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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
1
6
00:00
A
B
C
D
E
Difficulty:
55% (hard)
Question Stats:
64% (03:51) correct 36% (03:51) wrong based on 51 sessions
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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?
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
1
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 passports b+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 passports b+d+f+1 = c+e+f+1-2 b+d = c+e-2 --Eq (ii)
number of spies who carry Bolivian passports = twice the number of spies who carry both Bolivian and Chilean passports b+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 room b+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 passports d+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 passports d+1 = f+1 -2 d = f-2 Eq (vi)
From (v) & (vi) e+4 = f --(vii)
From (vi) & (iii) b+f-2 = f+1 b=3
From (ii) & (iii) & (vii) b+d = c+e-2 && b+d = f+1 f+1 = c+e-2 e+4+1 = c+e-2 c=7
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
_________________
I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯
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+1-2 or y+b=z+a-2
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=b-2
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+1-2 or b=c-2 or c= b+2
Now, putting the value of c in statement 3)
y+b= b+2+1 or y = 3
putting 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+b-4 or z = 7
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+1-2 or y+b=z+a-2
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=b-2
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+1-2 or b=c-2 or c= b+2
Now, putting the value of c in statement 3)
y+b= b+2+1 or y = 3
putting 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+b-4 or z = 7
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.
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.
Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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10 Sep 2015, 02:25
2
1
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
_________________
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.
Re: In a roomful of spies, some spies carry Argentinian passports, some ca [#permalink]
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10 Sep 2017, 19:38
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