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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55% = 1100
62% = 1240
37%= 740

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

860

option D

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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The survey consists of ppl who read 0,1 or 2 papers. Total surveyed: 2000
Total that read 2 papers: 37% x 2000 = 740.
Ppl that read atmost one => Ppl that read <=1 . That is 0 or 1 newspapers.
Total - Both = 2000 - 740 = 1260

Agree with Nikita. It says at most 1 "among A & B". Hence answer should be 860.

Please explain.