The Mathematics Club in a school held an open house on three afternoons. 115, 75 and 110 students
attended on the first, second and third afternoons respectively. 30 attended only on the first day, 35 on the first and second days. 70 attended the first and the third days and 35 attended on the second and third days. How many attended on all the three days?
(A) 15 (B) 25 (C) 21 (D) 20 (E) 30
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n(F) = n(F only) + n(F&S) + n(F&T) - n(F&S&T)
F --> First day.
S --> Second day.
T --> Third day.
[spoiler]115 = 30 + 35 + 70 - n(F&S&T)
n(F&S&T) = 135 - 115 = 20
Hence (D)[/spoiler]
I think info about n(S&T), n(S) and n(T) are not needed to solve this question.
F --> First day.
S --> Second day.
T --> Third day.
[spoiler]115 = 30 + 35 + 70 - n(F&S&T)
n(F&S&T) = 135 - 115 = 20
Hence (D)[/spoiler]
I think info about n(S&T), n(S) and n(T) are not needed to solve this question.
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chufus
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Everything seems correct to your approach. Wonder why the other info was mentioned in the question. Doesn't seem like a GMAT question !rppala90 wrote:n(F) = n(F only) + n(F&S) + n(F&T) - n(F&S&T)
F --> First day.
S --> Second day.
T --> Third day.
[spoiler]115 = 30 + 35 + 70 - n(F&S&T)
n(F&S&T) = 135 - 115 = 20
Hence (D)[/spoiler]
I think info about n(S&T), n(S) and n(T) are not needed to solve this question.
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krishna8143
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edvhou812
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This is a GMAT question.chufus wrote:Everything seems correct to your approach. Wonder why the other info was mentioned in the question. Doesn't seem like a GMAT question !rppala90 wrote:n(F) = n(F only) + n(F&S) + n(F&T) - n(F&S&T)
F --> First day.
S --> Second day.
T --> Third day.
[spoiler]115 = 30 + 35 + 70 - n(F&S&T)
n(F&S&T) = 135 - 115 = 20
Hence (D)[/spoiler]
I think info about n(S&T), n(S) and n(T) are not needed to solve this question.
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