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In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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30 Aug 2015, 10:56
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In a roomful of spies, some spies carry Argentinian passports, some carry Bolivian passports, and some carry Chilean passports. Every spy carries at least one passport, no spy carries more than one passport from a given country, and only one spy carries passports from all three countries. The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports, two less than the number of spies who carry Chilean passports, twice the number of spies who carry both Bolivian and Chilean passports, and half the total number of spies in the room. The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports and two less than the number of spies who carry both Bolivian and Chilean passports. If a spy is selected at random, what is the probability that he will carry an Argentinian passport? (A) 2/9 (B) 13/57 (C) 6/23 (D) 3/7 (E) 1/2 Kudos for a correct solution.
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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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30 Aug 2015, 19:26
Hi Bunuel, I wanted to know whether GMAT can present such a big question. Solving it will certainly take more than 34 mins.



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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31 Aug 2015, 00:21



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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Updated on: 03 Sep 2015, 00:52
IMO : D Consider the following Venn Diagram Attachment:
Capture.JPG [ 18.03 KiB  Viewed 1557 times ]
No. of ppl carrying all 3 passports = 1 No. of ppl carrying none = 0 number of spies who carry Bolivian passports = two more than the number of spies who carry Argentinian passportsb+d+f+1 = a+d+e+1+2 b+f = a+e+2 Eq (i)number of spies who carry Bolivian passports = two less than the number of spies who carry Chilean passportsb+d+f+1 = c+e+f+12 b+d = c+e2 Eq (ii)number of spies who carry Bolivian passports = twice the number of spies who carry both Bolivian and Chilean passportsb+d+f+1 = 2(f+1) b+d = f+1 Eq (iii)number of spies who carry Bolivian passports = half the total number of spies in the roomb+d+f+1 = 1/2(a+b+c+d+e+f+1) b+d+f+1 = a+e+c Eq (iv)number of spies who carry both Argentinian and Bolivian passports = two more than the number of spies who carry both Argentinian and Chilean passportsd+1 = e+1 +2 d=e+2  Eq (v) number of spies who carry both Argentinian and Bolivian passports = two less than the number of spies who carry both Bolivian and Chilean passportsd+1 = f+1 2 d = f2 Eq (vi) From (v) & (vi) e+4 = f (vii) From (vi) & (iii) b+f2 = f+1 b=3From (ii) & (iii) & (vii) b+d = c+e2 && b+d = f+1 f+1 = c+e2 e+4+1 = c+e2 c=7From (i) & (iv) b+f = a+e+2 && b+d+f+1 = a+e+c a+e+2+d+1 = a+e+c d+3=c d=4Sub b=3, d=4, c=7 in (ii) e=2From (vii) f=6sub b=3. c=7, d=4, e=2 , f=6 in (i) b+f = a+e+2 9 = a+4 a=5 On solving these equations we end with a = 5, b = 3, c = 7, d = 4, e = 2, f = 6Prob = \(\frac{a+d+e+1}{Total}\) = \(\frac{12}{28}\) = \(\frac{3}{7}\)
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Originally posted by VenoMfTw on 31 Aug 2015, 02:59.
Last edited by VenoMfTw on 03 Sep 2015, 00:52, edited 1 time in total.



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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01 Sep 2015, 19:52
IMO: D  3/7
The total number of spies is
T = A + B + C  (AB + AC + BC) + 1 (Spy with all three passports)  0 (Spy without any of the passports)
Now,
B = A + 2 = C  2 = 2BC = 0.5T AB = AC + 2 = BC  2
Thus,
T = B  2 + B + B + 2  (BC  2 + BC  4 + BC) + 1 T = 3B  (1.5B  6) + 1 = 1.5B + 7
2B = 1.5B + 7
B=14, A=12, C=16, BC=7, AB=5, AC=3, T=28
A/T=12/28=3/7



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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02 Sep 2015, 02:50
I can not attach pic. So I am using just labels. Let T  total AB  A and B holder AC  A and C holder BC  B and C holder a  only A holder b  only B holder c  only C holder. Given: A+2=B C2=B T=2B AB=AC+2 AB=BC2 From these we can derive: A+4=C AC+4=BC Further, if you look at the pic and differences between the numbers you will find out a=b+2 c=b+4 c=a+2 or you can derive these by making more substitutions. So, if you make substitution for all the numbers and express the whole table with one variable a, it will be like this. AB=a1 BC=a+1 AC=a3 , in order to make it positive integer a must be more than equal 4. Now, we can easily estimate following: A=3a3 B=3a1 C=3a+1 There is a problem here. a can be any number more than equal 4 and all these number will satisfy the whole relations. I started substitution from 4. a=4, A=9, B=11, C=13. Hence A/2B=9/22 a=5, A=12, B=14, C=16. Hence A/2B=3/7 This is in the answer choices. a=6, A=15, B=17, C=19. Hence A/2B=15/34 a=7, A=18, B=20, C=22. Hence A/2B=9/20 a=100, A=297, B=299, C=301. Hence blaa blaa blaa Is there max limit for a? I couldn't find



