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# Devil's Dozen!!!

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19 Mar 2012, 06:54
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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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03 Sep 2013, 14:40
Thanks Bunuel, my mistake I will memorize this formula for forever.
Really a silly mistake done by me
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09 Sep 2013, 21:11
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.

In B. If 45 have arachnophobia , and 32 have only arachnophobia, so # of patients common to both is 45-32=13.
So total number of patients are : 32 +13 + only acrophobia = 58.
Whats wrong in my approach ?
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10 Sep 2013, 01:31
ygdrasil24 wrote:
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.

In B. If 45 have arachnophobia , and 32 have only arachnophobia, so # of patients common to both is 45-32=13.
So total number of patients are : 32 +13 + only acrophobia = 58.
Whats wrong in my approach ?

Aren't you missing something? What about those who has neither? There is a matrix given in my solution, plug your values, will you get the answer?
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10 Sep 2013, 10:21
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.

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?

Hi Bunnel- Even I have same doubt. Is this question discussed in detail ? If so, it would be great if you can post the link. I know its pretty old now..but yet classic.
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11 Sep 2013, 03:00
suyash23n 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.

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?

Hi Bunnel- Even I have same doubt. Is this question discussed in detail ? If so, it would be great if you can post the link. I know its pretty old now..but yet classic.

Yes:
devil-s-dozen-129312-20.html#p1064472
devil-s-dozen-129312-20.html#p1064519
devil-s-dozen-129312-20.html#p1064618
devil-s-dozen-129312-40.html#p1075275

Hope this helps.
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07 Nov 2013, 04:55
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.

This might be a silly question, but i really don't understand how, if m < b, then 3m could replace 3b?
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07 Nov 2013, 06:43
sidvish 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.

This might be a silly question, but i really don't understand how, if m < b, then 3m could replace 3b?

The following post might help: devil-s-dozen-129312-80.html#p1246581
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23 May 2014, 00:29
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.

Hi Bunnel,

How we are getting 14/25

Thanks.
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23 May 2014, 01:15
PathFinder007 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.

Hi Bunnel,

How we are getting 14/25

Thanks.

Hi Pathfinder,
Suppose x = no. of people who do not listen to Bach , 0< x<=35

56% of x = x*(56/100) = x*(14/25) ..... on taking out 4(the common factor) both from Numerator and Denominator

Now, this 56% must be an integer implies x should be a multiple of 25
and x<=35 as given above
----------------> x =25, hence sufficient.
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23 May 2014, 04:54
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.

Hi Bunnel,

I just want to know why statement 2 is not sufficient. following is my logic

from question 11m+7b =15

st1- 7m+11b = 15

subtract st2 from 1

4m-4b = 0
m=b

so in question we can say 18b = 15
b=15/18=5/6. using this we can get the price for 27m and 27b

now same i can get from st2 then why this is not sufficient?

Thanks.
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23 May 2014, 05:14
PathFinder007 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.

Hi Bunnel,

I just want to know why statement 2 is not sufficient. following is my logic

from question 11m+7b =15

st1- 7m+11b = 15

subtract st2 from 1

4m-4b = 0
m=b

so in question we can say 18b = 15
b=15/18=5/6. using this we can get the price for 27m and 27b

now same i can get from st2 then why this is not sufficient?

Thanks.

There is a huge difference between $$11m+7b\leq{15}$$ and $$11m+7={15}$$. "Enough" should be translated as $$\leq$$ only.
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02 Jul 2014, 22:20
Bunuel wrote:
priyavenugopal wrote:
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.

Given,
From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..)
From (1) stmt - 7m + 11b = 15
From (2) stmt - 10m + 8b = 15
Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap)
Solving qn stem n stmt 2 eqns, got m and b as 5/6

Having m and b values, we can find whthr 45$is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D. Not even to analyze the rest of it, I must say that there is a huge difference between $$11m+7b\leq{15}$$ and $$11m+7={15}$$. You can not just write = sing instead of <= just because it's more convenient and "enough" should be translated as <= only. Bunuel the statement says 15$ is enough, so it implies <= 15$, consider max situation that it takes 15$ to buy 10m and 8b, which implies 45$is sufficient . Can we not imply that for any amount less than 15$ , the solution will hold?.

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03 Jul 2014, 05:57
ayushee01 wrote:
Bunuel wrote:
priyavenugopal wrote:
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.

Given,
From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..)
From (1) stmt - 7m + 11b = 15
From (2) stmt - 10m + 8b = 15
Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap)
Solving qn stem n stmt 2 eqns, got m and b as 5/6

Having m and b values, we can find whthr 45$is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D. Not even to analyze the rest of it, I must say that there is a huge difference between $$11m+7b\leq{15}$$ and $$11m+7={15}$$. You can not just write = sing instead of <= just because it's more convenient and "enough" should be translated as <= only. Bunuel the statement says 15$ is enough, so it implies <= 15$, consider max situation that it takes 15$ to buy 10m and 8b, which implies 45$is sufficient . Can we not imply that for any amount less than 15$ , the solution will hold?.

$15 is enough to buy 10 muffins and 8 brownies. 15*3=$45 is enough to buy 10*3=30 muffins and 8*3=24 brownies.

But this is not sufficient to say whether \$45 enough to buy 27 muffins and 27 brownies.
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03 Jul 2014, 06:08
correct , but suppose i consider the max value for 11m and 7b = 15 and max value for 10m + 8b as 15, we can solve and get 5/6. earlier you mentioned that we cannot consider less than as equal to , but can we consider the max situation ,i.e, the max value that 11m and 7b can be 15 (ques says 15 is enough)
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03 Jul 2014, 06:12
ayushee01 wrote:
correct , but suppose i consider the max value for 11m and 7b = 15 and max value for 10m + 8b as 15, we can solve and get 5/6. earlier you mentioned that we cannot consider less than as equal to , but can we consider the max situation ,i.e, the max value that 11m and 7b can be 15 (ques says 15 is enough)

______________
No we cannot.
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23 Jul 2014, 07:04
why there are no answers to some questions??
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23 Jul 2014, 09:18
HarvinderSaini wrote:
why there are no answers to some questions??

That's not true. Links to solutions are here: devil-s-dozen-129312.html#p1060761
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18 Aug 2014, 08:18
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:

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.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?
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18 Aug 2014, 10:08
mahendru1992 wrote:
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:

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.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?

There might be a group of patients who has neither arachnophobia nor acrophobia. So, it should be a + b + c + neither = 58.
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18 Aug 2014, 10:22
Bunuel wrote:
mahendru1992 wrote:
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:

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.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?

There might be a group of patients who has neither arachnophobia nor acrophobia. So, it should be a + b + c + neither = 58.

Okay but b+c=13, D can still be the answer? The question is asking us for b+c
Re: Devil's Dozen!!!   [#permalink] 18 Aug 2014, 10:22

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