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In a survey, 2000 executives were each asked whether they read newslet

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In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 03 Jul 2017, 03:34
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Question Stats:

31% (02:04) correct 69% (02:40) wrong based on 103 sessions

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In a survey, 2000 executives were each asked whether they read newsletter A or newsletter B. According to the survey, 55 percent of the executives read newsletter A, 62 percent read newsletter B, and 37 percent read both newsletter A and newsletter B. How many of the executives surveyed read at most one among newsletter A and newsletter B?

(A) 1600
(B) 1260
(C) 900
(D) 860
(E) 760

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In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 03 Jul 2017, 05:11
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Given data :
55 percent of the executives read newsletter A
62 percent read newsletter B
37 percent read both newsletter A and newsletter B
Total = 2000

From the data we can infer
P(A) = \(\frac{55}{100} * 2000 = 1100\)
P(B) = \(\frac{62}{100} * 2000 = 1240\)
P(Both) = \(\frac{37}{100} * 2000 = 740\)
P(Only A) = P(A) - P(Both) = 1100 - 740 = 360
P(Only B) = P(B) - P(Both) = 1240 - 740 = 500

P(Total) = P(Only A) + P(Only B) + P(Both) + P(Neither)
Hence, P(Neither) = P(Total) - {P(Only A) + P(Only B) + P(Both)}
Substituting values,
P(Neither) = 2000 - 360 - 500 - 740 = 400

P(At most 1) = P(Neither) + P(Only A) + P(Only B) = 400 + 500 + 360 = 1260(Option B)

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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 04 Jul 2017, 03:58
55% of 2000 =1100
62% of 2000=1240
37% of 2000=740
We need people who read only A and those who read only B, so using the double matrix method we can deduce the following,
1-people reading A but not B =1100-740 =360
2-people reading B but not A = 1240-740 =500
From 1 and 2 = 500+360 = 860
hence, the answer is : D

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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 07 Jul 2017, 04:34
55% = 1100
62% = 1240
37%= 740

Only A + Only B = (1100-740) + (1240-740)

860

option D
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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 10 Jul 2017, 20:43
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Total executives who read newsletter = A+B-AB = 55+62-37 = 80
Therefore, 80% of the total of 2000 executives read newsletter i.e 1600 this includes executives who read both A and B newsletters. We need no. of the executive who read ATMOST 1 newsletter.

Therefore, 1600-(2000*37%) = 860
Additionally, we also have 400 executives who neither read newsletter A nor B

So, 860+400= 1260

Answer: B
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In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post Updated on: 10 Jul 2017, 21:46
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resolved it in the same way as guys above, 860 which for sure makes no sense as we know at least 55% reads A, so 1100 reads newspaper. Missed the clue "at most"

P.S.
pushpitkc
P(B) = 62/100∗2000=1240 not 1210
so in your formula 430 should be corrected to 500
then
P(Neither) = 2000 - 360 - 500- 740 = 400
then
P(At most 1) = P(Neither) + P(Only A) + P(Only B) = 500 + 400 + 360 = 1260(Option B)

Originally posted by cbh on 10 Jul 2017, 21:35.
Last edited by cbh on 10 Jul 2017, 21:46, edited 1 time in total.
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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 10 Jul 2017, 21:41
I think this is pretty simple, only one calculation required. There are 37% who read both newsletters A and B.

So remaining 63% (of 2000) are those who read either only one newsletter (A or B), or neither. So all those 63% count as 'those who read at most one newsletter'. Thus answer = 63% of 2000 = 1260

Hence B answer
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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 11 Jul 2017, 02:02
cbh wrote:
resolved it in the same way as guys above, 860 which for sure makes no sense as we know at least 55% reads A, so 1100 reads newspaper. Missed the clue "at most"

P.S.
pushpitkc
P(B) = 62/100∗2000=1240 not 1210
so in your formula 430 should be corrected to 500
then
P(Neither) = 2000 - 360 - 500- 740 = 400
then
P(At most 1) = P(Neither) + P(Only A) + P(Only B) = 500 + 400 + 360 = 1260(Option B)


Thanks a ton for noticing and informing, cbh. Have made the necessary changes!
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Re: In a survey, 2000 executives were each asked whether they read newslet  [#permalink]

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New post 12 Jul 2017, 16:33
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Bunuel wrote:
In a survey, 2000 executives were each asked whether they read newsletter A or newsletter B. According to the survey, 55 percent of the executives read newsletter A, 62 percent read newsletter B, and 37 percent read both newsletter A and newsletter B. How many of the executives surveyed read at most one among newsletter A and newsletter B?

(A) 1600
(B) 1260
(C) 900
(D) 860
(E) 760


The phrase “at most one” means “one or fewer.” Since 2,000 x 0.37 = 740 executives read both newsletters A and B, there must be 2,000 - 740 = 1,260 executives who read at most one of two newsletters (which means they could read only newsletter A, or only newsletter B, or neither).

Answer: B
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Re: In a survey, 2000 executives were each asked whether they read newslet   [#permalink] 12 Jul 2017, 16:33
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