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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02 Sep 2015, 04:41
Marchewski wrote: IMO: D  3/7
The total number of spies is
T = A + B + C  (AB + AC + BC) + 1 (Spy with all three passports)  0 (Spy without any of the passports)
Now,
B = A + 2 = C  2 = 2BC = 0.5T AB = AC + 2 = BC  2
Thus,
T = B  2 + B + B + 2  (BC  2 + BC  4 + BC) + 1 T = 3B  (1.5B  6) + 1 = 1.5B + 7
2B = 1.5B + 7
B=14, A=12, C=16, BC=7, AB=5, AC=3, T=28
A/T=12/28=3/7 Hi, You have considered B = A+2 for (The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports) statement. The question didn't say who carry only Bolivian passports right? Correct me if i am wrong. Regards, Rakesh
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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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02 Sep 2015, 07:35
Bunuel wrote: In a roomful of spies, some spies carry Argentinian passports, some carry Bolivian passports, and some carry Chilean passports. Every spy carries at least one passport, no spy carries more than one passport from a given country, and only one spy carries passports from all three countries. The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports, two less than the number of spies who carry Chilean passports, twice the number of spies who carry both Bolivian and Chilean passports, and half the total number of spies in the room. The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports and two less than the number of spies who carry both Bolivian and Chilean passports. If a spy is selected at random, what is the probability that he will carry an Argentinian passport?
(A) 2/9 (B) 13/57 (C) 6/23 (D) 3/7 (E) 1/2
Kudos for a correct solution. This is going to be a big explanation. It is Better to do the question with Venn diagram. Let the number of only Argentinian spies be x the number of only Bolivian spies be y the number of only Chilean spies be z the number of only Argentinian and Chilean spies be a the number of only Argentinian and Bolivian spies be b the number of only Chilean and Bolivian spies be c the number of spies who carry all 3 passports is 1. the probability that the spy selected at random will carry Argentinian passport will be \(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\) Now, moving on to the conditions 1) The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports y+b+c+1 = x+a+b+1+2 or y+c=x+a+2 2) The number of spies who carry Bolivian passports is two less than the number of spies who carry Chilean passports y+b+c+1 =z+c+a+12 or y+b=z+a2 3) The number of spies who carry Bolivian passports is twice the number of spies who carry both Bolivian and Chilean passports y+b+c+1 = 2(c+1) or y+b = c+1 4) The number of spies who carry Bolivian passports is half the total number of spies in the room y+b+c+1 = 1/2(x+y+z+a+b+c+1) or 2y+2b+2c+2 = x+y+z+a+b+c+1 or y+b+c+1= x+z+a 5) The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports b+1 = a+1+2 or b=a+2 or a=b2 6) The number of spies who carry both Argentinian and Bolivian passports is two less than the number of spies who carry both Bolivian and Chilean passports b+1=c+12 or b=c2 or c= b+2 Now, putting the value of c in statement 3) y+b= b+2+1 or y = 3putting the value of a and c (in terms of b from statements 5 & 6 respectively) and y in statement 1) 3+b+2= x+b or x = 5 putting value of a (in terms of b from statement 5) and y in statement 2) 3+b= z+b4 or z = 7putting values of x,y,z,a and c in statement 4) 3+b+b+2+1=5+7+b2 or 2b+6 = b+10 or b = 4putting this value of b in statement 5) a=b2 or a = 2putting this value of b in statement 6) c=b+2 or c = 6so the probability will be \(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\) = \(\frac{(5+2+4+1)}{(5+3+7+2+4+6+1)}\) = \(\frac{12}{28}\)= \(\frac{3}{7}\) Answer: D



