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Devil's Dozen!!!

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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: http://gmatclub.com/forum/devil-s-dozen ... l#p1063846

2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!-n!
(2) p is a factor of (n+2)!/n!

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063847

3. 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: http://gmatclub.com/forum/devil-s-dozen ... l#p1063848

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.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063863

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.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063867

6. 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: http://gmatclub.com/forum/devil-s-dozen ... l#p1063871

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.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063874

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.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063884

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.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063886

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.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063888

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: http://gmatclub.com/forum/devil-s-dozen ... l#p1063892

12. If x>0 and xy=z, what is the value of yz?
(1) \(x^2*y=3\).
(2) \(\sqrt{x*y^2}=3\).

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063894

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.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063897
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Re: Devil's Dozen!!! [#permalink]

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New post 28 Sep 2015, 10:00
Bunuel wrote:
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.



Dear,
How do we know that there are only 2 snakes in the box? In the statement 2, it says " From any two snakes from Pandora's box at least one is a viper.". Doesn't it mean that when picking any 2 snakes out of more numbers of the snakes? Does it imply that there are only two snakes in the box?

Thanks!
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Re: Devil's Dozen!!! [#permalink]

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New post 28 Sep 2015, 21:55
andy2whang wrote:
Bunuel 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.



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.



Dear,
How do we know that there are only 2 snakes in the box? In the statement 2, it says " From any two snakes from Pandora's box at least one is a viper.". Doesn't it mean that when picking any 2 snakes out of more numbers of the snakes? Does it imply that there are only two snakes in the box?

Thanks!
Regards
Andy


We don't know whether there is only 2 snakes in the box. All we know from the second statement is that there must be one cobra and more than or equal to 1 vipers.
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Re: Devil's Dozen!!! [#permalink]

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New post 14 Dec 2015, 02:09
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 males who do NOT listen to Bach. Notice that \(0.56x=\frac{14}{25}x\) must be an integer. Hence x must be a multiple of 25: 25, 50, 75, ... But \(x\) (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So \(x\) can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient.

(2) Of the astronauts who listen to Bach 70% are female. Now, if we apply the same logic here we get that, if # of astronauts who listen to Bach is \(y\) then \(0.7y\) is # of females who listen to Bach: \(0.7y=\frac{7}{10}y\) must be an integer. Hence it must be a multiple of 10, but in this case it can take more than 1 value: 10, 20, 30. So, this statement is not sufficient.

Answer: A.


In your explanation you have quoted "But \(x\) (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So \(x\) can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient." but when we take x = 50 ([14/25]*50 = 28 < 35). Though your answer is logically right i suppose we might need a better explanation here. Please correct me if i am wrong.
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Re: Devil's Dozen!!! [#permalink]

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New post 14 Dec 2015, 07:27
sachinsrini wrote:
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 males who do NOT listen to Bach. Notice that \(0.56x=\frac{14}{25}x\) must be an integer. Hence x must be a multiple of 25: 25, 50, 75, ... But \(x\) (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So \(x\) can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient.

(2) Of the astronauts who listen to Bach 70% are female. Now, if we apply the same logic here we get that, if # of astronauts who listen to Bach is \(y\) then \(0.7y\) is # of females who listen to Bach: \(0.7y=\frac{7}{10}y\) must be an integer. Hence it must be a multiple of 10, but in this case it can take more than 1 value: 10, 20, 30. So, this statement is not sufficient.

Answer: A.


In your explanation you have quoted "But \(x\) (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So \(x\) can only be 25, which makes # of astronauts who do listen to Bach equal to 35-25=10. Sufficient." but when we take x = 50 ([14/25]*50 = 28 < 35). Though your answer is logically right i suppose we might need a better explanation here. Please correct me if i am wrong.


Please re-read the highlighted part above.
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Re: Devil's Dozen!!! [#permalink]

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New post 25 Dec 2015, 19:34
I was wondering the same thing...Do you randomly add 1 to both sides to get (a + 1)(b+1) = 35???


deeuk wrote:
Bunuel wrote:
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?

\(a+b+ab=34\) --> \((a+1)(b+1)=35\)



I was wondering if there was any other way of getting through this- i mean, without the classical form?
Thanks in advance
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Re: Devil's Dozen!!! [#permalink]

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New post 03 Jun 2017, 21:40
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.


HI Bunuel.

Thanks for the clarification. But I stiil have a doubt.

Here to calculate the Probability, the Numerator is 1 ( understandable, as only 1 way the numbers can be ascending).

But lets consider the Denominator.

Are you saying 12P5 = 20P5 = 75P5 ????

Means, are you saying, k doesnt matter here ?

So it means :-

The No of ways 5 numbers can be selected from 12 numbers = No of ways 5 numbers can be selected from 75 numbers.

I'm not clear here. Can you please explain ?

