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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.
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) 32 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.
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.
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 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.
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.
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.
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.
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.
We need to find the number of patients who have acrophobia. No matter what the value of x is, from the matrix (devil-s-dozen-129312.html#p1063863) we an get that the number of patients who have acrophobia is 13. _________________
We need to find the number of patients who have acrophobia. No matter what the value of x is, from the matrix (devil-s-dozen-129312.html#p1063863) we an get that the number of patients who have acrophobia is 13.
Right you are.
I see it now: was looking at the question rather being how many people have both acrophobia and arachnophobia. That would been a tougher question to ask, perhaps...and that answer would have been "C."
Re: Devil's Dozen!!! [#permalink]
16 Nov 2012, 01:28
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
Bunuel - All the sols are clear, except this one.
If for any 2 snakes there is 1 viper then
1 other snake + 1 cobra : 1 Viper 1 other snake + 1 cobra : 1 Viper or 1 cobra + 1 cobra : 1 Viper or 1 anysnake + 1 anysnake = 1 Viper.
Re: Devil's Dozen!!! [#permalink]
16 Nov 2012, 03:56
Expert's post
Jp27 wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
Bunuel - All the sols are clear, except this one.
If for any 2 snakes there is 1 viper then
1 other snake + 1 cobra : 1 Viper 1 other snake + 1 cobra : 1 Viper or 1 cobra + 1 cobra : 1 Viper or 1 anysnake + 1 anysnake = 1 Viper.
Can you please elaborate this 1 please.
Cheers
Sorry, but I don't understand your question at all. Can you please explain what does the text in red mean? _________________
Re: Devil's Dozen!!! [#permalink]
11 Feb 2013, 09:07
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.
hey,
I couldn't able to understand the parts which i have marked in red color. also the combined part i am not getting. ........ FTG
Re: Devil's Dozen!!! [#permalink]
11 Feb 2013, 09:27
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
we could also conclude that there is also one viper snake??
Re: Devil's Dozen!!! [#permalink]
13 Feb 2013, 00:46
Expert's post
FTG 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.
hey,
I couldn't able to understand the parts which i have marked in red color. also the combined part i am not getting. ........ FTG
Re: Devil's Dozen!!! [#permalink]
13 Feb 2013, 00:52
Expert's post
FTG wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
we could also conclude that there is also one viper snake??
No, that's not true. From the second statement we have that there must be one cobra in the box, but all we can say about vipers is that there must be at least one.
For example, there can be 1 cobra and 1 viper, or 1 cobra and 2 vipers, 1 cobra and 55 vipers... In all these cases from any two snakes from the box at least one will be viper.
Re: Devil's Dozen!!! [#permalink]
13 Feb 2013, 20:08
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.
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. . _________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Re: Devil's Dozen!!! [#permalink]
14 Feb 2013, 01:59
1
This post received KUDOS
Expert's post
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 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.
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.
Re: Devil's Dozen!!! [#permalink]
15 Feb 2013, 13:30
Expert's post
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. (2) p is a prime.
(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.
Dear BUNUEL, i remember a specific rule about this. it is if any prime number greater than or equal to 5 divided by 12 or 24 the division will leave the remainder as 1 always. now remainder when p^2/12 will be same as that of p/12 so i think all we need to know is whether p is prime greater than or equal to 5. Correct?
Re: Devil's Dozen!!! [#permalink]
29 Mar 2013, 16:12
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.
HI all,
in this below similar question:
A football team has 99 players. Each player has a uniform number from 1 to 99 and no two players share the same number. When football practice ends, all the players run off the field one-by-one in a completely random manner. What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72,etc) ?
(A) 1/64 (B) 1/48 (C) 1/36 (D) 1/24 (E) 1/16
No. of ways in which 4 people can leave the field i.e. No. of ways of arranging 4 people = 4! = 24
Out of these 24 ways, there is only one way in which they are in ascending order.
So, probability = 1/24
Answer is D.
in this above solution why there is no role of total players ,i.e., 99 ???
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.
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?
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:
The attachment Vertigo.png is no longer available
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.
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
Double-set matrix has all cases possible. The case you are talking about is red box below:
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.
isnt the answer D, because we know from statement A that there are 12 items in the set? kind regards
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.
isnt the answer D, because we know from statement A that there are 12 items in the set? kind regards
No, the answer is B. The size of the set is not enough, while the size of the subset is. _________________
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