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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02 Sep 2015, 12:54
kunal555 wrote: Bunuel wrote: In a roomful of spies, some spies carry Argentinian passports, some carry Bolivian passports, and some carry Chilean passports. Every spy carries at least one passport, no spy carries more than one passport from a given country, and only one spy carries passports from all three countries. The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports, two less than the number of spies who carry Chilean passports, twice the number of spies who carry both Bolivian and Chilean passports, and half the total number of spies in the room. The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports and two less than the number of spies who carry both Bolivian and Chilean passports. If a spy is selected at random, what is the probability that he will carry an Argentinian passport?
(A) 2/9 (B) 13/57 (C) 6/23 (D) 3/7 (E) 1/2
Kudos for a correct solution. This is going to be a big explanation. It is Better to do the question with Venn diagram. Let the number of only Argentinian spies be x the number of only Bolivian spies be y the number of only Chilean spies be z the number of only Argentinian and Chilean spies be a the number of only Argentinian and Bolivian spies be b the number of only Chilean and Bolivian spies be c the number of spies who carry all 3 passports is 1. the probability that the spy selected at random will carry Argentinian passport will be \(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\) Now, moving on to the conditions 1) The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports y+b+c+1 = x+a+b+1+2 or y+c=x+a+2 2) The number of spies who carry Bolivian passports is two less than the number of spies who carry Chilean passports y+b+c+1 =z+c+a+12 or y+b=z+a2 3) The number of spies who carry Bolivian passports is twice the number of spies who carry both Bolivian and Chilean passports y+b+c+1 = 2(c+1) or y+b = c+1 4) The number of spies who carry Bolivian passports is half the total number of spies in the room y+b+c+1 = 1/2(x+y+z+a+b+c+1) or 2y+2b+2c+2 = x+y+z+a+b+c+1 or y+b+c+1= x+z+a 5) The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports b+1 = a+1+2 or b=a+2 or a=b2 6) The number of spies who carry both Argentinian and Bolivian passports is two less than the number of spies who carry both Bolivian and Chilean passports b+1=c+12 or b=c2 or c= b+2 Now, putting the value of c in statement 3) y+b= b+2+1 or y = 3putting the value of a and c (in terms of b from statements 5 & 6 respectively) and y in statement 1) 3+b+2= x+b or x = 5 putting value of a (in terms of b from statement 5) and y in statement 2) 3+b= z+b4 or z = 7putting values of x,y,z,a and c in statement 4) 3+b+b+2+1=5+7+b2 or 2b+6 = b+10 or b = 4putting this value of b in statement 5) a=b2 or a = 2putting this value of b in statement 6) c=b+2 or c = 6so the probability will be \(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\) = \(\frac{(5+2+4+1)}{(5+3+7+2+4+6+1)}\) = \(\frac{12}{28}\)= \(\frac{3}{7}\) Answer: D Oh my God, careless me. Missed the fact that the number of spies who carry Bolivian passports is twice the number of spies who carry both Bolivian and Chilean passports. That is why I was struggling to find definite values.



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In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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08 Sep 2015, 17:59
VenoMfTw wrote: Marchewski wrote: IMO: D  3/7
The total number of spies is
T = A + B + C  (AB + AC + BC) + 1 (Spy with all three passports)  0 (Spy without any of the passports)
Now,
B = A + 2 = C  2 = 2BC = 0.5T AB = AC + 2 = BC  2
Thus,
T = B  2 + B + B + 2  (BC  2 + BC  4 + BC) + 1 T = 3B  (1.5B  6) + 1 = 1.5B + 7
2B = 1.5B + 7
B=14, A=12, C=16, BC=7, AB=5, AC=3, T=28
A/T=12/28=3/7 Hi, You have considered B = A+2 for (The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports) statement. The question didn't say who carry only Bolivian passports right? Correct me if i am wrong. Regards, Rakesh Sorry for my slow reponse. For the formula I used it is not necessecary to know the number of people who only carry Bolivian passports. B is just the number of spies that have Bolivian passports  no matter if they have more than just this one passport. Considering your Venn B = b + d + f + 1 = A + 2.



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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10 Sep 2015, 02:25
Bunuel wrote: In a roomful of spies, some spies carry Argentinian passports, some carry Bolivian passports, and some carry Chilean passports. Every spy carries at least one passport, no spy carries more than one passport from a given country, and only one spy carries passports from all three countries. The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports, two less than the number of spies who carry Chilean passports, twice the number of spies who carry both Bolivian and Chilean passports, and half the total number of spies in the room. The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports and two less than the number of spies who carry both Bolivian and Chilean passports. If a spy is selected at random, what is the probability that he will carry an Argentinian passport?
(A) 2/9 (B) 13/57 (C) 6/23 (D) 3/7 (E) 1/2
Kudos for a correct solution. The question is long but not hard. Let's try to stick with one variable. Note: "The number of spies who carry Bolivian passports is two more..., two less ..., twice ..., and half the total number of spies in the room. This is an interesting relation. Let's assume total number of spies = N. Every spy has at least one passport. Now from the given information, you can draw this venn diagram: Attachment:
Sets.jpg [ 1.19 MiB  Viewed 1183 times ]
Total = n(A) + n(B) + n(C)  n(A and B)  n(B and C)  n(C and A) + n(A and B and C) N = (N/2 2) +N/2 + (N/2 + 2)  (N/4  2)  N/4  (N/4  4) + 1 N = 28 n(A) = 12 Required Probability = 12/28 = 3/7
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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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10 Sep 2015, 02:41
You don't need to solve anything to arrive at the answer. Probability of picking up a spy with Bolivian passport = 1/2 And the number of spies who hold Argentinian passport is just 2 less than that of spies who hold Bolivian passports. So the probability here will be little less than 1/2 but not a lot. 3/7 is the only option that fits. However if the options were any closer you'll have to solve for the answer.



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Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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Re: In a roomful of spies, some spies carry Argentinian passports, some ca
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