Because as per my opinion, C should be the answer. To calculate Probability, we need to be sure about both Favorable Outcomes (Numerator) & Total Outcomes (Denominator).

Favorable Outcomes (Numerator) is 1 Here ( no doubt about that)

what about Total Outcomes (Denominator). ?? It definitely has dependency on the value of k. Since as k varies the value of KPn varies.

Kindly elaborate...

Thanks in advance..
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Re: Devil's Dozen!!! [#permalink]

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New post 04 Jun 2017, 04:35
Bunuel wrote:
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ question.

(1) a-b=15. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D.


hi Bunuel,
I didn't get your point mentioned in "For a given perimeter equilateral triangle has the largest area." because in statement B nothing is mentioned about type of triangle. Also please correct me if I am wrong for the statement "can't we do like this if we are assuming triangle as equilateral: (50= sqrt of 3 * a^2 ) / 4 and find what is a? Then a+a+a will give definite answer. So B is also sufficient to answer this question."

Thanks.
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Re: Devil's Dozen!!! [#permalink]

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New post 04 Jun 2017, 05:16
goalMBA1990 wrote:
Bunuel wrote:
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ question.

(1) a-b=15. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D.


hi Bunuel,
I didn't get your point mentioned in "For a given perimeter equilateral triangle has the largest area." because in statement B nothing is mentioned about type of triangle. Also please correct me if I am wrong for the statement "can't we do like this if we are assuming triangle as equilateral: (50= sqrt of 3 * a^2 ) / 4 and find what is a? Then a+a+a will give definite answer. So B is also sufficient to answer this question."

Thanks.


From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30. So, if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.
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Re: Devil's Dozen!!! [#permalink]

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New post 04 Jun 2017, 10:44
Bunuel wrote:
goalMBA1990 wrote:
Bunuel wrote:
6. Is the perimeter of triangle with the sides a, b and c greater than 30?

700+ question.

(1) a-b=15. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, a+b>c>15 --> a+b+c>30. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3}<50\). Since even equilateral triangle with perimeter of 30 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.

Answer: D.


hi Bunuel,
I didn't get your point mentioned in "For a given perimeter equilateral triangle has the largest area." because in statement B nothing is mentioned about type of triangle. Also please correct me if I am wrong for the statement "can't we do like this if we are assuming triangle as equilateral: (50= sqrt of 3 * a^2 ) / 4 and find what is a? Then a+a+a will give definite answer. So B is also sufficient to answer this question."

Thanks.


From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30. So, if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.


hi, I could not understand your explanation. Can you please explain in more simplified terms?
Let me put my doubts in sequence:
1. Why we are assuming it is equilateral triangle?
2. If we assume it equilateral triangle why we can't directly find side by using given area in statement 2?

Thanks.
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Re: Devil's Dozen!!! [#permalink]

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New post 04 Jun 2017, 10:52
HI RAN

I think I can elaborate you the details . At first i opted Option C. But its not correct. you said that while calculating the required probability numerator is one always. No its Not..Both numerator and denominator is changing, yet maintains fixed value of probability for a given n, irrespective of K value.

Let me explain by taking an example and evaluate the question from the beginning.
let K= {1,2,3,4} and n=2. as per question we need to take 2 numbers from K one by one and check if the numbers we pick are in ascending order to calculate the probaliity.
total number of selections possible = 4 x 3 = 12 (denominator value)
total number of favorable selection are = 3 (if we pick 1 first) or 2 (if we pick 2 first) or 1 (if pick 3 first) ..So totally there are 3+2+1 = 6 ways
so required probability = 6/(4x3) = 1/2.
now lets vary n value, say n =3
now total cases = 4x3x2 =24
no of favorable cases are = [2+1]+[1] =4. Note we get[2+1] when we assume the first number be 1 , then the next two number can be picked in 2+1 ways as explained in the previous example. similarly if we take 2 first then we have only one possible of taking (take 3 in second turn and 4 in 3rd turn)
so required probablity = 4/24 = 1/6
we can generalize that when n=4, favourable case is only 1.

from this argument we can conclude that n value is must and simply knowing K value will not be sufficient to calculate the req probability. So statement 1 is not sufficient.

Now lets look the other way round. Let fix n and vary the set K. Now set K has 5 numbers
K= {1,2,3,4,5} and n=2
then req probabilty = (4+3+2+1) / (5x4) = 1/2. same as when K=4, and N=2 right?
now let n=3
then numerator = (3+2+1) + (2+1) +(1) = 10
denominator = 5x4x3 = 60
req probability = 1/6. Surprised? we got the same value when k=4, and n=3 right?

so irrespective of K, n alone decides the value of req probability . Hence statement 2 alone is sufficient

Infact the numerator can be generalized as (1+2+.....n) + (1+2+3+....n-1) + (1+2+...n-2) + .........+ 1
and denominator can be generalized as = k x k-1 x k-2 x ........k-n+1.


