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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: devil-s-dozen-129312.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: devil-s-dozen-129312.html#p10638473. If x and y are integers, is y an even integer?(1) 4y^2+3x^2=x^4+y^4 (2) y=4-x^2 Solution: devil-s-dozen-129312.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: devil-s-dozen-129312.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: devil-s-dozen-129312.html#p10638676. Is the perimeter of triangle with the sides a, b and c greater than 30?(1) a-b=15. (2) The area of the triangle is 50. Solution: devil-s-dozen-129312.html#p10638717. 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. Solution: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.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: devil-s-dozen-129312-20.html#p1063897
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Re: Devil's Dozen!!! [#permalink]
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Bunuel wrote: 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. Thank you - Bunuel that was easy to understand
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Re: Devil's Dozen!!! [#permalink]
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28 Mar 2012, 07:39
I will go with option A as the answer. Its pretty straight forward 
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Re: Devil's Dozen!!! [#permalink]
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28 Mar 2012, 08:40
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Re: Devil's Dozen!!! [#permalink]
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04 Apr 2012, 13:43
Bunuel wrote: SOLUTIONS:
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.
Question: \(x-y=?\)
(1) In one year Jules earned $24 more than Jim from bond M. \(0.12x-0.12y=24\) --> \(0.12(x-y)=24\) --> \(x-y=200\). Sufficient.
(2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M. Basically the same type of information as above: \(0.2x-0.2y=40\) --> \(0.2(x-y)=40\) --> \(x-y=200\). Sufficient.
Answer: D.
Important note when two quantities are increased (decreased) by the same percent their difference also increase (decrease) by the same percent. For example if you increase 100 and 150 by 20% to 120 and 180 respectively, then their difference will also increase by the same 20% from 50 to 60. Hey Bunuel, I am bit confused on how to interpret statement 1. "In one year Jules earned $24 more than Jim from bond M." Does it mean : \(0.12x-0.12y=24\) Accounting only for the return on the investment? or \(1.12x-1.12y=24\) Accounting for the initial principal of the investment? In this question, I was lucky that its didn't matter. We could solve for x-y in both situations. But I can see how reading the question your way could have lead me to the wrong answer. Bunuel, your incredibly gifted dude!
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Re: Devil's Dozen!!! [#permalink]
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04 Apr 2012, 14:09
alphabeta1234 wrote: Bunuel wrote: SOLUTIONS:
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.
Question: \(x-y=?\)
(1) In one year Jules earned $24 more than Jim from bond M. \(0.12x-0.12y=24\) --> \(0.12(x-y)=24\) --> \(x-y=200\). Sufficient.
(2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M. Basically the same type of information as above: \(0.2x-0.2y=40\) --> \(0.2(x-y)=40\) --> \(x-y=200\). Sufficient.
Answer: D.
Important note when two quantities are increased (decreased) by the same percent their difference also increase (decrease) by the same percent. For example if you increase 100 and 150 by 20% to 120 and 180 respectively, then their difference will also increase by the same 20% from 50 to 60. Hey Bunuel, I am bit confused on how to interpret statement 1. "In one year Jules earned $24 more than Jim from bond M." Does it mean : \(0.12x-0.12y=24\) Accounting only for the return on the investment? or \(1.12x-1.12y=24\) Accounting for the initial principal of the investment? In this question, I was lucky that its didn't matter. We could solve for x-y in both situations. But I can see how reading the question your way could have lead me to the wrong answer. Bunuel, your incredibly gifted dude! The first reading is correct: if you invest $100 for one year in a bond which pays 12% simple interest annually then you earn 0.12*$100=$12 per year.
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Re: Devil's Dozen!!! [#permalink]
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04 Apr 2012, 17:28
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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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Re: Devil's Dozen!!! [#permalink]
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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!!! [#permalink]
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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!!! [#permalink]
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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!!! [#permalink]
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28 May 2012, 08:23
Bunuel wrote: 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?
Also very tricky.
(1) Of the astronauts who do NOT listen to Bach 56% are male --> if # of astronauts who do NOT listen to Bach is \(x\) then \(0.56x\) is # of females who listen to Bach.
Answer: A. Sorry for bringing up an old post here but I think there is a typo. Did you mean "the #males who Not listen to Bach instead"?Thanks.
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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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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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Re: Devil's Dozen!!! [#permalink]
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03 Jul 2012, 03:39
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.
I don't understand the logic of the following explanation -
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.
I see that you used the above logic to solve the two statements.....I don't get this.
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Re: Devil's Dozen!!! [#permalink]
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08 Jul 2012, 18:27
Hi Bunnel,
I didn't really understand the logic that you used for solving the muffin and brownies question, more specifically I have posted my question as above.
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Re: Devil's Dozen!!! [#permalink]
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14 Jul 2012, 03:21
for the first question i see either of the options individually giving the shares of the bond... is that wrong???
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Re: Devil's Dozen!!! [#permalink]
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14 Jul 2012, 03:23
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Re: Devil's Dozen!!! [#permalink]
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14 Jul 2012, 21:30
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Bunuel wrote: 3. If x and y are integers, is y an even integer?
(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4y)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4y)=odd*(odd-even)=odd*odd=odd\). Sufficient.
(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.
Answer: A. Regarding statement (1), I understood how we got \(x^2(3-x^2)=y^2(y^2-4)\). Later, we are examining \(y^2(y^2-4y)\) - 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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Re: Devil's Dozen!!! [#permalink]
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15 Jul 2012, 05:41
lateapp wrote: Bunuel wrote: 3. If x and y are integers, is y an even integer?
(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4y)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4y)=odd*(odd-even)=odd*odd=odd\). Sufficient.
(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.
Answer: A. Regarding statement (1), I understood how we got \(x^2(3-x^2)=y^2(y^2-4)\). Later, we are examining \(y^2(y^2-4y)\) - question: how did we get this extra 'y' here? Am getting a lot of DS questions wrong, would appreciate if you could explain. Thanks. Extra \(y\) there is clearly a typo, it should be \(y^2(y^2-4)\). Edited. Thank you.
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Re: Devil's Dozen!!! [#permalink]
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12 Sep 2012, 04:10
Bunuel wrote: 4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?Tricky question. (1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix: Attachment: Vertigo.png As you can see # of patients who has acrophobia is 58-45=13. Sufficient. (2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient. Answer: A. Can't statement 1 be insufficient if we can have the number of people who have both arachnophobia and acrophobia (which is the number of patients who have neither arachnophobia nor acrophobia) to be either zero or thirteen. Since we have not been told that at least one person has both arachnophobia and acrophobia, why do we rule out the possibility that no one has both? If that case applies, then the answer would be C, since statement 2 would help us eliminate the possibility that no one has both, leaving only thirteen as the only possible number of people who have both. Your thoughts, Bunuel. Cheers, Der alte Fritz.
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Re: Devil's Dozen!!!
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