, 5, 1/7, 1, 5  total 5 ways Probability of selecting numbers in increasing order = 1/6 Probability of selecting numbers in some other order = 5/6 Notice that it doesn't matter how many numbers there are in the original list. All we care about is arranging the n selected numbers. One arrangement will be in increasing order and the others will be nonincreasing. A good tricky question!
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Re: Devil's Dozen!!! [#permalink]
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27 Jun 2013, 05:33
Shouldn't the answer be D. How I approached the problem was that we are given 11m+7b=15 Statement 1: 7m+11b =15 If you solve both you get m=b i.e. cost of muffin is equal to brownie. and so I get cost of each is 15/18 (putting it back in given statement. and what we need is 27m +27b. so 54 x 15/18 = 45. Hence sufficient. Statement 2: 10m+8b = 15 Again solving it with given statement gets me m=b. Therefore, sufficient. Therefore D. What am I missing here? 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.



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Re: Devil's Dozen!!! [#permalink]
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27 Jun 2013, 07:21
poonamsharma83 wrote: Shouldn't the answer be D. How I approached the problem was that we are given 11m+7b =15 Statement 1: 7m+11b =15 If you solve both you get m=b i.e. cost of muffin is equal to brownie. and so I get cost of each is 15/18 (putting it back in given statement. and what we need is 27m +27b. so 54 x 15/18 = 45. Hence sufficient. Statement 2: 10m+8b = 15 Again solving it with given statement gets me m=b. Therefore, sufficient. Therefore D. What am I missing here? 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. In all cases we have "\(\leq\)" signs not "\(=\)" signs. The correct answer is A. Hope it's clear.
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15 Jul 2013, 09:32
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, Thanks for updating such elaborate answers ,I have one confusion regarding the answer posted above,which I have marked in red. If we can substitute one muffin with one brownie why can't we be sure to replace 2 brw with 2 muffins. Thanks in advance. Regards, Countdown



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15 Jul 2013, 10:00
Countdown 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, Thanks for updating such elaborate answers ,I have one confusion regarding the answer posted above,which I have marked in red. If we can substitute one muffin with one brownie why can't we be sure to replace 2 brw with 2 muffins. Thanks in advance. Regards, Countdown Say 11 muffins and 7 brownies cost $14 and the price of a muffin is less than the price of a brownie. 10 muffins and 8 brownies will cost more than $14 (since m<b). 9 muffins and 9 brownies will cost even more than that, so it could be more than $15. Hope it's clear.
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16 Aug 2013, 02:13
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 took gmatclub test and encountered the same question. The same explanation is provided in the solutions too. But , I am still not able to understand how you calculated this 4....4 . \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies. Please help Thanks
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16 Aug 2013, 02:19
TARGET730 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 took gmatclub test and encountered the same question. The same explanation is provided in the solutions too. But , I am still not able to understand how you calculated this 4....4 . \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies. Please help Thanks That's because both the question and the solution written by me. As for your question, following post might help: devilsdozen12931280.html#p1246581
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16 Aug 2013, 02:45
yes, got it now.. Thanks Alot...
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17 Aug 2013, 09:29
9) Is the OA  D? given 3 distinct +ve integers so let 0<A<B<C Also given ABC = C^2....i.e. AB= C  Equation 1 Case 1) (A+B+C)/3= 34/3 A + B + C = 34 From Eq 1 A + B + AB = 34 A + 1 + B(1+A) = 35 (1+B)(1+A) = 35  Eq (2) from Eq 2, A=4, B=6...so AB = 24 Case 2) C = 24 so AB = 24 So both statements are sufficient
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Re: Devil's Dozen!!! [#permalink]
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03 Sep 2013, 13:42
Bunuel, I have a doubt in 6th question part 2 explanation. 6. Is the perimeter of triangle with the sides a, b and c greater than 30? (1) ab=15. sufficient. (2) The area of the triangle is 50. There is another property of equilateral triangle which states that for a given area equilateral triangle has the lowest perimeter. Considering that : given area is 50=s^2 sqrt3/2 which gives us s=7.598 thus perimeter 3s = 22.79 Which means for a given area of 50 we can have a triangle with perimeter 22.79 which is less than 30. Thus (2) insufficient. Kindly help to verify. Thanks.
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Re: Devil's Dozen!!! [#permalink]
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03 Sep 2013, 14:35
PiyushK wrote: Bunuel, I have a doubt in 6th question part 2 explanation. 6. Is the perimeter of triangle with the sides a, b and c greater than 30? (1) ab=15. sufficient. (2) The area of the triangle is 50. There is another property of equilateral triangle which states that for a given area equilateral triangle has the lowest perimeter.
Considering that : given area is 50=s^2 sqrt3/2 which gives us s=7.598 thus perimeter 3s = 22.79
Which means for a given area of 50 we can have a triangle with perimeter 22.79 which is less than 30. Thus (2) insufficient.
Kindly help to verify. Thanks. The area of equilateral triangle is \(A=s^2*\frac{\sqrt{3}}{4}\), not \(A=s^2*\frac{\sqrt{3}}{2}\).
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