However during exam time we need to be fast enuf for these trail and error method or stick to standard derivation if required.

Hope this help
Thanks Bunuel for these wonderful questions.

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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?[/b]

(1) Set A consists of 12 even consecutive integers;
(2) n=5.


Hi,

The distinct number selected can have ONLY one way in which they are in ascending order or for that matter, one way in descending order.
1)Since n <=k and we are selecting n integers, selection can be n! Ways..

2) out of this, only one will be in ascending order.

So PROBABILITY is 1/n!

So we are looking for value of n..
Statement II gives us value of n, hence sufficient

Ans B
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Re: Devil's Dozen!!! [#permalink]

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New post 07 Jun 2017, 02:32
At first i opted Option C. But its not correct. you said that while calculating the required probability numerator is one always. No its Not..Both numerator and denominator is changing, yet maintains fixed value of probability for a given n, irrespective of K value.

Let me explain by taking an example and evaluate the question from the beginning.
let K= {1,2,3,4} and n=2. as per question we need to take 2 numbers from K one by one and check if the numbers we pick are in ascending order to calculate the probaliity.
total number of selections possible = 4 x 3 = 12 (denominator value)
total number of favorable selection are = 3 (if we pick 1 first) or 2 (if we pick 2 first) or 1 (if pick 3 first) ..So totally there are 3+2+1 = 6 ways
so required probability = 6/(4x3) = 1/2.
now lets vary n value, say n =3
now total cases = 4x3x2 =24
no of favorable cases are = [2+1]+[1] =4. Note we get[2+1] when we assume the first number be 1 , then the next two number can be picked in 2+1 ways as explained in the previous example. similarly if we take 2 first then we have only one possible of taking (take 3 in second turn and 4 in 3rd turn)
so required probablity = 4/24 = 1/6
we can generalize that when n=4, favourable case is only 1.

from this argument we can conclude that n value is must and simply knowing K value will not be sufficient to calculate the req probability. So statement 1 is not sufficient.

Now lets look the other way round. Let fix n and vary the set K. Now set K has 5 numbers
K= {1,2,3,4,5} and n=2
then req probabilty = (4+3+2+1) / (5x4) = 1/2. same as when K=4, and N=2 right?
now let n=3
then numerator = (3+2+1) + (2+1) +(1) = 10
denominator = 5x4x3 = 60
req probability = 1/6. Surprised? we got the same value when k=4, and n=3 right?

so irrespective of K, n alone decides the value of req probability . Hence statement 2 alone is sufficient

Infact the numerator can be generalized as (1+2+.....n) + (1+2+3+....n-1) + (1+2+...n-2) + .........+ 1
and denominator can be generalized as = k x k-1 x k-2 x ........k-n+1.


However during exam time we need to be fast enuf for these trail and error method or stick to standard derivation if required.

Hope this help
Thanks Bunuel for these wonderful questions.

Please Give me + 1 kudos if my post is valuable to you :)
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Re: Devil's Dozen!!! [#permalink]

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New post 30 Jun 2017, 04:17
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(1) Of the astronauts who do NOT listen to Bach, 56% are male. If # of astronauts who do NOT listen to Bach is xx, then 0.56x0.56x is # of males who do NOT listen to Bach. Notice that 0.56x=1425x0.56x=1425x must be an integer. Hence, xx must be a multiple of 25: 25, 50, 75, ... But xx (# of astronauts who do NOT listen to Bach) must also be less than (or equal to) 35. So xx can only be 25, which makes # of astronauts who do listen to Bach equal to 35−25=1035−25=10. Sufficient.

(2) Of the astronauts who listen to Bach, 70% are female. Now, if we apply the same logic here we get that, if # of astronauts who listen to Bach is yy, then 0.7y0.7y is # of females who listen to Bach: 0.7y=710y0.7y=710y must be an integer. Hence, it must be a multiple of 10, but in this case it can take more than 1 value: 10, 20, 30. So, this statement is not sufficient.
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Re: Devil's Dozen!!! [#permalink]

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New post 25 Jan 2018, 03:27
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HI

what is the solution for 8 and 9. the links are not working.

Thanks
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Re: Devil's Dozen!!! [#permalink]

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New post 25 Jan 2018, 04:34
coolnaren wrote:
HI

what is the solution for 8 and 9. the links are not working.

Thanks


Solutions are on pages 1 and 2.

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.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063884

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.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063886

Updated the links. Thank you.
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Re: Devil's Dozen!!! [#permalink]

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New post 27 Jan 2018, 10:51
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.

Why is there only 1 possible choice for ascending order?

If N = 3 and k = 10 then would 1,2,3; 2,3,4; ... 8,9,10; etc all be possible ascending order results?
Re: Devil's Dozen!!!   [#permalink] 27 Jan 2018, 10:51

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